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Signal Processing
An Introduction
Magnus Danielsen
NVDRit 2007:13
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Heiti / Title Signal Processing An Introduction
Høvundar / Authors Magnus Danielsen
Ritslag / Report Type Undirvísingartilfar/Teaching Material
NVDRit 2007:13 © Náttúruvísindadeildin og høvundurin
ISSN 1601-9741 Útgevari / Publisher Náttúruvísindadeildin, Fróðskaparsetur Føroya Bústaður / Address Nóatún 3, FO 100 Tórshavn, Føroyar (Faroe Islands) Postrúm / P.O. box 2109, FO 165 Argir, Føroyar (Faroe Islands)
@ +298 352550 +298 352551 [email protected]
Abstract: English
Signal Processing – An Introduction Fundamentals of signal processing and analysis is treated on the base of Fourier representations, Laplace transformation, and z-transformation. Continuous and discrete Fourier representations are defined, and treated in depth with emphasis on illustrating the connection, similarities and non-similarities between the four Fourier representations of 1) discrete time periodic, 2) continuous time periodic, 3) discrete time non-periodic, and 4) continuous time non-periodic signals. Discrete Fourier transform is introduced, and the fast Fourier transform algorithms are defined and illustrated. The Laplace transform and the z-transform are defined and treated in the light of being a kind of generalization of the continuous and discrete Fourier transformations respectively. On the base of the signal representations a number of applications are dealt with. Sampling and reconstruction of signals, frequency analysis, fundamental analogue and digital filter constructions are treated, and a number of specific application examples are lined out. Føroyskt: Signalfrøði – Ein inngangur Grundleggjandi signalfrøði – viðgerð og greining – er gjøgnumgingin við støði í Fourier røðum og transformum, Laplace-, og z-transformum. Fourier røð og transformatiónir eru lýst og viðgjørd við denti á sambandinum, líkheitini og ólíkheitini millum tær fýra Fourier umboðanirnar: 1) diskret tíð periodisk, 2) kontinuert tíð ikki periodisk, 3) diskret tíð periodisk, og 4) kontinuert tíð ikki periodisk signal. Diskret Fourier transform er tikið upp, og “Fast Fourier Transform” algoritmur eru lýstar. Laplace transform and the z-transform hættirnir eru lýstir og viðgjørdir í ljósinum av at vera eitt slag av algilding av kontinuertum og diskretum Fourier transformatiónum og røðum. Við støði í hesum signal umboðanunum eru nýtslur av hesum hættum tiknar upp. Sampling og endurskapan av signalum er viðgjørd, frekvensgreining, grundleggjandi analog and digital filtur eru viðgjørd, umframt ein røð av serstøkum nýtsluendamálum eru greinað.
Contents: 1 Introduction
1.1 Introduction to course 1.2 Elementary signals and basic operations
2 Linear time-invariant systems 2.1 Linear time invariant systems and convolution 2.2 Properties of impulse response of a linear system 2.3 Sinusoidal response, diff. equations, and block diagrams
3 Continuous and discrete time Fourier representation of periodic and non-periodic signals 3.1 Fourier representations of signals 3.2 Discrete time (DTFS) and continuous time (FS) Fourier series 3.3 Discrete time non – periodic signals 3.4 Continuous time nonperiodic signals 3.5 Basic properties of Fourier Representations (1) 3.6 Basic properties of Fourier Representations (2) 3.7 Fundamental Operations and System Properties of Fourier Representations 3.8 Technical Applications of Fourier Representations
4 Sampling and reconstruction of signals 4.1 Sampling of signals 4.2 Aliasing and reconstruction of sampled signals
5 Discrete Fourier Transform and Fast Fourier Transform 5.1 Discrete Fourier Transform (DTF) for numerical evaluation 5.2 Fast Fourier Transform (FFT) – an efficient algorithm for numerical evaluation of DFT
6 Laplace Transform representation og signals 6.1 Laplace transform principles 6.2 The Laplace transform – transfer functions
7 Z-transform representation of signals 7.1 Digital filters – overview, and z-transforms 7.2 The z-transform and its properties 7.3 The z-transform – transfer functions 7.4 The z-transform – transfer functions. Frequency response and block diagrammes
8 Analogue and Digital Filters 8.1 Analogue filters 8.2 Analogue Butterworth filters 8.3 Analogue Chebyshev filters and frequency transformations of filters 8.4 Analogue filters – practical constructions 8.5 Digital FIR filter 8.6 Digital IIR filter – Low-pass, and high-pass filters 8.7 Digital IIR filter – Bandpass, and bandstop filters
9 Applications of signal processing in systems – some examples 9.1 Acoustic spectrogramme example 9.2 Delays in systems 9.3 Seismic example
Signal ProcessingPart 1.1 Introduction to Course
Magnus Danielsen
Textbooks incl. Supplementary Books
• S.Haykin. B.V.Veen: ”Signal and systems”, John Wiley and Sons, ISBN:0-471-13820-7 (Course book)
• M.Danielsen: “Phase delay and group delay” NVDRit 2007:02, (Course note)
• J.W. Nilsson, S.A. Riedel: “Electric Circuits”, 7th Edition, (background knowledge)
• H. Baher. “Analog and Digital Signal Processing”, 2nd Edition, John Wiley and Sons, ISBN: 0471-62354-7 (supplementary)
• P.A. Lynn, W. Fuerst, Introductory Digital Signal Processing, 2nd Edition John Wiley and Sons, ISBN: 0-471-97631-8 (supplementary)
• E.W.Kamen, B.S.Heck:”Fundamentals of Signals and Systems”, Prentice Hall, 2nd Edition, ISBN 0-13-017293-6 (supplementary)
• J.H.McClellan, “Signal Processing First”, ISBN 0-13-1201265-0 (supplementary)
• J.G.Proakis, D.G.Manolakis, “Digital Signal Processing”, Prentice Hall, ISBN 0-13-373762-4 (Advanced for electronics)
• B.Burrkus, ”Spectral and Filter Theory”, Springer, ISBN 3-540-62674-3 (Advanced for geophysics)
Contents of Course (1)• Introduction – basic signals and operations –
application examples• Time-domain – linear time invariant systems• Energy signals and power signals• Fourier transforms:
– Discrete time periodic signals DTFS– Continuous time periodic signals CTFS = FS– Discrete time non-periodic signals DTFT– Continous time non-periodic signals CTFT = FT
• Discrete Fourier transform and Fast fourier transform DFT & FFT
• Convolution/deconvolution• Sampling and reconstruction of signals
Contents of Course (2)• Laplace transform and its relations to Fourier
transforms• Z-transform and its relations to Laplace transform
and Fourier transforms• Analogue Filters • Digital FIR filters• Digital IIR filters• Equalizers• Application examples:
– Seismic geophone signal– Communication modulation wave forms– Speech signal spectrograms – Control feedback system
• MATLAB as signal processing tool
Types of Signals -Phenomenologically:
• Analogue signals– Function f(t) – Continous magnitude range– Continous argument t
• Discrete signals– Function f(t)=f(nT) or f(n) – Continuous magnitude range – Discrete argument t=nT or n (n is an integer)
• Digital signals– Function f(t) or f(n)– Discrete magnitude range– Discrete argument t=nT or n (n is an integer)
Types of Signals – Physically:• Sound
– Speech (100-10000 Hz)– Music (20-20000 Hz) – Telephone(300-3400 Hz)– Seismics (1-1000 Hz)– Ultra sound (20-100 kHz →)
• Light– Communications (glass fibres)– Infrared temp.measurements
• Electrical current and voltage• Sea-waves and -currents • Communications systems
(data, speech, pictures)• Internet & Email• Radiowaves
– Radar– Broadcasting (sound, pictures)– Communication
• Biological signals – ECG (hearth signals)– EEG (brain signals)– Blood pressure– Temperature
• Weather – Temperature– Moisture– Windspeed and direction
• Visual signals– Television– Computer display
• Statistics– World market– Finance, price, stock– Demography
• Etc.
Signal Processing System
• Communication system• Control system• Remote sensing system• Biomedical system• Auditory system• Audio system• Seismic system
SYSTEMInput signal Output signal
Communication System
• Message: Speech, TV/video, data• System: Digital or analogue• Channel: Opt.fibre, coax, radiolink, satellite,
mobile tel./PC• Disturbance: Noise, distortion, interference• Transceiv.: Coding, compression reconstruction,
modulation (AM, FM, PM, ASK, FSK, PSK, QAM etc.)• Electronics: VLSI• Modes: Broadcasting, point-to-point
Transmitter Channel ReceiverMessage signal
Transmitted signal
Received signal
Estimated received message signal
Disturbance
Control System
• Examples: – aircraft, autopilots, automobil engine, machine tools, oil raffineries,
oildrilling, papermills, fish industry, electrical powerplants, nuclear reactors, robots
• Response: Output follows input reference, regulation• Robustness: Good regulation despite of disturbances• Closed-loop: Feed-back control system• SISO: Single input / single output• MIMO: Multiple input / multiple output• Analogue systems: Analogue electronics• Digital systems: Digital electronics using microprocessors
Controller
Sensor
Plant ++Reference
Disturbance
Output+
−
+
+
Remote Sensing System
• Examples of fields:– Acoustic (sound, seismic, ultrasound), electromagnetic, electric, magnetic, gravitational– Passive (e.g. Earth quake), active (seismic)
• Sensors and systems:– Radar: surface properties, topography, roughness, moisture, dielectric constant– Infrared: near surface termal properties– Visual/near infrared: Chemical composition– X- and γ-ray: radioactive properties– Seismics and seismology: oil and gass exploration, earth quakes
• Digital signal processing - examples– Synthetic aperture radar (SAR): Use of single antenna in place of array, and use of FFT– Ocean seismic tecnology using airgun/hydrophone streamer
Passive system:
Sensor Signal processing
Input Signal
Object
Detected Signal
Processed Signal
Disturbance
Active system:
Sensor Signal processing
Received Signal
Object
Detected Signal
Processed Signal
Trans-mitter
Transmitted Signal Signal
modulator
Disturbance
Biomedical System
• Extraction of information from biological signals• Biological signals trace back to electrical activity• Neurons • ECG (EKG)• EEG• Artifacts
– Instrumental : thermal noise, 50Hz powerline noise – Biological: EEG interfered by ECG– Signal analytical: roundoff, quantization
• Signal processing: – Filtering– Use of apriori known signal properties: frequency range, randomlike properties,etc.
EEG: ECG:
Auditory System
Audio System - Spectrogram
Seismic System
A. Ziolkowski et al: “The signature of an air gun array: Computation from nearfield measurements including interactions”,Geophysics,vol. 47, oct. 1982, P. 1413
Signal ProcessingPart 1.2 Elementary Signals and Basic Operations
Magnus Danielsen
Contents
• Types of signals• Classification• Basic operations• Elementary signals• Systems, response and properties• Operators and block diagrams
Limitation of Signals Applied in this Course
• One dimensional signals and variables• Single valued signals• Real valued signals• Complex valued signals• Real independent variable (time or space)
Analogue signal = continuous signal:
X(t)
t
Discrete signal:
n
X[n]
T
Digital Signal
t
f(t)
T
Analogue or Digital Signal Processing?
Analogue:– Real time– Simple circuits (often)– Physical in nature– Inflexible change of
system (rebuilding hardware)
– Suffers from parametrical variations
Digital:– Flexible– Software change of
system by changing programme
– Repeatibility– Complicated circuits
(VLSI)
Classification of Signals (1)1. Continuous / discrete time signals
– x(t)– x[n] with sampling time T
2. Even / odd signals– Even: x(-t)=x(t) x[-n]=x[n]– Odd: x(-t)=-x(t) x(-n)=-x[n]– Conjugate symmetric: x(-t)=x*(t) x[-n]=x*[n]
x(t)=a(t)+jb(t); a(-t)=a(t); b(-t)=-b(t) 3. Deterministic signals/
Indeterministic signals (random)
Classification of Signals (2)4. Periodic / non-periodic signals
– Periodic : x(t+T)=x(t) per.= T=1/f=(2π)/ωx[n+N]=x[n] per.= N=(2π)/Ω
– Non-per. : x(t) and x[n] not repeated periodically
5. Energy / power signals– Energy signals:
– Power signals:T/2 N
2 2
T N n NT/2
1 1lim x(t) dt lim x[n]T N→∞ →∞
=−−
< ∞ < ∞∑∫
2 2
n
x(t) dt x[n]∞ ∞
=−∞−∞
< ∞ < ∞∑∫
Basic Operations on Signals• Amplitude scaling
– y(t)=cx(t) y[n]=cx[n]• Addition
– y(t)=x1(t)+x2(t) y[n]=x1[n]+x2[n]• Multiplication
– y(t)=x1(t)·x2(t) y[n]=x1[n]·x2[n]• Differentiation / differensing
– y(t)=dx(t)/dt y[n]= x[n+1]-x[n]• Integration / summation
– y(t)=∫x(t)dt y[n]=Σx[n]
Basic Operation on the Independent Variable(1)
• Scaling– y(t)=x(at)
– Y[n]=x[kn]k = integer
• Reflection– y(t)=x(-t)
– y[n]=x[-n]
tx(t)
ty(t)
nx[n]
ny[n]
tx(t)
ty(t)
n
x[n] y[n]
n
Basic Operation on the Independent Variable
• Time shifting– y(t)=x(t-t0) – y[n]=x[n-m]
• Precedence operation– y(t)=x(at-b)– y[n]=x[pn-q]
ty(t)
t0t
x(t)
ty(t)
t0t
x(t)
Elementary Signals(1)• Exponential signal
– x(t)=B exp(at) a>0 growing signala<0 decaying signal
– x[n]=B rn r>1 growing; 0<r<1 decaying(B>0) r>0 positive x; r<0 alternating x
• Sinussoidal signal– x(t)=A cos(ωt+ϕ); x(t+T)=x(t); T=1/f=2π/ω
x[n]=A cos(Ωn+ϕ); – x[n+N]=x(t); Ω(n+N)+ ϕ = Ωn+2πm + ϕ
Periodicity requires: N=2πm/ Ω
Elementary Signals(2)
• Complex exponential and sinusoidal fct.– Eulers identities: ejθ = cos θ+jsin θ
cos θ = 1/2 (ejθ + ejθ)sin θ = 1/(2j) (ejθ - e-jθ)
– B = A ejθ
– A cos(ωt+θ) = ReB ejωt– A sin(ωt+θ) = ImB ejωt
Elementary Signals(3)
• Damped oscillation– Analogue: X(t) = A e-αt cos(ωt+θ) α > 0– Discrete: X[n] = B rn cos(Ωn+θ) 0 < r <1
• Step function– Analogue: u(t) = 0 t < 0
= 1 t ≥ 0– Discrete: u[n] = 0 t < 0
u[n] = 1 t ≥ 0 • Rectangular pulse
– p(t) = u(t + ½T) – u(t – ½T) p(t) = u(t) – u(t – T)– p[n] = u[n] – u[n – N]
Elementary Signals(4)• Impulse:
– Analogue:
– Discrete:
• Ramp: r(t) = t u(t) r[n] = n u[n]
0 0
0 t 0( t )
t 0
( t )dt 1 x ( t ) ( t t )dt x ( t )∞ ∞
−∞ −∞
≠δ = ∞ =
δ = δ − =∫ ∫
0 n 0( n )
1 n 0≠
δ = =
(t)δt
1
[n]δn
1
t0
n0
System and Operators Operator formulation:
Shift operator: x[n-k] = Skx[n] H = Sk (or St0)Differencing operator: y[n] = x[n-1] – x[n]
H = Sk – 1
Moving average operator: y[n] = 1/3·(x[n]+x[n-1]+x[n-2])
H = 1/3·(1+Sk+Sk 2)
Block diagram:
Hx(t)
x[n] y[n] = Hx[n]
y(t) = Hx(t)
Sk Sk
+
1/3y[n]
x[n]
+ Σ=
• StabilityBIBO: x ≤ Mx < ∞ ⇒ y ≤ My < ∞
• Memory: y(t0) depends on x for t<t0– memory: y[n] = 1/3·(x[n]+x[n-1]+x[n-2]) (Moving average )
i(t)=1/L ∫v(t)dt (Inductance)
– no memory: y[n]=x[n]2 (Square function)
v(t)=R i(t) (Ohms law)
• Causality:– y[n] depends only on x[m] for m ≤ n– y(t0) depends only on x(t) for t ≤ t0
Causal signal: y[n] = 1/3·(x[n]+x[n-1]+x[n-2])Noncausal sign.: y[n] = 1/3·(x[n+1]+x[n]+x[n-1])
Properties of System (1)Hx y
Properties of System (2)
Hxy=Hx
H-1
z=H-1H x=x
Invertibility – inverse system:– H-1H=I– H-1=inverse operator; I=identity operator
• Examples– Equalizer– Seismic signal detection– y(t)=1/L · ∫x(t)dt H= 1/L · ∫ H-1 =L·d/dt (inductance)– y=x2 x= ±y½ two solutions ⇒ non invertible
Properties of System (3)• Time invariance
Hxy=Hx
Sk
z= SkH x
Skxy=Hx
Hz=HSk x
Commutative rule: SkH = HSk
Properties of System (4)
• Linearity
i i
i i i i
Input : x a x
Output : y H x a y a H x
=
= = =
∑
∑ ∑
• Examples x= Σ ai xi– Linearity: y = nx = nΣ aixi = Σainxi = Σaiyi
– Non-linearity y = x(t)x(t-1) y = Σ ai xi(t)Σ aj xj(t-1) = Σ Σ ai aj xi(t) xj(t-1) ≠ Σ ai yi
Signal ProcessingPart 2.1 Linear Time Invariant Systems and Convolution
Magnus Danielsen
Contents of Part 2.1-2.3 • System• Impulse responce h(t) and h[n]• Convolution formulation for output: y=h∗x• Differential equations for analogue signals• Difference equations for discrete signals• Block diagram
– Scalar multiplication– Addition– Integration for analogue signals– Timeshift for discrete signals
• State variable description: – 1st order differential for analogue signals– 1st order difference equations for discrete signals
Impulse Response
Hx(t)=δ(t) y(t)=Hδ(t)=h(t)
Analogue signals
Hx[n]=δ[n] y[n]=Hδ[n]=h[n]
Discrete signals
x(t) y(t)t
x[n] n
Convolution for Discrete Signals
• Single pulse input: – Input signal: x[n]=x[k]·δ[n-k]– Output signal: y[n]=Hx[n]=h[n-k]
• Multi pulse input:– Input signal:
– Output signal:
k=
k=-
x[n]= x[k] · [n-k]∞
∞
δ∑
k= k=
k=- k=-
y[n]= x[k] · h[n-k]= x[n-k] · h[k] x[n] h[n]∞ ∞
∞ ∞
= ∗∑ ∑
Convolution for Analogue Signals
• Infiniticimal pulse input: – Input signal: dx=x(τ) · δ(t-τ)dτ– Output signal: dy=Hx(τ) · δ(t-τ)dτ=x(τ)h(t-τ)dτ
• Analogue function input:– Input signal:
– Output signal:
- -
y(t) = x( ) · h(t- )d = x(t- ) · h( )d x(t) h(t)∞ ∞
∞ ∞
τ τ τ τ τ τ = ∗∫ ∫
-
x(t)= x( ) · (t- )d∞
∞
τ δ τ τ∫
Discrete Convolution Graphically (1)x[n]=[-1 ½ 1 ½ -½] h[n]=[½ 1 ½ -½ ½ ¼ ]
nn
vk[n]=pk[n] h[n-k]=x[k] h[n-k]nk=-1
nk=0
nk=1
nk=2
nk=3
kk
k
y[n] p [n]h[n k]
x[k]h[n k]
∞
=−∞
∞
=−∞
= −
= −
∑
∑
pk[n]=x[k]δ[n-k]
n-4 -3 -2 -1 0 1 2 3 4 5 6 7 8
2
1
0
Discrete Convolution Graphically (2)x[n]=[-1 ½ 1 ½ -½] h[n]=[½ 1 ½ -½ ½ ¼ ]
n
k
k
k
k
k
k
k
k
k
k
k
k
h[-1-k]
h[-k]
h[1-k]
h[2-k]
h[3-k]
h[4-k]
h[5-k]
h[6-k]
h[7-k]
h[8-k]
h[9-k]
y(n) x[k] h[n k]∞
−∞
= ⋅ −∑n=2
n=1
Discrete Convolution Calculation1 1 1 1 1h[n] [ 1 ]2 2 2 2 4
= −1 1 1x[n] [ 1 1 ]2 2 2
= − −
0½h(-2-k)
......
0...¼ h(9-k)
+1/8 = 0.125...½¼ h(8-k)
+1/8-¼ = - 0.125 ...-½½¼ h(7-k)
¼+¼+¼ = 0.75 ...½-½½¼ h(6-k)
+1/8+½-¼-¼ = 0.125 ...1½-½½¼ h(5-k)
-¼+¼-½+¼-½ = 0.75...½1½-½½¼ h(4-k)
-½-¼+½+½-¼ = 0...½1½-½½¼ h(3-k)
½+¼+1+¼ = 2 ...½1½-½½h(2-k)
-½+½+½ = 0.5...½1½-½h(1-k)
-1+¼ = - 0.75 ...½1½h(-k)
-½ = - 0.5 ...½1h(-1-k)
...0-½½1½-10...h(n-k) x(k)
543210-1-2...k=
k
y[n] x[k]h[n k]∞
=−∞
= −∑
Example on Discrete Convolution
3 2 13 3 34 4 4y[3] ( ) ( ) ( ) 1 2.734= + + + =
.... k
.... kh[3 k]−
.... kh[5 k]−
n 1n 3k 43
4 30 4
1 ( )y[n] 1 ( )1
+−= ⋅ =
−∑
10 1n 310 43
4 30 4
1 ( )y[10] ( ) 3.8311
+−= = =
−∑
x[k] u[k]=
y[ 5] 0− =
5 1n 3k 43
4 30 4
1 ( )y[5] ( ) 3.2881
+−= = =
−∑
( )n 1 2 33 3 3 34 4 4 4x[n] u[n] h[n] u[n] [..0 0 1 ( ) ( ) ( ) ....]= = =
Convolutional Integral for Analogue Signals (1)
Hx(t) y(t)x[n] y[n]
x[n]=δ[n-k] ⇒ y[n]=h[n-k]
x[n]=Σx[k]δ[n-k] ⇒ y[n]=Σx[k]·h[n-k]
x(t)=δ(t-τ) ⇒ y(t)=h(t-τ)= Hδ(t-τ)
y(t)=x(t)∗h(t)=h(t)∗x(t)
- -
- -
x( ) (t- )d y(t)=H x( ) (t- )d
= x( )h(t- )d = x(t- )h( )d
∞ ∞
∞ ∞
∞ ∞
∞ ∞
τ δ τ τ ⇒ τ δ τ τ
τ τ τ τ τ τ
∫ ∫
∫ ∫
Convolutional Integral for Analogue Signals (1)Example:
x(t)=e-3t u(t) – u(t-2) h(t) = e-t u(t)
t
t t-3 -(t- ) -t -2 t 2t t 3t
0 02
-3 -(t- ) t 2 2 4
0
0 t 0
y(t) e e d e e d ½e [e 1] ½(e e ) 0 t 2
e e d ½e [e 1] ½(1 e ) t 2
τ τ τ − − − −
τ τ − − ⋅ −
<= ⋅ τ = τ = − = − ≤ ≤ ⋅ τ = − − = − >
∫ ∫
∫
x(τ)
Convolution of Discrete Step FunctionsNon - delayed step functions :
[ ] [ ] [ ] [ ]
( ) [ ]
n
k 0
u n u n u k u n k
n 1 u n=
∗ = ⋅ −
= + ⋅
∑ [ ] [ ] [ ] [ ]
( ) [ ]
2
1
n n
1 2 1 2k n
1 2 1 2
u n n u n n u k n u n n k
n n n 1 u n n n
−
=
− ∗ − = − ⋅ − −
= − − + ⋅ − −
∑
....
n
[ ] [ ]u n u n∗
.... k....u[n k]−
k=nk=0
n=0
.... k....u[k]
k=0
.... k....2u[n n k]− −
k=n-n2k=0
1u[k n ]−.... k....
k=n1k=0
n=0 n=n1+n2
....
n
[ ] [ ]1 2u n n u n n− ∗ −
....
Delayed step functions :
Moving Average System
[ ] [ ]N 1
k 0
1y n x n kN
where N is the averaging window width
−
=
= −∑Definition :
[ ]x n [ ]y n[ ]h n
[ ] [ ] [ ] [ ]
[ ] [ ]
[ ] [ ] [ ] [ ]( )
[ ] [ ]( )
N 1
k 0
x n n y n h n
1h n n kN1 n n 1 n 2 ... n N 1N1 u n u n NN
−
=
= δ ⇒ =
= δ −
= δ + δ − + δ − + + δ − +
= − −
∑
Impulse response :
Moving Average of Rectangular Pulse[ ] [ ] [ ]x n u n u n M= − −
Moving average of rectangular pulse of width M :
[ ] [ ] [ ] [ ] [ ]( ) [ ] [ ]( )
[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]( )
( ) [ ] ( ) [ ](( ) [ ] ( ) [ ])
1y n h n x n u n u n N u n u n MN
1 u n u n u n N u n u n u n M u n N u n MN1 n 1 u n n N 1 u n NN
n M 1 u n M n N M 1 u n N M
= ∗ = − − ∗ − − =
∗ − − ∗ − ∗ − + − ∗ − =
+ − − + −
− − + − + − − + − −
Rectangular pulse of width M :
Averaging window width N=4
Pulse width M=10
.... n
n=0
y[n]
n=N-1 n=M-1 n=N+M-1
1
¼
Convolution of Continuous Step Functions
( ) ( )
( ) ( )
( )
t
0
u t u t
u u t d
t u t
∗
= τ ⋅ − τ τ
= ⋅
∫
( ) ( )
( ) ( )
( )
2
1
1 2
t t
1 2t
1 2 1 2
u t t u t t
u t u t t d
(t t t ) u t t t
−
− ∗ −
= τ − ⋅ − − τ τ
= − − ⋅ − −
∫
Non - delayed stepfunctions : Delayed stepfunctions :
t
u(t)∗u(t)
t = 0
τ
u(t-τ)
τ = tτ = 0
τ
u(τ)
τ = 0τ
u(τ)
τ=t1τ = 0
τ
u(t-τ)
τ = t-t2τ = 0
t
u(t-t1)∗u(t-t2)
t = t1+t2t = 0
Signal ProcessingPart 2.2 Properties of Impulse Response
of a Linearsystem
Magnus Danielsen
Contents of Part 2.2
• Properties of impulse response– Parallel and cascade connections – Memory– Causality– Stability– Inversion– Step response
Properties of Impulse Respons (1)
h1
h2
x(t) +y(t)=x∗h1 + x∗h2
=x∗(h1 + h2)
Parallel connection: h = h1 + h2
Cascade connection: h = h1 ∗ h2
h1 h2x(t) y(t)=z∗h1 = x∗h1∗h2z(t)=
Exampley = - x∗h4 + x∗(h1 + h2)∗h3
= x∗(-h4 + (h1 + h2)∗h3)x
h1
h2
+ h3
h4
++
−
1h (n) u(n)= 2h (n) u(n 2) u(n)= + −
3h (n) (n 2)= δ − n4h (n) u(n)= α ⋅
[ ]
[ ] [ ][ ]
4 1 2 3
n
n
n
h n h (h h ) h
u(n) (u(n) u(n 2) u(n)) (n 2)u n u n
(1 ) u n
= − + + ∗
= −α ⋅ + + + − ∗δ −
= −α ⋅ +
= − α ⋅
Properties of Impulse Respons (2)
• Memoryless system
• Causal system
k=
k=-
y[n] h[n] x[n]= h[k]·x[n-k] c·x[n]
h[n] c [n]
∞
∞
= ∗ =
⇒ = ⋅ δ
∑
-
y(t) h ( t ) x ( t ) = h ( )·x(t- )d cx ( t )
h ( t ) c ( t )
∞
∞
= ∗ τ τ τ =
⇒ = ⋅ δ
∫
k = k =
k = - k = 0
y [n ] h [ n ] x [ n ]= h [k ] ·x [n -k ]= h [k ] ·x [n -k ]∞ ∞
∞
= ∗ ∑ ∑
- 0
y(t) h ( t ) x ( t ) = h ( )·x(t- )d = h( )·x(t- )d∞ ∞
∞
= ∗ τ τ τ τ τ τ∫ ∫
Properties of Impulse Respons (3)
• Stable systems: Bounded In – Bounded Out: BIBOx yx[n] M y[n] M≤ < ∞ ⇒ ≤ < ∞
k=
k=-
y[n]= h[k] · x[n-k]∞
∞∑
x yy[n] h[k] · x[n-k] M h[k] <M h[k] <≤ ≤ ⇒ ∞∑ ∑ ∑
x yx(t) M y(t) M≤ < ∞ ⇒ ≤ < ∞
y(n)= h( )·x(t- )d∞
−∞
τ τ τ∫
x yy(t) h( ) · x(t- ) d M h( ) d <M h( ) d <∞ ∞ ∞
−∞ −∞ −∞
≤ τ τ τ ≤ τ τ ⇒ τ τ ∞∫ ∫ ∫
Discrete
Analogue
Examples on Stability
n
n 2 1
2
h[n] a u[n 2]1For 1 a 1: h[n] a a a BIBO stable
1 a
For a 1: h[n] unstable
∞ ∞− −
−∞ −
∞
−∞
= +
− < < = = + + ⇒−
≥ = ∞ ⇒
∑ ∑
∑
Discrete system
Analogue systemat
ata
0
a
h(t) e u(t)
e1For a 0 (i.e. e 1) : h(t) dt dt BIBO stablea
For a 0 (i.e. e 1) : h(t) dt unstable
−
−∞ ∞
−
−∞
∞−
−∞
=
> < = = ⇒
≤ ≥ = ∞ ⇒
∫ ∫
∫
Properties of Impulse Respons (4)
• Invertible systems and deconvolution – Examples:
• Equalizer in telecommunication:
• Seismic exploration
h(t)x(t) y(t)=h(t)∗x(t)
h-1(t)y(t) x(t)=h-1(t)∗y(t)
Transmitterh(t)
Receiverh-1(t)
Channelx(t) y(t)= h-1(t) ∗ h(t) ∗ x(t) = x(t)
x[n] y[n]=h[n]∗x[n]
h[n] = y[n] ∗ x-1[n]= h[n] ∗ x[n] ∗ x-1[n]
Example on Invertible Systems
h[n]x[n] y[n]= x[n] +ax[n-1]
y[n]=h[n]∗x[n]=Σh[n-k]x[k]Impulse response: h[n] = [1 a]
h-1[n]∗h[n] = Σ h-1[n-k] h[k] = δ[n]
h-1[0]·1 + h-1[-1]·a = 1 ⇒ h-1[0] = 1
h-1[1]·1 + h-1[0]·a = 0 ⇒ h-1[1] = -a
h-1[2]·1 + h-1[1]·a = 0 ⇒ h-1[2] = (-a)2
Inverse impulse response: h-1[n] = (-a)n = [1 -a (-a)2 (-a)3 (-a)4... ]
Undesired echo on data:
Step Response
h[n]u[n] s[n]=h[n]∗u[n]=Σh[n-k]u[k]
s(t) = h(t)∗u(t)=∫h(t-τ)u(τ)dτu(t)
n
k kt
s[n ] h[n ] u[n ] h[k ] u[n k ] h[k ]
s( t ) h ( t ) u ( t ) h ( ) u ( t )d h ( )d
∞
= −∞ = −∞
∞
−∞ −∞
= ∗ = ⋅ − =
= ∗ = τ ⋅ − τ τ = τ τ
∑ ∑
∫ ∫
Step response found from impulse response:
h[n ] s[n ] s[n 1]dh ( t ) s( t )d t
= − −
=
Impulse response found from step response:
Examples on Step Response
0
t
dE RI V I C V(t)dt
d(t) V(t) V(t) RC 1dt
V(0 ) (t)dt 1
Impulse response : V(t) h(t) exp( t)u(t)
Step response : s(t) h(t)dt (1 exp( t)) u(t)
+
+
−∞
−∞
= + =
δ = τ + τ = =
= δ =
= = −
= = − − ⋅
∫
∫
Analogue signal:
Discrete signal:
n
nn
0
Impulse response: h[n] = (-a) u[n] a 1
1 ( a)Step response : s[n] h[n] u[n] (-a) u[n]1 a
∞ ∞
−∞
<
− −= = ⋅ = ⋅
+∑ ∑
E=δ(t) +−
R
C
+
V=h(t)−
I
Properties of the Impulse Response
Impulse response
Step response
h [n] ∗ hinv [n] = δ[n]h(t) ∗ hinv(t) = δ(t)Invertible
Stable
h[n]=0 for n<0h(t)=0 for t<0Causal
h[n]=c⋅δ[n]h(t)=c⋅δ(t)Memoryless
Discrete systemAnalogue systemProperty \ System
h(t) dt∞
−∞< ∞∫ n
h[n]∞
=−∞< ∞∑
n
ks[n ] h[k ]
= −∞= ∑
ts ( t ) h ( )d
−∞= τ τ∫
dh ( t ) s( t )d t
= h[n ] s[n ] s[n 1]= − −
Signal ProcessingPart 2.3 Sinusoidal Response, Diff. Equations, and Block Diagrams
Magnus Danielsen
Contents for Part 2.3
Sinussoidal response– Analogue systems– Discrete systems
• Differential equations • Difference equations• Block diagrams• State variables
hx[n]=ejΩn y[n]=h[n]∗x[n]=Σh[k]x[n-k]=Σh[k] ejΩ(n-k)=ejΩn · Σh[k] e-jΩk
y(t) = h(t)∗x(t)=∫h(τ)x(t-τ)dt= ∫h(τ) ejω(t- τ)dt= ejωt · ∫h(τ) e-jωτdτx(t) =ejωt
Discrete frequency response: H(ejΩ)=Σh[k] e-jΩk =H· ejϕ
y[n] = H(ejΩ)ejnΩ = H·x[n]
Analogue frequency response: H(ejω)=∫h(τ) e-jωτdτ =H· ejϕ
y(t) = H(ejΩ)ejωt = H·x(t)
Sinusoisal Steady State Response
Example on Sinusoidal Response
y(t)h
x[n]= A cos(Ωn+θ) y[n]
x(t) = A cos(ωt+θ)
Discrete frequency response:
x[n] = A cos(Ωn+θ) = ½A (ej(Ωn+θ) + e-j(Ωn+θ))
H = H(ejΩ) =H(ejΩ)· ejϕ(exp(jΩ)) =H· ejϕ
y[n] = H(ejΩ)· ejϕ(exp(jΩ)) ·½A ej(Ωn+θ) + H(e-jΩ)· ejϕ(exp(-jΩ)) ½A e-j(Ωn+θ)
= H· A cos(Ωn+ θ + ϕ)
Analogue frequency response:
x(t) = A cos(ωt+θ) = ½A (ej(ωt+θ) + e-j(ωt+θ))
H(jω) =H(jω)· ejϕ(jω)
y(t) = H(jω)· ejϕ(jω) ·½A ej(ωt+θ) + H(-jω)· e-jϕ(jω) ½A e-j(ωt+θ)
=H· A cos(ωt+ θ + ϕ)
If h is reel ⇒ H(ejΩ)= H(e-jΩ)* og H(jω)= H(-jω)*
Example on Sinussoidal Response Obtained from Impulse Response
• h1[n] = ½(δ[n] + δ[n-1])H1(ejΩ) = ½(1 + ejΩ) = ejΩ cos(Ω/2)
• h2[n] = ½(δ[n] - δ[n-1])H2(e-jΩ) = ½(1 - e-jΩ) = j e-jΩ sin(Ω/2)
Ω-π π
H1
Ω-π π
H2
∠H1
Ω-π π-π
0
∠H1
Ω-π π-π/2
0
π/2
Example: Analogue Sinusoidal ResponsetRC
jRC
1h(t) e u(t)RC
11 RCH( j ) e u(t)e d 1RC j
RC
−
∞ τ− − ωτ
−∞
=
ω = τ =+ ω
∫
22
1RCH( j )1RC
ω = + ω
angle(H) arctan RC= − ω
-5 0 50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-5 0 5-1.5
-1
-0.5
0
0.5
1
1.5
ω
ω
Example: Discrete Sinusoidal Responsen
j j n n j n
n n 0
jj
h[n] ( a) u[n]
H(e ) h[n]e ( a) e
1H(e )1 a e
∞ ∞Ω − Ω − Ω
=−∞ =
ΩΩ
= −
= = −
=+ ⋅
∑ ∑
j
2 2
1H(e )(1 (a cos ) (a sin )
a sinangle(H) arctan1 (a cos )
Ω =+ Ω + Ω
Ω= −
+ Ω
Differential Equationsk kN M
k kk kk 0 k 0
d da y(t) b x(t)dt dt= =
=∑ ∑
y(t)
x(t) +−
R
C
L
Example 1 t
2
2
dy 1L Ry ydt xdt C
d y dy 1 dxL R ydt dt C dt
−∞
+ + =
+ + =
∫
Example 2
km
x(t)f
2
2
d y dym f ky xdt dt
+ + =
y(t)
Difference EquationsN M
k kk 0 k 0
a y[n k] b x[n k]= =
− = −∑ ∑
y[n] + y[n-1] + ¼ y[n-2] = x[n] + x[n-1]
y[n] = - y[n-1] - ¼ y[n-2] + x[n] + x[n-1]
Example
N M
k kk 1 k 0
y[n] a y[n k] b x[n k]= =
= − − + −∑ ∑
Recursive formula for y[n]
Example Using Recursive Formula
Difference equation:
y[n] = – ½ y[n – 1] + ¼ y[n – 2] + x[n] – x[n – 5]
Initial output values: y[–1] = 1 y[–2] = 2Input: x[n] = u[n]
h[n]y[n]x[n]
Output solution:y[-2] = 2 = 2.00y[-1] = 1 = 1.00y[0] = –½ ⋅ 1 + ¼ ⋅ 2 + 1 – 0 = 1.00y[1] = –½ ⋅ 1 + ¼ ⋅ 1 + 1 – 0 = 0.75y[2] = –½ ⋅ 0.75 + ¼ ⋅ 1 + 1 – 0 = 0.89y[3] = –½ ⋅ 0.89 + ¼ ⋅ 0.75 + 1 – 0 = 0.75 y[4] = –½ ⋅ 0.75 + ¼ ⋅ 0.89 + 1 – 0 = 0.85 y[5] = –½ ⋅ 0.85 + ¼ ⋅ 0.75 + 1 – 1 = – 0.24y[6] = ½ ⋅ 0.23 + ¼ ⋅ 0.85 + 1 – 1 = 0.33 y[7] = –½ ⋅ 0.33 – ¼ ⋅ 0.23 + 1 – 1 = – 0.22y[8] = ½ ⋅ 0.22 + ¼ ⋅ 0.33 + 1 – 1 = 0.15y[9] = –½ ⋅ 0.15 – ¼ ⋅ 0.22 + 1 – 1 = – 0.13y[10]= ½ ⋅ 0.13 + ¼ ⋅ 0.15 + 1 – 1 = 0.10
y[n]
n
x[n]
.....
-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
n5 7 9
-3 -2 -1 0 1 2 3 4 6 8 10
Block Diagrams - Symbolscx y=cx
+x y=x+w
w
∫x(t) y(t)= ∫t-∞ x(t)dt
Sx[n] y[n]=x[n-1]
Skx[n] y[n]=x[n-k]
Σx y=x+w
w
Block Diagram – Discrete 2nd Order
y[n] = b0x[n] + b1x[n-1] + b2x[n-2] - a1y[n-1] – a2y[n-2]
S
S
+
+
-a2
-a1
y
S
S
+
+
b2
b1
x wb0
S
S
+
+
b2
b1
x
S
S-a2
-a1
y
+
b0
S
S
+
+
b2
b1
yb0
+
+
-a2
-a1
x
w
Block Diagram – Analogue 2nd Order
∫
∫
+
+
-a0
-a1
y
∫
∫
+
+
b0
b1
x wb2
∫
∫
+
+
b0
b1
x
∫
∫
-a0
-a1
y
+
b2
∫
∫
+
+
b0
b1
yb2
+
+
-a0
-a1
x
y = b0∫∫x + b1∫x + b2x – a0∫∫y’ – a1∫y
w
a0y + a1y’ + a2y’’ = b0x + b1x’ + b2x’’
a0∫∫y + a1∫y + a2y = b0∫∫x + b1∫x + b2x
Example –Block diagram for system with given difference equation
y[n] + 1/2 ⋅ y[n-1] – 1/3 ⋅ y[n-2] = 4 ⋅ x[n] – 3 ⋅ x[n-1] + 2 ⋅ x[n-2]
S
S
+
+
2
-3
y4
+
+
1/3
-1/2
x
State Variable Description
S
S
+
+
b2
b1
y[n]b0
+
+
-a2
-a1
x[n]
q1[n]
q2[n]
1 1 1 2 2
2 1
1 1 2 2 1 1 2 2
q [n 1] a q [n] a q [n] x[n]q [n 1] q [n]y[n] x[n] a q [n] a q [n] b q [n] b q [n]
+ = − − ++ == − − + +
1 11 2
2 2
11 1 2 2
2
q [n 1] q [n]a a 1x[n]
q [n 1] q [n]1 0 0
q [n]y[n] b a b a 1 x[n]
q [n]
q[n 1] A q[n] bx[n]
y c q Dx
+ − − = + +
= − − +
+ = ⋅ +
= ⋅ +
Signal ProcessingPart 3.1 Fourier Representations of Signals
Magnus Danielsen
Fourier Representations of Signals
• Inventor: Joseph Fourier 1768 – 1830• Theory includes: Any ”wellbehaved signal” can be
represented as a continous (integral) or discrete (summation) superposition of sinusoids
• Sinussoids can be complex (mostly applied) or reel. • Complex fourier representations can be transformed into
real representations and vice verse• Fourier methods have widespread applications in signal
systems like– Communication systems– Seismic exploration systems– Audio systems – etc.
Fourier Series using Real Sinusoids
0 n 0 n 0n 1 n 1
f (t) a a cosn t b sinn t∞ ∞
= =
= + ω + ω∑ ∑T
00
1a f (t)dtT
= ∫ ( )T
n 00
2b f (t)sin n t dtT
= ω∫( )T
n 00
2a f (t)cos n t dtT
= ω∫
2 2n n n
n n n
d a barctan(a / b )
= +
ϕ =
n n n
n n n
a d cosb d sin
= ϕ= ϕ
f(t) is a periodic time functions with period T
0 n 0 nn 1
f (t) a d cos(n t )∞
=
= + ω −ϕ∑
• Eulers equations:cos x = 1/2·(ejx + e-jx) sin x = 1/2j·(ejx – e-jx)ejx = cos x + j sin x e-jx = cos x – j sin x
Fourier Series using Complex Sinusoids
0jn tn
n
f (t) c e∞
ω
=−∞
= ∑ 0
Tjn t
n0
1c f (t)e dtT
− ω= ∫
( )0 0
n n n
c ac ½ a jb for n 0=
= − ≠ ( )n n n
n n n
a c c
b j c c
∗
∗
= +
= −
Definition of Eigenvalue Formulation
( ) ( )( )
Eigenvalue formulation for the operator :
t t
t eigenfunction = eigenvalue
ψ = λ ⋅ψ
ψ =
λ
H
H
k kk
k
k
Eigenvalue formulation for the matrix :
e e
e eigenvector= eigenvalue
⋅ = λ ⋅
=λ
A
A
Linear Time Invariant (LTI) Systems –Complex Sinusoids Eigenfunctions for Analogue Systems
k kj t j tk k k
k k
x(t) a e y(t) x(t) a H(j )e∞ ∞
ω ω
=−∞ =−∞
= ⇒ = = ω∑ ∑
For LTI systems applies :
H
j t
j t
h( )
= h( ) corresponding to convolution operation
: (t) = e selected as input signal
y(t) = (t) =
e = h( )
ω
ω
τ
τ ∗
ψ
ψ
τ ∗
Impulse response :
System operator : H
Eigenfunction
Output signal : H
H j t j ( t- ) j t -j j t
- -
-j
-
e h( ) e d e h( )e d H( j )e
: H( j ) h( )e d
∞ ∞ω ω τ ω ωτ ω
∞ ∞
∞ωτ
∞
= τ ⋅ τ = τ τ = ω
λ = ω = τ τ
∫ ∫
∫Eigenvalue
H y(t) = H x(t)x(t)General LTI system:
Hy(t) = H(jω) ejωtψ(t) = ejωt
Eigenpresentation of system:
Linear Time Invariant (LTI) Systems –Complex Sinusoids Eigenfunctions for Discrete Systems
k k kj n j j nk k
k k
x[n] a e y[n] x[n] a H(e )e∞ ∞
Ω Ω Ω
=−∞ =−∞
= ⇒ = =∑ ∑
For LTI systems applies :
H
[ ][ ]
[ ][ ] [ ]
[ ] [ ]
j n
j n j n j (
h n
= h n corresponding to convolution operation
: n = e selected as input signal
y n = n =
e = h n e h k e
Ω
Ω Ω Ω
∗
ψ
ψ
∗ = ⋅
Impulse response :
System operator : H
Eigenfunction
Output signal : H
H [ ]
[ ]
n k) j n j k j j n
k=- k=-
j j k
k=-
e h k e H(e )e
: H(e ) h k e
∞ ∞− Ω − Ω Ω Ω
∞ ∞
∞Ω − Ω
∞
= ⋅ =
λ = = ⋅
∑ ∑
∑Eigenvalue
H y[n]= H x[n]x[n]General LTI system:
Hy[n] = H(ejΩ)ejΩnψ[n] = ejΩn
Eigenpresentation of system:
Fourier Representation for 4 Signal Classes
DTFT =Discrete – Time Fourier Transform
DTFS (→ *DFT) =Discrete – Time Fourier Series
Discrete
FT = CTFT = Continuous – Time Fourier Transform
FS = CTFS =Continuous – Time Fourier Series
Continuous
NonperiodicPeriodicTime property
*DFT = Discrete Fourier Transform
[ ] [ ] [ ] [ ]0 0
N 1 N 1jk n jk n
k 0 n 0
1x n X k e X k x n eN
− −Ω − Ω
= =
= =∑ ∑Discrete Time Fourier Series = DTFS (→ DFT) :
0 0
Tjn t jn t
n nn 0
1f (t) c e c f (t)e dtT
∞ω − ω
=−∞
= =∑ ∫
Continuous Time Fourier Series = FS = CTFS:
[ ] ( ) ( ) [ ]j j n j j n
n
1x n X e e d X e x n e2
π ∞Ω Ω Ω − Ω
=−∞−π
= Ω =π ∑∫
Discrete Time Fourier Transform = DTFT:
( ) ( ) ( ) ( )j t j t1x t X j e d X j x t e dt2
π πω − ω
−π −π
= ω ω ω =π ∫ ∫
Continous Time Fourier Transform = FT=CTFT:
Definitions of Fourier Representations
Orthogonality
[ ]
[ ]
0
0 0 0
0
0
jk nk 0
N 1jk n jm n j( k m ) n
km k mN N n 0
j( k m ) N
j( k m )
D iscre te func tions:
n e N 2
I e e e
N for k m1 e N k m0 for k m1 e
Ω
−Ω − Ω − Ω∗
=
− Ω
− Ω
φ = Ω = π
= φ φ = =
= −= = = ⋅ δ − ≠−
∑ ∑ ∑
[ ]
0
0 0 0
0
jk tk 0
T Tjk t jm t j( k m ) t
km k mT t 0 t 0
Tj( k m ) t
0 0
Continuous functions:
( t ) e T 2
I ( t ) ( t )dt e e dt e dt
T for k me T k m0 for k mj(k m )
ω
ω − ω − ω∗
= =
− ω
φ = ω = π
= φ φ = =
= = = = ⋅ δ − ≠− ω
∫ ∫ ∫
Discrete -Time Periodic Signalswith Period N, and DTFS
[ ] [ ]
[ ] [ ] [ ] 0 0
0
jk n jk N
N
2x n x n N cyclic frequencyN
approximate x n by x n A k e e 1Ω Ω
π= + Ω =
= =∑
[ ] [ ]
[ ] [ ] [ ] [ ] [ ]0 0
2
N
jk n jk n
N N
1Mean Square Error : MSE x n x n 0 whenN
1A k X k x n e and x n X k eN
− Ω Ω
= − =
= = =
∑
∑ ∑
0DTFS,x[n] X[k]Ω←→
Proof of DTFS[ ] [ ]
0jk N0
: x n x n N2Cyclic frequency e 1N
Ω
= +
πΩ = =
Given periodic discrete function
[ ] [ ] 0jk n
N
x n X k e Ω= ∑Consequently :
[ ] [ ] [ ]
[ ] ( ) [ ] [ ] [ ]
0 0 0
0
N 1 N 1 N 1jk n jk n ' jk n
k 0 k 0 n ' 0N 1 N 1 N 1
jk n n '
n ' 0 k 0 n ' 0
1x n X k e x n ' e eN
1 1x n ' e x n ' N n n ' x nN N
− − −Ω − Ω Ω
= = =
− − −Ω −
= = =
= =
= = ⋅ ⋅ δ − =
∑ ∑∑
∑ ∑ ∑
Proof :
[ ] [ ] 0
N 1jk n '
n ' 0
1X k x n ' eN
−− Ω
=
= ∑Define :
Definition of DFT and FFT is most often defined
as a modification of DTFS, where the factor 1/N is moved from X[k] to x[n], to be used for a finite number N of data:
Discret Fourier Transform = DFT
is defined as specificallyefficient algoritms used to calculate Fast Fourier Transform = FFT
DFT
[ ] [ ]0 00
N 1 N 1jk n jk nDFT,
k 0 n 0
1x[n] X k e X[k] x n eN
− −Ω − ΩΩ
= =
= ←→ =∑ ∑
Continuous Periodic Signalswith Period T, and FS=CTFS( ) ( )
( ) ( ) [ ] 0
0
jk t
k
2x t x t T cyclic frequencyT
approximate x t by x t A k e∞
ω
=−∞
π= + ω =
= ∑
( ) ( )
[ ] [ ] ( ) ( ) ( )0 0
T 2
0T
jk t jk t
k0
1Mean Square Error : MSE x t x t dt 0 whenT
1A k X k x t e dt and x t X eT
∞− ω ω
=−∞
= − =
= = = ω
∫
∑∫
( ) 0FS,x t X[k]ω←→
Proof of FS( ) ( )
0jk T0
: x t x t T2Cyclic frequency e 1T
ω
= +
πω = =
Given periodic continuous function
( ) [ ] [ ] ( )0 0
Tjk t jk t
k 0
1ˆx t X k e and X k x t e dtT
∞ω − ω
=−∞
= =∑ ∫Consequently :
[ ] ( ) [ ]
[ ] ( ) [ ] [ ] [ ]
0 0 0
0
T Tjk t jk ' t jk t
k '0 0T
j k ' k t
k ' k '0
1 1X k x t e dt X k ' e e dtT T
1 1X k ' e dt X k ' T k ' k X kT T
∞− ω ω − ω
=−∞
∞ ∞− ω
=−∞ =−∞
= =
= ⋅ = ⋅ ⋅ δ − =
∑∫ ∫
∑ ∑∫
Proof :
[ ] ( ) [ ] 0jk t
k
X k such that x t X k e∞
ω
=−∞
= ∑Define :
Signal ProcessingPart 3.2 Discrete Time (DTFS) and Continuous Time (FS) Fourier Series
Magnus Danielsen
Example on DTFS[ ] 0
2x n cos( n ) Period N 16 cyclic frequency8 16 8π π π
= + ϕ = Ω = =
[ ] [ ]
[ ]
8j n j n jk nj j8 8 8
7
j
j
x n ½(e e e e ) X k e
½e for k 1X k ½e for k 1
0 for k 1, 1
π π π−ϕ − ϕ
−
− ϕ
ϕ
= + =
= −= = ≠ −
∑
-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18k
X[k]½
∠X[k]
-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18k
ϕ
-ϕ
Example on DTFS for Periodic Square WavePeriod = N cyclic frequency =Ω0
[ ] [ ] [ ]0 0 0
00
0
n M n M 2Mjk n jk M jk (n M)
n M n M 0
0jk (2M 1)
jk M 0jk
1 1X k x n e e x n eN N
1 sin(k (M ½) for k 0, N, 2N,...N sin(½k )1 1 ee
N 1 e 2M 1 for k 0, N, 2N,...N
= + =− Ω Ω − Ω +
=− + =
− Ω +Ω
− Ω
= =
Ω + ≠ ± ± Ω− = = − + ≠ ± ±
∑ ∑
n
x(n)
-M M
··· ··· ···N-M
···N
···
X[k]
k
N=64
M=5
Discrete Time Fourier Serieswith Harmonically Related Cosines – Derivation
[ ] [ ] [ ] [ ]0 0jk n jk n
N N
1x n X k e X k x n eN
Ω − Ω= =∑ ∑
[ ] [ ] [ ] [ ] [ ]
[ ]
[ ][ ]
[ ]
0 0 0
N N 12 2
jk n jk n jk n j n
N k 1k 12
N 12
0k 0
Nx n X k e X 0 X k e X k e X e2
B k cos(k n)
NX k for k 0,2B kN2X k for k 1,2,.., 12
−
Ω Ω − Ω − π
==− +
−
=
= = + + − +
= Ω
== = −
∑ ∑
∑
[ ] [ ] 0Even symmetry : X k X k Periodicity :N 2 N is even= − Ω = π
Discrete Time Fourier Series with Harmonically Related Cosines
[ ] [ ] 0Even symmetry : X k X k Periodicity :N 2 N is even= − Ω = π
[ ] [ ] [ ] [ ]0 0jk n jk n
N N
1x n X k e X k x n eN
Ω − Ω= =∑ ∑
[ ] [ ] [ ][ ]
[ ]
N 12
0k 0
NX k for k 0,2x n B k cos(k n) B kN2X k for k 1,2,.., 12
−
=
== Ω = = −
∑
[ ] [ ] 0Even symmetry : X k X k Periodicity :N 2 N is odd= − Ω = π
[ ] [ ] [ ][ ]
[ ]
N 12
0k 0
X k for k 0x n B k cos(k n) B k N 12X k for k 1,2,..,
2
−
=
== Ω = −
=∑
Discrete Time Fourier Series with Harmonically Related Sines
[ ] [ ] 0Odd symmetry : X k X k Periodicity :N 2 N is even= − − Ω = π
[ ] [ ] [ ] [ ]0 0jk n jk n
N N
1x n X k e X k x n eN
Ω − Ω= =∑ ∑
[ ] [ ] [ ] [ ]N 12
0k 1
Nx n B k sin(k n) B k 2 jX k for k 1,2,.., 12
−
=
= Ω = = −∑
[ ] [ ] 0Odd symmetry : X k X k Periodicity :N 2 N is odd= − − Ω = π
[ ] [ ] [ ] [ ]N 12
0k 1
N 1x n B k sin(k n) B k 2 jX k for k 1,2,..,2
−
=
−= Ω = =∑
Fourier Transform – FS
[ ] ( ) 0
Tjk t
0
1X k x t e dtT
− ω= ∫
( ) ( ) 02x t x t T cyclic frequencyTπ
= + ω =
( ) ( ) 0jk t
n
x t X k e∞
ω
=−∞
= ∑
( ) [ ]
[ ] [ ] [ ] [ ]
0 0 0
0
T Tjk t jm t jk t
m0 0T
j(m k) t
m m0
1 1Proof: x t e dt X m e e dtT T
1X m e dt X m m k X kT
∞− ω ω − ω
=−∞
∞ ∞− ω
=−∞ =−∞
=
= = δ − =
∑∫ ∫
∑ ∑∫
Example – Cosine Functionj j t j j t4 2 4 23 3x(t) 3cos( t ) e e e e
2 4 2 2
π π π π− −π π
= + = +
[ ]X k
k-1 10
32
32
k
[ ]X k∠
-1 10
4π
−
4π
Example – Sine Functions
j3 j2 t j3 j2 t j3(2 t ) j3(2 t )
x(t) 2sin(2 t 3) sin(6 t)j jje e je e e e2 2
− π − π π − π
= π − + π
= − + − +
[ ]X k
k-1 10 2 3-23
½
1
k
[ ]X k∠
-1 102π
−
32π
− −
1
32π
+
2π
−
-2-3
Example: FS of Square Wavex(t)
t-Ts Ts T T+TsT-Ts-T -T+Ts-T-Ts
[ ] ( )s
s
sT0 s s s s
T
Tsin(k2 )1 sin(k T ) 2T sin( u) 2T TTX k x t dt sinc(u) u=2T k k T u T T−
πω π= = = = =
π π π∫
- 5 0 - 4 0 - 3 0 - 2 0 - 1 0 0 1 0 2 0 3 0 4 0 5 0- 0 . 0 4
- 0 . 0 2
0
0 . 0 2
0 . 0 4
0 . 0 6
0 . 0 8
0 . 1
0 . 1 2
0 . 1 4
FS Approximation of Square Wave Using Finite Number of Terms
Haykin et al.
RC Circuit with Square Wave Input+
y(t)
−
x(t) +−
R
C
L
01/ RCH( j ) k
j 1/ RCω = ω = ω
ω+
[ ] [ ]0
s
0
X k
Tsin(k2
1/RCY kjk 1/RC
1/RCjk 1/RC
)T
kksin10
j2 k 10
( )2k
π
= ⋅ω +
= ⋅ω +
π
ππ
= ⋅+ π
x(t) ↔X[k] is a rectangular per.pulse train with amplitude = 1 RC=0.1 s Ts/T=1/4 ω0 = 2π
-1 -0 .5 0 0 .5 10
0 .2
0 .4
0 .6
0 .8
1
( )y t
Time (s)
-2 0 -1 5 -1 0 -5 0 5 1 0 1 5 2 0-3
-2
-1
0
1
2
3
[ ]( )a rg Y k
k
-2 0 -1 5 -1 0 -5 0 5 1 0 1 5 2 0-0 .2
0
0 .2
0 .4
0 .6
[ ]Y k
k
Signal ProcessingPart 3.3 Discrete Time Non – Periodic Signals
Magnus Danielsen
Discrete-time Nonperiodic Signals (1)DTFT
[ ][ ] [ ] [ ]
[ ] [ ]M
Non-periodic signal: x n for n
Define a periodic signal: x n x n+p N x n in period M n MLength of period : N 2M 1 (p is integer)
x n lim x n→∞
− ∞ ≤ ≤ ∞
= ⋅ = − ≤ ≤
= +
=
[ ] [ ]
[ ] [ ] [ ]
0
0 0
Mjk n
0M
M Mjk n jk n
M M
2DTFS for per. signal : x n X k e2M 1
1 1X k x n e x n e2M 1 2M 1
Ω
−
− Ω − Ω
− −
π= Ω =
+
= =+ +
∑
∑ ∑
[ ]Non-periodic x n
[ ]Periodic x n − M M
0n..... .....
0n..... .....
( ) [ ] [ ]0
0 0
Mjk nj j t
M M
2D efine : k d2 M 1
X e lim x n e x n e∞
− ΩΩ − Ω
→ ∞− −∞
πΩ = Ω Ω = Ω =
+
= =∑ ∑
[ ] [ ] [ ]
[ ]
( ) ( )
0
0 0
Mjk n
M M k MM M
jk n ' jk n
M k M n ' M
j j n j j n
x n lim x n lim X k e
1lim x n ' e e2M 1
d 1X e e X e e d2 2
Ω
→∞ →∞=−
− Ω Ω
→∞=− =−
π πΩ Ω Ω Ω
−π −π
= =
=+
Ω = = Ω π π
∑
∑ ∑
∫ ∫
Discrete-time Nonperiodic Signals (2) DTFT
DTFT Representation
[ ] ( ) ( ) [ ]j j n j j n1x n X e e d X e x n e2
π ∞Ω Ω Ω − Ω
−∞−π
= Ω =π ∑∫
( )DTFT, jx[n] X eΩ Ω←→
( )j
x[n] is non periodic and discrete in n
X e is periodic and continuous inΩ
• −
• Ω
Example on Exponential Sequence (1)
[ ] [ ]
( ) [ ]
n
j n j n j nj
0
x n u n
1X e u n e ( e ) for 11 e
∞ ∞Ω − Ω − Ω
− Ω−∞
= α
= α = α = α <− α∑ ∑
( )( )( ) ( )
j½ ½2 22 2
1 1X e1 2 cos1 cos sin
Ω = =α + − α Ω−α Ω + α Ω
( )j sinX e arctan1 cos
Ω α Ω ∠ = − − α Ω
( ) ( ) ( )jj X ej jX e X e eΩ∠Ω Ω=
Example on Exponential Sequence (2)
- 1 5 - 1 0 - 5 0 5 1 0 1 50
0 . 2
0 . 4
0 . 6
0 . 8
1
1 . 2
1 . 4
1 . 6
1 . 8
2
- 1 5 - 1 0 - 5 0 5 1 0 1 5- 0 . 8
- 0 . 6
- 0 . 4
- 0 . 2
0
0 . 2
0 . 4
0 . 6
2π-2π
2π-2π 0
0
( )
( )
jΩ
½2
X e
1=α +1-2αcosΩ
( )jX e
sinarctan1 cos
Ω∠
α Ω = − − α Ω
α = 0.5
α = 0.5
Example: Rectangular Pulse (1)
( )j ( 2 M 1 )
j MMj j n j n
M
s in ( 2 M 1)1 e 2e 0, 2 , 4 .....
X e 1 e 1 e sin2
2 M 1 0, 2 , 4 .....
− Ω +− Ω
Ω − Ω − Ω
−
Ω + − = Ω ≠ ± π ± πΩ= ⋅ = −
+ Ω = ± π ± π
∑
M = 10
n
[ ]1 for n 1
x n0 for n 1
≤= >
Example: Rectangular Pulse (2)
-1 5 -1 0 -5 0 5 1 0 1 5-5
0
5
1 0
1 5
2 0
2 5
M = 10( )jX e Ω
Ω
2π2− π 0
Example: Square Frequency Function – Discrete Sinc Time Function
-2 -1.5 -1 -0.5 0 0.5 1 1.5 20
0.2
0.4
0.6
0.8
1
1.2
1.4
Ω
2π
2π
− W− W
( )j 1 for WX e
0 for WΩ Ω ≤
= Ω >[ ] ( )sin Wn W Wnx n sinc( )
n= =
π π π
-40 -30 -20 -10 0 10 20 30 40-0.04
-0.02
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
n
π/W
DTFT for x[n]=δ[n] and x[n]=1
( )DTFT j j n
n
x[n] [n] X e [n]e 1∞
Ω Ω
=−∞
= δ ←→ = δ =∑
( ) ( ) ( )DTFTj j n1 1X e x[n] e d2 2
πΩ Ω
−π
= δ Ω ←→ = δ Ω Ω =π π∫
1n
1Ω
Ω
1
Ω
Example: Sinusoidal Spectrum for a Discrete Non-periodic Signal (1)
[ ] ( )
j
j n j2 j2 j n
j (2 n) j ( 2 n)
Discrete Time Fourier Transform : X(e ) 2cos2
1 1x n 2cos2 e d e e e d2 2
1 1 1e e2 j(2 n) j( 2 n)
1 for n 21 for n 20 for n 2
Ω
π πΩ Ω − Ω Ω
−π −π
πΩ + Ω − +
−π
= Ω
= Ω ⋅ Ω = + ⋅ Ωπ π
= + π + − +
= −= = + ≠ ±
∫ ∫
Example: Sinusoidal Spectrum for a Discrete Non-periodic Signal (2)
-5 0 50
0.5
1
1.5[ ]Discrete time function x n
n
-4 -3 -2 -1 0 1 2 3 40
0.5
1
1.5
2
2.5
3
3.5
jAngle : X(e )Ω∠
Ω
-4 -3 -2 -1 0 1 2 3 40
0.5
1
1.5
2
jMagnitude : X(e ) 2 cos2Ω = Ω
Ω
Signal ProcessingPart 3.4 Continuous – Time Nonperiodic Signals
Magnus Danielsen
Continuous-time Nonperiodic Signals (1)FT = CTFT
( )
( ) ( )
( ) ( )T
Non-periodic signal: x t for tT TDefine a periodic signal: x t x t period t2 2
x t lim x t→∞
− ∞ ≤ ≤ ∞
= − ≤ ≤
=
( ) [ ]
[ ] ( ) ( )
0
0 0
jk t0
T T2 2
jk t jk t
T T2 2
2FS for per. signal : x t X k eT
1 1X k x t e dt x t e dtT T
∞ω
−∞
− ω − ω
− −
π= ω =
= =
∑
∫ ∫
( )N on-periodic x t
t..... .....
( )Periodic x t − T/2 T/2
0t..... .....
( ) ( ) ( )0
0 0
T2
jk t j t
TT2
2D efine : k dT
X j lim x t e d t x t e d t∞
− ω − ω
→ ∞−∞−
πω = ω ω = ω =
ω = =∫ ∫
Discrete-time Nonperiodic Signals (2) DTFT
( ) ( ) [ ]
[ ]
( ) ( )
0
0 0
jk t
T T k
T / 2jk t ' jk t
T k T / 2
j t j t
x t lim x t lim X k e
1lim x t ' e dt ' eT
d 1X j e X j e d2 2
∞ω
→∞ →∞=−∞
∞− ω ω
→∞=−∞ −
∞ ∞ω ω
−∞ −∞
= =
=
ω = ω = ω ω π π
∑
∑ ∫
∫ ∫
FT Representation
[ ] ( )FT,x t X jω←→ ω
( )( )
x t is non periodic and continuous in t
X j is non periodic and continuous in
• −
• ω − ω
( ) ( ) ( ) ( )j t j t1x t X j e d X j x t e d t2
∞ ∞ω − ω
−∞ −∞
= ω ω ω =π ∫ ∫
Example on Exponential Decay
( ) ( )at j t 1X e u t e dta j
∞− − ω
−∞
ω = =+ ω∫
( )2 2
1X(a )
ω =+ ω
( )X j arctanaω
∠ ω = −
-1 0 1 2 3 4 50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
( ) ( )atx t e u t−=
Time t
a=
0.5
1
2-3 -2 -1 0 1 2 3
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
( )X ω
Cyclic frequency ω
a=0.5
1
2
-3 -2 -1 0 1 2 3-1.5
-1
-0.5
0
0.5
1
1.5
( )X j∠ ω
Cyclic frequency ω
a=
0.5
1
2
Rectangular Pulse
-20 -15 -10 -5 0 5 10 15 20-0.5
0
0.5
1
1.5
2
Cyclic frequency ω
Tπ
2Tπ 3
Tπ
Tπ
−
-3 -2 -1 0 1 2 30
0.2
0.4
0.6
0.8
1
1.2
1.4
Time t
-T T
T=1
( ) ( )TT
j t j t j t
T T
1 2 TX j x(t)e dt e dt e sin T 2Tsincj
∞− ω − ω − ω
−∞ − −
ω ω = = = = ω = − ω ω π ∫ ∫
Rectangular SpectrumWπ
( ) ( )WW
j t j t j t
W W
1 1 1 1 W Wtx(t) X j e dt e dt e sin Wt sinc2 2 j t
∞ω ω ω
−∞ − −
= ω = = = = π π ω π π π ∫ ∫
FT for an Impulse
FT for a DC signal
( ) ( )FT j tt t e dt 1∞
− ω
−∞
δ ←→ δ =∫
( ) ( ) ( )FT j t1 1x t e d2 2
∞ω
−∞
δ ω ←→ = δ ω ω =π π∫
( )FT1 2←→ πδ ω
FT of Ramp Function (1)
[ ]t t 1
x(t) t u(t 1) u(t 1)0 t 1
≤= = ⋅ + − − >
( ) ( )
11 1j t j t j t
1 1112
j t j t
1
2
1 1X( ) t e dt t e e dtj j
1 1t e ej j
2 2j cos j sin
− ω − ω − ω
− −−
− ω − ω
−
ω = ⋅ = ⋅ − − ω − ω
= ⋅ − − ω − ω
= ω − ωω ω
∫ ∫
x(t)
t1
-1
-1
1
FT of Ramp Function (2)
-20 -15 -10 -5 0 5 10 15 20-100
-80
-60
-40
-20
0Magnitude:
Cyclic frequency ω
20 lo
g |X
(ω)|
(dB
)
-20 -15 -10 -5 0 5 10 15 20-2
-1.5
-1
-0.5
0
0.5
1
1.5
2Phase:
∠X
(ra
dian
)
Cyclic frequency ω
( ) ( )2
X( )2 2j cos j sin
ω =
ω − ωω ω
Binary Phase Shift Keying (BPSK) Digital Communication Signal (1)
Single rectangular pulse shape:
Single raised cosine pulse shape:
0
0
½T2 2
r0 ½T
1Power : P x (t) dt AT −
= =∫
0
0
½T2 2
c0 ½T
1 3Power : P x (t) dt AT 8−
= =∫
t-½T0 ½T0
Axr(t)
-0 .5 0 0 .50
0 .5
1
t-½T0 ½T0
Axc(t)
0r
0
A t ½Tx (t)
0 t ½T≤
= ≥
0 0r
0
½A(1 cos(2 t / T )) t ½Tx (t)
0 t ½T+ π ≤
= ≥
Raised cosine pulse shaped BPSK
0 2 4 6 8 10-1
-0.5
0
0.5
1
t
1 0 1 1 0 1 0 0 1
T0
Rectangular pulse shaped BPSK1 0 1 1 0 1 0 0 1
tT0
Binary Phase Shift Keying (BPSK) Digital Communication Signal (2)
-10 -5 0 5 10-150
-100
-50
0Rectangular pulse, P=1 ⇒ A=1:
Raised cosine pulse, P=1 ⇒ A=(8/3)½ :
Power spectrum:
20⋅lo
g |X
(f)|
(d
B)
Rectangular pulse
Raised cosine pulse
Frequency f (unit = 1/T0)
P=1 ( )
( )
0
0
T / 2j2 ft
rT / 2
0
X ' jf 1 e dt
sin fTf
− π
−
= ⋅
π=
π
∫
( )
( )
( )( )( )( )( )( )
0
0
T / 2j2 ft
c 0T / 2
0
0 0
0
0 0
0
1 8X ' jf (1 cos(2 t / T )e dt2 3
sin fT23 f
sin f 1/ T T20.53 f 1/ T
sin f 1/ T T20.53 f 1/ T
− π
−
= + π
π=
ππ −
+ ⋅π −
π ++ ⋅
π +
∫
Signal ProcessingPart 3.5 Basic Properties of Fourier Representations (1)
Magnus Danielsen
Basic Properties of Fourier Representations
• 4 Fourier Representations• Periodicity• Linearity• Symmetry• Time-shift• Frequency-shift• Scaling• Differentiation/ differencing in time• Differentiation/ differencing in frequency• Integration/summation
Frequencyproperty
Continuous Discrete
Periodic
Non-periodic
DTFT =Discrete – Time Fourier Transform
DTFS = DFT =Discrete – Time Fourier Series
Discrete
FT = CTFT = Continuous – Time Fourier Transform
FS = CTFS = Continuous – Time Fourier Series
Conti-nuous
NonperiodicPeriodicTime property
[ ]0FS,x(t) X kω→ [ ]FTx(t) X j→ ω
[ ] [ ]0DTFS;x n X kΩ→ [ ] [ ]DTFTx n X k→
Linearityx X y Y↔ ↔
z ax by Z aX bY= + ↔ = +
[ ] 1Z k 3sin k sin k2k 4 2
π π = + π
[ ] 1Y k sin kk 4
π = π
[ ] 1X k sin kk 2
π = π
Example with Fourier series:
x(t)
t-1/8 1/8 11
y(t)
t1/4-1/4
z(t)=3/2·x(t)+1/2·y(t)
t
Linearity and Decomposition - FT
( ) ( )( )
( )
( )
( ) ( )( )
( )( )
M NM kpkp
k 0p 0N
q 0
B jf j for M Nb j A jB j
X jA j B ja j for M N
A j
−
==
=
ω⋅ ω + ≥ω ωω ω = = = ω ωω < ω
∑∑
∑
( ) ( )( )
N Pk,pk
k pk 1 p 2 fork k
P mult.poles
DB j CFor M N f 0 X j ( )A j j d ( j d )= =
−
ω< ⇒ = ω = = +
ω ω− ω−∑ ∑
( ) ( ) ( )k
FT FT FT kk
d dt 1 t j t ( j )dt dt
δ ←→ δ ←→ ω δ ←→ ω
( ) ( )k kd t d tFT FTpp
k k
1 (p 1)!x t e x t t ej d ( j d )
− − −= ←→ = ⋅ ←→
ω− ω−
Linearity and Decomposition - DTFT
( ) ( )( )
( )
( )
( ) ( )( )
( )( )
jM N kM jpj k jj p k 0p 0jNj jqj
qjq 0
B ef e for M Nb e A eB e
X eA e B ea e for M N
A e
− Ω−− Ω− Ω
− Ω− Ω==Ω
− Ω − Ω− Ω
− Ω=
⋅ + ≥
= = =
<
∑∑
∑
( ) ( )( )
j N Pk,pj k
k j j pjk 1 p 2 fork k
P mult.poles
B e DCFor M N f 0 X e ( )1 d e (1 d e )A e
− Ω− Ω
− Ω − Ω− Ω= =
−
< ⇒ = = = +− −∑ ∑
[ ] [ ] [ ] 0j nDTFT DTFT DTFTj0n 1 n 1 e n n e− Ω− Ωδ ←→ δ − ←→ δ − ←→
[ ] [ ] ( )
[ ] [ ] ( )
DTFTn jj
p 1DTFTp n j
j p
1x n d u n X e1 de(p 1)! ( d)x n n d u n X e(1 de )
Ω− Ω
−Ω
− Ω
= ←→ =−
− ⋅ −= ←→ =
−
Examples on Decomposition
( ) 2
1 2
j 1X j( j ) 5j 6C C 1 2j 2 j 3 j 2 j 3
ω+ω =
ω + ω+−
= + = +ω+ ω+ ω+ ω+
( ) j2t j3tx t e 2e− −= − +FT←→
FT with 2 different poles
( )j
j
j 2 j
1 2
j j
j j
5 e 56X e 1 1(e ) e 1
6 6C C1 11 e 1 e2 3
4 11 11 e 1 e2 3
− Ω
Ω
− Ω − Ω
− Ω − Ω
− Ω − Ω
− +=− + +
= ++ −
= ++ − [ ] [ ] [ ]
n n1 1x n 4 u n u n2 3
= − +
DTFT←→
DTFT with 2 different poles
Symmetry for Real Valued Time Signals
( ) ( ) j tFT for non periodic signal x(t) : X j x t e dt∞
− ω
−∞
− ω = ∫
( ) [ ] ( ) 0jk t0
T
2 1FSfor per.signal x t , : X k x t e dtT T
− ωπω = = ∫
[ ] ( ) [ ]j j nDTFT for non periodic signal x n : X e x n e∞
Ω − Ω
−∞
− =∑
[ ] [ ]X k X k∗ = −
[ ] [ ]X k X k∗ = −
j jX e X e∗ Ω − Ω =
[ ] [ ]X j X j∗ ω = − ω
[ ] [ ] [ ] 0jk n0
N
2 1DTFS for periodic x n , : X k x n eN N
− ΩπΩ = = ∑
Symmetry for Imaginary Valued Time Signals
( ) ( ) j tFT for non periodic signal x(t) : X j x t e dt∞
− ω
−∞
− ω = ∫
( ) [ ] ( ) 0jk t0
T
2 1FSfor per.signal x t , : X k x t e dtT T
− ωπω = = ∫
[ ] ( ) [ ]j j nDTFT for non periodic signal x n : X e x n e∞
Ω − Ω
−∞
− =∑
[ ] [ ]X k X k∗ = − −
[ ] [ ]X k X k∗ = − −
j jX e X e∗ Ω − Ω = −
[ ] [ ]X j X j∗ ω = − − ω
[ ] [ ] [ ] 0jk n0
N
2 1DTFS for periodic x n , : X k x n eN N
− ΩπΩ = = ∑
Even and Odd Real Valued Function Symmetries
( ) ( ) j tFT for non periodic signal x(t) : X j x t e dt∞
− ω
−∞
− ω = ∫
( ) [ ] ( ) 0jk t0
T
2 1FSfor per.signal x t , : X k x t e dtT T
− ωπω = = ∫
[ ] [ ] [ ] 0jk n0
N
2 1DTFS for periodic x n , : X k x n eN N
− ΩπΩ = = ∑
[ ] ( ) [ ]j j nDTFT for non periodic signal x n : X e x n e∞
Ω − Ω
−∞
− =∑
Even valued real functions x(-t)=x(t) and x[-n]= x[n]: X* = X ⇒ ImX=0
Odd valued real functions x(-t)=x(t) and x[-n]= -x[n]: X* = - X ⇒ ReX=0
Time - Shift
( ) ( )
( )
0
0 0
j t0 t 0
j ( t t ) j t
FT x(t t ) : X j x t t e dt
x t e dt e X( j )
∞− ω
−∞
∞− ω + − ω
−∞
− ω = −
= = ω
∫
∫
( ) [ ] ( )
( ) [ ]
0
0
0 0 0 0
jk t0 t 0
T
jk ( t t ) jk t
T
1FS x t t : X k x t t e dtT
1 x t e dt e X kT
− ω
− ω + − ω
− = −
= =
∫
∫
[ ] [ ] [ ]
[ ] [ ]
0
0
0 0 0 0
jk n0 n 0
N
jk ( n n ) jk n
N
1DTFS x n n : X k x n n eN
1 x n e e X kN
− Ω
− Ω + − Ω
− = −
= =
∑
∑
[ ] ( ) [ ]
[ ] ( )
0
0 0
j j n0 n 0
j ( n n ) j n j
DTFT x n n : X e x n n e
x n e e X e
∞Ω − Ω
−∞
∞− Ω + − Ω Ω
−∞
− = −
= =
∑
∑ ( )0j nDTFT j0x[n n ] e X e− Ω Ω− ←→
[ ]0 00 jk nDTFS,0x[n n ] e X k− ΩΩ− ←→
( ) [ ]0 00 jk tFS,0x t t e X k− ωω− ←→
( ) ( )0j tFT0x t t e X j− ω− ←→ ω
Example on Time Shift
0 2T
z(t)
t1
-T T
x(t)
t1
0
( ) FT 2x t sin( T)←→ ωω
( ) ( ) ( )FT j T
j T
z(t) x t T Z j e X j2e sin( T)
− ω
− ω
= − ←→ ω = ω
= ωω
- 4 - 3 - 2 - 1 0 1 2 3 4- 5
0
5
1 0
1 5
2 0
- 4 - 3 - 2 - 1 0 1 2 3 40
2
4
6
8
1 0
1 2
1 4
1 6
1 8
2 0
- 4 - 3 - 2 - 1 0 1 2 3 4- 4
- 3
- 2
- 1
0
1
2
3
4
- 4 - 3 - 2 - 1 0 1 2 3 40
0 . 5
1
1 . 5
2
2 . 5
3
3 . 5
X(ω) Z(ω)= X(ω)
∠ Z(ω)∠ X(ω)
π π
-π0Cyclic frequency ω Cyclic frequency ω
Signal ProcessingPart 3.6 Basic Properties of Fourier Representations (2)
Magnus Danielsen
Frequency Shift( )FTx(t) X j←→ ω
( ) ( )
( ) ( )
FT
j t j( ' ) t j t
Z j X j( )
1 1z(t) X j( ) e d X j ' e d ' e x(t)2 2
∞ ∞ω ω +γ γ
−∞ −∞
ω = ω− γ ←→
= ω− γ ω = ω ω =π π∫ ∫
[ ] [ ][ ]
[ ] ( )( )
0 0 0
0 0 0
jk n DTFS,0
jk t FS,0
DTFTj t j( )
FTj t
e x n X k k
e x(t) X k k
e x n X e
e x(t) X j( )
Ω Ω
ω ω
Γ Ω−Γ
γ
⋅ ←→ −
⋅ ←→ −
⋅ ←→
⋅ ←→ ω− γ
Example on Frequency Shift for FT
j10tj10te for t
z(t) e x(t)0 for t
≤ π = = > π
1 for tx(t)
0 for t≤ π
= > π
( ) ( ) ( )2X j sin 2 sincω = ωπ = π ωω
( ) ( )
( )
2Z j sin ( 10)10
2 sinc 10
ω = ω− πω−
= π ω−
-3 -2 -1 0 1 2 30
0 .2
0 .4
0 .6
0 .8
1
1 .2
1 .4
Time t-π π
1
0
x(t)
-1 5 -1 0 -5 0 5 1 0 1 5-2
-1
0
1
2
3
4
5
6
7
X(jω) Z(jω)
Cyclic frequency ω
-3 -2 -1 0 1 2 3-1 .5
-1
-0 .5
0
0 .5
1
1 .5
-π π
1
0Time t
Rez(t)
Example on Frequency Shift for DTFT
[ ] [ ] [ ]j n j nn4 4z n e u n e x nπ π
− −= α =
[ ] [ ] ( )DTFTn jj
1x n u n X e1 e
ΩΩ= α ←→ =
−α
( ) ( )j j( / 4)j( / 4)
1Z e X e1 e
Ω Ω+πΩ+π= =
− α
Scaling
( )jFT j t a1 1z(t) x(at) Z j x(at)e dt x( )e d X j
a a a
∞ ∞ τ− ω− ω
−∞ −∞
ω = ←→ ω = = τ τ = ∫ ∫
Example on rectangular pulse: a=½
-3 -2 -1 0 1 2 30
0 .2
0 .4
0 .6
0 .8
1
1 .2
1 .4
Time t-1 1
1
0
( ) ( )2X j sinω = ωω
( ) ( )1 2 2Z j sin sin 2½ /½ ½
ω ω = = ω ω ω
x(t)
- 3 - 2 - 1 0 1 2 30
0 . 2
0 . 4
0 . 6
0 . 8
1
1 . 2
1 . 4
Time t-2 2
1
0
z(t)=x(½t)
( )FTx(t) X j←→ ω
-20 -15 -10 -5 0 5 10 15 20-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
Scaling for Analogue Time Functions
( )FTx(t) X j←→ ω
( )jFT j t a1 1z(t) x(at) Z j x(at)e dt x( )e d X j
a a a
∞ ∞ τ− ω− ω
−∞ −∞
ω = ←→ ω = = τ τ = ∫ ∫
[ ] [ ]00 jka tFS,a
T / a
az(t) x(at) Z k x(at)e dt X kT
ωω= ←→ = =∫
a = any real number
[ ]0FS,x(t) X jω←→ ω
Continous Time Fourier Transform
Continous Time Fourier Series
Scaling for Discrete Time Signalsp = any integer number
[ ] [ ] [ ]
( )
DTFTz
j j nz
n
j n jp p
z zn
z n x p n x p n
Z e x [pn]e
x [n]e X (e )
∞Ω − Ω
=−∞
Ω Ω∞ − −
=−∞
= ⋅ = ⋅ ←→
=
= =
∑
∑
[ ] ( )DTFT jx n X e Ω←→
Discrete Time Fourier Transform
[ ] [ ] [ ] [ ] [ ]0DTFS,pz zz n x pn x pn Z k pX kΩ= = ←→ =
[ ] [ ]0DTFS,x n X kΩ←→Discrete Time Fourier Series
n0
z[n]
n
0
x[n]
0
nxz [n]
Differentiation in Time
( ) ( ) ( ) ( )j t j t1x t X j e d X j x t e dt2
∞ ∞ω − ω
−∞ −∞
= ω ω ω =π ∫ ∫
( ) ( ) ( )F Tj td 1x t X j ( j ) e d ( j ) X jd t 2
∞ω
−∞
= ω ⋅ ω ⋅ ω ←→ ω ωπ ∫
( ) ( )FTd x t ( j )X jdt
= ←→ ω ω
[ ] [ ]0 0
Tjk t jk t
n 0
1x(t) X k e X k x(t)e dtT
∞ω − ω
=−∞
= =∑ ∫
Continuous Time Fourier Series, FS
Continuous Time Fourier transform, FT
( ) [ ]0FS,0
d x t jk X kdt
ω= ←→ ω
Differencing in Time
[ ] [ ] [ ] [ ] [ ]n
kj j j j
y n x k x n y n y n 1
X(e ) Y(e ) e Y(e )=−∞
Ω Ω − Ω Ω
= = − −
= −
∑
[ ] DTFT j j jx n X(e ) (1 e )Y(e )Ω − Ω Ω←→ = −
Discrete Time Fourier transform, DTFT
[ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ]0
n
kjk
y n x k x n y n y n 1
X k Y k e Y k=−∞
− Ω
= = − −
= −
∑
[ ] [ ] [ ]00 jkDTFS,x n X k (1 e )Y k− ΩΩ←→ = −
Discrete Time Fourier Series, DTFS
Differentiation in Frequency
( ) ( ) ( ) ( )j t j t1x t X j e d X j x t e d t2
∞ ∞ω − ω
−∞ −∞
= ω ω ω =π ∫ ∫
( ) ( ) ( )FTj td X j x t ( jt)e dt jt x td
∞− ω
−∞
ω = − ←→ − ⋅ω ∫
( ) ( )FTd X j jt x td
ω ←→− ⋅ω
Continuous Time Fourier transform, FT
[ ] DTFT jdjn x n X(e )d
Ω− ⋅ ←→Ω
Discrete Time Fourier transform, DTFT
( ) [ ] ( ) [ ]j j n j j n
n n
dX e x n e X e jn x n ed
∞ ∞Ω − Ω Ω − Ω
=−∞ =−∞
= = − ⋅Ω∑ ∑
Differencing in Frequency
[ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ]0
k
mjn
Y k X k X k Y k X k 1
x n y n e y n=−∞
Ω
= = − −
= −
∑
[ ] [ ] [ ] [ ] [ ]0 0jn DTFS,x n (1 e )y n X k Y k X k 1Ω Ω= − ←→ = − −
Discrete Time Fourier Series, DTFS
Differencing in Frequency
[ ] [ ]0 0
Tjk t jk t
k 0
1x( t) X k e X k x( t)e dtT
∞ω − ω
=−∞
= =∑ ∫
Continuous Time Fourier Series, FS
[ ] [ ] [ ] [ ] [ ]
( ) ( ) ( )0
k
mj t
Y k X m X k Y k Y k 1
x t y t e y t=−∞
ω
= = − −
= −
∑
( ) [ ] [ ] [ ]0 0j t FS,x(t) (1 e )y t X k Y k Y k 1ω ω= − ←→ = − −
Integration
( ) ( ) ( ) ( ) ( )t
FT 1y(t) x d Y j X j X j0j
where first term equals zero for 0 −∞
= τ τ←→ ω = ω + π δ ωω
ω =
∫
( ) ( ) ( ) ( )
( ) ( ) ( )
t
tt tj t j t j t
t
dy(t) x d y t x(t) j Y j X jdt
1 1Y j x d e dt x d e x(t)e dtj j
1X( j0) ( ) X( j )j
−∞
=∞∞− ω − ω − ω
−∞ −∞ −∞ =−∞
= τ τ = → ω ω = ω
ω = τ τ = τ τ ⋅ − − ω − ω
= ⋅ πδ ω + ωω
∫
∫ ∫ ∫ ∫
Fourier Transform, FT
Example of Integration u(t) = ∫δ(t)dt
( )
( ) ( )
( )
t
Continuous impulse : t
Step function : u t t dt
1Fourier transform : U( j ) 1j
−∞
δ
= δ
ω = + π ⋅ ⋅ δ ωω
∫
Use of the integration formula :
( ) ( )
( )
( ) ( )
( )
FT
FT
1 1Step function : u t sgn t2 2
2 0jFourier transform : sgn t0 0
1 1 22 2
1Fourier transform : U( j )j
= +
ω = ω←→ ω =
←→ πδ ω = πδ ω
ω = + πδ ωω
Alternative calculation :
Summation[ ] [ ]
n
k
y n x k=−∞
= ∑
[ ] ( )DTFT j j j0j
DTFT1y n Y(e ) X(e ) X(e )
(1 e )for
where first term equals zero for 0
Ω Ω− Ω←→ = + π δ Ω
−− π ≤ Ω ≤ π
Ω =
[ ] [ ] [ ] DTFT j j jy n y n 1 x n X(e ) (1 e )Y(e )Ω − Ω Ω− − = ←→ = −
Signal ProcessingPart 3.7 Fundamental Operations and System Properties of
Fourier Representations
Magnus Danielsen
Fundamental Operations and System Properties of Fourier Representations
• Convolution• Modulation• Power spectrum and Parceval relationships• Duality• Time – bandwidth product• Auto – correlation • Cross – correlation
Non-periodic Convolution (FT)
( ) ( )
( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( )
j t
j t
j
y( t) h( t) x(t) h x t d
Y j h x t d e dt
h x t e dt d
h X j e d
H j X j
∞
−∞
∞ ∞− ω
−∞ −∞
∞ ∞− ω
−∞ −∞
∞− ωτ
−∞
= ∗ = τ − τ τ
ω = τ − τ τ
= τ − τ ⋅ τ
= τ ω τ
= ω ⋅ ω
∫
∫ ∫
∫ ∫
∫
( )FTy(t) Y j←→ ω( )FTx(t) X j←→ ω( )FTh(t) H j←→ ω
Non-periodic Convolution (DTFT)
[ ] [ ] [ ] [ ] [ ]
( ) [ ] [ ]
[ ] [ ]
( ) ( )
k
j j n
n k
j k j ( n k )
k n
j j
y n h n x n h k x n k
Y e h k x n k e
h k e x n k e
H e X e
∞
=−∞
∞ ∞Ω − Ω
=−∞ =−∞
∞ ∞− Ω − Ω −
=−∞ =−∞
Ω Ω
= ∗ = −
= − ⋅
= − ⋅
= ⋅
∑
∑ ∑
∑ ∑
[ ] ( )DTFT jy n Y e Ω←→[ ] ( )DTFT jx n X e Ω←→
[ ] ( )DTFT jh n H e Ω←→
Periodic Convolution (DTFS)
[ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ]N
y n h n x n h k x n k
Y k N H k X k
= ∗ = ⋅ −
= ⋅ ⋅
∑
[ ] [ ]02DTFS,Ny n Y kπ
Ω =←→[ ] [ ]0
2DTFS,Nx n X kπ
Ω =←→
[ ] [ ]02DTFS,Nh n H kπ
Ω =←→
Periodic Convolution (FS)
( ) ( )
[ ] [ ] [ ]T
y(t) h(t) x(t) h x t d
Y k T H k X k
= ∗ = τ − τ τ
= ⋅ ⋅
∫
[ ]02FS,Ty(t) Y kπ
ω =←→[ ]0
2FS,Tx(t) X kπ
ω =←→
[ ]02FS,Th(t) H kπ
ω =←→
Non-periodic Modulation (FT)
( ) ( )
( )( ) ( )
( )( ) ( )
( )( ) ( )
( ) ( ) ( )
j ' t j t
j ' t j t j( ' ) t2 2
j t2
j t j t
1 1y( t) z( t) x( t) Z j ' e d ' X j e d2 2
1 1Z j ' e X j e d ' d Z j ' X j e d ' d2 2
1 Z j ' X j( ' e d ' d2
1e Z j ' X j( ' d ' d e Z j X2
∞ ∞ω ω
−∞ −∞
∞ ∞ ∞ ∞ω ω ω +ω
−∞ −∞ −∞ −∞
∞ ∞λ
−∞ −∞
∞ ∞λ λ
−∞ −∞
= ⋅ = ω ω ⋅ ω ωπ π
= ω ω ω ω = ω ω ω ωπ π
= ω λ − ω ω λπ
= ω λ − ω ω λ = λ ∗ π
∫ ∫
∫ ∫ ∫ ∫
∫ ∫
∫ ∫ ( )j d∞
−∞
λ λ∫
( )FTy(t) x(t) z(t) Y j= ⋅ ←→ ω
( )FTz(t) Z j←→ ω
( )FTx(t) X j←→ ω⊗
( ) ( ) ( )FT 1y(t) z( t) x(t) Y j Z j X j2
= ⋅ ←→ ω = ω ∗ ωπ
Non-periodic Modulation (DTFT)
[ ] [ ] [ ]( )DTFT j
y n x n z n
Y e Ω
= ⋅
←→[ ] ( )DTFT jz n Z e Ω←→
[ ] ( )DTFT jx n X e Ω←→⊗
[ ] [ ] [ ] ( ) ( ) ( )DTFT j j j1y n z n x n Y e Z e X e2
Ω Ω Ω= ⋅ ←→ = ∗π
Parceval Relationships( ) [ ]FTx t X j←→ ω
Energy relations for nonperiodic signals:
[ ] ( )DTFT jx n X e Ω←→
( ) [ ]02FS,Tx t X kπ
ω =←→
[ ] [ ]02DTFS,Nx n X kπ
Ω =←→
( ) ( ) ( ) [ ] ( )
[ ] [ ] [ ]
2 j tx
2
1Pr oof : E x t dt x t x t dt X j e d x t dt2
1 1X j X j d X j d2 2
∞ ∞ ∞ ∞∗ ∗ − ω
−∞ −∞ −∞ −∞
∞ ∞∗
−∞ −∞
= = = ω ω π
= ω ω ω= ω ωπ π
∫ ∫ ∫ ∫
∫ ∫
( ) [ ]2 2x
1E x t dt X j d2
∞ ∞
−∞ −∞
= = ω ωπ∫ ∫
[ ] [ ]2 2x
N N
1P x n X kN
= =∑ ∑
( ) [ ]2 2x
T
1P x t dt X kT
∞
−∞
= = ∑∫
[ ] ( ) 22 jx
2
1E x n X e d2
∞Ω
−∞ π
= = Ωπ∑ ∫
Power relations for periodic signals:
Duality
( ) [ ] ( ) [ ]FT FTx t X j X jt 2 x←→ ω ⇔ ←→ π⋅ −ω
[ ] ( ) ( ) [ ]DTFT FS, 1j jtx n X e X e x kΩ←→ ⇔ ←→ −
[ ] [ ] [ ] [ ]0 02 2DTFS, DTFS,N N 1x n X k X n x k
N
π πΩ = Ω =
←→ ⇔ ←→ −
Time – Bandwidth Product
( ) ( )2X j sin Tω = ωω
( ) ( )1 2 T 2Z j sin sin 2T½ /½ ½
ω ω = = ω ω ω
-3 -2 -1 0 1 2 30
0 .2
0 .4
0 .6
0 .8
1
1 .2
1 .4
Time t-T T
1
0
x(t)
- 3 - 2 - 1 0 1 2 30
0 . 2
0 . 4
0 . 6
0 . 8
1
1 . 2
1 . 4
Time t-2T 2T
1
0
z(t)=x(½t)
x Xt 2T 2 4Tπ
∆ ⋅∆ω = ⋅ = π z Zt 4T 2 42Tπ
∆ ⋅∆ω = ⋅ = π
( )
( )
1222
d2
t x t dtT
x t dt
∞
−∞∞
−∞
=
∫
∫
( )
( )
1222
w2
X dB
X d
∞
−∞∞
−∞
ω ω ω
=
ω ω
∫
∫ w d1B T2
⋅ ≥
Example
General rule
-20 -15 -10 -5 0 5 10 15 20-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
Correlation Functions( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
FT
2FT *ff
FT FT
FT *fg
f t F j
R f t f t dt F j F j F j
f t F j g t G j
R f t g t dt F j G j
∞
−∞
∞
−∞
−
←→ ω
τ = + τ ←→ ω ω = ω
−
←→ ω ←→ ω
τ = + τ ←→ ω ω
∫
∫
Auto correlation :
Cross correlation :
Continuous signals:
[ ] ( )[ ] [ ] [ ] ( ) ( ) ( )
[ ] ( ) [ ] ( )[ ] [ ] [ ] ( ) ( )
DTFT j
2DTFT * j j jff
n
DTFT DTFTj j
DTFT * j jfg
n
f n F e
R m f n f n m F e F e F e
f n F e g n G e
R m f n g n m F e G e
Ω
∞Ω Ω Ω
=−∞
Ω Ω
∞Ω Ω
=−∞
−
←→
= + ←→ =
−
←→ ←→
= + ←→
∑
∑
Auto correlation :
Cross correlation :
Discrete signals:
Signal ProcessingPart 3.8 Technical Applications of Fourier representations
Magnus Danielsen
Technical Applications of Fourier representations
• Radar spectrum• Filters• Accelerometers• Geophones• Gauss Pulse• AM radio
Radar MeasurementsSpectrum of RF Pulse Train
0 2T0 T T+2T0 t
p(t)
s(t)
t
[ ] ( )0 0jkT 0 0sin kT
P k ek
− ω ω=
π
[ ] [ ] [ ]1 1S k k 500 k 5002j 2j
=− δ + + δ −
x(t)
t
[ ] [ ] [ ] [ ] [ ] ( )
( ) ( )
0 0
0 0 0 0
jkT 0 0
j( k 500)T j( k 500)T0 0 0 0
sin kT1 1X k S k P k k 500 k 500 e2 j 2 j k
sin (k 500)T sin (k 500)T1 1e e2 j (k 500) 2 j (k 500)
− ω
− + ω − − ω
ω = ∗ = − δ + + δ − ∗ π
+ ω − ω= − +
+ π − π
[ ] [ ] [ ] [ ] [ ]0 0 0Hint: k k Z k k ' k Z k k ' Z k kδ − ∗ = δ − − = −∑
FiltersLP=Low-pass filter
HP=High-pass filter
BP=Band-pass filter
BS=Band-stopp filter
ω-W W
H(jω)
ω-W W
H(jω)
ωW1 W2
H(jω)
-W2 -W1
ωW1 W2
H(jω)
-W2 W1
H(ejΩ)
H(ejΩ)Ω
-2π - π -W W π 2π
Ω-2π - π -W W π 2π
H(ejΩ)Ω
-2π - π W1 W2 π 2π
H(ejΩ)Ω
-2π - π W1 W2 π 2π
Analogue filter Digital filter
The Accelerometer Equation and the Geophone Equation
km
f
y
0u (t)
u0u
22nn 02
d y(t) dy(t) y(t) u (t)dt Q dt
ω+ + ω =
d 1y(t) v(t)dtv(t) voltage ingeophonecoil
= ⋅α
=
0 0du (t) u (t)dt
=
2 22nn 02 2
d v(t) dv(t) dv(t) u (t)dt Q dt dt
ω+ +ω = α
( ) ( )( )
( )( )
2
G2 2n0 0
n
V j Y j ( j ) ( j )H jU j U j /( j ) ( j ) j
Q
ω ω ⋅ ω α ωω = = =
ωω ω ω ω + ω + ω
( ) ( )( )A
2 2n0n
Y j 1H jU j ( j ) j
Q
ωω = =
ωω ω + ω+ ω
Accelerometer:
Geophone:
Accelerometer Characteristics
22nn2
d y(t) dy(t) y(t) u(t)dt Q dt
ω+ + ω =
( ) ( )( )A
2 2nn
Y jH j
U j1
( j ) jQ
ωω =
ω
=ωω + ω + ω
u(t) = acceleration to be measured y(t) = displacement
100
101
102
-150
-100
-50
0
100
101
102
-200
-150
-100
-50
0
Q=5
0.5
0.33
Q=5
0.5
0.33
Geophone
A. Barzilai, T Van Zandt, T. Pike, S. Manion, T. Kenny, Stanford University and Jet Propulsion Laboratory, AGU 1998)
PermanentMagnet
Spoli av kopartráði
Geofon húsiFjøður
Geophone Characteristics
100
101
102
-50
-40
-30
-20
-10
0
10
20
100
101
102
0
50
100
150
200
2 22n 0n2 2
d v(t) dv(t) d u (t)v(t)dt Q dt dt
ω+ +ω =
( ) ( )( )G
0
2
2 2nn
V jH j
U j
( j )
( j ) jQ
ωω =
ω
α ω=
ωω + ω + ω
Quality factor: Q 1Damping constant: ξ =
2Q
Q=5
0.5
0.33
Q=5
0.5
0.330u (t) = velocity to be measured
v(t) = measured coil voltage
Gauss Pulse
( ) ( )
( ) ( )
2tFT2
FT
d tg(t) e t g t j G jdt 2
djt g t G jd
−= − = − ⋅ ←→ ω⋅ ω
π
− ⋅ ←→ ωω
FTg(t) G( j )←→ ω
( ) ( )2
2
dDifferential equation : G j G jd
G( ) c eω
−
ω = −ω⋅ ωω
ω = ⋅
2 21g(t) dt 1 G( j ) d c 12
∞ ∞
−∞ −∞
= = ω ω → =π∫ ∫
-5 0 50
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.42
2t
21g(t) e2
−σ=
σ π
1/ et 2 2∆ = σ
-5 0 50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
12 2
2G( ) eσ ω
−ω = σ
1/ e2
∆ω =σ
2 2 2
2t
FT2 21GeneralGauss pulse : g(t) e G( ) e2
σ ω− −σ= ←→ ω = σ
σ π
1/ e 1/ e2Uncertainty relation for Gauss pulse: t 2 2 8∆ ⋅∆ω = σ⋅ =σ
Amplitude Modulated Radio System – AM
( ) ( ) ( ) ( ) ( )
( ) ( )( ) ( )( )
c cj t j tc
c c
1Transmitted signal : r t m t cos t m t e e2
1 1R j M j M j2 2
ω − ω= ω = +
ω = ω − ω + ω + ω
( ) ( ) ( )
( ) ( )( ) ( )( )
( ) ( ) ( )
FTc
c c
c c
Received signal: q t r t cos t
1 1Q j R j R j2 21 1 1M j( 2 ) M j M j( 2 )4 2 4
= ω ←→
ω = ω−ω + ω+ω
= ω− ω + ω + ω+ ω
Source Transmitter Channel Receiver
Usually applied AM – mode : m(t) = 1 + m⋅x(t) where x(t) is the source signal
m(t)
cos(ωct)
r(t)=m(t)·cos(ωct)⊗cos(ωct)
q(t)⊗ LP-filter
y(t)x(t)Source coding
AM Radio - Spectrum
( ) ( )( ) ( ) ( )
( ) ( )( ) ( )( )
( )( ) ( )( )( ) ( )( ) ( )( )( )
c c
c c c c
for positive frequencies for negativ frequencies
Usually applied AM mode: m t 1 m x t 0 m 1
M j j mX j1 1Transmitted signalspectrum : R j M j M j2 2
1 1j mX j j mX j2 2
− = + ⋅ ≤ ≤
ω = δ ω + ω
ω = ω − ω + ω + ω
= δ ω − ω + ω − ω + δ ω + ω + ω + ω
( ) ( ) ( ) ( )( ) ( )
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )
FT
FT
FT
x t X jω
Amplitude modulation : m t 1 m x t M j j +m X jω
Double-sideband modulation : m t x t M j X jω
Single-sideband m
←→
− = + ⋅ ←→ ω = δ ω ⋅
− = ←→ ω =
Source signal x t and spectrum X jω :
Source - coded signal m t spectrum M jω :
AM
DSB
( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )
FTˆodulation : m t x t jx t M j X jω sgn j X jωˆwhere x t is the Hilbert transform of x t .
AM, DSB, and SSB are all amplitude modulations, but frequently only AM is named amplidtude modulation. SSB requires mor
− = + ←→ ω = + ωSSB
e sophisticated modulation method than sketched here.
AM Radio Signals and Spectra
0 0.01 0.02 0.03 0.04 0.05-1
-0.5
0
0.5
1
0 0.01 0.02 0.03 0.04 0.05-2
-1
0
1
2
0 0.01 0.02 0.03 0.04 0.05-1
-0.5
0
0.5
1
0 0.01 0.02 0.03 0.04 0.05-2
-1
0
1
2
0 0.01 0.02 0.03 0.04 0.05-1
-0.5
0
0.5
1
0 0.01 0.02 0.03 0.04 0.05-2
-1
0
1
2
0 500 1000 1500 20000
0.5
0 500 1000 1500 20000
0.50 500 1000 1500 2000
0
0.5
1
0 500 1000 1500 20000
0.50 500 1000 1500 2000
0
0.2
0.4
0 500 1000 1500 20000
0.5
x(t)
r(t)
( )R jf
( )X jf
Sine signal Square signal Sawtooth signal
Spectrum (positive frequencies,and m=½ )
Time signal (m=½)
Signal ProcessingPart 4.1 Sampling of Signals
Magnus Danielsen
Sampling and Reconstruction of Analogue Signals
• Sampling • Relations between FT, FS, DTFT & DTFS• Impulse sampling method • Spectral properties of sampled signal• Aliasing • Examples of aliasing• Reconstruction of continuous-time signals• Applications
Sampling
( ) [ ]Sampling at times t nT: x nT x n= =
( )x t
tT
( )( ) ( )FT
Frequency property of x t :
x t X j←→ ω
[ ] [ ] ( )DTFT jFrequency property of x n : x n X e Ω←→
Practical sampling requires that:
• x(t) is described by a number of discrete values x[n] at discrete times t=nTs
• Number of samples N is finite
• x[n] can be periodised, and described by DTFS resulting in a discrete spectrum X[k]
• x(t) has a spectrum described by X(jω)
• The spectrum of x(t) is obtained by 4 relations between FT, FS, DTFT, and DTFS
FT to FS Relation( ) [ ]
( )( )0 0
FS
FT
jk t FT0 0
x t X k
1 2
e 2 kω
←→
←→ πδ ω
←→ πδ ω − ω
( ) [ ] ( ) [ ] ( )0jk t FT0
k k
x t X k e X j 2 X k k∞ ∞
ω
=−∞ =−∞
= ←→ ω = π δ ω − ω∑ ∑
k
0 0jk tFS for e ω
1
0kω
0 0jk tFT for e ω
0kω
2π
[ ] [ ]( )
( )0
DTFS
DTFT
jk n DTFT0
m
x n X k
1 2 for
e 2 2 mk∞
Ω
=−∞
←→
←→ πδ Ω − π < Ω < π
←→ π δ Ω πΩ −−∑
[ ] [ ] ( ) [ ] ( )0
N 1jk n DTFT j
0k 0 k m
x n X k e X e 2 X k 2k m− ∞ ∞
Ω Ω
= =−∞ =−∞
−= ←→ = π Ω πδ Ω−∑ ∑ ∑
k
0 0jk nDTFS for e Ω
1
0k 0N k+
NN−
0N k− +
... ...Ω
0 0jk nDTFT for e Ω
1
0 0k Ω 0 02 kπ + Ω
2π
0 02 k− π + Ω
... ...2− π
DTFT to DTFS Relation
FT to DTFT Relation[ ] ( )( )
( ) s
DTFT j
FT
j T nFTs
x n X e
t 1
t nT e
Ω
ω
←→
δ ←→
δ − ←→
( )
[ ] ( )
( )
( ) [ ] s
s
FTj T nj
s Tn n
x t X j
x n t nT X e x n e
δ δ
∞ ∞− ωΩ
Ω=ω=−∞ =−∞
= ω = ←→
δ − = ∑ ∑
s
s
T2Sampling interval: T
Ω = ωπ
=ω
Ω
( )jDTFT : X e Ω
2 π... ...
2− π
ω
( )FT : X jδ ω
s
2Tπ
... ...
s
2Tπ
−
t
( )x tδ
... ...
Ts
[ ]x n
... ...n
[ ] [ ]DTFSx n X k←→
[ ] [ ] ( ) [ ] ( )0
N 1jk n DTFT j
0k 0 k
x n X k e X e 2 X k k− ∞
Ω Ω
= =−∞
= ←→ = π δ Ω − Ω∑ ∑
s
s
T2Sampling interval: T
Ω = ωπ
=ω
( ) ( ) ( )
[ ]
s
FT j j
T
0s
k s
x t X e X e
2 X k T ( k )T
Ω Ωδ δ Ω=ω
∞
=−∞
←→ =
Ω= π δ ω−
∑
...
[ ]x n
... ...
NN−
n( )x tδ
...
sNTsNT−t
n
[ ]X k
... ...
NN−
k
( )X jδ ω
... ...
s
2Tπ
s
2Tπ
−
k
FT to DTFS Relation
Sampling – Impulse Sampling Method
( ) [ ]s sSampled signal at times t nT : x nT x n= =
( )x t
tTs
( )( ) ( )FT
Frequency property of x t :
x t X j←→ ω
[ ] [ ] ( )DTFT jFrequency property of x n : x n X e Ω←→
( ) ( ) ( )1X j X j P j2δ ω = ω ∗ ωπ
( ) ( )
( ) [ ] ( ) ( ) ( ) ( )
sn
s s sn n
p t t nT
x t x n t nT x(nT ) t nT = x t p t
∞
=−∞
∞ ∞
δ=−∞ =−∞
= δ −
= δ − = δ − ⋅
∑
∑ ∑Impulse sampled signal :
Spectrum of impulsesampled signal :
Spectrum for Sampled Signal (1)
( ) ( ) ( )1X j X j P j2δ ω = ω ∗ ωπ
( )x t
tTs
( ) ( ) ( )x t x t p tδ = ⋅
( )p t
tTs
......
( ) ( )
[ ] ( )s
0
s
FSs
nT2
jk t
Ts s2
p t t nT
1 1P k t e dtT T
∞
=−∞
− ω
−
= δ − ←→
= δ ⋅ =
∑
∫
( ) ( ) [ ]FTs
n k ks s s
2 2 2p t t nT P( j ) 2 P k k kT T T
∞ ∞ ∞
=−∞ =−∞ =−∞
π π π= δ − ←→ ω = π δ ω − = δ ω −
∑ ∑ ∑
( ) ( ) ( )
( )k ks s s s
1X j X j P j21 2 2 1 2X j k X k
2 T T T T
δ
∞ ∞
=−∞ =−∞
ω = ω ∗ ωπ
π π π= ω ∗ δ ω − = ω − π
∑ ∑
( )x t
tTs
( ) ( ) ( )x t x t p tδ =
( )p t
tTs
......
( ) ( ) ( )ks s
1 1 2X j X j P j X k2 T T
∞
δ=−∞
πω = ω ∗ ω = ω − π
∑
( )P jω
ω......
ss
2Tπ
ω =
ω-W W
X(jω)
ω-W W
Xδ (jω)
ss
2Tπ
ω = s2ωs−ωs2− ω
... ...
Spectrum for Sampled Signal (2)
Sampling with Rectangular Pulses (1)( )x t
tTs
( ) ( ) ( )sx t x t p t= ⋅
tTs
( )x tτ
( )p t
tTs
......τ
( )( ) ( )FT
Frequency property of x t :
x t X j←→ ω
( ) [ ] [ ]0jk t FS
k
p t c k e c k∞
ω
=−∞
= ←→∑
( )
( ) ( ) ( ) [ ] ( )
( ) ( )
[ ] ( )
[ ] ( ) ( )
[ ]
s
s
s
s
jk ts
k
FT j ts s
jk t j t
k
j k t
sk k
Frequency property of x t :
x t p t x t c k e x t
X j x t e dt
c k e x t e dt
c k x t e dtc k X( k )
∞ω
=−∞
∞− ω
−∞
∞ ∞ω − ω
=−∞−∞
∞− ω− ω∞ ∞
−∞=−∞ =−∞
= ⋅ = ⋅
←→ ω =
= ⋅ ⋅
⋅= = ω − ω
∑
∫
∑∫
∫∑ ∑
Spectrum for Rectangular Pulse Sampled Signal
ω-W Ws
s
2Tπ
ω = s2ωs−ωs2− ω... ...
|Xs(jω)|c[k]=c[ω=kωs]
( ) [ ]s sk
X j c k X( k )∞
=−∞
ω = ω − ω∑
ω-W W
|X(jω)|
The distribution of Xs(jω) on the ω-axis is the same as for the ideally sampled signal
( ) ( )FTx t X j←→ ω
Signal ProcessingPart 4.2 Aliasing and Reconstruction of Sampled Signals
Magnus Danielsen
Aliasing
Nyquist Sampling Theorem:
The condition for reconstruction of a sampled signal x(t) is that the maximum bandwidth W of the sampled signal x(t) shall be less than half of the sampling frequency, or W < ½·ωs = π/Ts.
ωs is called the Nyquist cyclic frequency,
and fs= ωs/2π is the Nyquist sampling frequency.(Geophysisists name fs = ωs/2π the Nyquist frequency)
Xδ (jω)
ω-W Ws
s
2Tπ
ω = s2ωs−ωs2− ω
... ...
s3− ω s3ω
sW2ω
>
Sampling with aliasing error
ω-W W
Xδ (jω)
ss
2Tπ
ω = s2ωs−ωs2− ω
... ... sW2ω
<
Sampling with no aliasing error
ω-W W
Xδ (jω)
ss
2Tπ
ω = s2ωs−ωs2− ω
... ...
s3ωs3− ωsW
2ω
=
Sampling on the limit to aliasing error
DTFT of Sampled Signal Spectrum
[ ] ( )s
DTFT j
T
A sampled signal having the frequency spectrum X (j ) corresponds to the time signal x (t). Therefore the sampling values are
x n X e X (j )
δ
δ
ΩΩδ ω=
ω
←→ = ω
sWT < πΩ-WTs WTs
2π 4π2− π4− π
... ...
( )jX e Ω
Ω
... ...
-WTs WTs2π 4π2− π4− π
( )jX e Ω
sWT = π
sWT > π
Ω
... ...
( )jX e Ω
-WTs WTs2π 4π2− π4− π
Example: Sampling a Sinusoidal Signal( ) ( )Signal: x t cos t= π sSampling time: T
- 4 - 3 - 2 - 1 0 1 2 3 4- 1
- 0 . 8
- 0 . 6
- 0 . 4
- 0 . 2
0
0 . 2
0 . 4
0 . 6
0 . 8
1 ( )x t
- 4 - 3 - 2 - 1 0 1 2 3 4- 1
- 0 . 8
- 0 . 6
- 0 . 4
- 0 . 2
0
0 . 2
0 . 4
0 . 6
0 . 8
1 ( )x t
• •• • •••
- 4 - 3 - 2 - 1 0 1 2 3 4- 1
- 0 . 8
- 0 . 6
- 0 . 4
- 0 . 2
0
0 . 2
0 . 4
0 . 6
0 . 8
1 ( )x t
- 4 - 3 - 2 - 1 0 1 2 3 4- 1
- 0 . 8
- 0 . 6
- 0 . 4
- 0 . 2
0
0 . 2
0 . 4
0 . 6
0 . 8
1 ( )x t
4π 8π8− π 4− π 0 π−π
X ( j )δ ω
ω
4π 8π8− π 4− π 0 π−π
X( j )ω
ω
s1T4
=
s 8ω = π
4π 8π8− π 4− π 0 π−π
X ( j )δ ω
ω
..... .....s
3T2
=
4s 3ω = π
4π 8π8− π 4− π 0 π−π
X ( j )δ ω
ωsT 1=
s 2ω = π
..... .....
s 8ω = π
Aliasing of Moving Wheel in Movies
(rad / s)ω
sTϕ = ω< π
sSampling time: T
(rad / s)ω
sTϕ = ω> π
No aliasing error:
Correct apparent rotation direction and angle frequency ωa = ω
Aliasing error :
Wrong apparent rotation direction and wrong apparent angle frequency ωa = (2π - ϕ)/Ts = 2π/Ts - ω
Reconstruction of Continuous-time Signal (1)
ω-W W
Xδ (jω)
ss
2Tπ
ω = s2ωs−ωs2− ω
... ... sW2ω
<
Sampling with no aliasing error
ω-W W
X(jω)
ω
H(jω)
BB−
sW B2ω
< <
( ) ( ) ( ) ( ) [ ] ( ) [ ] ( )
[ ]
r r s r s
ss
y t h t x t h t x n t nT x n h t nT
x n sinc (t nT )2
∞ ∞
δ−∞ −∞
∞
−∞
= ∗ = ∗ δ − = −
ω = − π
∑ ∑
∑
( ) ( )FTy t Y j←→ ω( ) ( )FTx t X jδ δ←→ ω ( ) ( )FTr rh t H j←→ ω
Filter with bandwidth B = ωs/2
( ) s s sr
Th t sin t sinc tt 2 2
ω ω = = π π ( )
ss
rs
T for2H j
0 for2
ω ω <ω = ω ω >
Reconstruction of Continuous-time Signal (2)
-4 -3 -2 -1 0 1 2 3-2
-1
0
1
2
3
4
5
Reconstruction of Continuous-time Signal (3)
Practical ReconstructionZero-order Hold
Zero-order hold
[ ]y n [ ]0y n
Anti-imaging filter
( )y t
[ ]y n
[ ]0y n( )x t
( )y t
Multipath Communication Channel- Reflections from Environment (1)
[ ] [ ] [ ]( ) ( ) ( )
DTFT
j j j
y n x n a x n 1
Y e X e 1 a eΩ Ω − Ω
= + ⋅ − ←→
= + ⋅
[ ] ( )DTFT jx n X e Ω←→[ ] ( )FT jh n H e Ω←→
Discrete-time model
( ) ( ) ( )( ) ( ) ( )diff
FTdiff
j T
y t x t x t T
Y j X j 1 e− ω
= + α ⋅ − ←→
ω = ω + α
( ) ( )FTx t X j←→ ω( ) ( )FTh t H j←→ ω
Two-path communication channel
( ) ( )( ) ( )diffj TY j
H j 1 eX j
− ωωω = = + α
ω
( ) ( )( ) ( )
jj j
j
Y eH e 1 a e
X e
ΩΩ − Ω
Ω= = + ⋅
( ) ( ) ( )( )
s
s
/ T2 2 2 2s
/ T
2 2
min
TMSE H j H j d a 12
MSE 1 for a
π
δ−π
= ω − ω ω = − αγ + α − γπ
= α − γ = αγ
∫
( ) ( ) ( )s
s
j Tj
TH j H e 1 a e− ωΩ
δ Ω=ωω = = + ⋅⇒
( ) ( )
( )
s
s
s
diff s
s
/ T2s
/ T
/ T2j T j Ts
/ T
2 2
2 2 2
TMSE H j H j d2
T e a e d2
a a * * a *
a 1
π
δ−π
π− ω − ω
−π
= ω − ω ωπ
= α − ⋅ ωπ
= α + − α γ − α γ
= − αγ + α − γ
∫
∫
( ) ( )sj TH j 1 a e− ωδ ω = + ⋅( ) ( )diffj TH j 1 e− ωω = + α
( )2 2
minMSE 1 for a= α − γ = αγ
s
diff s
s
/Tj (T T )s diff s
s/T
T T Te d sinc2 T
π− ω −
−π
−γ = ω= π
∫
0.5 1 1.50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
21− γ
Tdiff/Ts
Multipath Communication Channel− Reflections from Environment (2)
Basic Discrete-time Signal Processing System for Analogue Signals
Anti-aliasing filter
Sampling at interv.Ts
.
Discrete-time processing
Sample and hold
Anti-imaging filter
( )x t ( )ax t [ ]x n [ ]y n ( )0y t ( )y t
Equivalent continuous-time system G(jω)
( )x t ( )y t
Signal ProcessingPart 5.1 Discrete Fourier Transform (DTF)
for nNumerical Evaluation
Magnus Danielsen
Discret Forier Transform (DTF) and Fast Fourier Transform (FFT)
• DTFS is the only Fourier representation capable of exactnumerical evaluation
• DTFT related to DTFS• FS related to DTFS• FT related to DTFS• DFT = Discrete Fourier Transform – definition and reasons• FFT = Fast Fourier Transform
= efficient algorithms for calculating DTFS• FFT with decimation in time• FFT with decimation in frequency• FFT and IFFT use nearly same algoritme
Periodicing of Finite Signal Duration
n
M-1 N
[ ]x n
2Nperiod
[ ] [ ]
[ ] ( ) [ ]
[ ]
M 1DTFT j j n
n 0N 1
j n
n 0
Finite duration signal :
x n for 0 n M 1x n
0 for n 0 n M M N
x n X e x n e
x n e
−Ω − Ω
=
−− Ω
=
≤ < −=
< ∨ ≥ ≤
←→ =
=
∑
∑
[ ]x n
n
M-1 N
[ ] [ ] [ ]
[ ] [ ] [ ] 0
periodisedm
N 1jk nDTFS
n 0
Periodiced signal :
x n x n x n mN
1x n X k x n eN
∞
=−∞
−− Ω
=
= = +
←→ =
∑
∑
DTFT Related to DTFS
[ ] [ ] [ ]
[ ] [ ] [ ] 0
periodisedm
N 1jk nDTFS
n 0
Periodiced signal :
x n x n x n mN
1x n X k x n eN
∞
=−∞
−− Ω
=
= = +
←→ =
∑
∑
[ ] [ ]
[ ] ( ) ( ) [ ]N 1
DTFTj j n j j n
n 0
Finite duration signal :
x n for 0 n N 1x n
0 for n 0 n N
1x n X e e d X e x n e2
π −Ω Ω Ω − Ω
=−π
≤ < −=
< ∨ ≥
= Ω←→ =π ∑∫
[ ] ( ) ( )0
0
jkjk
1 1X k X e X eN N
ΩΩΩ= Ω= =
[ ] ( ) ( ) [ ]0 0 0
N 1 N 1j k j kn j knj j n
k 0 k 0
1 1x n X e e d X e e X k e2 N
π − −Ω − Ω − ΩΩ Ω
= =− π
= Ω ≅ =π ∑ ∑∫
0
0
Some relations :2dN
k
πΩ = Ω =
Ω = Ω
[ ] ( ) [ ] [ ] 0
N 1jk nDTFS
disc s discn 0
Discrete periodic signal (sampled signal) :
1x n x nT X k x n eN
−− Ω
=
= ←→ = ∑
( ) ( ) [ ] ( ) [ ]0
0 0 s
T Njk t jk nTFS
0 sn 00 00
Periodic continuous signal :
1 1x t x t T X k x t e dt x n e TT T
− ω − ω
=
= + ←→ = ≅ ∑∫
[ ] [ ]0 s 0
2disc TN
X k X k πω =Ω =
=
0 s 0 0 s s s 0 0 s 02Some relations : T NT T 2 T 2 N TNπ
= ω = π ω = π ω = ω ω = Ω =
FS Related to DTFS
[ ] ( ) [ ] [ ] 0
N 1jk nDTFS
disc s discn 0
Discrete periodic signal (sampled signal) :
1x n x nT X k x n eN
−− Ω
=
= ←→ = ∑
( ) ( )
( ) ( ) ( )0
0
0
TFT j t
0
Finite duration signal :
x t for 0 t Tx t
0 for t 0 t T
x t X j x t e dt− ω
≤ <= < ∨ ≥
←→ ω = ∫
[ ] [ ] [ ]0 s 0
s2disc TN 0 s
1 1X k X k X k X jkT NT N
πω =Ω =
ω = = =
( ) ( ) ( )
( ) [ ] ( )0
0
0periodisedmT
jk tFS
0 0
Periodiced signal :
x t x t x t mT
1x t X k x t e dtT
∞
=−∞
− ω
= = +
←→ =
∑
∫
0 s 0 0 s s s 0 0 s 02Some relations : T NT T 2 T 2 N TNπ
= ω = π ω = π ω = ω ω = Ω =
FT Related to DTFS
Discrete Fourier Transform (DFT) –Definition and Reasons
• DTFS is the only Fourier representation capable of exactnumerical evaluation
• DTFS (period N) relates directly to DTFT (finite time interval N)
• DTFT represents (nonperiodic) energy signals• DTFS represents (periodic) power signals• Nonperiodic signals restrictet to a time interval N, and
correspondingly periodiced signals with period N, energy E and power P are proportional (E=P·N). Correspondingly ”XDTFT=N · XDTFS”
• Resonably Discrete Fourier Transform (DFT) for a finite interval (N) signal is defined as a transform of an energy signal as N times the DTFS
• Consequently DFT is a numerical discrete method to calculate the DTFT for an energy signal.
DFT Formulation for Calculation of DTFT
[ ] [ ] [ ] [ ]0 0
N 1 N 1jk n jk nDTFS
k 0 n 0
Periodiced signal, and DTFS :
1x n X k e X k x n eN
− −− Ω − Ω
= =
= ←→ =∑ ∑
[ ] ( ) ( ) [ ]N 1
DTFTj j n j j n
n 0
Finite duration signal, and DTFT :
1x n X e e d X e x n e2
π −Ω Ω Ω − Ω
=−π
= Ω←→ =π ∑∫
[ ] [ ] [ ] [ ]0 0
N 1 N 1jk n jk nDFT
k 0 n 0
Discrete Fourier Transform(DFT)formulation
for calculation of DTFT of finite duration signal :
1x n X k e X k x n eN
− −Ω − Ω
= =
= ←→ =∑ ∑
Matrix Formulation of DFT (1)[ ] [ ] [ ] [ ]0 0
N 1 N 1jk n jk nDFT
k 0 n 0
1x n X k e X k x n eN
− −Ω − Ω
= =
= ←→ =∑ ∑
[ ][ ][ ][ ]
[ ]
[ ]
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
j j2 j3 jk j(N 1)
j2 j4 j6 j2k j2(N 1)
j3 j6 j9 n j3k j3(N 1)
1 1 1 1 ... 1 ... 1x 01 e e e ... e ... ex 11 e e e ... e ... ex 21 e e e ... e ... ex 3 1... ... ... ... ... ... ... ...... N1 ex n
...x N 1
Ω Ω Ω Ω − Ω
Ω Ω Ω Ω − Ω
Ω Ω Ω Ω − Ω
= ⋅ −
[ ][ ][ ][ ]
[ ]
[ ]
0 0 0 0 0
20 0 0 0 0
jn jn2 jn3 jnk jn(N 1)
j(N 1) j(N 1)2 j(N 1)3 j(N 1)k j(N 1)
e e ... e ... e... ... ... ... ... ... ... ...
1 e e e ... e ... e
Ω Ω Ω Ω − Ω
− Ω − Ω − Ω − Ω − Ω
X 0X 1X 2X 3...X k...
X N - 1
[ ][ ][ ][ ]
[ ]
[ ]
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
j j2 j3 jk j(N 1)
j2 j4 j6 j2k j2(N 1)
j3 j6 j9 n j3k j3(N 1)
1 1 1 1 ... 1 ... 11 e e e ... e ... e1 e e e ... e ... e1 e e e ... e ... e... ... ... ... ..
− Ω − Ω − Ω − Ω − − Ω
− Ω − Ω − Ω − Ω − − Ω
− Ω − Ω − Ω − Ω − − Ω
=
X 0X 1X 2X 3...X k...
X N - 1
[ ][ ][ ][ ]
[ ]
[ ]
0 0 0 0 0
20 0 0 0 0
jk j2k j3k jnk j(N 1)k
j(N 1) j2(N 1) j3(N 1) jn(N 1) j(N 1)
x 0x 1x 2x 3
. ... ... ... ...1 e e e ... e ... e x n... ... ... ... ... ... ... ... ...
x N 11 e e e ... e ... e
− Ω − Ω − Ω − Ω − − Ω
− − Ω − − Ω − − Ω − − Ω − − Ω
−
[ ] [ ]DFT
N
N
x n X k
X W x1x W XN
←→
= ⋅∗
= ⋅ ⋅
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
j j2 j3 jk j(N 1)
j2 j4 j6 j2k j2(N 1)
j3 j6 j9 n j3k j3(N 1)
N
jn jn2 jn3 jnk jn(N 1)
1 1 1 1 ... 1 ... 11 e e e ... e ... e1 e e e ... e ... e1 e e e ... e ... e
W ... ... ... ... ... ... ... ...1 e e e ... e ... e... ... ..
Ω Ω Ω Ω − Ω
Ω Ω Ω Ω − Ω
Ω Ω Ω Ω − Ω
Ω Ω Ω Ω − Ω
∗=
20 0 0 0 0j(N 1) j(N 1)2 j(N 1)3 j(N 1)k j(N 1)
. ... ... ... ... ...
1 e e e ... e ... e− Ω − Ω − Ω − Ω − Ω
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0
j j2 j3 jk j(N 1)
j2 j4 j6 j2k j2(N 1)
j3 j6 j9 n j3k j3(N 1)
N
jk j2k j3k jnk
1 1 1 1 ... 1 ... 11 e e e ... e ... e1 e e e ... e ... e1 e e e ... e ... e
W ... ... ... ... ... ... ... ...1 e e e ... e
− Ω − Ω − Ω − Ω − − Ω
− Ω − Ω − Ω − Ω − − Ω
− Ω − Ω − Ω − Ω − − Ω
− Ω − Ω − Ω − Ω
= 0 0
20 0 0 0 0
j(N 1)k
j(N 1) j2(N 1) j3(N 1) jn(N 1) j(N 1)
... e... ... ... ... ... ... ... ...
1 e e e ... e ... e
− − Ω
− − Ω − − Ω − − Ω − − Ω − − Ω
N N N1 W W UNunity matrix
∗ ⋅ ⋅ =
=
Matrix Formulation of DFT (2)
Matrix Formulation of DFT (3)[ ] [ ] [ ] [ ] [ ] [ ]0 0
N 1 N 1 N 1 N 1jk n jk nDFTkn kn
k 0 k 0 n 0 n 0
1 1x n X k e X k w X k x n e x n wN N
− − − −Ω − Ω−
= = = =
= = ←→ = =∑ ∑ ∑ ∑
[ ][ ][ ][ ]
[ ]
[ ]
( )N 11 2 3 k
2 4 6 2k 2(N 1)
3 6 9 3k 3(N 1)
n n 2 n 3 n k n(N 1)
1 1 1 1 ... 1 ... 1x 0x 1 1 w w w ... w ... wx 2 1 w w w ... w ... wx 3 1 w w w ... w ... w1... ... ... ... ... ... ... ... ...Nx n 1 w w w ... w ... w
....x N 1
− −− − − −
− − − − − −
− − − − − −
− − ⋅ − ⋅ − ⋅ − −
= ⋅ −
( )
[ ][ ][ ][ ]
[ ]
[ ]2N 1 (N 1) 2 (N 1) 3 (N 1) k (N 1)
.. ... ... ... ... ... ... ...
1 w w w ... w ... w− − − − ⋅ − − ⋅ − − ⋅ − −
X 0X 1X 2X 3...X k...
X N -1
[ ][ ][ ][ ]
[ ]
[ ]
1 2 3 n (N 1)
2 4 6 2n 2(N 1)
3 6 9 3n 3(N 1)
k k 2 k 3 k n k(N 1)
1 1 1 1 ... 1 ... 11 w w w ... w ... w1 w w w ... w ... w1 w w w ... w ... w... ... ... ... ... ... ... ...1 w w w ... w ... w... ... ... ... ... ... ... .
−
−
−
⋅ ⋅ ⋅ −
=
X 0X 1X 2X 3...X k...
X N - 1
[ ][ ][ ][ ]
[ ]
[ ]2(N 1) (N 1) 2 (N 1) 3 (N 1) n (N 1)
x 0x 1x 2x 3...x n
.. ...x N 11 w w w ... w ... w− − ⋅ − ⋅ − ⋅ −
−
0j
2jN
w e
e
− Ω
π−
=
=
[ ] [ ]DFT
N
N
x n X k
X W x1x W XN
←→
= ⋅∗
= ⋅
( )
( )
N 11 2 3 k
2 4 6 2k 2(N 1)
3 6 9 3k 3(N 1)
N
n n 2 n 3 n k n(N 1)
N 1 (N 1) 2 (N 1)
1 1 1 1 ... 1 ... 1
1 w w w ... w ... w1 w w w ... w ... w1 w w w ... w ... w
W... ... ... ... ... ... ... ...1 w w w ... w ... w... ... ... ... ... ... ... ...
1 w w w
− −− − − −
− − − − − −
− − − − − −
− − ⋅ − ⋅ − ⋅ − −
− − − − ⋅ − − ⋅
∗=
23 (N 1) k (N 1)... w ... w− − ⋅ − −
1 2 3 k (N 1)
2 4 6 2k 2(N 1)
3 6 9 3k 3(N 1)
N
n n 2 n 3 n k n(N 1)
(N 1) (N 1) 2 (N 1) 3 (N 1) k (N 1
1 1 1 1 ... 1 ... 11 w w w ... w ... w1 w w w ... w ... w1 w w w ... w ... w
W ... ... ... ... ... ... ... ...1 w w w ... w ... w... ... ... ... ... ... ... ...
1 w w w ... w ... w
−
−
−
⋅ ⋅ ⋅ −
− − ⋅ − ⋅ − ⋅ −
=
2)
1
N N
N
1 W WNU unity matrix
− ⋅ ⋅
= =
Matrix Formulation of DFT (4)
Signal ProcessingPart 5.2 Fast Fourier Transform (FFT) – an Efficient
Algorithm for Numerical Evaluation of DFT
Magnus Danielsen
Number of Addition and Multiplications
[ ] [ ] [ ] [ ]0 0
N 1 N 1jk n jk nDFT
k 0 n 0
1x n X k e X k x n eN
− −Ω − Ω
= =
= ←→ =∑ ∑
Using DTFS directly•Number of complex multiplication = (N-1)2
•Number of complex additions = N(N – 1)
Using FFT algorithms :•Number of complex multiplication = (N/2) log2(N/2)•Number of complex additions = N log2(N)
0
2jj NNw w e e
π−− Ω= = =
[ ][ ][ ][ ]
[ ][ ]
[ ]
( )
( )
( )
( )
( )
( )
( )
( )
0
1 2 3 (N 1)
2 4 6 2(N 1)
3(N 1)3 6 9
N / 2 1 N / 2 1 2 N / 2 1 3 N / 2 1 (N 1)
N / 2 N / 2 2 N / 2 3 N / 2 (N 1)
1 1 1 1 ... 11 w w w ... w1 w w w ... w1 w w w ... w... ... ... ... ... ...
1 w w w w...
1 w w w w... .
−
−
− Ω
− − − − −
−
=
X 0X 1X 2X 3...
X N/2 - 1
X N/2...
X N - 1
[ ][ ][ ][ ]
[ ][ ]
[ ]2(N 1) (N 1)2 (N 1)3 (N 1)
...
.. ... ... ... ... ...
1 w w w ... w
x 1
x
x 0
x 23
x N/2 -1
x N
x N/2
-1− − − −
N / 2 d N / 2
N / 2 d N / 2
W W W
W W
xxW
⋅ = ⋅
oddind
evenindN ex
xN e
/2X
X
Nx
N/2x
N NNw w 1= =
( )
1N / 2
2N / 2
3dN / 2
N / 2 1N / 2
1 0 0 0 ... 00 w 0 0 ... 00 0 w 0 ... 0
W 0 0 0 w ... 0... ... ... ... ... ...
0 0 0 0 ... w −
=
pnpnW w=
FFT Algorithm – Decimation in Time
2j 2 2N / 2N / 2 Nw e w w
π−
= = =
FFT Algorithme – Decimation in TimeFourier Transform X[k] for N=8
111
110
101
100
011
010
001
000
n =
111
011
101
001
110
010
100
000
k =
x[0]
x[4]
x[2]
x[6]
x[1]
x[5]
x[3]
x[7]
X[0]
X[1]
X[2]
X[3]
X[4]
X[5]
X[6]
X[7]
+
+
+
−
+
+
+
−
+
+
+
−
+
+
+
−
+
+
+
+
+
−
+
−
+
+
+
+
+
+
+
+
+
−
+
−
+
−
+
−
⊗
⊗
⊗
⊗1N / 2w
1N / 2w
+
+
+
+
+
−
+
−
⊗0Nw
⊗1Nw
⊗2Nw
⊗3Nw
0N / 2w
0N / 2w
0
2jj NNw w e e
π−− Ω= = =
N/2 N/2 d
N/2 N/2 d
W W W
W W W
xx
⋅ = ⋅
evenindex
Nodd
/2
e
N
ind xXX
Nx
N/2x
N NNw w 1= =
( )
1N/2
2N/ 2
3dN/ 2
N/ 2 1N/ 2
1 0 0 0 ... 00 w 0 0 ... 00 0 w 0 ... 0
W 0 0 0 w ... 0... ... ... ... ... ...
0 0 0 0 ... w −
=
pnpnW w=
FFT Algorithm – Decimation in Frequency
2j 2 2N / 2N / 2 Nw e w w
π−
= = =
[ ][ ][ ][ ]
[ ][ ]
[ ]
( )
( )
( )
1 (N/2 1) N/2 (N 1)
2 2 (N/2 1) 2 N/2 2(N 1)
3 3(N/2 1) 3 N/2 3(N 1)
N/2 1 N/2 1 (N/2 1)
N/2
1 1 1 1 ... 11 w w w ... w1 w w w ... w1 w w w ... w... ... .... ... ... ... ...
1 w w1 w
− −
⋅ − ⋅ −
⋅ − ⋅ −
− − ⋅ −
=
X 0
X 2
X
X 1
X 3
X ½N-1...
..X N-1
½N.
…( )
( )
( )
( )
( )
[ ][ ][ ][ ]
[ ][ ]
[ ]2
N/2 1 N/2 N/2 1 (N 1)
N/2 (N/2 1) N/2 (N/2) N/2 (N 1)
(N 1) (N 1) (N/2 1) (N 1) N/2 (N 1)
x 0x 1x 2
3...
w w x ½N -1...
x ½Nw w w... ... ... ... ... ... ...
x N -11 w w w ... w
− ⋅ − −
⋅ − ⋅ −
− − ⋅ − − ⋅ −
FFT Algorithme – Decimation in FrequencyFourier Transform X[k] for N=8
111 111
x[0]
x[1]
x[2]
x[3]
x[4]
x[5]
x[6]
x[7]
X[0]
X[4]
X[2]
X[6]
X[1]
X[5]
X[3]
X[7]
110
101
100
011
010
001
000
n =
011
101
001
110
010
100
000
k =
+
+
+
+
+
+
+
+
+
−
+
−
+
−
+
−
⊗0Nw
⊗1Nw⊗
2Nw
⊗3Nw
+
+
+
+
+
−
+
−
+
+
+
+
+
−
+
−
+
+
+
−
+
+
+
−
+
+
+
−
+
+
+
−⊗1N / 2w
⊗0N / 2w
⊗0N / 2w
⊗1N / 2w
0
2jj NNw w e e
π−− Ω= = =
[ ][ ][ ][ ]
[ ][ ]
[ ]
( )
( )
( )
( )
( )
( )
0
1 2 3 (N 1)
2 4 6 2(N 1)
j3(N 1)3 6 9
N / 2 1 N / 2 1 2 N / 2 1 3
N / 2 N / 2 2 N / 2 3
1 1 1 1 ... 1x 01 w w w ... wx 11 w w w ... wx 21 w w w ... wx 3
1 ... ... ... ... ... ......N 1 w w w wx N / 2 1
...1x N / 2 w w w
...x N 1
− − − − −
− − − − −
− − Ω− − −
− − − − − − −
− − −
= ⋅ − −
( )
( )
[ ][ ][ ][ ]
[ ][ ]
[ ]2
N / 2 1 (N 1)
N / 2 (N 1)
(N 1) (N 1)2 (N 1)3 (N 1)
w... ... ... ... ... ...
1 w w w ... w
− −
− −
− − − − − − − −
X 1
X 3
X N/2 - 1
X N -
X 0
X 2
X
.
N/
..
1
2...
N/2 d N/2
N/2 d
N 2
NN/2
/
x
W W W1N
W W
x
W
∗ ∗ ∗ ⋅ = ⋅
∗ ∗ ∗ ⋅
evenin
oddindex
dex
XX
Nx
N / 2x
N NNw w 1= =
( )
1N / 2
2N / 2
3dN / 2
N / 2 1N / 2
1 0 0 0 ... 00 w 0 0 ... 00 0 w 0 ... 0
W 0 0 0 w ... 0... ... ... ... ... ...
0 0 0 0 ... w
−
−
−
− −
∗ =
pnpnW w=
IFFT Algorithm – Decimation in Frequency
2j 2 2N / 2N / 2 Nw e w w
π−
= = =
IFFT Algorithme – Decimation in FrequencyTime Function x[n] for N=8
111
110
101
100
011
010
001
000
n =
111
011
101
001
110
010
100
000
k =
X[0]
X[4]
X[2]
X[6]
X[1]
X[5]
X[3]
X[7]
x[0]
x[1]
x[2]
x[3]
x[4]
x[5]
x[6]
x[7]
+
+
+
−
+
+
+
−
+
+
+
−
+
+
+
−
+
+
+
+
+
−
+
−
+
+
+
+
+
+
+
+
+
−
+
−
+
−
+
−
⊗
⊗
⊗
⊗1
N / 2w−
1N / 2w−
+
+
+
+
⊗
+
−
+
−
⊗
⊗
⊗
0Nw
1Nw−
2Nw−
3Nw−
0N / 2w
0N / 2w
⊗1N⊗
1N
⊗1N⊗
1N
⊗1N
⊗1N
⊗1N
⊗1N
0
2jj NNw w e e
π−− Ω= = =
oddindex
N/2 N/2 d
N/2
evenindex
N/2 d
W W W1N
Wx
W
x
W
∗ ∗ ∗ ⋅ = ⋅
∗ ∗ ∗ ⋅
N/2
N
X
X
NX
N/2X
N NNw w 1= =
( )
1N/2
2N/ 2
3dN/ 2
N/2 1N/2
1 0 0 0 ... 00 w 0 0 ... 00 0 w 0 ... 0
W 0 0 0 w ... 0... ... ... ... ... ...
0 0 0 0 ... w
−
−
−
− −
∗ =
pnpnW w=
IFFT Algorithm – Decimation in Time
2j 2 2N / 2N / 2 Nw e w w
π−
= = =
[ ][ ][ ][ ]
[ ][ ]
[ ]
( )
( )
1 (N/2 1) N/2 (N 1)
2 2 (N/2 1) 2 N/2 2(N 1)
3 3(N/2 1) 3 N/2 3(N 1)
N/2 1
N/2
x 1
x
x 0
x 2
x ½
3
x ½N 1
N
1 1 1 1 ... 11 w w w ... w1 w w w ... w1 w w w ... w
1 ... ... .... ... ... ... ......N 1 w
1 w..
N.
x 1
− − − − − −
− − ⋅ − − ⋅ − −
− − ⋅ − − ⋅ − −
− −
−
=
−
−
( )
( )
( )
( )
( )
( )
[ ][ ][ ][ ]
[ ][ ]
[ ]2
N/2 1 (N/2 1) N/2 1 N/2 N/2 1 (N 1)
N/2 (N/2 1) N/2 (N/2) N/2 (N 1)
(N 1) (N 1) (N/2 1) (N 1) N/2 (N 1)
w w w...
w w w... ... ... ... ... ...
1 w w w ... w
− − ⋅ − − − ⋅ − − −
− ⋅ − − ⋅ − −
− − − − ⋅ − − − ⋅ − −
X 0X 1X 2X 3...X ½N-1
X ½N...
X N-1
…
IFFT Algorithme – Decimation in TimeTime Function x[n] for N=8
X[0]
X[1]
X[2]
X[3]
X[4]
X[5]
X[6]
X[7]
x[0]
x[4]
x[2]
x[6]
x[1]
x[5]
x[3]
x[7]
n =
111
011
101
001
110
010
100
000
111
110
101
100
011
010
001
000
k =
+
+
+
+
+
+
+
+
+
−
+
−
+
−
+
−
⊗0Nw
⊗1
Nw−
⊗2
Nw−
⊗3
Nw−
+
+
+
+
+
−
+
−
+
+
+
+
+
−
+
−
+
+
+
−
+
+
+
−
+
+
+
−
+
+
+
−⊗1
N / 2w−
⊗0N / 2w
⊗0N / 2w
⊗1
N / 2w−
⊗1N⊗
1N
⊗1N⊗
1N
⊗1N
⊗1N
⊗1N
⊗1N
Matlab for Calculation of DFT (FFT)• Matlab Fourier representation is based on
DTF (FFT) calculations• Spectrum calculation of x(t):
– Define sampling interval Ts– Define t-range T for sampling (e.g. t=[0:Ts:T])– Number of samples N=T/Ts– Define corresponding frequency range (e.g.
f=[0:1/T:1/Ts] (or cyclic frequency ω=2πf)– Matlab command X=fft(x) calculates N DFT-samples– Spectrum of x is obtained:
• Magnitude spectrum: plot(f,abs(X))• Phase spectrum: plot(f,angle(X))
• Time function calculation of X(f):– Matlab command x=ifft(X) calculates N samples of
IDTF– x is generally a complex number, but usually very
near real, when x(t) represents a physical quantity – Time function is therefore found by x(t)=real(x)– A measure of approximation error is the imaginary
part denoted by imag(x)
Example of fft calculation:Ts=0.01;T=20;N=T/Ts;t=0:Ts:T;f=0:1/T:1/Ts;w=2*pi*f;x=exp(-t);X=fft(x);subplot(2,1,1);plot(f(1:50),abs(X(1:50)));subplot(2,1,2);plot(f(1:50),angle(X(1:50)));
Useful Matlab Commands
For vectors of sampled values x and y of time dependent physical functions x(t) and y(t) we can calculate:
• Convolution: x∗y=conv(x,y)• Crosscorrelation: Rxy=xcorr(x,y)• Autocorrelation: Rxx=xcorr(x,x)• p-multiple zero padding when FFT: fft(x,p)• p-multiple zero padding when IFFT: ifft(x,p)• Performing DTFS: x[n]=N·ifft(X)↔X[k]=1/N·fft(x)
Signal ProcessingPart 6.1 Laplace Transform Principles
Magnus Danielsen
Laplace Transform ( ) ( ) ( )
( ) ( ) ( )
L st
L st
0
Bilateral (double sided) : x t X s x t e dt
Unilateral (single sided) x t X s x t e dt
∞−
−∞
∞−
←→ =
←→ =
∫
∫
( ) ( )0
0
jst
j
Inverse Laplace Transform x t X s e dtσ + ∞
σ − ∞
= ∫
s-plane
σσ0
jω s=σ+jω
Integration loop for evaluation of inverse Laplace transform
Fourier transform: s = jω ⇒ X(s) = X(jω)
Eigenvalue Property of est
( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( )
( ) ( )
s t st s st
st
s
Operator formulation : y t H x t
Convolution : y t H x t h t x t h x t d
h e d e h e d e H s
Eigenfunction : x t eEigenvalue : H(s)
Transfer function eigenvalue : H s h e d
∞
−∞
∞ ∞−τ − τ
−∞ −∞
∞− τ
−∞
=
= = ∗ = τ − τ τ
= τ τ = τ τ =
=
= = τ τ
∫
∫ ∫
∫
H( ) stx t e= ( ) ( ) ( )sty t H x t e H s= =
Region of Convergence - ROC
( ) ( ) ( )L stt 0
0
Causal signal : x(t) 0 x t X s x t e dt∞
−<= ⇒ ←→ = ∫
ROC: Re(s)>σ0
σ
jω s-plane
( ) ( ) ( )Lat at st
0
Example: Causal exponential signal:
1x t e u t e u t e dts a
ROC : Re(s) a
∞−
−
= ←→ =−
>
∫
( ) ( ) ( )0
L stt 0
Anti causal signal : x(t) 0 x t X s x t e dt−>
−∞
− = ⇒ ←→ = ∫
ROC: Re(s)<σ0
σ
jω s-plane
( ) ( ) ( )0
Lat at st
Example: Anti-causal exponential signal:
1x t e u t e u t e dts a
ROC : Re(s) a
−
−∞
= − ←→ − =−
<
∫
Unilateral Laplace Transforms Relations for Causal Signals
( ) ( ) ( ) ( )
( )
( ) ( )( ) ( )
( ) ( ) ( ) ( )
0
L
L
L as
s t L0
L
• Linearity ax t by t aX s bY s
1 s• Scaling x at Xa a
• Time shift x t a e X s
• s-domain shift e x t X s s
• Convolution x t y t X s Y s
• Diffe
−
+ ←→ +
←→
− ←→ ⋅
←→⋅ −
∗ ←→ ⋅
( ) ( )
( ) ( ) ( )
( ) ( )
( ) ( )
L
L
tL
-
s
drentiation in s-domain t x t X sds
d• Differentiation in time x t s X s x 0dt
1• Integration x t dt X ss
• Initial value theorem lim (s X s ) x 0
• Final val
−
∞
+→∞
− ⋅ ←→
←→ ⋅ −
←→
⋅ =
∫
( ) ( )s 0
ue theorem lim (s X s ) x→
⋅ = ∞
( ) ( )( ) ( )
L
L
x t X s
y t Y s
←→
←→
Given Laplace transforms :
Some Basic Transform Pairs and use of Integral Relation
( )
( )
( )
( )
L
tL
-t
L2
-t
L23
-n-1
Ln
: (t) 1
1: u(t) t dts
1: t u(t) u t dts
1 1t u(t) t u t dt2 s
t 1u(t)n - 1 ! s
∞
∞
∞
δ ←→
= δ ←→
⋅ = ←→
⋅ = ⋅ ←→
⋅ ←→
∫
∫
∫
Delta function
Step function
Ramp function
Inverse Laplace Transforms of a Rational Function
( )( )
k kMM M 1 m
kM M 1 m 0 k 1NN N 1 n
N N 1 n 0 kk 1
in engineering
is of the form of a rational function w ith real coefficients a and b :
s cb s b s ... b s bX (s)a s a s ... a s a s d
−− =
−− =
−+ + + += =
+ + + + −∏∏
C om m only occuring Laplace transform
( )
( )
N 1 N 1' k ' kM N M Nk kk kk 0 k 0
k kN Nk 0 k 0kk kk 0 k 1
M N R N Pk k,rk rk 0 r 1 k 1
k
with real and complex poles into a sum of simpler function:
b s b sX(s) s s
a s s d
es where R is the mu
s d
− −− −= == =
= =
− −
= = =
= α + = α +−
= α +−
∑ ∑∑ ∑∑ ∏
∑ ∑ ∑
Decomposition of X(s)
ltiplicity (order) of the poles.
is obtained using
inverse Laplace transform for each of these simpler terms
The inverse Laplace transform x(t) of X(s)
k
k
M re a l o r com p lex ze ros c , w h e re th e n om in a tor is z e ro
N rea l o r c om p lex p o le s d , w h e re th e d en om in a to r is z e ro C om p lex p o le s an d ze ros occu r in com p lex p a ir w h en p re sen t
•
••
X (s) h a s :
Alternative Decomposition With Complex Pol-pairs Present
N 1 ' kM N M Nkk kk 0
k kNk 0 k 0kkk 0
with real and complex poles, into a sum of simpler function
The complex poles are paired giving quadric forms in the denominator.
b sX(s) s s
a s
−− −== =
=
= α + = α∑∑ ∑∑
Decomposition of X(s)
( ) ( )
( ) ( )
N 1 ' kkk 0
2k k k
real complexpoles pole pairs
R RM N k k,r k,r k,r
k r rk 0 2r 1 k 1 r 1 k 1k k kand real and complex
poles only pole pairs only
b s
s d s s
e f s gs
s d s s
where R is the multiplicity (order) of the poles.
−
=
−
−
== = = =
−
+− ⋅ + γ + ζ
+= α + +
− + γ + ζ
∑∏ ∏
∑ ∑ ∑ ∑ ∑
Th is obtained using
inverse Laplace transform for each of these simpler terms
e inverse Laplace transform of X(s)
( )( )
( )( )
( ) ( )( )
( ) ( )( )( )
( ) ( )( )( )
L22 2
2 2L
22 2
2L
22 2
L-a t22 2
2 2L-a t
22 2
-a t
- :
2 st sin ts
st cos ts
1 2t cos t sin ts
-
2 s ae t sin t
s a
s ae t cos t
s a
e
⋅
⋅
⋅
ω⋅ ω ←→
+ ω
− ω⋅ ω ←→
+ ω
− ω⋅ ω − ω ←→
ω + ω
ω +⋅ ⋅ ω ←→
+ + ω
+ − ω⋅ ⋅ ω ←→
+ + ω
2 m ultiple com plex pole pair = ±jω
2 m ultiple com plex pole pair = -a ± jω :
( ) ( )
( )( )
L
2
22 2
1t cos t sin t
2
s a
⋅ ⋅ ω − ω ←→ ω − ω
+ + ω
( )
( ) ( )
( )
( )
Ld t
Ld t2
n 1Ld t
n
L2 2
L2 2
1e u ( t )s d
1t e u ( t )s d
t 1e u ( t )n 1 ! s d
sin ts
scos ts
− ⋅
− ⋅
−− ⋅
⋅ ←→−
⋅ ⋅ ←→−
⋅ ⋅ ←→− −
ωω ←→
+ ω
ω ←→+ ω
Single pole, real or com plex = d :
2 - m ultiple pole = d :
n - m ultiple pole = d :
C om plex pole pair = ±jω :
C om plex pole pair = -a ± jω :
( )( )
( ) ( )( )
La t2 2
La t2 2
e sin ts a
s ae cos t
s a
− ⋅
− ⋅
ω⋅ ω ←→
+ + ω
+⋅ ω ←→
+ + ω
Laplace Transform Pairs of some Basic Functions
Examples of Decomposition of Laplace Transforms:
( ) ( ) ( ) ( ) ( ) ( )( ) t 2t 3t
3s 5 1 1 2: X(s)s 1 s 2 s 3 s 1 s 2 s 3
x t e e 2 e− − −
+= = + −
+ + + + + +
= + − ⋅
Three single real poles
( ) ( ) ( ) ( )
( )( )( )( ) ( )
( )( )( ) ( )( )
( ) ( ) ( )
2 2
32
2 2 2
t ½t ½t
s 2 1 s 1 X(s)s 1s 1 s s 1 s s 1
s ½ s ½1 1 ¾3s 1 s 1s ½ ¾ s ½ ¾ s ½ ¾
x t e e cos ¾t e sin ¾t− − −
+ −= = −
++ + + + +
+ − += − = − +
+ ++ + + + + +
= − +
1 real pole and 1 complex pole - pair :
( ) ( ) ( ) ( ) ( )( )
2 2
2t 4t 4t
3s 10 1 1 1 X(s)s 2 s 4s 2 s 4 s 4
x t e e t e− − −
+= = − +
+ ++ + +
= − + ⋅
Multiple real poles :
Solving Differential Equations with Initial Conditions
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( )
( ) ( ) ( ) ( )( ) ( )2 t 5t
d 1 1y t y t x tdt RC RCd y t 5y t 5x tdtsY s y 0 5Y s 5X s
3 15 25X(s) y(0 ) 5 s 2Y ss 5 s 5
2s 1 1 3s 5 s 2 s 2 s 5
y t (e 3e )u t
−
−
− −
+ =
+ =
− + =
−+ += =+ +
− += = −
+ + + +
= −
+
x(t)
-
R=1k
C=200µ
+
y(t)
-( )
( )
L2t3 3 1x t e5 5 s 2
RC 1k 200µy 0 2
−
−
= ←→+
τ = = ⋅
= −
- 1 - 0 . 5 0 0 . 5 1 1 . 5 20
0 . 1
0 . 2
0 . 3
0 . 4
0 . 5
0 . 6
0 . 7
00
0.6x(t)
- 1 - 0 . 5 0 0 . 5 1 1 . 5 2- 2
- 1 . 5
- 1
- 0 . 5
0
0 . 5
0
-2 0
y(t)
Laplace Transforms in Circuit Analysis - Impedance and Initial Generators
• Resistance
• CapacitanceInitial voltage = VC(0-)
• InductanceInitial current = IL(0-)
R
+ VR (s) –
IR (s)
+ VC (s) –
IC (s) 1sC
+ −
Cv (0 )s
−
sL
+ VL (s) –
IL (s)
+−
LLI (0 )−
+ VC (s) –
IC (s)
1sC
CC v (0 )−⋅
+ VL (s) –
IL (s)
Li (0 )s
−
sL
R
+ VR (s) –
IR (s)
Example of Laplace Transformed Circuit(Low-pass Filter)
1
1sC
+−
1Cv (0 ) 0.5s s
− =
2
1sC
+−
2Cv (0 ) 0.3s s
− =
R1R2
+
VR(s)
–
sL
+−
LI (0 ) L 0.1 L− ⋅ = ⋅
+−( )
1V ss
+
=
−
Bilateral Laplace Transform Properties
• Definition: x(t) for − ∞ < t < ∞• ROC must be given for the transform• x(t) found from X(s) depends of given ROC• Calculation rules same as for unilateral exept
differentiation, where x(0-) does not appear: dx(t)/dt has the transform sX(s)
• Care must be taken to ROC.
Example of Bilateral Laplace Transformand Inverted Time Function
( )( ) ( ) ( ) ( ) ( )3
2 2 2 2
4s 26s 1 1 1 1: X(s)s 2 s 2 s 3 s 3s 2 s 3
( ) : 2 Re(s) 2
−= = + + +
+ − + −− −
− < < +
Laplace transform
Regionof convergence ROC
2t 2t 3t 3tx(t) e u(t) e u( t) e u(t) e u( t)− −= + − + + −
Inverted Laplace transform time function :
First and third term are causal Second and forth term are anti-causal
x x x x
ROC
3-2 2-3
x = poless-plane
Signal ProcessingPart 6.2 The Laplace Transform – Transfer Functions
Magnus Danielsen
Transfer Function• Definition of transfer function H(s)=Y(s)/X(s), where Y(s)
and X(s) are output and input signals respectively• Initial conditions are zero• Causality – unilateral transforms• Stability – all poles are in ”negative” half of s-plane• Minimum phase – all zeros are in ”negative” halfplane• Inverse system transfer function H-1(s) of H(s) such that
H-1(s)·H(s)=1• Frequency response determined by poles and zeros• Bode diagrammes• Decibel – definition:
– A=20 logH ,where H is a transfer function. – A=10 log(P2/P1), where P2 and P1 are the output and input powers
respectively.
Transfer Function of System Definition and Differential Equation
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )M M 1 N N 1
M M 1 0 N N 1 1 0M M 1 N N 1
d d d d d db y t b y t ... b y t b y t a x t a x t ... a x t a x tdt dt dt dt dt dt
− −
−− −+ + + + = + + + +
Differential equation :
( )( )
MM M 1 mkM M 1 m 0 k 1
NN N 1 nN N 1 n 0 kk 1
s cY(s) b s b s ... b s bH (s) KX(s) a s a s ... a s a s d
−− =
−− =
−+ + + += = = ⋅
+ + + + −∏∏
Definition of transferfunction :
( )M 1 N 1 N 2
M 1 N 1 N 2
d d d d d For it applies that: y ... y y t 0 and x x ... x x 0dt dt dt dt dt
− − −
− − − −= = = = = = = = =
Initial conditions :
t = 0
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )M M 1 N N 1M M 1 1 0 N N 1 1 0b s Y s b s Y s ... b sY s b Y s a s X s a s X s ... a sX s a X s− −
− −+ + + + = + + + +
Laplace transform of differential equation :
H(s) Lx(t) X(s)←→ Ly(t) Y(s)←→
Transfer Function and Impulse Response
( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )
L
Y s H s X s
y t h t x t , where h t H s
x t t y t h t t h t
=
= ∗ ←→
= δ ⇒ = ∗δ =
Laplace transform of output signal :
Output signal time function :
Impulse response h(t) :
Lx(t) X(s)←→ Ly(t) Y(s)←→( ) ( )Lh t H s←→
( )
( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )
L
1Example on impulse response : H ss a
x t t X s 1
Y s H s X s H s
y t h t exp at
=+
= δ ←→ =
= =
= = −
Stability of a SystemFor a transfer function H(s) with impulse response h(t) holds
• Stable systems:– H(s) contains poles in negative half s-plane only– h(t) contains only terms decaying exponentially with time*
– The system is stable• Oscillating unstable systems:
– H(s) contains poles on the imaginary axis of the s-plane– h(t) contains oscillating terms versus time t– The system is oscillating unstable
• Unstable systems:– H(s) contains poles in positive half s-plane – h(t) contains exponentially increasing terms versus time t– The system is unstable
(*impulses can also be present)
Transfer Function -Inverted Transfer Function
( ) ( )Lh t H s←→ ( ) ( )Ly t Y s←→( ) ( )Lx t X s←→
( ) ( )( )
( )
( )
M Mk
k kk 0 k 1N N
kk k
k 0 k 1
b s s cY sH s K
X s a s s d
= =
= =
−= = = ⋅
−
∑ ∏
∑ ∏
( ) ( )k kN M
k kk kk 0 k 0
d da y t b x tdt dt= =
=∑ ∑ ( ) ( ) ( ) ( )k k
L Lk kk k
All initial values are zero for x(t), y(t), and their derivatives d dy t s Y s x t s X sdt dt
⇒
←→ ←→
Transfer function: Inverted transfer function:
( ) ( )( )
( )
( )
N Nk
k k1 k 0 k 1
M Mk
k kk 0 k 1
a s s dX s 1H sY s Kb s s c
− = =
= =
−= = = ⋅
−
∑ ∏
∑ ∏
The inverted transfer function H-1(s) has:• Poles equal to zeros of the transfer function H(s)• Zeros equal to poles of the transfer function H(s)
Inversion, Example 1
( ) ( )Lh t H s←→ ( ) ( )Ly t Y s←→( ) ( )Lx t X s←→
( ) ( )( )
( ) ( ) ( ) ( )
2
3t
Y s s s 2 4Transfer function : H s s 2X s s 3 s 3dImpulse response: h t t 2 t 4e u tdt
−
+ −= = = − +
+ +
= δ − δ +
( ) ( )( )
12
1 t 2t
X s s 3 1 1Inverse transfer function : H sY s s s 2 s 1 s 2
Impulse response : h (t) u(t)e u(t)etwo poles : s 1, 2Inverse system is unstable.
−
− −
+= = = −
+ − − +
= −= −
( ) ( ) ( ) ( ) ( )2
2
d d dy t 3y t x t x t 2x tdt dt dt
+ = + −
Inversion, Example 2
( ) ( )Lh t H s←→ ( ) ( )Ly t Y s←→( ) ( )Lx t X s←→
( ) ( )( )
( ) ( ) ( ) ( )
1
-1 7t
X s (s 3)(s 2) 50Inverse Transfer function : H s s 8Y s s 7 s 7
dImpulse response: h t t 8 t 50e u tdt
− + −= = = + +
− −
= δ + δ +
( ) ( )( )
( ) ( ) ( )3t 2t
Y s 2 1 s 7Transfer function : H sX s s 3 s 2 (s 3)(s 2)
Impulse response: h t 2e u t e u t−
−= = − =
+ − + −
= −
Magnitude Response (Amplitude Characteristics)Phase Response (Phase Characteristics)
( ) ( )Lh t H s←→ ( ) ( )Ly t Y s←→( ) ( )Lx t X s←→
( ) ( )( )
( )
( )
MM
kk 1 kk 1
N N
kk 1 k 1 k
s 1s cY s cFactorised transfer function : H s K K '
X s ss d 1d
==
= =
−−
= = ⋅ = ⋅ − −
∏∏
∏ ∏
We apply for usually occuring systems M ≤ N
( )M M
k 1 k 1k k
Definition of magnitude response :
j jA 20log H j 20log K ' 20log 1 20log 1c d= =
ω ω= ω = + + − +∑ ∑
( ) ( )( )( )( )
M N
k 1 k 1k k
Definition of phase response :
Im H jH j arctan K ' arctan arctan
Re H j c d= =
ω ω ωϕ = ∠ ω = = ∠ + − ω
∑ ∑
Bode Diagram of First-order Real Pole Factor
10-2
10-1
100
101
102
-40
-35
-30
-25
-20
-15
-10
-5
0
10-2
10-1
100
101
102
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
nNormalized cyclic frequency ω/ω
nNormalized cyclic frequency ω/ω
Mag
nitu
de re
spon
se A
(ω)
(dB)
Phas
e re
spon
se ϕ
(ω)
(radi
an)
Positive real pole (unstable H)
-π
-π/2
10-2
10-1
100
101
102
-40
-35
-30
-25
-20
-15
-10
-5
0
10-2
10-1
100
101
102
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
Phas
e re
spon
se ϕ
(ω)
(radi
an)
nNormalized cyclic frequency ω/ω
nNormalized cyclic frequency ω/ω
Mag
nitu
de re
spon
se A
(ω)
(dB)
Negative real pole
-π/2
-π
Bode Diagram of First-order Real Zero Factor
10-2
10-1
100
101
102
0
5
10
15
20
25
30
35
40
10-2
10-1
100
101
102
0
0.5
1
1.5
2
2.5
3
3.5
nNormalized cyclic frequency ω/ω
nNormalized cyclic frequency ω/ω
Mag
nitu
de re
spon
se A
(ω)
(dB)
Phas
e re
spon
se ϕ
(ω)
(radi
an)
π
π/2
Negative real zero
10-2
10-1
100
101
102
0
5
10
15
20
25
30
35
40
10-2
10-1
100
101
102
0
0.5
1
1.5
2
2.5
3
3.5
nNormalized cyclic frequency ω/ω
nNormalized cyclic frequency ω/ω
Mag
nitu
de re
spon
se A
(ω)
(dB)
Phas
e re
spon
se ϕ
(ω)
(radi
an)
π
π/2
Positive real zero
Example on a Transfer Function (1)
( ) ( )Lh t H s←→ ( ) ( )Ly t Y s←→( ) ( )Lx t X s←→
( ) ( )( ) ( )22
Y s s+2 s+2H sX s s 2s 1 s 1
= = =+ + +( ) ( ) ( ) ( ) ( )
2
2
d d dy t 2 y t y t x t 2x tdt dt dt
+ + = +
Frequency response: s = jω
( ) ( ) ( )( ) ( ) ( )2 2s j
Y j j + 2 j( / 2 )+1H j H s 2X j j 2 j 1 j 1= ω
ω ω ωω = = = =
ω ω + ω + ω +
( ) ( )
( )( ) ( ) ( )½ ½ ½2 2 2
A 20log H j
20log 2 20log / 2 1 20log 1 20log 1
ω = ω =
+ ω + − ω + − ω +
( ) ( ) ( ) ( )H 0 arctan / 2 arctan arctanϕ = ∠ ω = + −ω − ω − ω
Amplitude characteristics:
Phase characteristics:
( ) ( )
( )( ) ( ) ( )2 ½½2 2½
20log 2 20
Amplitude characteris
20log 1 20lolog / 2 1
tics : A 20 g H j
g 1
lo
−+ ω + − ωω
=
+
ω
+
ω =
Example on a Transfer Function(2)
110− 010 110 210 310
20log 2 10
10−
20−
( )A(dB)
ω
30−
40−
20
ω
( ) ( ) ( ) ( )arctan / 2 arctaH n a ctan0 r+ −ϕ = ∠ ω = ω − ω − ω
Example on a Transfer Function(3)Phase characteristics
110− 010 110 210 310
2π
( )Hϕ = ∠ ω
2π
−
−π
ω
Example on Transfer Function with Complex Poles
( ) ( ) ( ) ( )( ) 2s j
n n
Y s 1H j H s H sX s j j2 1
= ωω = = = =
ω ω+ ζ + ω ω
( ) ( )( ) 2
n n
Y s 1H sX s s s2 1
= =
+ ζ + ω ω
( ) ( )
12 22 2
2
n n
A 20log H j 20log 1 4 ω ω ω = ω = − − + ζ ω ω
( ) ( ) n2s j
n
2H j H s arctan
1= ω
ω ζω ϕ = ∠ ω = = − ω − ω
Frequency response: s = jω
Amplitude characteristics:
Phase characteristics:
Bode Diagram of Second-order Pole Factor
10-1
100
101
102
-100
-80
-60
-40
-20
0
20 ζ = 0.1
0.2
0.5
1.0
nNormalized cyclic frequency ω/ω
Mag
nitu
de re
spon
se A
(ω)
(dB
)
10-1
100
101
102
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0ζ = 0.1
0.2
0.5
1.0
nNormalized cyclic frequency ω/ω
Phas
e re
spon
se ϕ
(ω)
(rad
ian)
Complex pole-pair with negative real value
Bode Diagram of Second-order Zero Factor
10-1
100
101
102
-20
0
20
40
60
80
100
ζ = 1.0
0.5
0.2
0.1
( )A ω
Amplitude characteristics:
Complex zeros-pair in either left half or right half of the s-plane results in same amplitude characteristics
Am
plitu
de (
dB)
nNormalized cyclic frequency ω/ω
1 0-1
1 00
1 01
1 02
0
0 .5
1
1 .5
2
2 .5
3
3 .5
ζ = 0.1
0.2
0.5
1.0( )ϕ ω
Phase characteristics:
Pha
se (
radi
an)
Complex zeros-pair in left half of the s-plane
nNormalized cyclic frequency ω/ω
10-1
100
101
102
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0Phase characteristics:
Pha
se (
radi
an)
nNormalized cyclic frequency ω/ω
Complex zeros-pair in right half of the s-plane
ζ = -1.0- 0.5- 0.2
- 0.1
Minimum, Maximum and Mixed Phase Systems• Transfer functions H(s) with zeros in right and left half s-plane
with identical real parts results in:– Identical maginitude response | H(s) |– Different phase response ∠H(s)– Different impulse response h(t)
• Minimum phase systems:– Zeros in negative half s-plane only– Narrowest impulse respons h(t)– Inverse system H-1(s) containes poles in negative half s-plane only– Inverse system is stable
• Maximum phase systems:– Zeros in positive half s-plane only– Impulse respons h(t) is not the narrowest possible– Inverse system H-1(s) containes poles in positive half s-plane only– Inverse system is unstable
• Mixed phase systems:– Zeros in positive and negative half s-plane– Impulse respons h(t) is not the narrowest possible– Inverse system H-1(s) containes poles in positive half s-plane– Inverse system is unstable
Minimum and Maximum Phase Analogue Signals
Zero: s = αs=jω
s-plane
ϕ
Minimum phase: zero in s < 0
When 0 ≤ ω ≤ ∞:- ϕ0 ≤ ϕ = ∠(jω − α) ≤ π/2
ϕ0
s-plane
Zero: s = α
s=jω ϕ
Maximum phase: zero in s > 0
When 0 ≤ ω ≤ ∞:π/2 ≤ ϕ = ∠(jω − α) ≤ π + ϕ0
ϕ0
Maximum and Minimum Phase Example on Analogue Signal
( ) ( )s 3Maximum phase : H
s 4 s 5−
=+ +
( ) ( )s 3Minimum phase : H
s 4 s 5+
=+ +
5t 4t
Maximum phase impulse response :h(t) 8e 7e− −= −
5t 4t
Minimum phase impulse response :h(t) 2e e− −= −
0 0.5 1 1.5 2 2.5-0.5
0
0.5
1Maximum phase
time (sec)
impu
lsre
spon
se (a
rb.u
nits
)
0 0.5 1 1.5 2 2.5-0.5
0
0.5
1Minimum phase
time (sec)
impu
lsre
spon
se (a
rb.u
nits
)
0 10 20 30 40 50-2
0
2
4Maximum phase
cyclic frequency (rad/sec)
Pha
se (r
adia
n)
0 10 20 30 40 50-1.5
-1
-0.5
0Minimum phase
cyclic frequency (rad/sec)
Pha
se (r
adia
n)
∠H = arctan(ω/3) – arctan(ω/4) – arctan(ω/4)
∠H = π – arctan(ω/3) – arctan(ω/4) – arctan(ω/4)
Signal ProcessingPart 7.1 Digital Filters – Overview
and z-Transforms
Magnus Danielsen
Introductory Remarks
• FIR – filters:requires finite numbers of terms in a DFTF
• IIR – filters:requires infinite numbers of terms in a DFTF
• Equalization:requires inverted transfer functions
• Generalized method for construction of digital transfer function constructions:
z – transforms – a generalisation of DTFTz – transform propertiesz – transform applications
Elementary FIR Filter: FIR – filter
h[n]↔ H(ejΩ)
y[n]↔Y(ejΩ)x[n]↔X(ejΩ)
0 1 2 3 4 5 6 7-60
-50
-40
-30
-20
-10
0
10
0 0.5 1 1.5 2 2.5 3 3.50
0.2
0.4
0.6
0.8
1
1.2
1.4
-30 -20 -10 0 10 20 30-0.05
0
0.05
0.1
0.15
0.2
hd[n]↔ Hd(ejΩ)
hd[n]=Ωc/π sinc[(Ω/π(n-M/2)]
Hd(ejΩ)=ejMΩ/2 for Ω≤Ωc=0 for Ω>Ωc
-40 -30 -20 -10 0 10 20 30 40-0.05
0
0.05
0.1
0.15
0.2
h[n] = hd[n] for n≤6= 0 for n>6
20logHd (dB)
Desired filter
Approximated filter of M+1=13th order
Infinite Impulse Response or IIR – Filters
• Transfer function is a DTFT corresponding to an infinite impulse response series
• Realisation requires feedback in the block diagram
• Best realized by generalization of DTFT by introduction of z – transform
• Method for filter construction will be based on bilinear transform, and converting of analogue transfer functions to digital transfer functions
z – Transform Definitionj
n n n jn n n
Define a complex number: z r e r cos jr sinSpecifically for z we obtain : z r e r cos n jr sin n
Ω
± ± ± ± Ω ± ±
= ⋅ = Ω + Ω
= ⋅ = Ω ± Ω
[ ]
[ ] [ ] n
n
For a discrete signal: x n
we define z-transform: X z x n z∞
−
=−∞
= ∑±±±
[ ] ( ) [ ]z n
n
z transformation : x n X z = x n z∞
−
=−∞
− ←→ ∑
[ ] ( ) [ ]
j
DTFT j jn
n
DTFT transformation for comparison, replacing z by e :
x n X e = x n e
Ω
∞Ω − Ω
=−∞
−
←→ ∑
Inverse z – Transform [ ] ( ) [ ]z p
p
z-transform: x n X z = x p z∞
−
=−∞
←→ ∑
jz re Ω=
jdz rd je z jdΩ= Ω ⋅ = ⋅ Ωz-plane( )
[ ] [ ]
[ ] [ ]
n 1
c
p n 1 n p 1
p pc c
n p j(n p)
p c
Determine the integral X z z dz
x p z z dz x p z z jd
jx p r e d jx n 2
−
∞ ∞− − − −
=−∞ =−∞
∞− −
=−∞
= = ⋅ Ω
= Ω = ⋅ π
∫
∑ ∑∫ ∫
∑ ∫
[ ] ( ) n 1
c
1Inverse z-transform: x n X z z dz2 j
−=π ∫
Convergence - ROC
( ) [ ] [ ] [ ]n n n
n n n
X z = x n z converges when x n z x n r∞ ∞ ∞
− − −
=−∞ =−∞ =−∞
≤ < ∞∑ ∑ ∑
Definition for region of convergencearea in z plane, where sum converges
== −ROC
[ ] [ ]n
n n 1
n 0
Example with exponential discrete signal x n u n 1:
r ROC : z r∞
− −
=
= α α >
α < ∞ → = > α∑
Unit Circle in z – Plane is Transformed into the Frequency Axis for DTFT
j
unit circle :z r 1 and z e Ω= = =
jz re Ω=z-plane
rΩ
( ) [ ]
[ ]
jj
nz e
n z e
jn j
n
X z = x n z
x n e X e
ΩΩ
∞−
==−∞ =
∞− Ω Ω
=−∞
= =
∑
∑
[ ] [ ]
( )( )
1 2
j j j j2
x n ...0 0 0 1 2 1 1 0 0 0...
z transform : X z z 2 z z
DTFT transform : X e e 2 e e
− −
Ω Ω − Ω − Ω
= −
↑
− = + − +
= + − +
Example comparing z – transform and DTFT
Difference Equation and z – Transform
[ ] ( ) [ ]
[ ] [ ] [ ] ( )0 0
z n
n
(n n ) nz n0 0
n n
x n X z = x n z
x n-n x n-n z x n z z X z
∞−
=−∞
∞ ∞− + −−
=−∞ =−∞
←→
←→ = = ⋅
∑
∑ ∑
Time shift:
[ ] ( ) [ ] ( )[ ] [ ] ( ) ( )
z z
z
x n X z y n Y z
a x n b y n a X z b Y z
←→ ←→
⋅ + ⋅ ←→ ⋅ + ⋅
Linearity:
N M
k kk 0 k 0
a y[n k] b x[n k]= =
− = −∑ ∑
Difference equation:
( ) ( )N M
k kk k
k 0 k 0
a Y z z b X z z− −
= =
=∑ ∑z-transformed equation:
( ) ( )( )
Mk
kk 0N
kk
k 0
b zY zH z
X z a z
−
=
−
=
= =∑
∑Transfer Function :
Transfer FunctionPoles and Zeros
x[n] y[n][ ] ( )zh n H z←→( )X z ( )Y z
( ) ( )( )
Mk
kk 0
Nk
kk 01 M
0 1 M1 N
0 1 NM
1k
0 k 1N
10k
k 1
b zY zH z
X z a z
b b z ... b za a z ... a z
(1 c z )ba (1 d z )
−
=
−
=
− −
− −
−
=
−
=
= =
+ + +=
+ + +
−= ⋅
−
∑
∑
∏
∏
Transfer Function :
Poles: dk where k=1 ... NZeros: ck where k=1 ... M
Causality and Anticausality – Two Examples
[ ] [ ] ( )zn n n1
n 0
Causal expon
1x n u n X z z ROC : z
ential signal:
1 z−
−=
= α ←→ = α = > α− α∑
0 α Re(z)
Im(z)
ROC
0 α Re(z)
Im(z)
ROC
[ ] [ ] ( )1
zn n n
n
n n 11 1
n 1
x n u n
Anticausal e
1 X z z
1 1z z ROC : z
xponential
1 z
si
1 z
gnal:−
−
=−∞
∞− −
− −=
= −α − − ←→ = −α
= −α = −α = < α− α − α
∑
∑
Two – sided SignalExample
[ ] [ ] [ ]
( )
n
z
1ROC:z 11ROC: z2
1x n u n 1 u n2
1 1X z 1 11 z 1 z2
−<
>
= − − − +
←→ = − + − −
( )( )
3z z2X z
11 z z2
1ROC : z 12
− = − −
< <
Re(z)
Im(z)
ROC
0 ½ 1
Signal ProcessingPart 7.2 The z-Transform and its Properties
Magnus Danielsen
Properties of Region of Convergence (ROC)
N-n
n 00
-n
n N
ROC contains no polesROC = tot. z-plane, except z 0,z
Causal x[n]: X(z) x[n]z has only a pole in z 0
Anticausal x[n]: X(z) x[n]z has only a pole in z
=
=−
•• = = ∞
− = =
− = = ∞
•
∑
∑
Finite duration signal x[n] :
Infinite duration si
( )1
-n -n -n
n n n 0
ROC = anular region in z-plane
X z x[n]z x[n]z x[n]z I I∞ − ∞
− +=−∞ =−∞ =
= = + = +∑ ∑ ∑
gnal x[n] :
Re(z)
Im(z)
ROC of I+
Causal signal part
Im(z)
Re(z)
ROC of I−
Anticausal signal part
Re(z)
Im(z)
ROC of X(z)
Two-sided signal
Examples of Two-sided ROC[ ] [ ] [ ] ( )
n nz
n n0n n
1 10 ROC: z2 1ROC: z
4
1 1x n u n 2 u n X z2 4
1 1 1 2z 2 z 12 4 1 2z 1 z4
∞− −
−−∞ <
>
= − − + ←→ =
− + = + + −∑ ∑ 1 1ROC : < z <
4 2
[ ] [ ] [ ] ( )n n
z
n nn n
1 10 01 1ROC: z ROC: z2 4
1 1y n u n 2 u n Y z2 4
1 1 1 2z 2 z 1 12 4 1 z 1 z2 4
∞ ∞− −
− −
> >
= − + ←→ =
− + = + + −
∑ ∑ 1ROC : z >2
[ ] [ ] [ ] ( )n n
z
n n0 0n n
1 1ROC: z ROC: z2 4
1 1w n u n 2 u n W z2 4
1 1 1 2z 2 z2 4 1 2z 1 4z
− −
−∞ −∞ < <
= − − + − ←→ =
− + = + + − ∑ ∑ 1ROC : z <
4
Ex.1
Ex. 2
Ex. 3
Definition of ROC
x
x
ROC, named R is defined by r z r1 1 1ROC, named is defined by z
R r r
+ −
− +
< <
< <
ROC of a right-sided signal is of the form z r causal signal
ROC of a left-sided signal is of the form z r anti causal signal
ROC of a two-sided signal is of the form: r z r non causal signal
+
−
+ −
> ⇒
< ⇒ −
< < ⇒ −
mixed causal and anti causal signal= −
Properties and Calculation Rules[ ] ( ) [ ] ( )
[ ] [ ] ( ) ( )
[ ]
[ ] ( )
[ ]
0
zx y
z
xnz
0 x
znx
: :
Linearity : ax n by n aX z bY z ROC: R R
1 1Time reversal : x n X ROC:z R
Time shift : x n n z X z ROC: R
zMultiplication exp. sequence: x n X ROC: R
Convolution
−
←→ ←→
+ ←→ +
− ←→
− ←→
α ←→ α α
z zx yx n X z ROC R y n Y z ROC R
∩
[ ] [ ] ( ) ( )
[ ] ( )
zx y
zx
: x n y n X z Y z ROC: R R
dDifferentiation in z domain: nx n z X z ROC: Rdz
∗ ←→
←→−
∩
Example: Application of Calculation Rules
[ ] [ ] [ ] [ ]
x w y
z
1ROC R R R : z 42
1 z 42x n w n y n X z 1 z 4z2 = < <
− −= ∗ ←→ = ⋅ − +
∩
[ ] [ ] [ ] [ ] [ ] [ ] [ ]n n1 1w n n u n y n u n
2 4
− − ∗ = ⋅ = − x n = w n y n
[ ]
[ ] [ ]
[ ] [ ]
nz
nz
1ROC z2
nz
ROC z 4
1 zu n 12 z2
1 z1 d z 2w n n u n z 1 12 dz z z2 2
1 4y n u n4 z 4
>
−
<
− ←→ ⇒ +
− − = ⋅ ←→− = + +
− = − ←→ −
Examples on Basic z – Transforms (1)[ ] [ ] ( )z n
10 ROC z 1
1 x n u n X z z1 z
∞−
−>
= ←→ = =−∑Discrete step signal :
[ ] [ ] ( ) ( )zn n n
10
ROC z a
1 x n a u n X z a z1 az
∞−
−
>
= ←→ = =−∑Exponential signal :
[ ] [ ] ( )zROC total z plane
(the only function with ROC = total z plane):
x n n X z 1−
−
= δ ←→ =
Discrete impulse signal
[ ] [ ] ( )z k
ROC z 0x n n k X z z−
>= δ − ←→ =Discrete impulse signal :
Examples on Basic z – Transforms (2)
[ ] [ ] [ ]
( )
( ) ( ) ( )( ) ( )
0 0
0 0
0 0
0 0
0 0
j n j n zn n n0
j n j nn n n nj j1 1
0 0
j j1 1
j j1 1
1 1x n a cos( n)u n a e a e u n2 2
1 1 1 1 1X z a e z a e z2 2 2 1 ae z 1 ae z
1 ae z 1 ae z12 1 ae z 1 ae z
Ω − Ω
∞ ∞Ω − Ω− −
Ω − Ω− −
Ω − Ω− −
Ω − Ω− −
= Ω = + ←→
= + = + − −
− + − ⋅= =
⋅ ⋅− −
∑ ∑-1
0-1 2
0
1 - acosΩ zX z1 - 2a cosΩ z + a
ROC z a>
-2z
Damped cosine function
[ ] [ ] [ ]
( )
( ) ( ) ( )( ) ( )
0 0
0 0
0 0
0 0
0 0
j n j n zn n n0
j n j nn n n nj j1 1
0 0
j j1 1
j j1 1
1 1x n a sin( n)u n a e a e u n2 j 2 j
1 1 1 1 1X z a e z a e z2 j 2 j 2 j 1 ae z 1 ae z
1 ae z 1 ae z12 j 1 ae z 1 ae z
Ω − Ω
∞ ∞Ω − Ω− −
Ω − Ω− −
− Ω Ω− −
Ω − Ω− −
= Ω = − ←→
= − = − − −
− − − ⋅ ⋅= =
⋅ ⋅− −
∑ ∑-1
0
0
a sinΩ zX z1 - 2a cosΩ z
ROC z a>
-1 2 -2+ a z
Damped sine function
Inversion of z – TransformsPartial Fraction Method
( ) ( )( )
( )( )
Mk
1 Mk M Nkk 0 0 1 M
kN 1 Nk k 00 1 N
kk 0
M N Nk k
k 1k 0 k 0 k
b zB z B zb b z ... b zX z f zA z a a z ... a z A za z
Af z (no multiple poles)1 d z
−− − −
−=− −
− =
=
−−
−= =
+ + += = = = +
+ + +
= +−
∑∑
∑
∑ ∑
[ ] zn kk k k 1
k
ACausal solution ROC z d : A d u n1 d z−> ←→−
[ ] zn kk k k 1
k
AAnti causalsolution ROC z d : A d u n 11 d z−− < − − − ←→−
Example 1: Inversion of z – TransformsPartial Fraction Method
( )( ) ( )
( ) ( ) ( )
1 2
1 1 1
1 2 31 1
1
1 z zX z with ROC 1 z 211 z 1 2z 1 z2
A A AX z1 1 2z 1 z1 z2
− −
− − −
− −−
− += < < − − −
= + + − −−
[ ] [ ] [ ] [ ]n
n1x n u n 2 2 u n 1 2 u n2
= + ⋅ − − − ⋅
( ) ( ) ( )1
21
1z 2
2 3
1 1 2 2A 1 z X z 12 1 2 2 1 2
Similarily : A 2 A 2−
−
=
− + = − = = − ⋅ − = = −
Re(z)
Im(z)
ROC of X(z)
½ 1 2
Poles and ROC
Example 2: Inversion of z – TransformsPartial Fraction Method
( )
( )11 1
3 2
2
1 2 3
1 2
3 2 1
2 1
11
2 1
z 10z 4z 4X z2z 2z 4
z 1 10z 4z 4z2 1 z 2zz 4z 4z 10z 12 2z z 1z 5z 22z 32 2z z 1
z z2 2
1 32z 3 W z1 z 1 2z
− − −
− −
− − −
− −
−−
− −
−− −
− − +=
− −− − +
= ⋅− −− − +
= ⋅− −
− + + − + −
+ − −
= ⋅ − + + − − +
⋅ =
= ⋅
[ ] [ ] [ ] ( ) [ ] [ ]
[ ] [ ] [ ] ( ) [ ] [ ]n 1
n n
n
w n 2 n 1 3 n 1 u n 13 1x n n n 1 1
3
u n 2 3 2 u n 2
2 1
2 2
u n+= −δ + δ + − − − − +
= − δ − + δ − − − − + ⋅ − −
⋅ − −
Re(z)
Im(z)
-1 +2
Poles and ROC: |z|<1
ROC of X(z)
Inversion of z – TransformsPower Series Expansion Method
( )1
1
2 z 1X z ROC z1 21 z2
−
−
+= ⋅ >
−Example1 :
( )
[ ] [ ] [ ] [ ] [ ]
11 2 3
1
2 z 1X z 2 2z z z .....1 21 z2
1x n 2 n 2 n 1 n 2 n 3 .........2
−− − −
−
+= = + + + +
−
= δ + δ − + δ − + δ − +
( ) 2zX z e ROC z= ≠ ∞Example2 : ( )
[ ]
20
z 2k n
k 0 n 2kn even
1 1X z e z znk! !
20 for n 0 or n odd
1x n for n 0 and n evenn !
2
∞−
= =− =−∞
= = =−
>= ≤ −
∑ ∑
Signal ProcessingPart 7.3 The z-Transform – Transfer Functions
Magnus Danielsen
Transfer Function• Definition of transfer function: H(z)=Y(z)/X(z), where Y(z)
and X(z) are output and input signals respectively• Initial conditions are zero• Causality – unilateral transforms• Stability – all poles are inside the unit circle of the z-plane• Minimum phase – all zeros are inside the unit circle of the
z-plane• Inverse system transfer function H-1(z) of H(z) such that
H-1(z)·H(z)=1• Frequency response is determined by poles and zeros• Decibel – definition:
– A=20 logH ,where H is a transfer function. – A=10 log(P2/P1), where P2 and P1 are the output and input powers
respectively.
Transfer Function
( ) ( )( )
Mk
kk 0N
kk
k 0
b zY zH z
X z a z
−
=
−
=
= =∑
∑Transfer Function :
x[n] y[n][ ] ( )zh n H z←→( )X z ( )Y z
[ ] [ ] ( )
[ ] ( ) [ ] [ ] ( )
( )
nz
x1
nn z
11
y1 1
1 1 1x n u n X z ROC R z13 31 z3
1 3 1y n 3 1 u n u n Y z 13 1 z 1 z3
4 ROC R z 111 z 1 z3
−
−−
− −
= − ←→ = > +
= − + ←→ = + + −
= > + −
( )( ) ( )
1
y x y1 1
14 1 zY z 3H(z) ROC R R R : z 11X z 1 z 1 z3
−
− −
+ = = = >
+ −
∩
Example on a transfer function
Transfer Function and Impulse Response
( ) ( ) ( )[ ] [ ] [ ] [ ] ( )[ ] [ ] [ ] [ ] [ ] [ ]
L
Y z H z X z
y n h n x n , where h n H z
x n n y n h n n h n
=
= ∗ ←→
= δ ⇒ = ∗ δ =
Z transform of output signal :
Output signal time function :
Impulse response h(t) :
[ ] ( )zx n X z←→ [ ] ( )zy n Y z←→[ ] ( )zh n H z←→
[ ]
[ ] [ ] ( )( ) ( ) ( ) ( )[ ] [ ] [ ]
1
z
n
1Example on impulse response : H z1 az
x n n X z 1
Y z H z X z H z
y n h n a u n
−=+= δ ←→ =
= =
= =
Causal System • Causality requires that h[n]=0 for n<0• This requires rightsided transfer function
Stability for a Causal Signal• A pole |dk|<1 inside the unit circle, gives an
exponentially decaying term • A pole |dk|>1 outside the unit circle, gives
an exponentially increasing term • A pole |dk|=1 on the unit circle, gives an
complex sinusoid term (i.e. Constant ampl.)
Anti-causal System • Anti-causality requires that h[n]=0 for n≥0• This requires leftsided transfer function
Stability for a Anti-causal Signal• A pole |dk|<1 side the unit circle, gives an
exponentially decaying term for n → +∞• A pole |dk|>1 outside the unit circle, gives
an exponentially increasing term for n → -∞• A pole |dk|=1 on the unit circle, gives an
complex sinusoid term (i.e. Constant ampl.)
Example on Causality and Stability( )
[ ] [ ] [ ] ( ) [ ]
( ) [ ]
1j j1 14 4
j j4 4
n nj j n4 4
n
2 2 3H z1 2z1 0.9e z 1 0.9e z
The system shall be either or .
Poles z 0.9e z 0.9e z 2
Stable system : h n 2 0.9e u n 2 0.9e u n 3 2 u n 1
4 0.9 cos n u n 34
π π −−− −
π π−
π π−
= + ++
− −
= = = −
= ⋅ + ⋅ + − − −
π = + −
stabel causalSolution :
( ) [ ]
[ ] [ ] [ ] ( ) [ ]
( ) [ ] ( ) [ ]
n
n nj j n4 4
n n
2 u n 1
Causal system : h n 2 0.9e u n 2 0.9e u n 3 2 u n
4 0.9 cos n u n 3 2 u n4
π π−
− −
= ⋅ + ⋅ + −
π = + −
Inverse Systemsx[n] y[n][ ] ( )zh n H z←→( )X z ( )Y z
[ ] ( )zinv invh n H z←→ x’[n]
( )X ' z
[ ] [ ][ ] [ ] [ ]( ) ( )
inv
inv
x ' n x n requires that
h n h n n
H z H z 1
=
∗ = δ
⋅ =
( ) ( )( ) ( )( ) ( )
inv
inv
inv
1H zH z
Poles in H z are converted to zeros in H z
Zeros in H z are converted to poles in H z
=
Example on a Stable and Causal Inverse System
1 1 1y[n] y[n 1] y[n 2] x[n] x[n 1] x[n 2]4 4 8
− − + − = + − − −Difference equation :
x[n] y[n][ ] ( )zh n H z←→( )X z ( )Y z
[ ] ( )zinv invh n H z←→ x’[n]
( )X ' z
( ) ( )( )
( ) ( )
1 2 1 1
1 2 1 2
1 1 1 11 z z (1 z )(1 z )Y z 4 8 4 2H z 1 1X z 1 z z (1 z )4 2
1 1 1z and z z (double pole)4 2 2
− − − −
− − −
+ − − += = =
− + −
= = − =
Transfer function :
Zeros for H z : Poles for H z :
( ) ( )( )
( ) ( )
1 2
inv
1 1
1(1 z )Y z 2H z 1 1X z (1 z )(1 z )4 2
1 1 1z (double zero) z and z2 4 2
Both poles are inside unit circle stable inverted systemBoth zeroes are inside
−
− −
−= =
− +
= = = −
⇒
inv inv
Inverse Transfer function :
Zeros for H z : Poles for H z :
unit circle minimum phase inverted system⇒
Example on Multipath Communication System
y[n] x[n] ax[n k]= + −Difference equation :
( )
( ) ( ) ( )
k
r phase(a)1 1 jkk k
H z 1 az
z a a e where r 1,3,...,2k 1
−
π +
= +
= − = = −
Transfer function:
Zeros for H z : No polesfor H z
( ) ( )( )
( ) ( ) ( )
invk
r phase(a)1 1 jkk k
Y z 1H zX z 1 az
z a a e where r 1,3,..., 2k 1
All poles are inside unit circle if a <1 stable inverted system
−
π +
= =−
= − = = −
⇒
inv inv
Inverse Transfer function :
Poles for H z : No zeros for H z
Sampling time = TsTime difference between direct and reflected signal: ∆T=kTs
Direct signal
Reflected signal
Transmitter Receiver
Mountain
Unilateral z – Transform used for Causal Systems
[ ]
[ ] [ ]
[ ] [ ]
[ ]
uz n
n 0
z n
n
Definition : x n is defined for n
unilateral z transform : x n x n z ,
and bilateral z transform : x n x n z
are identical for causal signals, i.e. if x n 0 for n 0
∞−
=
∞−
=−∞
− ∞ < < ∞
− ←→
− ←→
= <
∑
∑
[ ] [ ]uz n
n 0
x n x n z∞
−
=
←→∑ [ ] [ ] [ ] [ ]uz n 1 n
n 0 n 0
x n-1 x n-1 z x -1 z x n z∞ ∞
− − −
= =
←→ = +∑ ∑
[ ] [ ] [ ] [ ] [ ] [ ]uz n 1 k 1 k n
n 0 n 0
x n-k x n-k z x -k x -k+1 z ...x -1 z z x n z∞ ∞
− − − + − −
= =
←→ = + + +∑ ∑
Difference Equation with Initial Condition
N M
k kk 0 k 0
a y[n k] b x[n k]= =
− = −∑ ∑Difference equation:
z-transformed equation:
( ) ( ) ( ) ( ) ( )A z Y z C z B z X z+ =
( )N
kk
k 0
A z a z−
=
= ∑ ( )M
kk
k 0
B z b z−
=
= ∑ ( ) [ ]N 1 N
kk
m 0 k m 1
C z a y k m z−
−
= = +
= − +∑ ∑
Initial conditions for y[n]: y[-1], y[-2] , y[-3] , y[-4] , .... , y[-N+1]
Initial conditions for x[n]: 0, 0, 0, ... i.e. X[n] is taken to be causal
Signal ProcessingPart 7.4 The z-Transform – Transfer Functions Frequency Response and Block Diagrammes
Magnus Danielsen
Frequency Response
( ) ( )j
Mj
kj N M k 1
Nz ej
kk 1
(e c )with H z K e
(e d )Ω
Ω
Ω − ==
Ω
=
−= ⋅
−
∏
∏jΩTransfer Function z = e :
( ) ( )( )
Mk
1 Mk0 1 Mk 0
N 1 Nk 0 1 N
kk 0
M M1
k kN Mk 1 k 1
N N1
k kk 1 k 1
b zY z b b z ... b zH zX z a a z ... a za z
(1 c z ) (z c )K K z
(1 d z ) (z d )
−− −
=− −
−
=
−
−= =
−
= =
+ + += = =
+ + +
− −= ⋅ = ⋅
− −
∑
∑
∏ ∏
∏ ∏
Transfer Function :
jΩegjΩIllustration of the factor : e - g
jΩe - g
Ω
Example: Amplitude and Phase Characteristics
-4 -3 -2 -1 0 1 2 3 4-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
-4 -3 -2 -1 0 1 2 3 40
2
4
6
8
10
12
14
|H|
∠H
Ω
Ω
( ) ( )1 1
1 2j j1 14 4
1 z 1 zH z1 0.9 2z 0.81z1 0.9e z 1 0.9e z
− −
π π − −−− −
+ += = − +− −
Transfer function :
( )
( ) ( )
j
j
z e j 2 j
2 2
:
1 eH z
1 0.9 2e 0.81e
2 2cos
1 0.9 2 cos 0.81cos2 0.9 2 sin 0.81sin 2
Ω
− Ω
= − Ω − Ω
+=
− +
+ Ω=
− Ω + Ω + Ω + Ω
Amplitude characteristic
( )
( )( )
( )( )
jz e
:
H z
sin( ) sin( )sin 4 4arctan arctan arctan1 cos 1 cos 1 cos
4 4sin 0.9 2 sin 0.81sin 2arctan arctan1 cos 1 0.9 2 cos 0.81cos2
Ω=∠ =
π π−Ω + −Ω −−Ω
− −π π+ −Ω + −Ω + + −Ω −
−Ω Ω + Ω
= −+ −Ω − Ω + Ω
Phase characteristic
Minimum, Maximum and Mixed Phase Systems
• Transfer functions H(z) with zeros inside the unit-circle and outside the unitcircle in the z-plane with reciprocal values result in:
– Identical maginitude response | H(z) |– Different phase response ∠H(z)– Different impulse response h[n]
• Minimum phase systems:– Zeros inside the unit circle in the z-plane only– Narrowest impulse respons h[n]– Inverse system H-1(z) containes poles inside the unit circle in the z-plane only– Inverse system is stable
• Maximum phase systems:– Zeros outside the unit circle in the z-plane only– Impulse respons h[n] is not the narrowest possible– Inverse system H-1(s) containes poles outside the unit circle in the z-plane only– Inverse system is unstable
• Mixed phase systems:– Zeros inside and outside the unit circle in the z-plane– Impulse respons h[n] is not the narrowest possible– Inverse system H-1(s) containes poles outside the unit circle in the z-plane– Inverse system is unstable
Minimum and Maximum Phase Discrete Signals( ) 0j1H z 1 az Zero : z a | a | e ϕ−= − = =
Minimum phase: Zero z=a inside unit circle |z| < 1
When – π ≤ Ω ≤ π:- Arcsin|a| ≤ ϕ = ∠(1-az-1) ≤ Arcsin|a|
z=e-jΩ
z-plane
ϕ
ϕ0
z=a- Ω
z=ae-jΩ
1–ae-jΩ
Maximum phase: Zero z=a outside unit circle |z| >1
When – π ≤ Ω ≤ π:– π ≤ ϕ = ∠(1-az-1) ≤ π
z=e-jΩ
z-planeϕ
ϕ0
z=a- Ω
z=ae-jΩ
1–ae-jΩ
Maximum and Minimum Phase Example on Discrete Signal
1
1 1
Maximum phase :1 2zH
1 11 z 1 z3 4
−
− −
+= + +
1
1 1
Minimum phase :2 zH
1 11 z 1 z3 4
−
− −
+= + +
-4 -2 0 2 4-4
-3
-2
-1
0
1
2
3
4Maximum phase
cyclic frequency (rad/sec)
Pha
se (r
adia
n)
-4 -2 0 2 4-4
-3
-2
-1
0
1
2
3
4Minimum phase
cyclic frequency (rad/sec)
Pha
se (r
adia
n)
-2 0 2 4 6 8 1 0-1 .5
-1
-0 .5
0
0 .5
1M a xim um p ha se
n
impu
lsre
spon
se
-2 0 2 4 6 8 1 0-0 .5
0
0 .5
1
1 .5
2M inim um p ha se
n
impu
lsre
spon
se
[ ] [ ] [ ]n n
Maximum phase impulse response :
1 1h n 20 u n 21 u n3 4
= − +
[ ] [ ] [ ]n n
Minimum phase impulse response :
1 1h n 4 u n 6 u n3 4
= − +
Difference EquationsN M
k kk 0 k 0
a y[n k] b x[n k]= =
− = −∑ ∑
M N
k kk 0 k 1
y[n] b x[n k] a y[n k]= =
= − − −∑ ∑
Recursive formula for y[n]
( ) ( ) ( )M N
k kk k
k 0 k 1
Y z b z X z a z Y z− −
= =
= −∑ ∑
z-transform Y(z) for y[n]
Block Diagram – Discrete 2nd ordery[n] = b0x[n] + b1x[n-1] + b2x[n-2] - a1y[n-1] – a2y[n-2]
w
Y(z) = b0X(z) + b1z-1X(z) + b2z-2X(z) - a1z-1 Y(z) – a2z-2Y(z)
W(z)
W(z)X(z)
z-1
+
+
b2
b1
b0
z-1
z-1
+
+
-a2
-a1
Y(z)
z-1
z-1
+
+
b2
b1
X(z)
-a2
-a1
Y(z)
+
b0
z-1
z-1 z-1
z-1
+
+
b2
b1
Y(z)b0+
+
-a2
-a1
X(z)
z-1
z-1
( ) ( )( )
Y zH z
X z=
Block Diagram – Discrete N’th ordery[n] = b0x[n] + b1x[n-1] + b2x[n-2]+ ...+ bMx[n-M] – a1y[n-1] – a2y[n-2] – .... – aNy[n-N]
wY(z) = b0X(z) + b1z-1X(z) + b2z-2X(z)+...+ bMX(z) – a1z-1 Y(z) – a2z-2Y(z) –...– aNz-2Y(z)
W(z)
( ) ( )( )
Mk
kk 0N
kk
k 0
b zY zTransfer function : H z
X z a z
−
=
−
=
= =∑
∑
+
+
b2
b1
Y(z)b0+
+
-a2
-a1
X(z)
z-1
z-1
z-1
bM-aN
++
Cascade-form Implementation
+
+
b’2
b’1
Y(z)b’0+
+
-a’2
-a’1
W(z)
z-1
z-1
+
+
b2
b1
b0+
+
-a2
-a1
X(z)
z-1
z-1
( ) ( )( )
( )( )
( )( ) ( ) ( )1 2
Y z W z Y zH z H z H z
X z X z W z= = ⋅ = ⋅
Example: Cascade-form Implementation
( ) ( )( )( )1 1
1j j1 1
1
2j j1 18 84 4
1 jz 1 jzH
1 11 e z 1 e z
1 zH
3 31 e z 1 e z42 2 4
H z− −
π π
−
π π−− − −−−
+ −
− −
+
− −
= =
⋅
( ) ( )( )
2
11 2
W z1 zH z1 X Z1 cos z z
4 4
−
− −
+= =
π − +
( ) ( ) ( )( )
1
21 2
1 z Y zH z
3 9 W z1 cos z z2 8 16
−
− −
+= =
π − +
+
1
1+
+
-1/4
cos(π/4)
X(z)
z-1
z-1
+
1
Y(z)1+
+
-9/16
3/4⋅cos(π/8)
z-1
z-1
W(z)
Parallel-form Implementation+
+
b’2
b’1
Y’(z)b’0+
+
-a’2
-a’1
X(z)
z-1
z-1
+
+
b2
b1
Y’’(z)b0+
+
-a2
-a1
X(z)
z-1
z-1
X(z) + Y(z)=Y’(z)+Y’’(z)=[H’(z)+H’’(z)]X(z)
Signal ProcessingPart 8.1 Analogue Filters
Magnus Danielsen
Filter Types• Ideal filters:
– Constant amplitude in passband
– Linearly varying phase versus frequency
• Low-pass filters (LP)• High-pass filters (HP)• Band-pass filters (BP)• Band-stop filters (BS)• Characteristics (amplitude and
phase):– Butterworth filters– Chebyshev filters
• Frequency transformations– LP to HP transformation– LP to BP transformation– LP to BS transformation
• Analogue filters– Passive– Active filters– Electronic circuits designed
filters• Digital filters
– FIR– IIR– Processor based designed
filters
Distortion free Ideal Transmission of Analogue Signal
( )
( )
0j ts j
s j
0
Y(s)H( j ) H s CeX(s)
h(t) C t t
− ω= ω
= ω
ω = = =
= δ −
( ) ( )Lh t H s←→ ( ) 0y t Cx(t t )= −( )x t0stY(s) CX(s)e−=
H(jω)
ω
C
Amplitude characteristics
ϕ=∠ H(jω)
dϕ/d ω= - t0
ω
Phase characteristics
Ideal transmission: constant amplitude amplification C, and group delay t0 for all frequencies composing the input signal x(t)
Ideal Low-pass Filters( ) ( )Lh t H s←→ ( )y t( )x t
0j tc
c
e forH( j )
0 for
− ω ω ≤ ωω = ω > ω( )
cc 00
c c
j ( t t )j t j t
0
c 0 c c0
0
1 1 eh(t) e e d2 2 j(t t )
sin (t t )sinc (t t )
(t t )
ωω ω −− ω ω
−ω −ω
= ω = π π −
ω − ω ω = = − π − π π
∫
H(jω)
ω
C
− ωc ωc
ϕ=∠ H(jω)ω
− ωc ωc
-3 0 -2 0 -1 0 0 1 0 2 0 3 0-0 .1
-0 .0 5
0
0 .0 5
0 .1
0 .1 5
0 .2
0 .2 5
0 .3
0 .3 5
t0
h(t)
Transmission of Rectangular Time Pulse Gibbs Phenomenon
0
0
T1 for t2x(t)T0 for t2
≤= >
( ) ( )Lh t H s←→ ( )y t( )x t
( )c 0 c c0
0
sin (t t )h(t) sinc (t t )
(t t )ω − ω ω = = − π − π π
( ) ( ) ( )0
0
T2
c 0
T 02
0 0c 0 c 0
sin (t t )y(t) x h t d d
(t t )
1 T TSi (t t ) Si (t t )2 2
∞
−∞ −
ω − − τ= τ − τ τ = τ
π − − τ
= ω − + − ω − − π
∫ ∫
( )u
0
sinSi(u) d
λ= λ
λ∫
Rectangular input pulse: Ideal low-pass filter impulse response :
Output pulse response:
Sine Integral Function Si(u)
-40 -30 -20 -10 0 10 20 30 40-2
-1.5
-1
-0.5
0
0.5
1
1.5
2Sine integral
u (radian)
Si(u
)
( )u
0
sinSi(u) d
λ= λ
λ∫
maxSi(u) 1.1790
2π
≅ ⋅
/ 2π
/ 2−π
Gibbs PhenomenonRectangular Pulse filtred in Ideal Bandpass Filter
-0.5 0 0.5 1 1.5-0.2
0
0.2
0.4
0.6
0.8
1
1.2
Time (sec)
Out
put s
igna
l ωcT0 =200
-0.5 0 0.5 1 1.5-0.2
0
0.2
0.4
0.6
0.8
1
1.2
Time (sec)
Out
put s
igna
l ωcT0 =20
-0.5 0 0.5 1 1.5-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (sec)
Out
put s
igna
l
-0.5 0 0.5 1 1.5-0.2
0
0.2
0.4
0.6
0.8
1
1.2
Time (sec)
Out
put s
igna
l ωcT0 =4ωcT0 = 2
T0 T0
T0T0
maxy 1.09
maxy maxy
maxy
Design of Filters
• Prescription of frequency response of amplitude and phase characteristics
• Function selected must be causal and stable
• Realization of approximating transfer function by a physical system
• Analogue approach
• Analogue to digital approach
• Direct digital approach
Amplitude characteristic
0 ωp ωs
1
δ
Passband Stoppband
ω
Transition band
|H(jω)|2
−
2
1 ε1=
1+ γ
Fundamental Filter Parameters
Definition of filter passband parameters:• Passband tolerance parameter ε• Passband maximum attenuation αp = − 10⋅log(1 − ε) dB• Passband cutoff cyclic frequency ωp radian/sec• Stopband cutoff frequency fp = ωp/(2π) Hz
Definition of filter passband parameters:• Stopband tolerance parameter δ• Stopband minimum attenuation αs = − 10⋅ log(δ) dB• Stopband cutoff cyclic frequency ωs radian/sec• Stopband cutoff frequency fs Hz
• Maximally flat responsedkH(jω)/dωk=0 for ω=0 and ω=∞ :
– Butterworth filter characteristics
• Equiripple magnitude response– In passband, Chebyshev I filter characteristics– In stopband, Chebyshev II filter characteristics– In both passband and stopband, Elliptic filter
characteristics
Approximating Functions
Magnitude Response Description for Low-pass Filters
( )
( ) ( )2 22s j
Transfer function for analogue filter: H s1Magnitude response: H s H j
1 F ( )= ω= ω =
+ ω
n
c
nc
cn
Functions to be selected for filter design:
Butterworth filters F( ) maximally flat filter
Chebyshev I filters F( ) T equiripple filter in passband:
1Chebyshev II filters F( )T
ωω = ω
ωω = γ ω
ω =ω
γω
( )
n
n n
T is the Chebyshev polynomium equiripple filter in stopband :
Elliptic filters F( ) R equiripple filter in both passband and stopband: R is the elliptic function
ω = ω
Digital Filters used for Analogue Signals
• Signal is samples with sampling time Ts• Transfer function is described by
H(ejΩ), where -π < Ω < π• Normalized cyclic frequency = Ω• Cyclic frequency ω = Ω/Ts• Transfer function in cyclic frequency
domain:H = H(ejωTs) where -π/Ts< ω < π/Ts
• H(ejωTs) is matemathically periodic in ω
Signal ProcessingPart 8.2 Analogue Butterworth Filters
Magnus Danielsen
Butterworth Filters( ) 2
2K
c
1H j K is an integer number
1
ω = ω
+ ω
0 ωp ωs
11-ε
δ
Passband Stoppband
ω
Transition band
Amplitude characteristic
H(jω)2
12K
p c
12K
s c
1
1
ε ω = ω − ε
− δ ω = ω δ
αp = − 10⋅log(1 − ε) dBαs = − 10⋅ log(δ) dB
Butterworth Transfer Function( ) ( ) ( )
( ) ( )
2
2Ks j
c
2K
c
1H s H s H jj1j
1H s H ss1
j
= ω− = ω =
ω+ ω
− =
+ ω
( ) ( )1 (2k 1)j
2K 2Kc c
1 (2k 1)j ( )2 2K
c
Poles of H s H s :
s j ( 1) j e
e , where k 1,2,...,2K
−π
−π +
−
= ω − = ω
= ω =
Stability for H(s) requires the poles of H(s) to be placed in left half-plane
⋅
c
Poles of H(s) H(-s) are distributed with equal angular distance on a circle with radius = ω :
cω
Butterworth Transfer Function( ) ( ) 2K
c
1H s H ss1
j
− =
+ ω
( ) ( )1 (2k 1)j
2K 2Kc c
1 (2k 1)j ( )2 2K
c
Poles of H s H s :
s j ( 1) j e
e ,where k 0,1,...,2K 1
+π
+π +
−
= ω − = ω
=ω = −
K= 4Poles for H(s) Poles for H(-s)
K=3Poles for H(s) Poles for H(-s)
c2j3
c
se
π±
−ω= ω
c
j3
c
se
π±
ω= ω
5j8
c7j8
c
es
e
π±
π±
ω= ω
j8
c3j8
c
es
e
π±
π±
ω=
ω
Prototype transfer functions are defined by ωc=1
Design of Buttereworth LP-FiltersDesign a Butterworth (prototype) LP-filter with ωc=1, and K=3
Poles: s= −1,
s=e-j2π/3=-½+j½√3 and s=e-j2π/3=-½-j½√3
( )( )
3 2
1 1H ss 2s 2s 11 3 1 3s 1 s s
2 2 2 2
= =+ + +
+ + − + +
Change of bandwidth from ωc = 1 to ωc = ωc :
Substitution of s by s/ωc results in:
( ) 3 2
c c c c c c
1 1H ss s 1 3 s 1 3 s s s1 2 2 12 2 2 2
= = + + − + + + + + ω ω ω ω ω ω
Butterworth Filter Prototype Transfer Functions (ωc=1)
( )
( )
( )
( )
( )
( )
2
3 2
4 3 2
5 4 3 2
6 5 4 3 2
1K 1: H ss 1
1K 2 : H ss 2s 1
1K 3 : H ss 2s 2s 1
1K 4 : H ss 2.6131s 3.4142s 2.6131s 1
1K 5 : H ss 3.2361s 5.2361s 5.2361s 3.2361s 1
1K 6 : H ss 3.8637s 7.4641s 9.1416s 7.4641s 3.8637s 1
= =+
= =+ +
= =+ + +
= =+ + + +
= =+ + + + +
= =+ + + + + +
Butterworth Polynomials (denominator of H(s))
General FormulasEven polynomials of K’th order:
Examples on odd Butterworth polynomials:B1 = s+1B3 = (s+1) ⋅ (s2 + 2s cos(4/6 π) + 1)B5 = (s+1) ⋅ (s2 – 2 s cos(6/10 π) + 1) ⋅ (s2 –2 s cos(8/10 π) + 1)B7 = (s+1) ⋅ (s2 – 2 s cos(8/14 π) + 1) ⋅ (s2 –2 s cos(10/14 π) + 1) ⋅ (s2 –2 s cos(12/14 π) + 1)
K / 22
Kn 1
2n K 1B (s) s 2s cos 12K=
+ − = − π + ∏
( )( )K 1 / 2
2K
n 1
2n K 1B (s) s 1 s 2scos 12K
−
=
+ − = + − π + ∏
Examples on even Butterworth polynomials:B2 = s2 – 2 s cos(3/4 π) + 1B4 = (s2 – 2 s cos(5/8 π) + 1) ⋅ (s2 –2 s cos(7/8 π) + 1)B6 = (s2 – 2 s cos(7/12 π) + 1) ⋅ (s2 –2 s cos(9/12 π) + 1) ⋅ (s2 –2 s cos(11/12 π) + 1)B16 = (s2 – 2 s cos(17/32 π) + 1) ⋅ (s2 –2 s cos(19/32 π) + 1) ⋅ (s2 –2 s cos(21/32 π) + 1) ⋅
(s2 – 2 s cos(23/32 π) + 1) ⋅ (s2 –2 s cos(25/32 π) + 1) ⋅ (s2 –2 s cos(27/32 π) + 1) ⋅(s2 – 2 s cos(29/32 π) + 1) ⋅ (s2 –2 s cos(31/32 π) + 1)
Odd polynomials of K’th order:
10-1
100
101
-120
-100
-80
-60
-40
-20
0
20
Butterworth Amplitude Characteristics for LP-filter
K =
1
2
3
4
5
6
Slope = K·20 dB/decade
ω/ωc
10-1
100
101
-4
-3
-2
-1
0
1
2
3
4
Butterworth Phase Characteristics for LP-filter
ω/ωc
K =
3
4
5
1
6
2
Pha
se
(ra
dian
)
Design of Butterworth LP-filters( ) 2
2K
c
1Amplitude characteristics : A 10log H j 10log
1
= ω = ω+ ω
p p
s s
Parameters K, ,Requirements: passband 0 A 20 log(1 ) dB
stopband A 20 log dB
ε δ≤ ω ≤ ω ≤ −α = − ε
ω ≤ ω ≤ ∞ ≥ −α = δ
( ) ( )( )( )
ps 0.10.1
s p
log 10 1 / 10 1K
2 log /
αα − −≥
ω ω
( )p
pc 1
0.1 2K10 1α
ωω =
−( )s
10.1 2K
s c 10 1αω = ω −
Example: Butterworth LP filter (1)Given: Passband attenuation: αp = 2 dB for 0 < f < fp = 8 kHz
Stopband attenuation: αp = 15 dB for fs = 8 kHz < f < ∞
( ) ( )( )( )
( ) ( )( )( )
ps 0.10.1 1.5 0.2
s p
log 10 1 / 10 1 log 10 1 / 10 1K 4.88 5
2 log 12 / 82 log /
αα − − − −≥ = = ≅
⋅⋅ ω ω
( ) 5 4 3 2
3 3 3 3 3
1H ss s s s s3.2361 5.2361 5.2361 3.2361 1
53 10 53 10 53 10 53 10 53 10
= + + + + + ⋅ ⋅ ⋅ ⋅ ⋅
( )p
pc 1
0.1 2K
ff 8.44 kHz
10 1α
= =−
( )s
10.1 2K
s cf f 10 1α= −
3 3c p2 f 2 8.44 10 53.0 10 rad / secω = π ⋅ = π ⋅ ⋅ = ⋅
Example: Butterworth LP filter (2)
10-1
100
101
102
-120
-100
-80
-60
-40
-20
0
10-1
100
101
102
-500
-400
-300
-200
-100
0
Frequency (kHz)
Frequency (kHz)
Pha
se (
degr
ees)
Mag
nitu
e (
dB)
Slope =20 K dB/dec- 120 dB/dec
K⋅90o=450o
K=5 fc = 8.44 kHz
Signal ProcessingPart 8.3 Analogue Chebyshev Filters and
Frequency Transformations of Filters
Magnus Danielsen
Chebyshev-I Filters( ) 2
2 2K
p
1H j K is an integer number1 T
ω = ω
+ γ ω
0 ωp ωs
11-ε
δ
Passband Stoppband
ω
Transition band
Amplitude characteristic
H(jω)
------ for odd K
for even K
Chebyshev Polynomials( )( )( )( )( )( )
0
1
22
33
4 24
5 35
T 1
T
T 2 1
T 4 3
T 8 8 1
T 16 20 5
ω =
ω = ω
ω = ω −
ω = ω − ω
ω = ω − ω −
ω = ω − ω − ω
( ) ( )( ) ( )
( ) ( ) ( )
1K
1K
K 1 K K 1
T cos K cos 1
T cosh K cosh 1
Recursion (generation) formula:T 2 T T
−
−
+ −
ω = ω ω ≤
ω = ω ω ≥
ω = ω ω − ω
Design of Chebyshev-I LP filters
p
s
Parameters K, ,Requirements: passband A 20 log dB
stopband A 20 log dB
ε δ− ≤ α = − ε
− ≥ α = − δ
( ) ( )( )( )
ps 0.10.11
1s p
cosh 10 1 / 10 1K
cosh /
αα−
−
− −≥
ω ω
( ) 2
2 2K
p
1H j K is an integer number1 T
ω = ω
+ γ ω
( ) ( )2 2
Kp
1H s H ss1 T
j
− =
+ γ ω
Chebyshev Polynomials and Chebyshev-I Amplitude Characteristics
-1.5 -1 -0.5 0 0.5 1 1.5-10
-5
0
5
10
15
20
Chebyshev polynomium
TK(ω/ωp)
K=1,2,3,4,5, and 6
ω/ωc
1
-1
-1.5 -1 -0.5 0 0.5 1 1.5-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
0
ω/ωc
A(ω/ωp)=
-10 log (1+γ2TK2(ω/ωp))
Chebyshev ampl.char.
K=1,2,3,4,5, and 6
ε=0.106 , and γ=0.5
Poles of Chebyshev-I LP-filter
( )1 11 1 1sinh sinh or sinh K sinhK
− − η = = η γ γ
Definition of parameter :
( ) ( )( ) ( ) ( )2 2 1
K K
Poles of H s H s will be found from :
T 1 and T cos K cos−−
γ ω = − ω = ω
( ) ( )1 2
r r rc
r
:
s jcos(sin j ) sin j 1 cos
(2r 1)where where r 1,..., 2K2K
−
−
= − η+ θ = −η θ − + η θω
−θ = π =
Poles for H s H s
( ) ( )2 2
Kc
1H s H ss1 T
j
− =
+ γ ω
Stability for H(s) requires the poles of H(s) to be placed in left half-plane
⋅ −2
Poles of H(s) H( s) are placed on an ellipse
with the axes ηand 1 + η :
21+ η
η
Chebyshev-I Prototype LP Transfer Function( )
( )K
2r r
r 1
1H s Cs [ sin j 1 cos ]
C must be determined from |H(j )| for ω in passband =
=+ η θ + + η θ
ω
∏
K = 3 γ = 0.5 η = 0.5 ωc = 1
Poles for H(s) Poles for H(-s)
c
1 0.97 j4
s 12
1 0.97 j4
− += −ω − −
c
1 0.97 j4
s 12
1 0.97 j4
−= ω +
( )cPrototype LP transfer function 1:
0.5H s1 1 1s s 0.97j s 0.97j2 4 4
ω =
= + + − + +
Prototype transfer functions are defined by ωc=1
( )H s 1 for s 0 C 0.5= = ⇒ =
Chebyshev I Transfer FunctionK=4 γ=0.5 ωc=1
c
0.1411 j0.98470.3407 j0.4079s0.3407 j0.40790.1411 j0.9847
− +− += − −ω − −
( ) ( ) ( ) ( ) ( )
cPr ototypeLP transfer function1H s C
s 0.1411 j0.9847 s 0.3407 j0.4079 s 0.3407 j0.4079 s 0.1411 j0.9847
ω=ω
=+ + + + + − + −
1 21 1sinh sinh 0.3688 1 1.0658K
− η = = + η = γ
Poles for H(s) Poles for H(-s)
( )H s 1 for s 0 C 0.2500= = ⇒ =
c
0.1411 j0.98470.3407 j0.4079s0.3407 j0.40790.1411 j0.9847
− −− −= − +ω − +
Example: Chebyshev I LP Filter Given: Passband attenuation: αp = 2 dB for 0 < f < fp = 8 kHz
Stopband attenuation: αp = 15 dB for fs = 12 kHz < f < ∞
( ) ( )( )
( ) ( )( )
ps 0.10.11 1 1.5 0.2
11s p
cosh 10 1 / 10 1 cosh 10 1 / 10 1K 2.77 3
cosh 12 / 8cosh /
αα− −
−−
− − − −≥ = = ≅
ω ω
c pf f 8 kHz= =
3c p
3
2 f 2 8 10
50 10 rad / sec
ω = π ⋅ = π ⋅ ⋅
= ⋅
1 ss p
p
Resulting stopband frequency:
f2.77f ' f cosh cosh 11.373 f
− = =
( )c
3 3 3
1LP transfer function 50 : H s Cs s 1 s 11 0.97j 0.97j
50 10 50 10 4 50 10 4
ω = = ⋅ + + − + + ⋅ ⋅ ⋅
Example: Chebyshev I LP FilterK=3 fc = 8 kHz
100
101
102
-80
-60
-40
-20
0
20
100
101
102
-300
-250
-200
-150
-100
-50
0
Frequency (kHz)
Frequency (kHz)
Pha
se (
degr
ees)
Mag
nitu
e (
dB)
Slope =20 K dB/dec- 60 dB/dec
K⋅90o=270o
fc
Example: Chebyshev I LP Filter K=4 fc = 8 kHz
100
101
102
-100
-80
-60
-40
-20
0
20
100
101
102
-400
-300
-200
-100
0
Frequency (kHz)
Frequency (kHz)
Pha
se (
degr
ees)
Mag
nitu
e (
dB)
Slope =20 K dB/dec- 80 dB/dec
K⋅90o=360o
fc
Frequency Transformations: Low-pass Prototype to Low-pass
(shift of cut-off frequency)
c
c
sLow-pass prototype to low-pass transformation: s
1 1ss d d
→ω
→− −
ω
( ) ( )
( ) ( )
c
LP ,prot 2
3c
LP 2 2 2c c c
c c c
sExample : Butterworth filter : s
1H (s)s 1 s s 1
1H (s)s s ss s s1 1
→ω
= →+ + +
ω= =
+ ω + ω + ω + + + ω ω ω
Frequency Transformations: Low-pass Prototype to High-pass
c
c c
Low-pass to high-pass transformation: ss
s1 1 d
s d d ss d
ω→
−→ =
ω ω− − −
( ) ( ) ( ) ( )
cc
3
LP,prot HP2 2
2
Example : Butterworth filter : s , 1s
1 1 sH (s) H (s)1 1 1s 1 s s 1 s 1 s s 11 1s s s
ω→ ω =
= → = = + + + + + ++ + +
Frequency Transformations:Low-pass Prototype to Band-pass
2 20 0 0
0
s sLow-pass to band-pass transformation: sBs B s
+ω ω ω→ = + ω
( ) ( )
( )
( )
1 20 0
0
2 20
1,22 2
0
1 1 Bss d s p s ps d
B s
½Bd ½Bdp
½Bd ½Bd
→ =− − − ω ω+ − ω
+ − ω= − − ω
Example: Butterworth Band-pass Filter0 0
0
sTransformation : sB s ω ω
→ + ω
( ) ( )LP,prot 2
2c LP,prototype
1Prototype filter: H (s)s 1 s s 1
with 3-dB cut-off frequency 1, where 10 log(|H | ) 3dB
=+ + +
ω = ω = ⋅ =
( )
BP
0 0 0 0 0 0
0 0 0
0 0
0 s=j
1,2 ( )
Band pass filter :1H (s)
s s s1 1B s B s B s
swith 3-dB cut-off frequencies found from putting =1 B s
giving 4 solutions ½ B ½ B
ω
+−
−
= ω ω ω ω ω ω + + + + + + ω ω ω
ω ω+ ω
ω = ± 2 20
2 1
, two of them positive.
The 3 dB bandwidth B
+ ω
− ω − ω =
112 20 0
0 0
s B sLow-pass to stop-band transformation: sBs s
−− +ω ω
→ = + ω ω
( ) ( )( ) ( )
0
0 1 20 01
1 20 00
00 0
ss s z s z1 1
s d Bd Bd s p s ps 1B s d B s ds
−
ω+ ω − −ω ω → = − = −− − − ω ω ω + −+ − ωω ω
Frequency Transformations: Low-pass Prototype to Stop-band
( )
( )
21 1 20
1,221 1 2
0
Poles :
½Bd ½Bdp
½Bd ½Bd
− −
− −
+ − ω= − − ω
01,2
0
Zeros :j
zjω
= − ω
LP and BP Butterworth FilterBP
0 0 0 0 0 0
0 0 0
1H (s)s s s1 1
B s B s B s
= ω ω ω ω ω ω + + + + + + ω ω ω
LP 2
c c c
1H (s)s s s1 1
= + + + ω ω ω
100
101
102
-120
-100
-80
-60
-40
-20
0
100
101
102
-700
-600
-500
-400
-300
-200
-100
0
100
101
102
-70
-60
-50
-40
-30
-20
-10
0
100
101
102
-300
-250
-200
-150
-100
-50
0
Pha
se (
degr
ees)
Mag
nitu
e (d
B)
Pha
se (
degr
ees)
Mag
nitu
e (d
B)
LP: K=3 fc=ω/2π= 8kHz BP : K=3 fc=ω/2π= 8kHz ∆f = =B/2π 2kHz
Frequency (kHz) Frequency (kHz)
LP and BS Butterworth Filter
100
101
102
-70
-60
-50
-40
-30
-20
-10
0
100
101
102
-300
-250
-200
-150
-100
-50
0
Pha
se (
degr
ees)
Mag
nitu
e (d
B)
LP: K=3 fc=ω/2π= 8kHz
LP 2
c c c
1H (s)s s s1 1
= + + + ω ω ω
100
101
102
-140
-120
-100
-80
-60
-40
-20
0
100
101
102
-800
-700
-600
-500
-400
-300
-200
-100
0P
hase
(de
gree
s)M
agni
tue
(dB
)
BS : K=3 f0=ω/2π= 8kHz ∆f = =B/2π 2kHz
BS 21 1 1
0 0 0
0 0 0 0 0 0
1H (s)B s B s B s1 1
s s s
− − −=
ω ω ω + + + + + + ω ω ω ω ω ω
Frequency (kHz) Frequency (kHz)
Signal ProcessingPart 8.4 Analogue Filters – Practical Constructions
Magnus Danielsen
Passive Butterworth LP-Filter of Third OrderV1
1 01 1 1
out2
1 1
1 1 1sC EsL R sL VNode equations : R
V1 1 1 0sCsL sL R
+ + − = − + +
0
1out
1 21 1 1 1
out
0 1 1 2 3 2
1 2 1 1 1 2 1 1 2
3 2 2 31 1 2 c c c
3 2
c c c
E1sL RV
1 1 1 1 1 1sC sCsL R sL R sL sL
V 1 1E RL C C 1 1 1 1 2s s s
RC RC L C L C L C C R1 1
RL C C s 2s 2s1K
s s s2 2
=
+ + + + − − −
= ⋅
+ + + + +
= ⋅+ ω + ω + ω
= ⋅
+ + ω ω ω 1+
2 3c c c 3
1 2 1 1 1 2 1 1 2 1 1 2 c
1 1 1 1 2 1 12 2 KRC RC L C L C L C C R RL C C 2
ω = + ω = + ω = = =ω
c c 1 1 2Example : R 50 f 159 Hz 1000 rad/sec L 50 mH C 68.3uF C 11.7 uF= Ω = ω = = = =
Passive Butterworth LP-Filter of Third Order (2)
Amplitude characteristics
Phase characteristics
Butterworth LP-filter Characteristics of 3rd Order
Passive Butterworth BP-Filter of 3rd Order
20 0 c c 0
c0
c2 2
c 0 0
Ls 1 1Inductances: sL L s L sL'B s B B s sC'
B 1Serial coupling of L' and C' L' L C'B L L'
ω ω ω ω ω → ω + = + = + ⇒ ω
ω= = =
ω ω ω
out 3 2
1LP-prototype filter: Vs 2s 2s 1
=+ + +
0 0
0
sPrototype Low-pass to band-pass transformation: sB s ω ω
→ + ω
c
0 0 0 0c
c 0 0
Low-pass to band-pass transformation cutoff frequency = :
s s sor sB s B s
ω
ω ω ω ω→ + →ω + ω ω ω
20 0 c c 0
c0
c2 2
c 0 0
Cs 1 1Capitances: sC C s C sC''B s B B s sL''
B 1Parallel coupling of C'' and L'' C'' C L''B C C''
ω ω ω ω ω → ω + = + = + ⇒ ω
ω= = =
ω ω ω
Center frequency:f0 = ω0/(2π)
Band width:∆f = B/(2π)
Passive Butterworth BP Filter of 3rd Order
Amplitude characteristics
Phase characteristics ∆f = 800 Hzf0 = 1000Hz
Active LP-filter – Single Real Pole
1
11out 1
in 3 3 1 1
1 1
1 sCRV R 1H(s)
V R R 1 sR C
1Real pole: sR C
−
+ = = − = −
+
= −
Active LP-filter – Single Real PoleFrequency Response
Magnitude response
Phase response
Example: LP to BP Transformation
0
0 0
0
2
1Prototype LP-filter: H(s)s 1
1 B 0.1s 1s ( ) 10(s )
B s s1 0.1sBandpass filter : H(s) 1 s 0.1s 110(s ) 1
s
=+
ω = =ω ω
→ + = +ω
= =+ ++ +
Poles : s 0.05 j0.998 0.05 jZero : s 0
= − ± ≅ − ±=
Transfer Function of BP-filter
1 0- 1
1 00
1 01
- 4 0
- 3 5
- 3 0
- 2 5
- 2 0
- 1 5
- 1 0
- 5
0
1 0- 1
1 00
1 01
- 2
- 1 . 5
- 1
- 0 . 5
0
0 . 5
1
1 . 5
2
Amplitude characteristics
Phase characteristics
2
0.1sH(s)s 0.1s 1
=+ +
Active BP-filter Two Complex Poles, and a Zero
1
11 1out
in 3
1
2 131 1
1
2
1 1 1 1 1 1
1 1sCR sLVH(s)
V RL s
LR s L C s 1R
1 1 1s j2R C L C 2R C
s 0
−
+ + = = −
= − ⋅+ +
= − ± −
=
Transfer function :
Complex poles :
Zero :
Active BP-filter – Two Complex Poles, and a ZeroMagnitude and Phase
1 1
Center frequency:
1 1f 1007Hz2 L C
= =π
1
2 131 1
1
H(s)L s
LR s L C s 1R
=
− ⋅+ +
Phase Characteristic
Amplitude Characterlistic
Signal ProcessingPart 8.5 Digital FIR Filter
Magnus Danielsen
FIR-Filter Properties and Applications
• Impulse response:h[n] has finite length
• Transfer function:H(z) has finite number of terms: H(z)=a0z0+a1z-1+a2z-2+...+aMz-M
• Difference equation:y[n]=a0x[n] + a1x[n-1] +a2x[n-2] +...+aMx[n-M]
• Linear phase⇒ group (signal) delay is
constant
• Desired freq. Response:– Hd(ejΩ)
• Practical freq. response:– H(ejΩ)
• Realization: – Mean-square error– Window functions– Implementation of FIR filters
with block diagrammes
• Applications– Discrete time differentiator– Types of filters– Practical examples of filters
Desired Filter and Mean Square Error
( ) ( ) [ ] [ ]
[ ] [ ]
2 2j jd d
1 2 2d d
M 1
1E H e H e d h n h n2
h n h n
π ∞Ω Ω
−∞−π
− ∞
−∞ +
= − Ω = −π
≤ +
∑∫
∑ ∑
Mean square error :
[ ]
[ ] [ ]
d
d
h n n
0 for n 0h n h n for 0 n M
0 for M n 0
− ∞ < < ∞
<= ≤ ≤ < <
Desired impulse response
Practical impulse response
Window Functions
[ ] [ ]
( ) ( )( )
rec
Mj jn jM / 2
n 0
0 for n 0: w n w n 1 for 0 n M
0 for M n 0
sin (M 1) / 2W e e e
sin / 2Ω Ω − Ω
=
<= = ≤ ≤ < <
Ω += = − π < Ω ≤ π
Ω∑
Rectangular Window
[ ] [ ] [ ]
( ) ( ) ( ) ( ) ( )
d
j j j j j( )d d
h n w n h n
1H e W e H e W e H e d2
πΩ Ω Ω Λ Ω−Λ
−π
=
= ∗ = Λπ ∫
Practical truncated impulse response :
Transfer function :
[ ]Hamming,M
:
n0.54 0.46cos 2 for 0 n Mw n M
0 for n 0 and n M
− π ≤ ≤ = < >
Hamming window
Example: Rectangular and Hamming Windows
0 0.5 1 1.5 2 2.5 3-70
-60
-50
-40
-30
-20
-10
0
10
20
Ω
dB
0 5 10 15 20 250
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Rectangular window
n
0 5 10 15 20 250
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Hammingwindow
n
M=11
M=11
Rectangular window
Hamming window
Implementation of FIR Filters with Rectangular Window(1)
( ) [ ]Mj2 DTFTj c
d d
c
:
e forH e h n0 for
− ΩΩ
Ω ≤ Ω= ←→Ω < Ω ≤ π
Desired frequency response
[ ] ( )c
c
Mj nj jn 2 c c
d d
:
1 1 Mh n H e e d
for 0 n <
e d sinc n2 2 2
Ω π − Ω Ω Ω
−π −Ω
Ω Ω = Ω = Ω = − π π π π
∞
≤
∫ ∫
Desired impulse response
[ ] [ ] [ ]rectc
,c
dM
:for
Mh n h n sinc n and 0 otherwise.
0 n
w n2
M
Ω Ω = ⋅ = − π π
≤
<
Frequency response with rectangular window
Implementation of FIR Filters with Rectangular Window(2)
0 5 10 15 20 25 30 35 40 45 50-0.05
0
0.05
0.1
0.15
0.2
0 5 10 15 20 25 30 35 40 45 500
0.2
0.4
0.6
0.8
1
0 5 10 15 20 25 30 35 40 45 50-0.05
0
0.05
0.1
0.15
0.2
0 1 2 3 4 5 6 7-40
-30
-20
-10
0
10
0 1 2 3 4 5 6 7-30
-20
-10
0
10
20
30
0 1 2 3 4 5 6 7-80
-60
-40
-20
0
20
M=12
hd
wrec
h
|Hd|
|Wrec|
|H|
dB
dB
dB
Ωn
Implementation of FIR Filters with Hamming Window(1)
( ) [ ]Mj2 DTFTj c
d d
:
e forH e h n0 for 0
ΩΩ
Ω ≤ Ω= ←→< Ω ≤ π
Desired frequency response
[ ]c c
c c
MM j nj jn 2 c c2d
for 0 n< :
1 1 Mh n e e d e d sinc n2 2 2
Ω Ω − Ω− Ω Ω
−Ω −Ω
Ω Ω = Ω = Ω = − π π π π
≤ ∞
∫ ∫
Desired impulse response
[ ] [ ] [ ]Hammic c
ng,M d
:
Mh n h n
for 0 n M
nw n 0.54 0.46c sinc n2
and 0 otherwise
o 2M
.
s
≤ <
− π Ω Ω = ⋅ = ⋅ − π π
Frequency response with Hamming window
Implementation of FIR Filters with Hamming Window(2)
0 5 10 15 20 25 30 35 40 45 50-0.05
0
0.05
0.1
0.15
0.2
0 5 10 15 20 25 30 35 40 45 500
0.2
0.4
0.6
0.8
1
0 5 10 15 20 25 30 35 40 45 50-0.05
0
0.05
0.1
0.15
0.2
0 1 2 3 4 5 6 7-40
-30
-20
-10
0
10
0 1 2 3 4 5 6 7-60
-40
-20
0
20
0 1 2 3 4 5 6 7-100
-80
-60
-40
-20
0
hd
wHam
h
|Hd|
|Wham|
|H|
M=12
FIR Filter Block Diagram Implementation
z-transformed output signal Y(z)= h[0] z0X(z)+ h[1] z-1X(z)+ h[2] z-2X(z)+...+h[n] z-MX(z)
++ Y(z)++
X(z)z-1 z-1 z-1
h[0] h[1] h[2] h[M-1] h[M]
Impulse response: h[n] = [... 0 0 h[0] h[1] h[2] ... h[M] 0 0 ...]
Input signal: x[n]
Output signal: y[n] = h[n]∗x[n] = h[0]x[n] + h[1]x[n-1] + h[2]x[n-2] +...+ h[M]x[n-M]
Discrete Time Differentiatorx yh H←→
( ) ( ) ( ) ( ) ( ) ( ) ( )
[ ] [ ] [ ] ( ) ( ) ( )
( ) sT j T j
s
z
z e e e
dy t x t Y s H s X s sX s H s s jdt
y n h n x n Y z H z X z
H z j jTω Ω= = =
= ←→ = = = = ω
= ∗ ←→ =
Ω= ω =
Differentiator in analogue system :
Differentiator in digital system :
( )
[ ]
Mjj 2d
Mj jn2d
Mth order
Desired transfer function : H e j e
Desired impulse response :
M Mcos n sin n2 2h n j e e d
M Mn s n2 2
− ΩΩ
π− Ω Ω
−π
−
= Ω − π < Ω ≤ π
π − π − = Ω Ω = − − π −
∫
Normalized digital differentiator filter :
2 for 0 n≤ < ∞
”Windowed” Discrete Time Differentiator
[ ]
[ ] 2
1 for 0 n Mw n
0 for n 0 and for M n 0
M Mcos n sin n2 2h n for 0 n M
M Mn s n2 2
≤ ≤= < < <
π − π − = − ≤ < − π −
Rectangular Window
[ ] [ ] [ ]( ) ( ) ( )
d
j j jd
Mth orderh n w n h n
H e W e H eΩ Ω Ω
−
=
= ∗
Windowed digital differentiator filter :Windowed impulse response :
Windowed transfer function :
[ ]
[ ] 2
n0.54 0.46cos 2 for 0 n Mw n M
0 for n 0 and for M n 0
M Mcos n sin nn2 2h n 0.54 0.46cos 2 for 0 n M
M MMn s n2 2
− π ≤ ≤ = < < <
π − π − = − ⋅ − π ≤ < − π −
Hamming Window
Discrete Time DifferentiatorRectangular Window
[ ]
[ ] 2
1 for 0 n Mw n
0 for n 0 and for M n 0
M Mcos n sin n2 2h n for 0 n M
M Mn s n2 2
≤ ≤= < < <
π − π − = − ≤ < − π −
Rectangular Window
-20 -15 -10 -5 0 5 10 15 20-1
-0.5
0
0.5
1
1.5
0 1 2 3 4 5 6 70
0.5
1
1.5
2
2.5
3
3.5
h[n] |H(ejΩ)|
n Ωπ 2π
M=12
Discrete Time DifferentiatorHamming Window[ ]
[ ] 2
n0.54 0.46cos 2 for 0 n Mw n M
0 for n 0 and for M n 0
M Mcos n sin nn2 2h n 0.54 0.46cos 2 for 0 n M
M MMn s n2 2
− π ≤ ≤ = < < <
π − π − = − ⋅ − π ≤ < − π −
Hamming Window
-20 -15 -10 -5 0 5 10 15 20-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5 6 70
0.5
1
1.5
2
2.5
h[n] |H(ejΩ)|
n Ωπ 2π
M=12
Filtering of Speech Signals• Speech must be processed by precise time alignement• FIR-filters with w[n], 0≤n≤M, symmetric or antisymmetric
around M/2 or (M+1)/2 have linear characteristics• Length M+1 of FIR is usually large (fx~100)• Time delay Tsignal=½MTs (group delay)• LP used to remove high frequency noise• LP used to freq.band limit signal • HP used to remove low frequency noise • BP used to limit frequency range, used in speech
processing equipment (fx. telephone 300 < f < 3100Hz)• BS used to remove noise at specific frequencies (fx. 50Hz)
Typical Examples on FIR-filter Characteristics for Filtering of Speech Signals
[ ][ ]
[ ]
[ ]
1 2 n1 cos 0 n M ( Hanning or raised cosine window,2 M
0 otherwise with symmetric h n )
M2 ( n )2sin 0 n M (with antisymmetric h[n])
M
0 otherwise
0
LP filter h n
BP filter h n
HP filter h n
π − ≤ ≤
π − ≤ ≤
− = − =
− = ( )cos n 0 n M (with antisymmetric h[n])
otherwise
g s
g signal s
d dTime delay (= group delay), defined by T Td d
for = T T ½ MT
π ≤ ≤
= − ϕ = − ϕω Ω
= =phase linear FIR filters
LP – FIR filter, An Example
0 1 2 3 4 5 6 70
1
2
3
4
5
6
-20 -15 -10 -5 0 5 10 15 200
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
[ ]1 2 n1 cos 0 n M2 M
0 otherwiseh n
π − ≤ ≤
=
n Ω
h[n] |H(ejΩ)|
M=12
BP – FIR filter, An Example
-20 -15 -10 -5 0 5 10 15 20-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5 6 70
1
2
3
4
5
6
7
M=12
[ ]
M2 ( n )2sin 0 n M
M
0 otherwiseh n
π − ≤ ≤
=
h[n] |H(ejΩ)|
Signal ProcessingPart 8.6 Digital IIR Filter
Lowpass and Highpass Filters
Magnus Danielsen
IIR-Filter Properties and Applications• Difference equation:
y[n]+a0y[n-1] +... aNy[N] =b0x[n] + b1x[n-1] +...+ bMx[n-M]• Impulse response:
h[n]=h[0], h[1], h[2], ... , h[n], ..... has infinite length• Transfer function:
H(z)=h[0]z0+h[1]z-1+...+h[n]z-n+..... has infinite no. of terms• Causality:
h[n]=0 for n<0 usually applied• Construction of filters by bilinear transform• Butterworth characteristics• Chebyshev characteristics• Elliptic characteristics• LP-filters• HP-filters• BP-filters• BS-filters• Linear distortion – transfer function dependence on frequency
Amplidtude distortionPhase distortion - non-linear phase ⇒ non-constant group (signal) delay
• Equivalizers• Implementation of IIR filters with block diagrammes
Difference Equation for IIR Filters Transfer Function and Infinite Impulse Response
( )1 2 M
0 1 2 M1 2 M
0 1 2 M1 2 n
b b z b z ... b zH za a z a z ... a z
h[0] h[1]z h[2]z .... h[n]z ........
− − −
− − −
− − −
+ + + +=
+ + + +
= + + + + +
Transfer function :
Difference equation:y[n]+a0y[n-1] +... aNy[N] =b0x[n] + b1x[n-1] +...+ bMx[n-M]
z-transformed difference equation:Y(z)+a0z-1Y(z) +... aNz-NY(z) =b0X(z) + b1z-1X(z) +...+ bMz-MX(z)
[ ] [ ] [ ] [ ] [ ] [ ]h n ....0 0 h 0 h 1 h 2 h 3 ... h n .....= Impulse response :
Bilinear Transform
s
ssamp s
sTj
s
2 2Sampling time : T
Relation between z and s : z re es jCyclic frequency :Frequency f /(2 )Normalized cyclic frequency : T
Ω
π π= =ω ω
= == σ + ω
ω= ω πΩ = ω
Definitions :
Analogue filter characteristic: H(s)Wanted digital filter characteristic: H(z)H(s) and H(z) shall have approximately the same properties
s
s
sTs
sTe -1 sTs j = tanhe +1 2
= σ + ω→ =
Bilinear transform :z - 1λ = = Σ' + jΩ'z + 1
samp
s
ω is often
and will bewritten as ωwhich howevercan be confusedwith the stopbandcyclic frequency
Transformation ofs – plane → z – plane → λ - plane
s j= σ + ω ssT' j ' tanh2
λ = Σ + Ω =ssTjz re eΩ= =
Negative s-plane strip: σ<0, -½ωs<ω< ½ωs is immaged in interiour of |z|=1 circle, and negative λ-plane
The frequency axis s=jω interval: σ=0, -½ωs< ω<½ωs is immaged on the |z|=1 circleand the imaginary λ-axis
Positiv s-plane strip: σ>0, -½ωs< ω <½ωs is immaged in exteriour of |z|=1 circle, and positive λ-plane
jω
s-plane
½ωs
−½ωs
σ
j 'Ω
'Σ
λ-plane
Ω
z-plane
Re(z)
Im(z)jz e Ω=
r 1=
Frequency Axis - Transforms s
s
s
j T Ts j ' j ' tanh j tan2 2
T' tan tan tan2 2
ω ω= ω→ λ = Σ + Ω = =
ω ω ΩΩ = = π = ω
The bilinear transformed frequency axis :
( ) ( )( ) ( )( ) ( )
Trigonometric relations :sinh j x j sin x sin j x j sinh x
cosh j x cos x cos j x cosh x
tanh j x j tan x tan j x j tanh x
⋅ = ⋅ ⋅ = ⋅
⋅ = ⋅ =
⋅ = ⋅ ⋅ = ⋅
0
0Ω’ →∞
πΩ
Ω’
IIR LP-filter with Butterworth Response( ) ( )( )
( )ps 0.10.1
s p
log 10 1 / 10 1K
2log /
αα − −≥
ω ω
Analogue LP-filter:Prototype filter is defined as LP-filter with ωc=1
( ) rK
rr 1 c
1 2r 1H s r 1,2,3,...,K2Ks jexp( j )
=
−= θ = π =
− θ ω
∏
Digital IIR LP-filter:Prototype filter is defined as LP-filter with Ω’c=1
( ) rK
rr 1 c
1 2r 1H r 1,2,3,...,K2K
jexp( j )'=
−λ = θ = π =
λ− θ Ω
∏
( ) ( )( )( )
ps 0.10.1
s p
log 10 1 / 10 1K
2log ' / '
αα − −≥
Ω Ω
( ) 2
2K
c
1H j ''1'
Ω = Ω+ Ω
( ) 2
2 K
c
1H j
1
ω = ω+ ω
1
1
z 1 1 zz 1 1 z
−
−
− −λ = =
+ +
Transfer Function ofLP IIR-filter with Butterworth Response
( ) rK
rr 1 c
1 2r 1H r 1,2,3,..., K2K
jexp( j )'=
−λ = θ = π =
λ− θ Ω
∏
( )
( )
r1K
r1r 1 c
K1
K1
r rr 1 c c
1 2r 1H z r 1,2,3,..., K2K1 1 z jexp( j )
' 1 z
1 z
1 1jexp( j ) jexp( j ) z' '
−
−=
−
−
=
−= θ = π =
−− θ Ω +
+=
− θ − + θ Ω Ω
∏
∏
1
1
z 1 1 zz 1 1 z
−
−
− −λ = =
+ +c s c
cs
T' tan tan2
ω ω Ω = = π ω
Poles and Zeros Butterworth LP IIR-filters
( ) ( )K1
rK1
r rr 1 c c
1 z 2r 1H z r 1, 2,3,..., K2K1 1jexp( j ) jexp( j ) z
' '
−
−
=
+ −= θ = π =
− θ − + θ Ω Ω
∏
rc
pole
rc
1 jexp( j )'z for r 1, 2,..., K1 jexp( j )'
+ θΩ
= =− θ
Ω
Poles :
zeroz 1 K-multiple zero= −Zeros :Ω
z-plane
Re(z)
Im(z)
jz e Ω=
r 1=
5 multpl. zero−⊗
c
K 5' 0.2=
Ω =
Example :
⊗
⊗
⊗
⊗
5 poles
K=5fc=200 Hzfs=1000 Hz
( ) 15
r1r 1 c
1H z1 1 z jexp( j )' 1 z
−
−=
= −
− θ Ω + ∏
r2r 1 1 3 5 7 9, , , ,2K 10 10 10 10 10−
θ = π = π π π π π
c s cc
s
T f' tan tan 0.7272 f
ω Ω = = π =
0 200 400 600 800 1000-100
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
10
0 200 400 600 800 1000-4
-3
-2
-1
0
1
2
3
4
A(f)=10log|H|
ϕ(f)=∠HFrequency f
Frequency f
Butterworth LP IIR-filter
Block – Diagram IIR Butterworth Filter
( ) ( ) ( )( ) ( )
1
1 2
1 2
1
0.0181 1 z
1 0.509
1 zH z
1 1.2505z53 0.5 7zz 45
−
− −
−
−
+=
+
+
−−
+
+2
Y(z)+
+
-0.5457
1.2505
z-1
z-1
++
0.50953
X(z)
z-1
0.0181
( )
( ) ( )
c
2
c c
2
c
Analogue LP filter: with 0.32491H ss s 1
s 0.03249s 0.10
s 1
0.0343s . 24 560 3 9
ω =
= + + ω ω
+ ω
=++
+
Ts=1.93 sec
LP IIR-filter with Chebyshev Response
( ) 2
2 2K
c
1H j1 T
ω = ω+ γ ω
Digital LP-filter:Prototype filter is defined as LP-filter with Ωc=1
( )( )
rK2
r rr 1
1 2r 1H C r 1,2,3,...,K2K[ sin j 1 cos ]
=
−λ = θ = π =
λ + η θ + +η θ∏
( ) 2
2 2K
c
1H j ''1 T'
Ω = Ω
+ γ Ω
( )( )
K2
r rr 1
1H s Cs [ sin j 1 cos ]
=
=+ η θ + + η θ∏
( ) ( )( )( )
ps 0.10.11
1s p
cosh 10 1 / 10 1K
cosh /
αα−
−
− −≥
ω ω
( ) ( )( )( )
ps 0.10.11
1s p
cosh 10 1 / 10 1K
cosh ' / '
αα−
−
− −≥
Ω Ω
Analogue LP-filter:Prototype filter is defined as LP-filter with ωc=1
For Chebyshev filter: Ωc=Ωp
For Chebyshev filter: ωc= ωp
Poles and Zeros of Chebyshev LP IIR-filter
( )1 11 1 1sinh sinh or sinh K sinhK
− − η = = η γ γ
D efinition of param eter η :
1r
c
2r r r
2r r
2r r
j cos(sin j ),'
(2r 1)sin j 1 cos where2K
1 sin j 1 cos1z1 1 sin j 1 cos
−λ= − η + θ
Ω−
= −η θ − + η θ θ = π
− η θ − + η θ+ λ= =
− λ + η θ + + η θ
Poles:
1z 1 is a K-multiple zero1+ λ
λ = ∞ ⇒ = = −− λ
Zeros :
LP Butterworth IIR FiltersT' tan tan
2 2ω Ω
Ω = =
( ) rK
rr 1 c
1 2r 1H r 1,2,3,...,K2K
jexp( j )'=
−λ = θ = π =
λ− θ Ω
∏
( ) ( )( )( )
ps 0.10.1
s p
log 10 1 / 10 1K
2log ' / '
αα − −≥
Ω Ω
( ) 2
2K
c
1H j ''1'
Ω = Ω+ Ω
z 1z 1−
λ =+
( ) ( )p s
p sc 1 1
0.1 0.12K 2K
' ''10 110 1α α
Ω ΩΩ = =
−−
HP Butterworth IIR Filter:
sT' tan tan2 2ω Ω
Ω = =
( ) rKc
rr 1
1 2r 1H r 1,2,3,...,K' 2Kjexp( j )
=
−λ = θ = π =
Ω − θ λ ∏
( ) ( )( )( )
ps 0.10.1
p s
log 10 1 / 10 1K
2 log ' / '
αα − −≥
Ω Ω
( ) 2
2Kc
1H j ''1'
Ω =Ω + Ω
z 1z 1−
λ =+
( ) ( )p s
1 10.1 0.12K 2K
c p s' ' 10 1 ' 10 1α αΩ = Ω − = Ω −
LP-filtur HP-filtur
c c
c c
' '' ' ' '
Ω Ωλ Ω→ →
Ω λ Ω ΩFrequency transformation :
PolesHP=PolesLP* (i.e. same poles)ZerosHP: λ=0 ⇒ z=1 (K-multiple)
K=5fc=200 Hzfs=1000 Hz
( ) 15
c r1r 1
1H z1 z' jexp( j )1 z
−
−=
= +Ω − θ −
∏
r2r 1 1 3 5 7 9, , , ,2K 10 10 10 10 10−
θ = π = π π π π π
c s cc
s
T f' tan tan 0.7272 f
ω Ω = = π =
Butterworth HP IIR Filter
0 200 400 600 800 1000-4
-3
-2
-1
0
1
2
3
4
ϕ(f)=∠H
½fsFrequency f
0 200 400 600 800 1000-100
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
10
A(f)=10log|H|
Frequency f½fs
HP Chebyshev IIR Filter:
sT' tan tan2 2ω Ω
Ω = =
1
1
1 z1 z
−
−
−λ =
+c pFor Chebyshev filters ' 'Ω = Ω
LP-filtur HP-filtur
c c
c c
' '' ' ' '
Ω Ωλ Ω→ →
Ω λ Ω ΩFrequency transformation :
PolesHP=PolesLP* (i.e. same poles)ZerosHP: λ=0 ⇒ z=1 (K-multiple)
( ) ( )( )( )
ps 0.10.11
1p s
cosh 10 1 / 10 1K
cosh ' / '
αα−
−
− −≥
Ω Ω
( ) rK2c
r rr 1
1 2r 1H C r 1,2,3,..., K' 2K[ sin j 1 cos ]
=
−λ = θ = π =
Ω + η θ + + η θ λ ∏
( ) 2
2 2 cK
1H j ''1 T'
Ω =Ω + γ Ω
LP and HP Chebyshev Filters of Order K=5
0 100 200 300 400 500 600 700 800 900 1000-100
-80
-60
-40
-20
0
0 100 200 300 400 500 600 700 800 900 1000-4
-3
-2
-1
0
1
2
3
4
0 100 200 300 400 500 600 700 800 900 1000-100
-80
-60
-40
-20
0
0 100 200 300 400 500 600 700 800 900 1000-4
-3
-2
-1
0
1
2
3
4
LP filter HP filter
fp =200 Hz fs = 1000Hz αp = 3dB γ = 1 η = 0.1775
Frequency (Hz) Frequency (Hz)
Pha
se (r
adia
n)
Pha
se (r
adia
n)M
agni
tude
(dB
)
Mag
nitu
de (
dB)
Signal ProcessingPart 8.7 Digital IIR Filter
Bandpass, and Bandstop Filters
Magnus Danielsen
Bandpass IIR Filter Frequency Transformation
0 0 0 0
c 0 c 0
' ' ' '' ' ' B' ' ' B' ' '
Ω Ω Ω Ωλ λ Ω Ω→ + → − Ω Ω λ Ω Ω Ω
Frequency transformation :
LP-filtur BP-filtur
p2'Ωp1'Ω s2'Ωs1'Ωs'Ωp'Ω 'Ω 'Ω
s
1j T j s
1
Tz 1 1 z z e e ' tan tanz 1 1 z 2 2
−ω Ω
−
ω− − Ωλ = = = = ⇒ Ω = =
+ +p1 s p2 s
p1 p2
T T' tan ' tan
2 2ω ω
Ω = Ω =
0 s0 p1 p2
T' tan ' '2
ωΩ = = Ω ⋅Ω
s1 s s2 ss1 s2
T T' tan ' tan2 2
ω ωΩ = Ω =
22
0 0
c 0
' '' 'H j H j' B ' ' '
Ω ΩΩ Ω→ − Ω Ω Ω
0 0
c 0
' 'H H' B '
Ω Ωλ λ→ + Ω Ω λ
⇒
p2 p1B ' ' '= Ω −Ω
BP Butterworth IIR Filter
( ) ( )( )ps 0.10.1
p1,2s1,2 0 0
0 s1,2 0 p1,2
log 10 1 / 10 1K
'' ' '2 log /' ' ' '
αα − −≥
ΩΩ Ω Ω− − Ω Ω Ω Ω
( ) rK0 0
rr 1 0
1 2r 1H r 1, 2,3,..., K2K' ' jexp( j )
B ' '=
−λ = θ = π =
Ω Ωλ+ − θ Ω λ
∏
( ) 2
2 K
0 0
0
1H j '' ''1
B ' ' '
Ω = Ω ΩΩ
+ − Ω Ω
( ) ( ) ( )( ) ( ) ( )
K K1 1
2 2 2K0 0 01 2
r rr 1
1 z 1 zH z
' 2 ' '1 2 1jexp( j ) z z jexp( j )B' B' B' B' B' B'
− −
− −
=
+ −=
Ω Ω Ω + − θ + − + + + + θ
∏
10 0 0 0
1c 0 c 0
' ' ' '' ' z 1 1 z' B' ' ' B' ' ' z 1 1 z
−
−
Ω Ω Ω Ωλ λ Ω Ω − −→ + → − λ = = Ω Ω λ Ω Ω Ω + +
LP-filtur BP-filtur
Bandpass IIR Butterworth Filter of Order K=5
0 100 200 300 400 500 600 700 800 900 1000-100
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
10
½fs
0 100 200 300 400 500 600 700 800 900 1000-30
-25
-20
-15
-10
-5
0
5π/2
-9π/2
-7π/2-8π/2
½fs
Frequency f (Hz)
Pha
se (
radi
an)
Mag
nitu
de (
dB)
fp1=150 Hz fp2=250 Hzfs=1000 Hz αp=3dB
( ) ( )( )
( )( )
50 0
rr 1 0
H z H z
1z' ' jexp( j )
B ' ' z=
= λ =
λΩ Ω+ − θ Ω λ
∏
r2r 1 1 3 5 7 9, , , ,2K 10 10 10 10 10−
θ = π = π π π π π
( )1
1
1 zz1 z
−
−
−λ = λ =
+
p1 s p2 sp1 p2
p2 p1
0 p1 p2
T T' tan ' tan
2 2B' ' '
' ' '
ω ωΩ = Ω =
=Ω −Ω
Ω = Ω ⋅Ω
Bandpass IIR Chebyshev I Filter of Order K=5
αp=3dB fp1=150 Hz fp2=250 Hzfs=1000 Hz αp=3dB
( ) ( )( )
( )( )
50 0
rr 1 0
H z H z
1z' ' jexp( j )
B ' ' z=
= λ =
λΩ Ω+ − θ Ω λ
∏
r2r 1 1 3 5 7 9, , , ,2K 10 10 10 10 10−
θ = π = π π π π π
( )1
1
1 zz1 z
−
−
−λ = λ =
+
p1 s p2 sp1 p2
p2 p1
0 p1 p2
T T' tan ' tan
2 2B' ' '
' ' '
ω ωΩ = Ω =
=Ω −Ω
Ω = Ω ⋅Ω
0 100 200 300 400 500 600 700 800 900 1000-100
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
10
0 100 200 300 400 500 600 700 800 900 1000-30
-25
-20
-15
-10
-5
0
½fs
-π/2
-11π/2
-9π/2-10π/2
½fs
Frequency f (Hz)
Pha
se (
radi
an)
Mag
nitu
de (
dB)
Bandstop IIR Filter Frequency Transformation
s
1j T j s
1
Tz 1 1 z z e e ' tan tanz 1 1 z 2 2
−ω Ω
−
ω− − Ωλ = = = = ⇒ Ω = =
+ +p1 s p2 s
p1 p2
T T' tan ' tan
2 2ω ω
Ω = Ω = p2 p1B' ' '= Ω −Ωs1 s s2 ss1 s2
T T' tan ' tan2 2
ω ωΩ = Ω =
1 1
0 0
c 0 0 c 0 0
' 'B' ' B' ' ' ' ' ' ' ' '
− − Ω Ωλ λ Ω Ω
→ + → − Ω Ω Ω λ Ω Ω Ω Ω Frequency transformation :
LP-filtur
s'Ωp'Ω 'Ω
BS-filtur
s2'Ωs1'Ω p2'Ωp1'Ω 'Ω
⇒
22 1
0
c 0 0
'' B ' 'H j H j' ' ' '
− ΩΩ Ω → − Ω Ω Ω Ω 1
0
c 0 0
'B 'H H' ' '
− Ωλ λ → + Ω Ω Ω λ
0 s0 p1 p2
T' tan ' '2
ωΩ = = Ω Ω
Band Stop (BS) Butterworth IIR Filter
( ) ( )( )ps 0.10.1
p1,2 s1,20 0
0 p1,2 0 s1,2
log 10 1 / 10 1K
' '' '2 log /' ' ' '
αα − −≥
Ω ΩΩ Ω− − Ω Ω Ω Ω
( ) r1K0
rr 1 0 0
1 2r 1H r 1, 2,3,..., K2K'B ' jexp( j )
' '
−
=
−λ = θ = π =
Ωλ + − θ Ω Ω λ ∏
( ) 2
2K1
0
0 0
1H j ''B ' '1
' ' '
−Ω =
ΩΩ + − Ω Ω Ω
( )( ) ( ) ( )
K
1 20 0 0K
0 0 002 2 2K
0 0 01 2r r r
r 1
1 1 1' 2 ' z ' z' ' ''H z
B' ' ' '1 1 11 jexp( j ) z 2 jexp( j ) z 1 jexp( j )B' B' B' B' B' B'
− −
− −
=
Ω + + Ω − + Ω + Ω Ω ΩΩ = Ω Ω Ω − + θ + ⋅ ⋅ − θ − + + θ
∏
1 1 10 0
1c 0 0 c 0 0
' 'B' ' B' ' z 1 1 z' ' ' ' ' ' ' z 1 1 z
− − −
−
Ω Ωλ λ Ω Ω − −→ + → − λ = = Ω Ω Ω λ Ω Ω Ω Ω + +
LP-filtur BS-filtur
Bandstop IIR Butterworth Filter of Order K=5
fp1=150 Hz fp2=250 Hzfsamp=1000 Hz αp=3dB
( ) ( )( )
150
rr 1 0 0
H z H z
1
'B ' jexp( j )' '
−
=
= λ =
Ωλ + − θ Ω Ω λ ∏
r2r 1 1 3 5 7 9, , , ,2K 10 10 10 10 10−
θ = π = π π π π π
( )1
1
1 zz1 z
−
−
−λ = λ =
+
s1 s s2 ss1 s2
s2 s1
0 s1 s2
T T' tan ' tan2 2
B' ' '
' ' '
ω ωΩ = Ω =
=Ω −Ω
Ω = Ω ⋅Ω 0 100 200 300 400 500 600 700 800 900 1000-30
-25
-20
-15
-10
-5
0
-3π/2
-4π
½fs
Frequency f (Hz)
Pha
se (
radi
an) -5π/2
0 100 200 300 400 500 600 700 800 900 1000-100
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
10
½fs
Mag
nitu
de (
dB)
Bandstop IIR Chebyshev I Filter of Order K=5
fs1=150 Hz fs2=250 Hzfsamp=1000 Hz αp=3dB
( ) ( )( )
150
rr 1 0 0
H z H z
1
'B ' jexp( j )' '
−
=
= λ =
Ωλ + − θ Ω Ω λ ∏
r2r 1 1 3 5 7 9, , , ,2K 10 10 10 10 10−
θ = π = π π π π π
( )1
1
1 zz1 z
−
−
−λ = λ =
+
s1 s s2 ss1 s2
s2 s1
0 s1 s2
T T' tan ' tan2 2
B' ' '
' ' '
ω ωΩ = Ω =
=Ω −Ω
Ω = Ω ⋅Ω
0 100 200 300 400 500 600 700 800 900 1000-100
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
10
0 100 200 300 400 500 600 700 800 900 1000-25
-20
-15
-10
-5
0
5π
-3π/2
½fs
Frequency f (Hz)
Pha
se (
radi
an) -π/2
½fs
Mag
nitu
de (
dB)
-3π
Poles and Zeros IIR Filters of Order K
• LP filter: K complex polesK-multiple zero z = -1
• HP filter: K complex polesK-multiple zero z = 1
• BP filter: 2K complex polesK-multiple zero z = -1K-multiple zero z = +1
• BS filter: 2K complex polesK-multiple complex zero pairs
Resumé: IIR responses of filters of K’th order for LP-, HP-, BP-, and BS-filters
• Same transformations for the LP-, HP-, BP-, and BS- filters are used for Butterworth type response and Chebyshev type response.
• Same transformations are used for other filter types transfer functions as well
Used variables in:
LP filter: λ/Ω’c
HP filter: Ω’c/ λ
BP filter: Ω’0/B’⋅(λ/Ω’0+ Ω’0/ λ) B’=Ω’p2 - Ω’p1= transformed bandwidth of passband
BS filter: B’/Ω’0⋅(λ/Ω’0+ Ω’0/λ)-1 B’=Ω’p2 - Ω’p1=transformed bandwidth of stopband
EqualizersHc(jω) Heq(jω)
( ) ( ) ( ) ( )0j t
FTeq d eq d
c
eh (t) h t H j H jH j
ω
←→ ω ω =ω
( )
( ) ( ) ( ) ( )s s
FT,d d s
nM
jn T jn T,d d s d s ,eq
n n 0
h (t) h (t) t nT
H j h nT e h nT e H j
∞
δ=−∞
∞− ω − ω
δ δ=−∞ =
= δ − ←→
ω = = ω
∑
∑ ∑
( ) [ ] ( ) [ ]
[ ] [ ] [ ] ( ) [ ]
DTFT j jnd s d d, d s
nM
DTFT j jneq d d,eq d s
n 0
h nT h n H e h n e T
h n w n h n H e h n e T
∞Ω − Ω
δ=−∞
Ω − Ω
=
= ←→ = Ω = ω
= ←→ = Ω = ω
∑
∑
Signal ProcessingPart 9.1: Acoustic Spectrogramme Example
Magnus Danielsen
Speech Production Mechanism
J.R.Deller Jr,J.G.Proakis, J.H.L.Hansen: Discrete-Time Processing of Speech Signals, Prentice Hall, New Jersey, 1987
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-5
0
5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.70
0.5
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-5
0
5
Speech:Continuous: f(t)Discrete: f[n]
Window function:Continuous: w(t-t0)Discrete: w[n-n0]
Windowed speech:Continuous: f(t)w(t-t0)Discrete: f[n]·w[n-n0]
Time (s)
[ ] [ ][ ] [ ]
0
0
0
j tFT0
j nFT j j0
j nDFT j j20 kN
00
s s s
Spectrum :Continuous : f (t) w(t t ) F( ) W( )e
Discrete DTFT : f n w n n F(e ) W(e )e
Discrete DFT : f n w n n F(e ) W(e )e
ttn n 0 k N 1T T T
ω
ΩΩ Ω
ΩΩ Ωπ
Ω=
⋅ − ←→ ω ∗ ω
⋅ − ←→ ∗
⋅ − ←→ ∗
Ω= = ω = ≤ ≤ −
Windowing of Speech Signal
Measurement System for Audio and Speech Spectrogrammes
Ampl. Sampl.Freq.= fs
Quant.Micro-phone
Small voltage
Amplified voltage
Sampledvalues
Sampl. data f[n] (numbers)
f[n]·w[n-n0]
LP analogue filter. B = fs/2
Filtred voltage
w[n-n0]Discrete time generator n0
FFTN = nfft
Data loggermemory
F(ejΩ,n0) F·F*
f = fs·Ω /(2π)Ω
Time
Freq
uenc
y
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450
500
1000
1500
2000
2500
3000
3500 Frequency f
Power to color or intensity transformation
n0
Displayt = n0/fs
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-3
-2
-1
0
1
2
3
4
Time
Freq
uenc
y
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450
500
1000
1500
2000
2500
3000
3500
Time (s)
Spectrogramme: specgram(mtlb,512,Fs,kaiser(500,5),475)
SignalNumber of samples: 4001Sampling frequency: 7418 HzOverlap: 475Window: Kaiser(500,5)FFT length: 512
0 100 200 300 400 5000
0.2
0.4
0.6
0.8
1
Spectrum
0 0.05 0.1 0.15 0.2 0.25 ... ...
No overlap
Time
Freq
uenc
y
0 0.1 0.2 0.3 0.4 0.50
500
1000
1500
2000
2500
3000
3500
Time
Freq
uenc
y
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450
500
1000
1500
2000
2500
3000
3500
specgram(mtlb,512,Fs,kaiser(500,5),0) specgram(mtlb,512,Fs,kaiser(500,5),475)
0 0.05 0.1 0.15 0.2 0.25 ... ...
Overlap: 475
FFT length = 515, Kaiser window, window lenth = 500, β=5
Influence of Overlap
Influence of Length of Window
Time
Freq
uenc
y
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450
500
1000
1500
2000
2500
3000
3500
specgram(mtlb,512,Fs,kaiser(500,5),475)Window length = 500
~ 0.0634 secFFT length = 512
Time
Freq
uenc
y
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40
500
1000
1500
2000
2500
3000
3500
specgram(mtlb,1024,Fs,kaiser(1000,5),975)
Window length = 1000~ 0.1287 sec
FFT length = 1024
Time
Freq
uenc
y
0 0.1 0.2 0.3 0.4 0.50
500
1000
1500
2000
2500
3000
3500
specgram(mtlb,512,Fs,kaiser(200,5),175)
Window length = 200 ~ 0.02696 sec
FFT length = 512
Time
Freq
uenc
y
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450
500
1000
1500
2000
2500
3000
3500
specgram(mtlb,512,Fs,kaiser(500,5),475)
Influence of Window Shape
Time
Freq
uenc
y
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450
500
1000
1500
2000
2500
3000
3500
specgram(mtlb,512,Fs,kaiser(500,0),475)
⇒-100 0 100 200 300 400 500 6000
0.2
0.4
0.6
0.8
1
1.2
1.4Rect.window = kaiser(500,0)
-100 0 100 200 300 400 500 6000
0.2
0.4
0.6
0.8
1
1.2
1.4Kaiser.window = kaiser(500,5)
Window length=500
Time
Freq
uenc
y
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450
500
1000
1500
2000
2500
3000
3500
specgram(mtlb,512,Fs,kaiser(500,10),475)
Time
Freq
uenc
y
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450
500
1000
1500
2000
2500
3000
3500
specgram(mtlb,512,Fs,kaiser(500,5),475)
0 100 200 300 400 5000
0.2
0.4
0.6
0.8
1
0 100 200 300 400 5000
0.2
0.4
0.6
0.8
1
⇒
⇒
Time
Freq
uenc
y
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450
500
1000
1500
2000
2500
3000
3500
specgram(mtlb,512,Fs,hamming(500),475)
Time
Freq
uenc
y
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.450
500
1000
1500
2000
2500
3000
3500
specgram(mtlb,512,Fs,hanning(500),475)
0 100 200 300 400 5000
0.2
0.4
0.6
0.8
1
0 100 200 300 400 5000
0.2
0.4
0.6
0.8
1
⇒
⇒
Influence of Window Shape
Signal ProcessingPart 9.2: Delays in Systems
Magnus Danielsen
Delay in Transmission Systems
SystemNo magnitude distortionTransfer function H(ω)
xphys(t) yphys(t)
Definition of system:•Type of system, e.g.
•Communication system•Measurement system•Audio Hi-Fi system•Television system
•Input signal: xphys(t)•Output signal: yphys(t)•Group delay = signal delay: Tgr•Phase delay: Tph•No magnitude distortion in the transmission band
Group Delay and Phase Delay
-3 -2 -1 0 1 2 3-1
-0 .8
-0 .6
-0 .4
-0 .2
0
0 .2
0 .4
0 .6
0 .8
1
Time (arb.units)
xphys(t)
x0(t)x0(t-Tgr)
Tgr
yphys(t)Sig
nal m
agni
tude
(ar
b.un
its)
Tph
( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( )
( ) ( ) ( )
phys 0 0
0 0 0 0
0 0
0 0
x t x t cos j t
½x t exp j t ½x t exp j t
Re x t exp j t
x t x t exp j t
= ω
= ω + − ω
= ω
= ω
Input signal :
( ) ( ) ( )phys 0 gr 0 phy t x t T cos j (t T )= − ω −
Output signal = delayed input signal :
Definition of System Transfer Function and Input Signal
( ) ( )( ) ( ) ( ) ( )
FT0 0
FT0 0 0 0
x t X
x t x t exp j t X
←→ ω
= ω ←→ ω−ω
Envelope :
Input signal :
( ) ( )( )
( ) ( ) ( ) ( )0
0 0
0 0 0 0 0
with no magnitude distortion :
H H exp j H constant
gives the linear frequency variation :
d'
dω=ω
ω = Ψ ω =
Ψ ωΨ ω = Ψ + ω− ω = Ψ + Ψ ⋅ ω− ω
ω
0
Transfer function
Phase - first order Taylor approx. around ω
Definition of Group Delay and Phase Delay
( ) ( ) ( )( )( )( )( )
( )( ) ( )
( ) ( )( ) ( )
0 0 0
0
0 0 0
0 0 0
0 0 0 0
0 0 0 0
0
0 0 0 0 0
0 0 0
0
0
Y H j X( j ) H exp j X ( )
H exp j ' X ( )
y
X ( ) exp j '
x (t ') exp j (t '
x (t ')
H exp j '
H exp j
exp
t '
j (tH
Ψ −Ψ ⋅ω
ω = ω ⋅ ω = Ψ ω ⋅ ω− ω
= Ψ + Ψ ⋅ ω− ω ⋅ ω− ω
= ⋅
= ⋅
ω− ω ⋅ Ψ ⋅ω
+Ψ − Ψ ⋅ω Ψ ⋅ ω
+
+ Ψ
+ Ψ ω⋅ Ψ= ⋅
Output signal Fourier transform :
Output signal :
( )( ) ( )phy
0
000 0s 0 0
)
x (t ')y cos (H )/t
/
t+ Ψ ω + Ψ⋅ ⋅ ω=
ω
0ph
0
T Ψ= −
ω
Phase delay :
0
gr 0
:
dT 'd ω=ω
Ψ= −Ψ = −
ω
Group delay
H(jω) X(jω) Y(jω)
Phase and Group Delay Based on Transfer Function
( ) ( )( )( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( )
H j H exp j
ln H j ln H j j
½ ln H j ½ ln H j j
½ ln H j ½ ln H j j
∗
ω = Ψ ω
ω = ω + Ψ ω = ω + ω + Ψ ω
= ω + − ω + Ψ ω
Transfer function in Fourier transform formulation :
( ) ( )( )
( ) ( )( )
( )( )gr
H jj ln2 H j
d H jdT ½ lnd d j H j
ωΨ ω = − − ω
Ψ ω ωω = − = − ω ω − ω
Phase :
Group delay :
Spectrum of input pulse
0 0.5 1 1.5 2
x 10-3
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
X Axis
Y A
xis
X Y P lot
Out
put p
ulse
mag
nitu
de
Time (sec)
0 0.5 1 1.5 2
x 10-3
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
X Axis
Y A
xis
X Y P lot
Inpu
t pul
se m
agni
tude
Time (sec)t1=0.5ms
t2=0.757 ms
Group delay=t2 - t1=0.257ms
XY Graph1
XY Graphs -3*10^4s+10^102
s +3*10^4s+10^102
Transfer Fcn2
s -3*10^4s+10^102
s +3*10^4s+10^102
Transfer Fcn1
Time
Sine Wave1
Sine Wave - envelope
PulseGenerator
Product1Product
BufferedFFT Scope
4th order all-pass filter
Simulation of Group Delay of Pulse in4th Order All-pass Filter by Simulink
Group delay for ω=105 rad/sec:
(4π)/ ∆ω = 0.257 ms
( )2 4
2 4
:
2s 3 10 s 1H ss 3 10 s 1
− ⋅ += + ⋅ +
Transfer function4 5
4 5
:
Double complex pole-pairs: s 1.5 10 j 10Double complex zero-pairs: s 1.5 10 j 10
≅ − ⋅ ± ⋅
≅ + ⋅ ± ⋅
Poles and zeros
4
2
3 104 Arctan1
− ⋅ ωϕ = − ⋅ − ω
Phase :
4th Order All-pass Filter
0 0.5 1 1.5 2 2.5 3 3.5
x 105
-14
-12
-10
-8
-6
-4
-2
0
4π
Pha
se (
radi
an)
Cyclic frequency (rad/sec)
∆ω=0.489·105 rad/sec
2π
Phase delay for ω=105 rad/sec:
(2π)/ω = 0.0628 ms
( ) ( ):
y t (1 m x(t)) cos t= + ⋅ ⋅ ω
Amplitude modulated signals -3*10^4s+10^102
s +3*10^4s+10^102
Transfer Fcn2
s -3*10^4s+10^102
s +3*10^4s+10^102
Transfer Fcn1Sum2
Sum
Sine Wave1
Sine Wave - envelope1
Scope1
Product1 Mux
Mux
3
Constant21
Constant1
BufferedFFT Scope
( )x(t)
cos t=
− ω
y(t) gry(t T )−
Delay in AM-system Simulink Simulation
0 5 10 15
x 104
0
100
200
300
400
500
Mag
nitu
de
Frequency (rad/sec)
Spectrum of AM-modulated signal
Tgr=0.25 ms
Time s
y(t)
gry(t T )−
Signal ProcessingPart 9.3: Seismic Example
Magnus Danielsen
Seismic Ship and Streamer Configuration
B K Galloway and S L Klemperer (Dept.Geoph. Stanford Univ., CA); J R Childs (Menlo Park, CA); Bering-Chukchi Working Group (Inst.of the Lithosphere and Inst.of Oceanology, Russian Academy of Sciences, Western Washington University, Lamont-Doherty Earth Observatory, Columbia University)
Wing-kei Yu, Rice University, Houston, Texas
SeismicSource -Airgun
Hydrophone cableHydrophones
Convolutional Model
Source Receiver
Sea surface
Seismic wavelet: h[k] = s[k]∗f[k]Source signal: s0[k]
”Source” signature:s[k]=s0[k]∗p[k]
Reverberationfunction: f[k]
Received signal: y[k] = h[k]∗ε[k] + n[k] = s0[k]∗p[k]∗f[k]∗ε[k] + n[k]
Noise: n[k]
Received signal: y[k]
Sea floor
Reflectivity function: ε[k]
Undergroundhorizons
Seismic Source - Airgun
A. Ziolkowski, J. R. Underhill, R.G.K. Johnston Wavelets, well ties, and the searchfor subtle stratigraphic trapsGEOPHYSICS, VOL. 63, 1998; P. 297–313
Maximum and Minimum Phase Example on Discrete Signal
1
1 1
Maximum phase :1 2zH
1 11 z 1 z3 4
−
− −
+= + +
1
1 1
Minimum phase :2 zH
1 11 z 1 z3 4
−
− −
+= + +
-4 -2 0 2 4-4
-3
-2
-1
0
1
2
3
4Maximum phase
cyclic frequency (rad/sec)
Pha
se (r
adia
n)
-4 -2 0 2 4-4
-3
-2
-1
0
1
2
3
4Minimum phase
cyclic frequency (rad/sec)
Pha
se (r
adia
n)
-2 0 2 4 6 8 1 0-1 .5
-1
-0 .5
0
0 .5
1M a xim um p ha se
n
impu
lsre
spon
se
-2 0 2 4 6 8 1 0-0 .5
0
0 .5
1
1 .5
2M inim um p ha se
n
impu
lsre
spon
se
[ ] [ ] [ ]n n
Maximum phase impulse response :
1 1h n 20 u n 21 u n3 4
= − +
[ ] [ ] [ ]n n
Minimum phase impulse response :
1 1h n 4 u n 6 u n3 4
= − +
All-pass Filter for Converting Max-phase to Min-phase – An Example
1
max1 1
Maximum phase :1 2zH
1 11 z 1 z3 4
−
− −
+= + +
1
min1 1
Minimum phase :2 zH
1 11 z 1 z3 4
−
− −
+= + +
- 4 - 2 0 2 4- 4
- 3
- 2
- 1
0
1
2
3
4M a x i m u m p h a s e
c y c li c fr e q u e n c y ( r a d /s e c )
Pha
se (r
adia
n)
- 4 - 2 0 2 4- 4
- 3
- 2
- 1
0
1
2
3
4M i n i m u m p h a s e
c y c li c fr e q u e n c y ( r a d /s e c )
Pha
se (r
adia
n)
-2 0 2 4 6 8 10-1 .5
-1
-0 .5
0
0 .5
1M aximum phase
n
impu
lsre
spon
se
-2 0 2 4 6 8 10-0 .5
0
0 .5
1
1 .5
2M inimum phase
n
impu
lsre
spon
se
Phase: Impulse response
( ) ( )( ) ( )
1min j
1max
All pass filter transfer function :H z 2 zH z H e 1H z 1 2z
−Ω
−
−
+= = =
+
Properties of FunctionsReceived signal: y[k] = h[k]∗ε[k] = s0[k]∗p[k]∗f[k]∗ε[k] + n[k]
Source signal: s0[k] Minimum phase signal
Source signature: s[k] = s0[k]∗p[k] Mixed phase signal
Spike filter response: p[k] All pass response
Reverberation function: f[k] Results from multiple reflections between bottom and surface
Reflectivity function: ε[k] Results from object to be measured
Reflectivity autocorr.fct.: Rεε[k]=σ2δ[k] ε as a random-like property
Noise: n[k] Can be random or coherent
Signature Deconvolution (1)
Source Receiver
Sea surface
Seismic wavelet: h[k] = s[k]∗f[k]Source signature: s[k]=s0[k]∗p[k]Reverberation function: f[k]
( )0
DTFT j
DTFT-1
Source signal s [k] is a
p[k] P e is an
and can be compensated for by convolution with an
inverse filter with the response q[k] = p [k]
Ω←→
← →
minimum phase signal
all - pass response function
( )( ) ( )
[ ] ( ) ( )[ ]
j
2j j ( ) j
jj
Q e
Properties: P e e P e =1 (Constant energy spectrum)
1 q[k] p[k]= k Q eP e
p k is stable and causal q[k] is anticausal
Ω
Ω θ Ω Ω
ΩΩ
=
∗ δ =
→
Signature Deconvolution (2)
Received signal: y[k] = h[k]∗ε[k] = s0[k]∗p[k]∗f[k]∗ε[k] + n[k]
Noise will be ignored (this is not always possible):n[k]∼0
Inverse source signature must be found: s-1[k]=s0
-1∗p-1[k]
Processing is now performed finding:s-1[k] ∗y[k] = Σn s-1[n] ∗y[n-k] =f[k]∗ε[k]
Reflection and Transmission Coefficients from Horizons with Perpendicular Incidents
Medium 1
Medium 2
Density of mass: ρ1Velocity of sound: ν1Impedance: Z1=ρ1ν1
Density of mass: ρ2Velocity of sound: ν2Impedance: Z2=ρ2ν2
1 c
t
Reflecting surface
( )
2 1
2 12
2 2 1
2 1
2
2 1
21 1 22
2 2 1
Z Z: cZ Z
Z Z: cZ Z
2Z: t 1 cZ Z
Z 4Z Z: tZ Z Z
−=
+
−= +
= + =+
=+
Reflection coefficient
Power reflection factor
Transmission coefficient
Power transmission factor
Reverberation Deconvolution (1) (deterministic)Seismic trace after signature deconvolution:u[k] = f[k]∗ε[k]
Reverberation function: f[k] Reflectivity function: ε[k]
Sea surface
Sea floor
1 -c (-c)2 (-c)3 (-c)3 (-c)3 .....
.....
Down going signal: g[k] = 1 0 0 .... 0 -c 0 0 .... (-c)2 0 0 ... 0 (-c)3 0 ...k = 0 ................ N ............. 2N ................. 3N .......
Reverberation Deconvolution (2) (deterministic)
( ) ( ) ( )
[ ]
2 3 4N 2 N 3N 4 NN
1 N
1
1G(z) 1 cz c z c z c z .....
1 czInverse z transform : G (z) 1 czInverse impulse response : g k 1 0 0 0...0 c 0.........
k 0 1 2 3...... N ...........
− − − −−
− −
−
= − + − + − + − =+
− = +
=
=
z - transform of downgoing signal :
Down going signal: g[k] = 1 0 0 .... 0 -c 0 0 .... (-c)2 0 0 ... 0 (-c)3 0 ...k = 0 ................ N ............. 2N ................. 3N .......
( )( )
( )[ ]
2 N 2 2 N 3 3N 4 4 N2N
21 N N 2 2 N
1
1F(z) G(z) 1 2cz 3c z 4c z 5c z .....
1 cz
Inverse z transform : F (z) 1 cz 1 2cz c z
Inverse impulse response : f k 1 0 0 0.
− − − −
−
− − − −
−
= = = − + − ++
− = + = + +
=
z - transform of two - way transmission of signal :
FIR filter
FIR filter 2..0 2c 0...0 c 0 0 0 0........k 0 1 2 3...... N .........2N....................=
Reverberation Deconvolution (3)
Sea surface
Sea floor
Target horizon
Direct signalSea surface
Sea floor
Target horizon
1st order multiple signal
Sea surface
Sea floor
Target horizon
2nd order multiple signal
N 2 2 N 3 3N
F(z)1 2cz 3c z 4c z ...− − −
=
− + − +
z - transform of two - waytransmission of signal :