handouts for dsp

8/10/2019 Handouts for DSP

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DSP and Digital Filters (2014-5289)

DSP & Digital Filters

Mike Brookes


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1: Introduction

1: IntroductionOrganization

Signals

ProcessingSyllabus

Sequences

Time Scaling

-TransformRegion ofConvergence

-Transform examples

Rational z-Transforms

Rational example

nverse z-TransformMATLAB routines

Summary



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Organization

1: Introduction

OrganizationSignals

ProcessingSyllabus

Sequences

Time Scaling


-Transform examples


Rational example


Summary


18 lectures: feel free to ask questions

Textbooks: (a) Mitra Digital Signal Processing ISBN

most of the course except for some of t (b) Harris Multirate Signal Processing IS

covers multirate material in more detailMitra

Lecture slides available via Blackboard or on myhttp://www.ee.ic.ac.uk/hp/staff/dmb/courses/d

quite dense - ensure you understand each l email me if you dont understand or dont a

Prerequisites: 3rd year DSP - attend lectures if

Exam + Formula Sheet (past exam papers + so

Problems: Mitra textbook contains many problechapter and also MATLAB exercises


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Signals

1: Introduction

Organization

SignalsProcessingSyllabus

Sequences

Time Scaling


-Transform examples


Rational example


Summary


A signal is a numerical quantity that is a functioindependent variables such as time or position.

Real-world signals are analog and vary continuocontinuous values.

Digital signals are sampled at discrete times andfinite number of discrete values

We will mostly consider one-dimensionsal real-vregular sample instants; except in a few places, quantization.

Extension to multiple dimensions and compis mostly straighforward.

Examples:


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Processing

1: Introduction

Organization

Signals

ProcessingSyllabusSequences

Time Scaling


-Transform examples


Rational example


Summary


Aims to improve a signal in some way or extfrom it

Examples:

Modulation/demodulation

Coding and decoding

Interference rejection and noise suppress

Signal detection, feature extraction

We are concerned with linear, time-invariant p


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Syllabus

1: Introduction

Organization

Signals

Processing SyllabusSequences

Time Scaling


-Transform examples


Rational example


Summary


Main topics:

Introduction/Revision

Transforms

Discrete Time Systems

Filter Design

FIR Filter Design

IIR Filter Design

Multirate systems

Multirate Fundamentals

Multirate Filters Subband processing


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Sequences

1: Introduction

Organization

Signals

ProcessingSyllabus

SequencesTime Scaling


-Transform examples


Rational example


Summary


We denote the nth sample of a signal as x[n] wherethe entire sequence as {x[n]} although we will ofte

Special sequences:

Unit step: u[n] =

1 n 0

0 otherwise

Unit impulse: [n] = 1 n= 0

0 otherwise

Condition: condition[n] =

1 condition is true

0 otherwise

Right-sided: x[n] = 0 forn < Nmin Left-sided: x[n] = 0 forn > Nmax Finite length: x[n] = 0 forn / [Nmin, Nmax]

Causal: x[n] = 0 forn


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Time Scaling

1: Introduction

Organization

Signals

ProcessingSyllabus

Sequences

Time Scaling-Transform

Region ofConvergence

-Transform examples


Rational example


Summary


For sampled signals, the nth sample is at time t=n

is the sample frequency.Often easiest to scale time so that fs = 1 Hz. E.g. low-pass filter for fs = 44.1kHz we can design a 0.fs = 1 Hz.

To scale back to real-world values: Every quantityomultiplied by Tn (or equivalently by fns ). Thus alT and all frequenciesand angular frequenciesbyT

We use for real angular frequencies and for nfrequency. The units of are radians per sample.

Warning: Several MATLAB routines scale time so tnon-standard and irritating.


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z-Transform

1: Introduction

Organization

Signals

ProcessingSyllabus

Sequences

Time Scaling

z-TransformRegion ofConvergence

-Transform examples


Rational example


Summary


The z-transform converts a sequence, {x[n]}, into aarbitrary complex-valued variable z.

Why do it?

Complex functions are easier to manipulate than

Useful operations on sequences correspond to sithe z-transform:

addition, multiplication, scalar multiplicatioconvolution

Definition: X(z) =+

n= x[n]zn


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Region of Convergence

1: Introduction

Organization

Signals

ProcessingSyllabus

Sequences

Time Scaling

-Transform

Region ofConvergence

-Transform examples


Rational example


Summary


The set ofz for which X(z)converges is its Region

Complex analysis: the ROC of a power series (if an annular region of the form 0 Rmin < |z| < Rm

X(z)will alwaysconverge absolutelyinside the ROsome, all, or none of the boundary.

converge absolutely

+n= |x[n]z

n

finite Rmin = 0, Rmax = ROC may included either, both or none of

absolutely summable X(z)converges for |z|

right-sided Rmax = causal X()converges

left-sided Rmin = 0 anticausal X(0)converges


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z-Transform examples


The sample at n= 0 is indicated by an open circle.

u[n] 11z1 1

x[n] 2z2 + 2 + z1 0

x[n 3] z3

2z2 + 2 + z1

0

nu[n]=0.81

1z1

nu[n 1] 11z1 0

nu[n] z1

12z1+z2 1

sin(n)u[n]=0.5z1 sin()

12z1 cos()+z2 1

cos(n)u[n]=0.51z1 cos()

12z1 cos()+z2 1

Note: Examples 4 and 5 have the same z-transform but different R

Geometric Progression:r

n=qn


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1: Introduction

Organization

Signals

ProcessingSyllabus

Sequences

Time Scaling


-Transform examples

Rationalz-Transforms

Rational example

nverse z-Transform

MATLAB routines

Summary


Most z-transforms that we will meet arerational pocoefficients, usually one polynomial in z1 divided b

G(z) =gMm=1(1zmz

1)Kk=1(1pkz

1) =gzKM

Mm=1(zzm)Kk1(zpk)

Completely defined by thepoles,zerosandgain.

Theabsolute valuesof the poles define the ROCs:R+ 1 different ROCs

where R is the number of distinct pole ma

Note: There are K M zeros or M Kpoles at


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Rational example

1: Introduction

Organization

Signals

ProcessingSyllabus

Sequences

Time Scaling


-Transform examples


Rational examplenverse z-Transform

MATLAB routines

Summary


G(z) = 82z1

44z13z2

Poles/Zeros: G(z) = 2z(z0.25))(z+0.5)(z1.5)Polesat z = {0.5, +1.5)},

Zerosat z = {0, +0.25}

Partial Fractions: G(z) = 0.751+0.5z1 + 1.2511.5z1

ROC ROC 0.751+0.5z11.25

11.5z1

a 0 |z|


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Inverse z-Transform

1: Introduction

Organization

Signals

ProcessingSyllabus

Sequences

Time Scaling


-Transform examples


Rational example

Inversez-TransformMATLAB routines

Summary


g[n] = 12j G(z)zn1dz where the integral is anti-

circle within the ROC, z=Rej

.Proof:1

2j

G(z)zn1dz= 12j

m= g[m]z

m

zn

(i)=

m= g[m] 12j

znm1d

(ii)= m= g[m][n m]=g[n]

(i) depends on the circle with radius R lying within

(ii) Cauchys theorem: 12j

zk1dz=[k] forz =dzd

=jRej 12j

zk1dz= 12j2=0 R

k1e

= Rk

2 2=0 ej=Rk(k)=

In practice use a combination of partial fractions an


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MATLAB routines

1: Introduction

Organization

Signals

ProcessingSyllabus

Sequences

Time Scaling


-Transform examples


Rational example

nverse z-Transform MATLAB routinesSummary


tf2zp,zp2tf b(z1)

a(z1) {zm, pk, g}

residuez b(z1)

a(z1)

krk

1pkz1

tf2sos,sos2tf b(z1)

a(z1)

l

b0,l+b1,lz1+b2,lz

1+a1,lz1+a2,lz2

zp2sos,sos2zp {zm, pk, g}

l

b0,l+b1,lz1+b2,l

1+a1,lz1+a2,l

zp2ss,ss2zp {zm, pk, g} x =Ax + Bu

y=Cx + Du

tf2ss,ss2tf b(z1)

a(z1)

x =Ax + Bu

y=Cx + Du


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Summary

1: Introduction

Organization

Signals

ProcessingSyllabus

Sequences

Time Scaling


-Transform examples


Rational example


Summary


Time scaling: assume fs = 1 Hz so <

z-transform: X(z) =+

n= x[n]n

ROC: 0 Rmin < |z| < Rmax Causal: ROC Absolutely summable: |z| = 1 ROC

Inversez-transform: g[n] = 12j G(z)zn1dz

Not uniqueunless ROC is specified Usepartial fractionsand/or atable

For further


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2: Three Different Fourier T

2: Three DifferentFourier Transforms

Fourier Transforms

Convergence ofDTFT

DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines



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Fourier Transforms


Fourier TransformsConvergence ofDTFT

DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


Three different Fourier Transforms:

Continuous-Time Fourier Transform (CTFT): Discrete-Time Fourier Transform (DTFT):x[ Discrete Fourier Transform (DFT):x[n] X

Forward Transform Inve

CTFT X(j) = x(t)e

jtdt x(t) = 12

DTFT X(e

j

) = x[n]e

jn

x[n] =

1

2

DFT X[k] =N1

0 x[n]ej2 kn

N x[n] = 1N

We use forreal and = T fornormalized aNyquist frequency is = 2 fs2 =

T

and =.

For power signals (energy time interval), CTFT

unbounded.Fix this by normalizing:

X(j) = limA12A

AA x(t)e

jtdt

X(ej) = limA12A

AA x[n]e

jn


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Convergence of DTFT


Fourier Transforms

Convergence ofDTFTDTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


DTFT:X(ej) = x[n]e

jn does not conver

Consider the finite sum: XK(ej

) =

K

Kx[n]ej

Strong Convergence:x[n] absolutely summable X(ej) converges

|x[n]| < supX(ej) XK(ej)

Weaker convergence:

x[n] finite energy X(ej

) converges inmean |x[n]|

2< 12

X(ej) XK(e

Example: x[n] = sin0.5nn

0 0.1 0.2 0.3 0.4 0.5-0.2

0

0.20.4

0.6

0.8

1

/2 (rad/sample)

K

j

K=5

0 0.1 0.2 0.3 0.4 0.5-0.2

0

0.20.4

0.6

0.8

1

/2 (rad/sample)

K

j

K=20

0 0.1 0.2-0.2

0

0.20.4

0.6

0.8

1

/2 (r

K

j

Gibbs phenomenon:Converges at each as K but peak error d


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[DTFT Convergence Proofs]

DSP and Digital Filters (2014-5289) Fourier

(1) Strong Convergence: [these

We are given that

|x[n]| < >0, Nsuch that

|n|>N

For K N, supX(ej ) XK(ej)= sup |n|>Kx[n]ejn

sup

|n|>K

x[n]ejn(2) Weak Convergence:

We are given that

|x[n]|2 < >0, Nsuch that|n|>N

Define y[K][n] =

0 |n| K

x[n] |n| > Kso that its DTFT is, Y[K](ej ) =

We see that X(ej ) XK(e

j ) = x[n]e

jn KKx[n]e

j

=|n|>Kx[n]e

jn = y

[K][n

From Parsevals theorem,

y[K][n]2 = 12

Y[K](ej )2 d= 1

2

X(ej) XK(eHence for K N, 1

2

X(ej ) XK(ej)2 d = y[K][n]


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DTFT Properties


Fourier Transforms

Convergence ofDTFT

DTFT PropertiesDFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


DTFT:X(ej) = x[n]e

jn

DTFTisperiodicin : X(e

j(+2m

) =X(e

j

) DTFTis the z-Transformevaluated at the poin

X(z) = x[n]z

n

DTFT converges iff the ROC includes |z| = 1.

DTFTis the same as the CTFT of a signal comthe sample times(dirac delta functions) :

x(t) =

x[n](t nT)=x(t) (

Equivalent to multiplying a continuousx(t)by a

Proof: X(ej) =

x[n]ejn

n= x[n]

(t nT)e

j tT

(i)

=n= x[n](t nT)e

j

(ii)=

x(t)e

jtdt

(i) OK if |x[n]| < . (ii) use = T


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DFT Properties


Fourier Transforms

Convergence ofDTFT

DTFT Properties

DFT PropertiesSymmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


DFT:X[k] =N1

0 x[n]ej2 kn

N

DTFT:X(ej) = x[n]ejn

Case 1: x[n] = 0 forn / [0, N 1]

DFT is the same as DTFT at k = 2N

k.

The {k} are uniformly spaced from = 0 to

DFT is the z-Transform evaluated at N equaaround the unit circle beginning at z= 1.

Case 2: x[n] is periodic with period N

DFT equals the normalized DTFT

X[k] = limKN

2K+1 XK(ejk

)

where XK(ej) =

KKx[n]e

jn


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Symmetries


Fourier Transforms

Convergence ofDTFT

DTFT Properties

DFT Properties

SymmetriesParsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


Ifx[n] has a special property then X(ej)and X[k]properties as shown in the table (and vice versa):

One domain Other d

Discrete PerioSymmetric Symm

Antisymmetric Antisym

Real Conjugate SImaginary Conjugate An

Real + Symmetric Real + SyReal + Antisymmetric Imaginary + A

Symmetric: x[n] =x[n]

X(ej) =X(ej)X[k] =X[(k)mod N] =X[N k] fo

Conjugate Symmetric: x[n] =x[n]Conjugate Antisymmetric: x[n] = x[n]


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Parsevals Theorem


Fourier Transforms

Convergence ofDTFT

DTFT Properties

DFT Properties

Symmetries

ParsevalsTheorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


Fourier transforms preserveenergy

CTFT

|x(t)|2

dt= 12

|X(j)|

2

dDTFT

|x[n]|

2= 12

X(ej)

2 d

DFT N1

0 |x[n]|2

= 1N

N10 |X[k]|

2

More generally, they actually preservecomplex inne

N10 x[n]y[n] = 1N

N10 X[k]Y[k]

Unitary matrix viewpoint for DFT:

If we regard x and Xas vectors, then X = Fxwa symmetric matrix defined by fk+1,n+1 =e

j2

The inverse DFT matrix is F1 = 1NFH

equivalently, G = 1NF is aunitary matrixwith


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Convolution


Fourier Transforms

Convergence ofDTFT

DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

ConvolutionSampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


DTFT:Convolution Productx[n] =g[n] h[n]=

k=g[k]h[n k]

X(ej) =G(ej)H(ej)

DFT:Circular convolutionProductx[n] =g[n] Nh[n]=

N1k=0 g[k]h[(n k)mo

X[k] =G[k]H[k]

DTFT:Product Circular Convolution 2

y[n] =g[n]h[n] Y(ej) = 12G(e

j) H(ej) = 12G(

DFT:ProductCircular ConvolutionNy[n] =g[n]h[n]

X[k] = 1N

G[k] NH[k]

g[n] : h[n] : g[n] h[n] : g


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Sampling Process


Time Time

Analog CT

Low PassFilter

* CT

Sample DT

Window DT

DFT DF


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Phase Unwrapping


Fourier Transforms

Convergence ofDTFT

DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase UnwrappingSummary

MATLAB routines


Phase of a DTFT is only defined to within an integ

-20

2

4

6

(

x[n] |X

-2 0 2

-2

0

2

(rad/sample)-2

-40

-20

0

(

X[k] X[k]

Phase unwrapping adds multiples of2 onto each phase as continuous as possible.


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Summary


Fourier Transforms

Convergence ofDTFT

DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


Three types: CTFT, DTFT, DFT

DTFT = CTFT of continuous signal

DFT = DTFT of periodic or finite suppo

DFT is a scaled unitary transform

DTFT: Convolution Product; Product

DFT: Product Circular Convolution

DFT: Zero Padding Denser freq sampling

Phase is only defined to within a multiple of2

Whenever you integrate frequency component

12 for CTFT and DTFT or

1N

for DFT

Inverse transform, Parseval, frequency do

For further details see Mitra: 3 & 5.


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MATLAB routines


Fourier Transforms

Convergence ofDTFT

DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


fft, ifft DFT with optional zero-padd

fftshift swap the two halves of a vecconv convolution or polynomial multipliccircular)

x[n]y[n] real(ifft(fft(x).*fft(y)))unwrap remove 2 jumps from phase spe


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3: Discrete Cosine Tra

3: Discrete CosineTransform

DFT Problems

DCTDCT of sine waveDCT/DFTEquivalence

DCT Properties

DCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform

MDCT (ModifiedDCT)MDCT & TDAC

Summary

MATLAB routines



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DFT Problems


DFT ProblemsDCTDCT of sine waveDCT/DFTEquivalence

DCT Properties

DCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform


Summary

MATLAB routines


For processing 1-D or 2-D signals (especially codingto divide the signal into frames and then apply aneach frame that compresses the information into few

The DFT has some problems when used for this pu

N real x[n] N complex X[k] : 2 real, N2

DFT the DTFT of a periodic signal formed Spurious frequency components from boun

N=20f=0.08

The Discrete Cosine Transform (DCT) overcomes t


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DCT


DFT Problems


DCT Properties

DCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform


Summary

MATLAB routines


To form the Discrete Cosine Transform (DCT), replreverse order and insert a zero between each pair of

0 12

y[r]

Take the DFT of length 4Nreal symmetric sequenc

Result is real, symmetric and anti-periodic: only ne

012

23

Y[k]

Forward DCT: XC[k] =N1

n=0 x[n]cos2(2n+1)k

4N fo

Compare DFT: XF[k] =N1

n=0 x[n]expj2(4n+0)

4N


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DCT of sine wave


DFT Problems

DCT DCT of sine waveDCT/DFTEquivalence

DCT Properties

DCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform


Summary

MATLAB routines


DCT: XC[k] =N1

n=0 x[n]cos2(2n+1)k

4N

f= mN f= mN

x[n]N=20f=0.10

N=20f=0.08

|XF[k]|

|XC[k]|

DFT: RealComplex; Freq range [0, 1]; Poorly unless f= m

N; |XF[k]| k

1 forNf < kDCT: RealReal; Freq range [0, 0.5]; Well loca

|XC[k]| k2 for2Nf < k < N


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DCT/DFT Equivalence


DFT Problems

DCTDCT of sine wave

DCT/DFTEquivalence

DCT Properties

DCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform


Summary

MATLAB routines


XC[k] =N1n=0 x[n]cos

2(2n+1)k4N

Define y[r] =

0 r even

xr12

r= 1 : 2 : 2N 1

x4N1r

2

r= 2N+ 1 : 2 : 4N

YF[k] = 4N1r=0 y[r]W

kr4N where WM =e

j2M

(i)=2N1

n=0 y[2n + 1]W(2n+1)k4N

(ii)=N1

n=0 y[2n+ 1]W(2n+1)k4N

+N1

m=0y[4N 2m 1]W(4N2m4N

(iii)=x[n]W

(2n+1)k4N +

N1m=0x[m]W

(2m+4N

= 2N1

n=0 x[n]cos2(2n+1)k

4N = 2XC[k]

(i) odd r only: r= 2n + 1, (ii) reverse order for n

(iii) substitute y definition & W4Nk4N =ej2 4Nk

4N


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DCT Properties


DFT Problems


DCT PropertiesDCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform


Summary

MATLAB routines


Definition: X[k] =N1

n=0 x[n]cos2(2n+1)k

4N

Linear: x[n] + y[n] X[k] + Y[k] DFTConvolutionMultiplication propert Symmetric: X[k] =X[k] sincecos k= c Anti-periodic: X[k+ 2N] = X[k] because:

cos(+ ) = cos 2(2n + 1)(k+ 2N) = 2(2n + 1)k+ 8

X[N] = 0 since X[N] =X[N] = X[N Periodic: X[k+ 4N] = X[k+ 2N] =X[k]

DCT basis functions:

0:

3:

1: 4:

2: 5:


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IDCT


DFT Problems


DCT Properties

IDCTEnergy Conservation

Energy Compaction

Frame-based coding

Lapped Transform


Summary

MATLAB routines


y[r] = 14N4N1

k=0 Y[k]Wrk4N =

12N

4N1k=0 X[k]W

x[n] =y[2n + 1] = 12N4N1

k=0 X[k]W(2n+1)k

4N (i)= 12N

2N1k=0 X[k]W

(2n+1)k4N

12N2N1

l=0 X[l]W(2n+1)(l+2N)4N

(ii)= 1

N2N1k=0 X[k]W

(2n+1)k4N

(iii)= 1N

X[0] + 1N

N1k=1 X[k]W

(2n+1)k4N +

1N

X

+ 1N

N1r=1 X[2N r]W

(2n+1)(24N

(iv)= 1

NX[0] + 1

N

N1k=1 X[k]W

(2n+1)k4N

+ 1N

N1r=1 X[r]W

(2n+1)r+2N4N

x[n] = 1

N

X[0] + 2

N

N1

k=1

X[k]cos 2(2n+1)k

4N

= Inv

(i) k=l+ 2N fork 2N and X[k+ 2N] = X[

(ii) (2n+1)(l+2N)4N = (2n+1)l

4N + n + 12

(iii) k= 2N r fork > N (iv) X[N] = 0 and


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Energy Conservation


DFT Problems


DCT Properties

DCT

EnergyConservation

Energy Compaction

Frame-based coding

Lapped Transform

MDCT (ModifiedDCT)

MDCT & TDAC

Summary

MATLAB routines


DCT: X[k] =N1

n=0 x[n]cos2(2n+1)k

4N

IDCT: x[n] = 1NX[0] +

2NN1

k=1 X[k]cos2(2n+1)

4N

rep

0 12 23

y[r]

DFT 0

12

Y[k]

Energy: E=N1n=0 |x[n]|

2: E 2E 8N E 0

E= 1N

|X|2

[0] + 2N

N1n=1 |X|

2[n]

Orthogonal DCT(preserves energy)

Define: c[k] =

2kN

c[0] =

1N

, c[k= 0

ODCT: X[k] =c[k]N1n=0 x[n]cos

2(2n+1)k4N

IODCT: x[n] =N1

k=0 c[k]X[k]cos2(2n+1)k

4N

Note: MATLAB dct


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Frame-based coding


DFT Problems


DCT Properties

DCT

Energy Conservation

Energy Compaction

Frame-basedcoding

Lapped Transform

MDCT (ModifiedDCT)

MDCT & TDAC

Summary

MATLAB routines


Divide continuous signalinto frames

Apply DCT to each frame

Encode DCT

e.g. keep only 30X[k]

Apply IDCT y[n]

x[n]

X[k]

y[n]

y[n]-x[n]

Problem: Coding may create discontinuities at frame.g. JPEG, MPEG use 8 8 pixel blocks

8.3 kB (PNG) 1.6 kB (JPEG) 0


41/211

Lapped Transform


DFT Problems


DCT Properties

DCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped TransformMDCT (ModifiedDCT)MDCT & TDAC

Summary

MATLAB routines


Modified Discrete Cosine Transform(MDCT): over

x[0 : 2N 1]MDCT

X0[0 :N 1]IMDCT

y0[0 : 2N 1]

x[N : 3N 1]MDCT

X1[N : 2N 1]

IMDCT y1[N : 3N 1]

x[2N : 4N 1]MDCT

X2[2N : 3N 1]IMDCT

y2[2N : 3N 1]

y[n] =y0[n] + y1[n] + y2[n]

X0[k]

y0[n]

X1[k]

y1[n]

X

y

y[n]

y[n]-x[n] = error

x[n]

0 N

N

MDCT: 2N N coefficients,IMDCT: N 2N saAddyi[n] together to get y[n]. Only two non-zero tErrors cancel exactly: Time-domain alias cancellatio


42/211

MDCT (Modified DCT)


DFT Problems


DCT Properties

DCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform

MDCT (ModifiedDCT)

MDCT & TDAC

Summary

MATLAB routines


MDCT: XM[k] =2N1

n=0 x[n]cos2(2n+1+N)(

8N

IMDCT: yA,B[n] = 1N

N1k=0 XM[k]cos 2

(2n+

MDCT from a DCT of length 2N:

Split x[n] into four N2-vectors:x=

a b c d

Form the N-vectoru=

d c a

b

wherec is in reverse order

Form v=

u u

Take 2N DCT: v VC

Symmetry VC[2k] = 0

Set XM[k] =VC[2k+ 1]

a

-c 100.

Example: M= 103, N= 104:

Direct: 2M N= 2 107with DFT: 12 (M+ N)log2(M+ N) = 1.8

But: (a) DFT may be very long ifN is large(b) No output samples until all x[n]has been i


57/211


58/211

Overlap Save

4: Linear Timenvariant Systems

LTI Systems

ConvolutionProperties

BIBO Stability

Frequency Response

Causality

Passive and Lossless

ConvolutionComplexity

Circular ConvolutionFrequency-domain

onvolution

Overlap Add Overlap SaveSummary

MATLAB routines


Alternative method:(1)chop x[n] into N

K

overlappingchunks of length K+ M 1

(2) K+M1 each chunk with h[n](3)discard first M 1 from each chunk(4)concatenate to make y[n]

The first M 1 points of each output chunk are in

Operations: slightly less than overlap-add because ncreate y[n]

Advantages:same as overlap add

Strangely, rather less popular than overlap-add


59/211

Summary


LTI Systems


BIBO Stability

Frequency Response

Causality




onvolution

Overlap AddOverlap Save

SummaryMATLAB routines


LTI systems: impulse response, frequency respo

BIBO stable, Causal, Passive, Lossless systems Convolution and circular convolution properties

Efficient methods for convolution single DFT overlap-add and overlap-save

For further


60/211

MATLAB routines


LTI Systems


BIBO Stability

Frequency Response

Causality




onvolution

Overlap AddOverlap Save

Summary

MATLAB routines


fftfilt Convolution using overlap ax

[n

]y

[n

] real(ifft(fft(x).*fft(y)))


61/211

5: Filters

5: FiltersDifference Equations

FIR Filters

FIR SymmetriesIR Frequency

Response

Negating z

Cubing z

Scaling z

IR Impulse responseexamples

IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase

Linear Phase Filters

Summary

MATLAB routines



62/211

Difference Equations

5: Filters

DifferenceEquations

FIR FiltersFIR Symmetries

IR FrequencyResponse

Negating z

Cubing z

Scaling z


IR Impulse response

Other BIBO

esponsesStability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


Most useful LTI systems can be described by a diffe

y[n] =Mr=0 b[r]x[n r]

N

r

Nr=0 a[r]y[n r] =Mr=0 b[r]x[n r]

a[n] y[n] =b[n] x[n] Y(z) = B(z)

A(z)X(z)

Y(ej) = B(ej)A(ej)X(e

j)

(1)Alwayscausal.(2)Orderof system ismax(M, N), the highest r w(3)We assume that a[0] = 1; if not, divide A(z)an(4)Filter is BIBO stable iff roots ofA(z)all lie wit

Notenegative signin first equation.Authors in some SP fields reverse the sign of the a[


63/211

FIR Filters

5: Filters


FIR FiltersFIR SymmetriesIR Frequency

Response

Negating z

Cubing z

Scaling z


IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


A(z) = 1: Finite Impulse Response(FIR) filter: Y(Impulse response is b[n]and is of length M+

Frequency responseis B(ej) and is the DTFT ofbComprises Mcomplex sinusoids + const:

b[0] +b[1]ej + +b[M]ejM

Small Mresponse contains only low quefrencies

Symmetrical b[n]H(ej

)e

jM

2 consists of M

2 cosin

M=4 M=14

0 1 2 30

0.5

1

0 1 2 30

0.5

1

Rule of thumb: Fastest possible transition 2M


64/211

FIR Symmetries

5: Filters


FIR Filters

FIR SymmetriesIR FrequencyResponse

Negating z

Cubing z

Scaling z


IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


B(ej)is determined by the zeros ofzMB(z) =

Real b[n] conjugate zeSymmetric: b[n] =b[M n] reciprocal zero

Real + Symmetric b[n] conjugate+or else pairs

Real: Symmetric: [1, 1.28, 0.64] [1, 1.64 + 0.27j, 1]

-1 0 1

-1

-0.5

0

0.5

1

z-1 0 1

-1

-0.5

0

0.5

1

z

-2 0 20

1

2

(rad/sample)-2 0 2

0

1

2

3

(rad/sample)


65/211

[FIR Symmetry Proofs]


In all of the proofs below, we assume that z=z0 is a root ofB(z)so that B

and then we prove that this implies that other values of z also satisfy B(z)

(1) Real b[n]

B(z0) =M

r=0b[r]

z0r

=M

r=0b[r]

z0r

since b[r] is real

= Mr=0b[r]zr0

take complex conjugate

= 0 = 0 since B(z0) = 0

(2) Symmetric: b[n] = b[M n]

B(z10 ) =M

r=0b[r]zr0

=

Mn=0b[M n]z

Mn0 substitute r= M n

=zM

0M

n=0b[M n]z

n

0 take out zM

0 factor=zM0

Mn=0b[n]z

n0 since b[M n] = b[n]

=zM0 0 = 0 since B(z0) = 0


66/211

IIR Frequency Response

5: Filters


FIR Filters

FIR Symmetries

IIR FrequencyResponse

Negating z

Cubing z

Scaling z


IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


Factorize H(z) = B(z)A(z) =

b[0]M

i=1(1qiz1)

Ni=1(1piz

1)

Roots ofA(z)and B(z) are thepoles{pi} andzAlso an additional N Mzeros at the origin (affecH(ej)= |b[0]||z

M|Mi=1|zqi||zN|

Ni=1|zpi|

forz=ej

Example:

H(z) = 2+2.4z110.96z1+0.64z2 = 2(1+1.2z

1

(1(0.480.64j)z1)(1(0

At= 1.3:H(ej)= 21.761.620.39= 5.6H(ej) = (0.6 + 1.3) (1.7 + 2.2) =

0 1 2 30

5

10

-1 0 1

-1

-0.5

0

0.5

1

1.76 1.62

0.39

(z)


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Negating z

5: Filters


FIR Filters


Response

Negating zCubing z

Scaling z


IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


Given a filter H(z) we can form a new one HR(z) =Negate all odd powers ofz, i.e. negate alterna

Example: H(z) = 2+2.4z1

10.96z1+0.64z2

-1 0 1-1

-0.5

0

0.5

1

(z)

0 1 2 30

5

10

Negate z: HR(z) = 22.4z1

1+0.96z1+0.64z2 Negate o

-1 0 1

-1

-0.5

0

0.5

1

(z)

0 1 2 30

5

10

Pole and zero positions arenegated, frequency resp


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Cubing z

5: Filters


FIR Filters


Response

Negating z

Cubing zScaling z


IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


Given a filter H(z) we can form a new one HC(z) =Insert two zeros between each a[n]and b[n] te


10.96z1+0.64z2

-1 0 1-1

-0.5

0

0.5

1

(z)

-2 0 20

5

10

(rad/sample)

Cube z: HC(z) = 2+2.4z3

10.96z3+0.64z6 Insert 2 ze

-1 0 1

-1

-0.5

0

0.5

1

z-2 0 2

0

5

10

(rad/sample)

C

Pole and zero positions arereplicated, magnitude re


69/211

Scaling z

5: Filters


FIR Filters


Response

Negating z

Cubing z

Scaling zIR Impulse response

examples

IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


Given a filter H(z) we can form a new one HS(z) =Multiply a[n]and b[n] byn


10.96z1+0.64z2

-1 0 1

-1

-0.5

0

0.5

1

(z)

0 1 2 30

5

10

Scale z: HS(z) =H( z1.1 ) =

2+2.64z1

11.056z1+0.7744z2

-1 0 1

-1

-0.5

0

0.5

1

z0 1 2 30

5

10

15

20

(rad/s)

Pole and zero positions aremultiplied by , >1 Pole at z =pgives peak bandwidth 2 |log |p||For pole near unit circle,decrease bandwidthby


70/211

IIR Impulse response examples

5: Filters


FIR Filters


Response

Negating z

Cubing z

Scaling z

IIR Impulseresponse examples

IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


To find the impulse response of B(z)A(z) , use partial fra

Example 1: M < N

B(z)A(z) =

13.4z1

10.3z10.4z2 = 3

1+0.5z1 + 210.8z1

h[n] = (3 (0.5)n 2 (0.8)n) u[n]Example 2: M NB(z)A(z) = 25

.7z1

+0.2z2

+0.8z3

10.3z10.4z2

= 1 2z1 + 13.4z110.3z10.4z2= 1 2z1 + 31+0.5z1 + 210.8z1

h[n] =[n] 2[n 1] + (3 (0.5)n 2 (0.8)n

Example 3: repeated pole at z = 0.5B(z)A(z) =

35z11.3z21+0.3z10.55z20.2z3

= 21+0.5z1 + 3

(1+0.5z1)2+ 210.8z1

h[n] = (2 (0.5)n + 3(n + 1) (0.5)n 2 (0.8


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IIR Impulse response

5: Filters


FIR Filters


Response

Negating z

Cubing z

Scaling z


IIR Impulseresponse

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


H(z) = B(z)A(z) we want h[n], the inverse z-Transform

(1) Assume M < Nelse, do a long division: H(z)

(2) Factorize A(z) =Npi=1

1 piz1

Ki (Np=(3) Then

B(z)A(z) =

Npi=1

Kik=1

ri,k

(1piz1)k (K

where ri,k =

(pi)kKi

(Kik)!

dKik

d(z1)Kik B(z)A(z) (1 p

For the common case Ki = 1 i(e.g. all poles B(z)A(z) =

Npi=1

ri1piz1

where ri=B(A(

(4) Theimpulse responseis given by

h[n] =Npi=1

Kik=1 Cn+k1k1 ri,kpni forn 0

where C = !!()! is a binomial coefficient

For the common case Ki = 1 i: h[n] =


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[Impulse Response: Proof of (3)]


[You do not ne

We assume thatB(z)

A(z) =Np

j=1Kj

l=1

rj,l

(1pjz1)land we want to determin

(1) Multiply both sides by

1 piz1Ki :

B(z)A(z)

1 piz

1Ki = Npj=1Kjl=1 rj,l(1piz1)Ki(1pjz1)l

now separate out the term with j =i:

B(z)A(z)

1 piz

1

Ki =

Kil=1ri,l

1 piz

1

Kil +

j=i

Kjl=1

rj,l(

(1

(2) DifferentiateKi k times with respect to z1 : d

Kik

d(z1)Kikand evalua

1 k Ki Ki k < Ki so we are differentiating < Ki times.

Hence all terms in

j=i

Kjl=1

rj,l(1piz1)Ki

(1pjz1)l will still include a fact

numerator and will equal zero at z = pi.

Similarly, terms inKil=1ri,l 1 piz1Kil with l < k will also vaniAlso, terms in

Kil=1ri,l

1 piz1

Kil with l > k will also different

The only term remaining is: (p)Kik (Ki k)!ri,k from which the re

[Similarly dn

dxn(xm)

x=0

is zero unless m= n i


73/211

[Impulse Response: Proof of (4)]


[You do not ne

We use induction on k to show that 1 pz1k = n=0

Cn+k1

k1

pnz

(1) Suppose the formula is true for k and differentiate both sides with respe

dd(z1)

1 pz1

k=

n=0Cn+k1k1 p

nzn

This gives:

pk

1 pz1

(k+1)

=

n=0nC

n+k1k1 p

nz(n1) =

n=1nC

where we have omitted the term with n= 0 because it equals zero.

Now substitute, m= n 1:

pk

1 pz1(k+1)

=

m=0(m+ 1)Cm+kk1 p

m+1zm

1 pz1(k+1)

=

m=0m+1k

Cm+kk1 p

mzm =

m=0Cmk

which proves the result for k+ 1.

(2) To start the induction at k = 1:

1 pz1

1 =

n=0Cn0pnzn =

n=0pnzn

This is true from the standard geometric progression formula. (Note th


74/211

Other BIBO responses

5: Filters


FIR Filters


Response

Negating z

Cubing z

Scaling z


IR Impulse response

Other BIBOresponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


All LTI systems described by difference equations hexponentially or exponentially times a power ofn.

It is perfectly possible to have BIBO stable LTI systresponses decay more slowly than this.

Example: h[n] = 11+n2 forn 0BIBO stable since

|h[n]| = 2.077... < Hence H(e

j

)exists.However, no finite difference equation can have t

0 1 0

0.5

1

1.5

2


75/211

Stability Triangle

5: Filters


FIR Filters


Response

Negating z

Cubing z

Scaling z


IR Impulse response

Other BIBOesponses

Stability TriangleLow-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


Not normally easy to tell if poles lie within

|z

|= 1, but

easy test for a 2nd order filter:

Suppose A(z) = 1 + a[1]z1 + a[2]z2

The roots are p1,2 = a[1]

a[1]24a[2]

2

Want the conditions that|p1,2| < KCase 1: Real roots:

(P) a[1]2

4a[2](Q)2K < a[1] a[1]2 4a[2]


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Low-pass filter

5: Filters


FIR Filters


Response

Negating z

Cubing z

Scaling z


IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filterAllpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


1st order low pass filter: extremely commony[n] = (1

p)x[n] +py[n

1]

H(z) = 1

1pz

Impulse response:h[n] = (1 p)pn = (1 p)en

where = 1 lnp is the time constant in sa

Magnitude response:

H(ej)

= 1p

12p cos+p2

Low-pass filter with DC gain of unity.3 dB frequency is 3dB = cos1

1 (1p)22p

Compare continuous time: HC(j) = 11+j

Indistinguishable for low but H(ej) is perio

-1 0 1

-1

-0.5

0

0.5

1

(z)

p=0.80

0.01

-30

-20

-10

0


77/211

Allpass filters

5: Filters


FIR Filters


Response

Negating z

Cubing z

Scaling z


IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filtersGroup Delay

Minimum Phase


Summary

MATLAB routines


Ifa[n] =b[M n] we have anallpass filter

H(ej) =M

r=0b[r]ejr

Mr=0b[r]e

j(Mr) =ejMM

r=0b[r]eMr=0b[r]

The two sums are complex conjugatesthey haveHence

H(ej)= 1 allpassHowever phase isnotconstant: H(ej) =M

1st order allpass: H(z) = p+z1

1pz1= p1p1z1

1pz1

Pole at p and zero at p1: reflected in unit ci

Constant distance ratio:ej p= |p|

ej

0 1 2 30

0.2

0.4

0.6

0.8

1

0 1 2 3

-3

-2

-1

0

In an allpass filter, thezeros are the poles reflected


78/211

Group Delay

5: Filters


FIR Filters


Response

Negating z

Cubing z

Scaling z


IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group DelayMinimum Phase


Summary

MATLAB routines


Group delay: H(ej) = dH(ej)

d = delay of the

Trick to get at phase: ln H(ej) = lnH(ej)+j

H= d((lnH(ej)))

d =

1H(ej)

dH(ej)d

=

H

H(ej) =n=0 h[n]e

jn= F(h[n])dH(ej)d

=n=0

jnh[n]ejn=

jF(nh[n]

H= 1H(ej)

dH(ej)d

=

jF(nh[n])F(h[n])

=

F(F

Example: H(z) = 11pz1 H= [1 p]=

-1 0 1

-1

-0.5

0

0.5

1

(z)

0 1 2 3

-0.8

-0.6

-0.4

-0.2

0

p=0.80

0 1

0

1

2

3 p=0.80

H

Average group delay (over ) = (# poles # zerosZeros on the unit circle count


79/211

[Group Delay Properties]


The group delay of a filter H(z)at a frequency , measured in seconds or sam

of the envelope of a modulated sine wave at a frequency

. It is defined aFor example, H(z) = zk defines a filter that delays its input by k samplegroup delay by evaluating

H(ej ) =

dH(ej)

d=

d

d

ejk

=

d

d(k

which tells us that this filter has a constant group delay of k samples that is

The average value of H equals the total change in

H(ej

) as goes f2. If you imagine an elastic string connecting a pole or zero to the point z goes from to+ the string will wind once around the pole or zero if it not if it is outside. Thus, the total change in H(ej) is equal to 2 times the unit circle minus the number of zeros inside the unit circle. A zero that icounts 1

2since there is a sudden discontinuity of in H(ej) as passes

When you multiply or divide complex numbers, their phases add or subtracyou multiply or divide transfer functions their group delays will add or sub

the group delay of an IIR filter, H(z) = B(z)A(z)

, is given by H = B Acan determine the group delay of a factorized transfer function by summinindividual factors.


80/211


81/211

[Group Delay Example]


As an example, suppose we want to determine the group delay of : H(z) =

ifH(z) = B(z)A(z) , then H = B A. In this case B = 0 so H = [1 p

Using equation (2) gives H = F([0 p])F([1 p])

since nh[n] = [0 1] [1

Applying the definition of the DTFT, we get

H =

pej

1 pej

=

p

ej p

=

p

ej p

(ej p) (ej p)=

As demonstrated above, the average value ofHis zero for this filter becauszero inside the unit circle.


82/211

Minimum Phase

5: Filters


FIR Filters


Response

Negating z

Cubing z

Scaling z


IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum PhaseLinear Phase Filters

Summary

MATLAB routines


Average group delay (over ) = (# poles # zeros

zeros on the unit circle count Reflecting an interior zero to the exteriormultiplies

H(ej) by a constant butincreases average group delayby 1 sample.

-1 0 1

-1

-0.5

0

0.5

1

(z)

0

-10

-5

0

-1 0 1

-1

-0.5

0

0.5

1

(z)

0 -10

0

10

20

30

H

A filter with all zeros inside the unit circle is amini Lowest possible group delayfor a given magn Energy in h[n] isconcentrated towards n= 0


83/211


5: Filters


FIR Filters


Response

Negating z

Cubing z

Scaling z


IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase

Linear PhaseFilters

Summary

MATLAB routines


The phase of alinear phasefilter is: H(ej) =0

Equivalentlyconstant group delay: H= dH(ej

d

A filter has linear phase iffh[n] issymmetricoranth[n] =h[M n]nor else h[n] = h[M

Mcan be even ( mid point) or odd (mid Proof

:

2H(ej) =M0 h[n]e

jn +M0 h[M n]ej

=ejM2

M0 h[n]e

j(nM2) + h[Mh[n]symmetric:

2H(ej) = 2ejM2

M0 h[n]cos

n M2

h[n]anti-symmetric:2H(ej) = 2jejM2 M0 h[n]sin

n M2

= 2ej(2+

M2)M

0 h[n]sin

n M2


84/211

Summary

5: Filters


FIR Filters


Response

Negating z

Cubing z

Scaling z


IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase




Useful filters havedifference equations:

Freq response determined by pole/zero pos N Mzeros at origin (or M N poles) Geometric construction of|H(ej)|

Pole bandwidth 2 |log |p|| 2 (1 | Stable if poles have|p|


85/211

MATLAB routines

5: Filters


FIR Filters


Response

Negating z

Cubing z

Scaling z


IR Impulse response

Other BIBOesponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


filter filter a signal

impz Impulse responseresiduez partial fraction expansiongrpdelay Group Delay

freqz Calculate filter frequency resp


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6: Window Filter De

6: Window FilterDesign

nverse DTFT

Rectangular windowDirichlet Kernel

Window relationships

Common Windows

Uncertainty principle

Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines



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Inverse DTFT


Inverse DTFTRectangular windowDirichlet Kernel


Common Windows


Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


For any BIBO stable filter, H(ej) is the DTFT of

H(ej) =

h[n]ejn h[n] = 12H

If we know H(ej) exactly, the IDTFT gives t

Example: Ideal Lowpass filter

H(ej) = 1 || 00 || > 0

h[n] = sin0nn

-2 0 20

0.5

1

20

0

2/

Note: Width in is 20, width in n is 20

: productSadly h[n] is infiniteandnon-causal. Solution: mul


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Rectangular window


nverse DTFT

RectangularwindowDirichlet Kernel


Common Windows


Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


Truncate to M2 to make finite; h1[n] is now of len

MSE Optimality:Define mean square error (MSE) in frequency do

E= 12

H(ej)H1(ej)2 d= 12

H(ej)

M2

M2

h1[n]ejn

2

d

Minimum E is when h1[n] = (h[n])Proof: Differentiate w.r.t. h1[r]and set to zero

However: 9%overshoot at a discontinuity even for

0

h1[n]

M=14

0 1 0

0.5

1

M=14

M=28

Normal to delay by M2 to make causal. Multiplies H


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Dirichlet Kernel


nverse DTFT

Rectangular window Dirichlet KernelWindow relationships

Common Windows


Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


TruncationMultiply h[n] by a rectangular windo

Circular Convolution HM+1(ej

) =

1

2

W(ej) =M

2

M2

ejn(i)= 1 + 2

0.5M1 cos(n)

(ii)=

Proof: (i) ej(n) + ej(+n) = 2 cos (n) (ii) Su

Effect: convolve ideal freq response withDirichlet k

-2 0 2

0

0.5

1

-2 0 2

0

0.5

1

4/(M+1)

-

0

0.5

1

-2 0 2

0

0.5

1M=14

Provided that 4M+1


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[Dirichlet Kernel]


Other properties ofW(ej):

The DTFT of a symmetric rectangular window of length M + 1 is W(esmall x we can approximate sin x x; the error is < 1% for x < 0.25. W(ej) 21 sin0.5(M+ 1).

The peak value is at = 0 and equals M+ 1;this means that the peak gr

be M+12

.

The minimum value of W(ej ) is approximately equal to the minimuum which is when sin0.5(M + 1) = 1 i.e. = 1.5

0.5(M+1) = 3

M+1

min21 sin0.5(M+ 1) = M+11.5 .

Passband and Stopband ripple:

The ripple in W(ej) = sin 0.5(+1)

sin 0.5 has a period of = 2

0.5(+1) =

M

ripple with this period in both the passband and stopband ofHM+1(ej).

However the stopband ripple takes the value of HM+1(ej) alternately po

plot the magnitude response,HM+1(ej ) then this ripple will be full-wave frequency so its period will now be 2

M+1.


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nverse DTFT


Windowrelationships

Common Windows


Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


When you multiply an impulse response by a windoHM+1(e

j) = 1

2

H(ej) W(ej)

-2 0 20

0.5

1

-2 0 2

0

10

20

M=20

0

0.5

1

1

(a) passband gain w[0]; peak w[0]2 + 0.52

main

rectangular window: passband gain = 1; pe

(b) transitionbandwidth, = width of the matransitionamplitude, H= integral of main

rectangular window: = 4M+1 , H 1

(c) stopband gain is an integral over oscillating srect window:

min H(ej)= 0.09 min(d) features narrower than the main lobe will be

attenuated


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Common Windows


nverse DTFT



Common WindowsUncertainty principle

Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


Rectangular: w[n] 1dont use

Hanning: 0.5 + 0.5c1ck = cos

2knM+1

rapid sidelobe decay

Hamming: 0.54 + 0.46c1best peak sidelobe

Blackman-Harris3-term:0.42 + 0.5c1+ 0.08c2

best peak sidelobe

Kaiser:I0

1( 2nM)

2

I0()

controls width v sidelobesGood compromise:

Widthvsidelobevdecay


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nverse DTFT



Common Windows

Uncertaintyprinciple

Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


CTFT uncertainty principle: t2|x(t)|2dt

|x(t)|2dt 1

2

2|X

|XThe first term measures thewidth ofx(t)aro

It is like if|x(t)|2

was a zero-mean probThe second term is similarly thewidth ofX(

A signalcannot be concentrated in both time and fSo a short window cannot have a narrow main

Proof:Assume

|x(t)|

2dt= 1

|X(j)|

2d= 2

Set v(t) = dxdt V(j) =jX(j) [by parts]

Now

txdxdt

dt= 12 tx2t=

12x

2dt= 12 [

So 14 = txdxdt

dt2 t

2x2dt dxdt

2dt

=

t2x2dt

|v(t)|2 dt

=

t2x2dt

12

=

t2x2dt

12

2 |X(j)|

2d

No exact equivalent for DTFT/DFT but a similar e


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[Uncertainty Principle Proof Steps]


(1)Suppose v(t) = dxdt

. Then integrating the CTFT definition by parts w.r.t

X(j) = x(t)e

jt

dt=1j x(t)e

jt +

1j

dx(t)

dt ej

(2)Since ddt

12

x2

= xdxdt

, we can apply integration by parts to get

tx dx

dtdt=

t 1

2x2

t=

dtdt 1

2x2dt= 1

2

x2dt=

It follows that

tx dx

dtdt

2=

1

2

2= 1

4which we will use below.

(3)The Cauchy-Schwarz inequality is that in a complex inner product space|u v|2 (u u) (v v). For the inner-product space of real-valued squa

this becomes

u(t)v(t)dt2

u2(t)dt

v2(t)dt. We app

and v(t) = dx(t)dt

to get

14

=

tx dxdt

dt

2 t2x2dt

dxdt

2dt

=

t2x2dt

(4)From Parsevals theorem for the CTFT,

v2

(t)dt= 12

|V(j|

2

d. Fsubstitute V(j) =jX(j)to obtain

v2(t)dt= 1

2

2 |X(j|2 d

in (3) gives14

t2x2dt

v2(t)dt

=

t2x2dt

12

2 |X(j|2 d


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Order Estimation


nverse DTFT



Common Windows


Order EstimationExample Design

Frequency sampling

Summary

MATLAB routines


Several formulae estimate the required order of a fil

E.g. for lowpass filterEstimated order is

M 5.64.3 log

10()

21 820 log

10

2.2

Required Mincreases as either thetransition width, 2 1, or the gain

tolerances and get smaller.Only approximate.

Example:Transition band: f1 = 1.8 kHz, f2 = 2.0 kHz,

1 = 2f1fs

= 0.943, 2 = 2f2fs

= 1.0

Ripple: 20 log10(1 + ) = 0.1 dB, 20 log10 == 10

0.1

20 1 = 0.0116, = 1035

20 =

M 5.64.3 log

10(2104)1.0470.943 =

10.250.105 = 98 o


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Example Design


nverse DTFT



Common Windows


Order Estimation

Example DesignFrequency sampling

Summary

MATLAB routines


Specifications:Bandpass: 1 = 0.5, 2 = 1Transition bandwidth: = 0.1Ripple: == 0.02

20 log10 = 34 dB20 log10(1 + ) = 0.17 dB

Order:

M

5.64.3 log10()

21 = 92Ideal Impulse Response:

Difference of two lowpass filtersh[n] = sin2n

n sin1n

n

Kaiser Window: = 2.5

0

M=92

0

0

-

-

-


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Frequency sampling


nverse DTFT



Common Windows


Order Estimation

Example Design

Frequencysampling

Summary

MATLAB routines


Take M+ 1 uniform samples ofH(ej); take IDFT

Advantage:exact match at sample points

Disadvantage:poor intermediate approximation if spectrum is

Solutions:

(1) make the filter transitions smooth over (2) oversample and do least squares fit (cant u(3) use non-uniform points with more near tran

-2 0 20

0.5

1M+1=93

0 1

0

0.5

1


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Summary


nverse DTFT



Common Windows


Order Estimation

Example Design

Frequency sampling



Make an FIR filter by windowing the IDTFT of idIdeal lowpass has h[n] = sin0n

nAdd/subtract lowpass filters to make any response

Ideal response is with window DTFTRectangular window (Dirichlet kernel) has

and is always a bad idea

Hamming, Blackman-Harris are good; KaUncertainty principle: cannot be concentrated in

Frequency sampling: IDFT of uniform frequency

For further details see Mitra: 7, 10.


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MATLAB routines


nverse DTFT



Common Windows


Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


diric(x,n) Dirichlet kernel: sin0.5sin0.

hanninghammingkaiser

Window functions(Note periodic optio

kaiserord Estimate required filter orde


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7: Optimal FIR filt

7: Optimal FIRfilters

Optimal Filters

Alternation TheoremChebyshevPolynomials

Maximal ErrorLocationsRemez ExchangeAlgorithm

Determine Polynomial

Example Design

FIR Pros and Cons

Summary

MATLAB routines



101/211

Optimal Filters

7: Optimal FIR filters

Optimal FiltersAlternation Theorem

ChebyshevPolynomials



Example Design

FIR Pros and Cons

Summary

MATLAB routines


We restrict ourselves to zero-phase filters of odd lenaround h[0], i.e. h[n] =h[n].

H() =H(ej) =M

2

M2

h[n]ejn=h[0] + 2

H() is real but not necessarily positive (unlik

Weighted error: e() =s()

H()d()

whereChoose s() to control the error variation with

Example: lowpass filter

d() =

1 01

0 2

s() = 1 01

1 2

e() =1 when H() lies at the edge of the spe

Minimax criterion: h[n] = arg minh[n]max|e()|:


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Alternation Theorem


Optimal Filters

Alternation

TheoremChebyshevPolynomials



Example Design

FIR Pros and Cons

Summary

MATLAB routines


Want to find the best fit line: with the smallest max

Best fit line always attains themaximal error three times withalternate signs

2 42

4

6

8

Proof:Assume the first maximal deviation from the line is

There must be an equally large positive deviation; odownwards to reduce the maximal deviation.This must be followed by another maximal negativecan rotate the line and reduce the deviations.

Alternation Theorem:A polynomial fit of degree n to a bounded set of po

only if it attains its maximal error at n+ 2 points wThere may be additional maximal error points.Fitting to a continuous function is the same as to apoints.


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Chebyshev Polynomials


Optimal Filters

Alternation Theorem

ChebyshevPolynomialsMaximal ErrorLocationsRemez ExchangeAlgorithm


Example Design

FIR Pros and Cons

Summary

MATLAB routines


H() =H(ej) =h[0] + 2M2

1 h[n]cos n

But cos n=Tn(cos ): Chebyshev polynomialof

cos2= 2 cos2 1 = T2(cos )cos3= 4 cos3 3 cos = T3(cos )

Recurrence Relation:Tn+1(x) = 2xTn(x) Tn1(x)with T0(x) = 1

Proof: cos(n+) + cos (n) = 2 cos

So H() is an M2 order polynomial in cos : altern

Example: Low-pass filter of length 5 (M= 4)

0 0.5 1 1.5 2 2.5 3 3.5

0

0.2

0.4

0.6

0.8

1

1.2

1.4

|g|

M=4

0 0.5 1 1.5 2 2.5 3

-15

-10

-5

0

|g|(dB)

M=4

-1

0

0.5

1

g


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Maximal Error Locations


Optimal Filters

Alternation Theorem


Maximal ErrorLocations

Remez ExchangeAlgorithm


Example Design

FIR Pros and Cons

Summary

MATLAB routines


Maximal error locations occur either at bandedges or when dH

d

= 0

H() =h[0] + 2M

2

1 h[n]cos n=P(cos )

where P(x)is a polynomial of order M2 .

|H|

dH

d

=P(cos )sin

= 0 at = 0, and at most M2 1 zeros of p

With two bands, we have at most M2 + 3 maxim

We require M2 + 2 of alternating signs for the optim

Only three possibilitiesexist (try them all):

(a) = 0 + two band edges + M2 1 zeros ofP(b) = + two band edges + M2 1 zeros ofP

(c) ={0 and } + two band edges + M2 2 z


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Remez Exchange Algorithm


Optimal Filters

Alternation Theorem


Maximal ErrorLocations

Remez ExchangeAlgorithm


Example Design

FIR Pros and Cons

Summary

MATLAB routines


1. Guessthe positions of the M2 + 2 maximal error alternating signs to the errors (e.g. choose evenl

2. Determinethe error magnitude, , and the M2 + the polynomial that passes through the maximal

3. Find the local maximaof the error function by eve() =s()

H() d()

on a dense set of

4. Update the maximal error frequenciesto be an athe local maxima + band edges + {0 and/or }

If maximum error is > , go back to step 2. (t

5. Evaluate H() on M+ 1 evenly spaced and d

0 0.5 1 1.5 2 2.5 3

-0.5

0

0.5

1

1.5

|g|

M = 4Iteration 1

0 0.5 1 1.5 2 2.5 3

0

0.5

1

|g|

M = 4Iteration 2

0

0

0.5

1

|g|


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Optimal Filters

Alternation Theorem



DeterminePolynomial

Example Design

FIR Pros and Cons

Summary

MATLAB routines


For each extremal frequency, i for1i M2 + 2

d(i) =H(i) + (1)i

s(i) =h[0] + 2

M2

n=1h[n]co

Method 1: (Computation time M3)Solve M2 + 2 equations in

M2 + 2 unknowns for h

In step 3, evaluate H() =h[0] + 2M2

n=1h[n]co

Method 2: Dont calculate h[n] explicitly (ComputaMultiply equations by ci =

j=i

1cosicosj

andM2+2

i=1 ci

h[0] + 2

M2

n=1h[n]cos n+ (1)is(i)

=

All terms involving h[n]sum to zero leaving

M2 +2i=1

(1)ici

s(i) = M2 +2

i=1 c

id(

i)

Solve for thencalculate the H(i)then useLag

H() =P(cos ) =M

2+2

i=1 H(i)

j=icoscosi

M2 + 1

-polynomial going through all the H(


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Example Design


Optimal Filters

Alternation Theorem




Example DesignFIR Pros and Cons

Summary

MATLAB routines


Filter Specifications:Bandpass = [0.5, 1], Transition widths: Stopband Attenuation: 25 dB and 15 dBPassband Ripple: 0.3 dB

Determinegain tolerancesfor each band:25 dB= 0.056, 0.3 dB= 1 0.034, 15 d

Predicted order: M= 36M2 + 2 extremal frequencies are distributed bet

Filter meets specs ; clearer on a decibel scaleMost zeros are on the unit circle + threereciprocal

Reciprocal pairs give a linear phase shift

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

|H|

M=36

0 0.5 1 1.5 2 2.5 3

-30

-25

-20

-15

-10

-5

0

|H|(d

B)

M=36

-1

-0.5

0

0.5

1


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FIR Pros and Cons


Optimal Filters

Alternation Theorem




Example Design

FIR Pros and ConsSummary

MATLAB routines


Can havelinear phase no envelope distortion, all frequencies ha symmetric or antisymmetric: h[n] =h[ antisymmetric filters haveH(ej0) =H(e symmetry means you only need M2 + 1 m

to implement the filter.

Alwaysstable

Low coefficient sensitivity

Optimal design methodfast and robust

Normally needshigher orderthan an IIR filter Filter orderM dBatten3.5 where is th

Filtering complexityMfs

dBatten

3.5 f2s for a given specification in unscaled


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110/211

MATLAB routines


Optimal Filters

Alternation Theorem




Example Design

FIR Pros and Cons

Summary

MATLAB routines


firpm optimal FIR filter design

firpmord estimate require order for fircfirpm arbitrary-response filter desigremez [obsolete] optimal FIR filter de


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8: IIR Filter Transform

8: IIR FilterTransformations

Continuous TimeFilters

Bilinear Mapping

Continuous TimeFiltersMapping Poles andZerosSpectralTransformationsConstantinidesTransformations

mpulse Invariance

Summary

MATLAB routines



112/211

Continuous Time Filters

8: IIR FilterTransformations


Bilinear Mapping


mpulse Invariance

Summary

MATLAB routines


Classical continuous-time filters optimize tradeoff:passband ripple v stopband ripple v transition widthThere are explicit formulae for pole/zero positions.

Butterworth:G2() = H(j)2 = 11+2N Monotonic

G() = 1 12

2N + 384N +

Maximally flat: 2N 1 derivatives are zeroChebyshev:G2() = 1

1+2T2N()

where polynomial TN(cos x) = cos N x passband equiripple + very flat at

Inverse Chebyshev:G2() = 1

1+(2

T2N(

1

))1

stopband equiripple + very flat at0

Elliptic: [no nice formula] Very steep + equiripple in pass and stop band


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Bilinear Mapping

8: IIR FilterTransformationsContinuous TimeFilters

Bilinear MappingContinuous TimeFiltersMapping Poles andZerosSpectralTransformationsConstantinidesTransformations

mpulse Invariance

Summary

MATLAB routines


Change variable: z= +ss s=

z1z+1 : a one-to-o

axis (s) axis (z)

axis (s) Unit circle (z)

Proof: z=ejs= ej1ej+1 =

ej2ej

2

ej2+ej

2

=

Left half plane(s)inside of unit circle (z)

Proof: s=x +jy |z|2 = |(+x)+jy|2

|(x)jy|2

= 2+2x+x2+y

22x+x2+y

x


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Continuous Time Filters


Bilinear Mapping


Mapping Poles andZerosSpectralTransformationsConstantinidesTransformations

mpulse Invariance

Summary

MATLAB routines


Take

H(s) = 1

s2+0.2s+4 and choose = 1

Substitute: s= z1z+1 [extra zeros at z=

H(z) = 1( z1z+1 )

2+0.2 z1

z+1+4

= (z+1)2

(z1)2+0.2(z1)(z+1)+4(z+1)2

= z2+2z+15.2z2+6z+4.8 = 0.19

1+2z1+z2

1+1.15z1+0.92z2

Frequency response is identical(both magnitude anphase) but with a distorted frequency axis:

Frequency mapping: = 2 tan1

=

2 3 4 5

= 1.6 2.2 2.5 2.65 2.75 Choosing: Set = 0

tan 120

to map 0 0

Set = 2fs= 2T

to map low frequen


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Mapping Poles and Zeros


Bilinear Mapping


Mapping Poles andZeros

SpectralTransformationsConstantinidesTransformations

mpulse Invariance

Summary

MATLAB routines


Alternative method:

H(s) = 1

s2+0.2s+4

Find the poles and zeros: ps= 0.1 2jMap using z =1+s1s pz = 0.58 0.77j

After the transformation we will always end up withthesame number of poles as zeros:

Add extra poles or zeros at z = 1

H(z) =g (1+z1

)

2

(1+(0.580.77j)z1)(1+(0.58+0.77j)z1)

=g 1+2z1+z2

1+1.15z1+0.92z2

Choose overallscale factor, g,to give the same gainat any convenient pair of mapped frequencies:

At0 = 0 s0 = 0 H(s0)

= 0.25

0 = 2 tan1 0 = 0 z0 =ej0 = 1

|H(z0)| =g 43.08 = 0.25 g= 0.19

H(z) = 0.19 1+2z1+z2

1+1.15z1+0.92z2


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Spectral Transformations


Bilinear Mapping

Continuous TimeFiltersMapping Poles andZeros

SpectralTransformations

ConstantinidesTransformations

mpulse Invariance

Summary

MATLAB routines


We can transform the z-plane to change the cutofffrequency by substituting

z= z1z z= z+1+z

Frequency Mapping:

Ifz =ej, then z =z 1+z1

1+z has modulus 1since the numerator and denominator are

complex conjugates.

Hence theunit circle is preserved. ej = e

j+1+ej

Some algebra gives: tan 2 =1+1

tan 2

Equivalent to:z s= z1

z+1 s= 11+s z=

1+s1s

Lowpass Filter example:Inverse Chebyshev

0 = 2 = 1.57

=0.6 0 = 0.49


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Impulse Invariance


Bilinear Mapping


Impulse InvarianceSummary

MATLAB routines


Bilinear transform works well for a lowpass filter bucompression of the frequency distorts any other resp

Alternative method:H(s) L1 h(t) sample h[n] ZExpressH(s)as a sum of partial fractionsH(s) Impulse response is h(t) =u(t)

Ni=1 gie

pit

Digital filter H(z) =N

i=1gi

1epiTz1has identic

Poles ofH(z) arepi=epiT

(where T = 1fs is sa

Zeros do not map in a simple way

Properties: Impulse response correct. No distort Frequency response is aliased.

Example: Standard telephone filter - 300 to 3400 H

0 5 10 15 20 250

0.5

1

Frequency (krad/s)

|H|

Analog Filter

0 0.5 1 1.5 2 2.5 30

0.5

1

(rad/sample)

|H|

Bilinear (fs= 8 kHz)

Matched at 3.4 kHz

0 0.5 1 1.5

0.2

0.4

0.6

0.8

1

(rad/s

|H|

Impulse Invarian


119/211

Summary


Bilinear Mapping


mpulse Invariance



Classical filtershave optimal tradeoffs in cont Order transition widthpass ripple Monotonic passband and/or stopband

Bilinear mapping Exact preservation of frequency response non-linear frequency axis distortion can choose to map 0 0 for one s

Spectral transformations lowpasslowpass, highpass, bandpass o bandpass and bandstop double the filter

Impulse Invariance Aliassing distortion of frequency response preserves frequency axis and impulse resp

For furt


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MATLAB routines


Bilinear Mapping


mpulse Invariance

Summary

MATLAB routines


bilinear Bilinear mapping

impinvar Impulse invariancebutterbutterord

Analog or digitalButterworth filter

cheby1cheby1ord

Analog or digitalChebyshev filter

cheby2cheby2ord

Analog or digitalInverse Chebyshev filter

ellipellipord

Analog or digitalElliptic filter


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9: Optimal IIR Des

9: Optimal IIRDesign

Error choices

Linear Least SquaresFrequency Sampling

terative Solution

Newton-Raphson

Magnitude-onlySpecification

Hilbert RelationsMagnitude PhaseRelation

Summary

MATLAB routines



122/211

Error choices

9: Optimal IIRDesign

Error choicesLinear Least SquaresFrequency Sampling

terative Solution

Newton-Raphson

Magnitude-onlySpecification

Hilbert RelationsMagnitude PhaseRelation

Summary

MATLAB routines


We want to find a filter H(ej) = B(ej)

A(ej) that appr

responseD(). Assume A is order N and B is ordTwo possible error measures:

Solution Error: ES() =WS()B(ej)A(ej) D(

Equation Error: EE() =WE() B(ej) D

We may know D()completely or else only |D

We minimize

|E()|p

d

where p= 2 (least squares)

Weight functions W() are chos

handouts for dsp

Documents