fourier series (for periodic signals)
TRANSCRIPT
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EECE 301
Signals & Systems
Prof. Mark Fowler
Note Set #13
C-T Signals: Fourier Series (for Periodic Signals) Reading Assignment: Section 3.2 & 3.3 of Kamen and Heck
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Ch. 1 Intro
C-T Signal Model
Functions on Real Line
D-T Signal Model
Functions on Integers
System Properties
LTICausal
Etc
Ch. 2 Diff EqsC-T System Model
Differential Equations
D-T Signal ModelDifference Equations
Zero-State Response
Zero-Input Response
Characteristic Eq.
Ch. 2 Convolution
C-T System Model
Convolution Integral
D-T Signal Model
Convolution Sum
Ch. 3: CT FourierSignalModels
Fourier Series
Periodic Signals
Fourier Transform (CTFT)
Non-Periodic Signals
New System Model
New Signal
Models
Ch. 5: CT FourierSystem Models
Frequency Response
Based on Fourier Transform
New System Model
Ch. 4: DT Fourier
SignalModels
DTFT
(for Hand Analysis)DFT & FFT
(for Computer Analysis)
New Signal
Model
Powerful
Analysis Tool
Ch. 6 & 8: LaplaceModels for CT
Signals & Systems
Transfer Function
New System Model
Ch. 7: Z Trans.
Models for DT
Signals & Systems
Transfer Function
New System
Model
Ch. 5: DT Fourier
System Models
Freq. Response for DT
Based on DTFT
New System Model
Course Flow DiagramThe arrows here show conceptual flow between ideas. Note the parallel structure between
the pink blocks (C-T Freq. Analysis) and the blue blocks (D-T Freq. Analysis).
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3.2 & 3.3 Fourier Series
In the last set of notes we looked at building signals using:
=
=N
Nk
tjk
kectx0)(
We saw that these build periodic signals.
Q: Can we getany periodic signal this way?
A: No! There are some periodic signals that need an infinite number of terms:
=
++=N
k
kk tkAAtx1
00 )cos()(
=
=k
tjk
kectx0)(
Fourier Series
(Complex Exp. Form)
kare integers
=
++=1
00 )cos()( k kk tkAAtx
kare integers
Fourier Series
(Trig. Form)
These are two
different
forms of the
same tool!!
There is a 3rd
form that
well see later.
Sect. 3.3
There is a related
Form in Sect. 3.2
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Q: Does this now let us get any periodic signal?
A: No! Although Fourier thought so!
Dirichlet showed that there are some that cant bewritten in terms of a FS!
But those will never show up in practice!
See top of
p. 155
So what??!! Here is what!!
We can now break virtually any periodic signal into a sum of simple things
and we already understand how these simple things travel through an LTI system!
So, instead of:
)(th)(tx )()()( thtxty =
We breakx(t) into its FS components and find how each component goes
through. (See chapter 5)
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To do this kind of convolution-evading analysis we need to be able to solve the
following:
Given time-domainsignal modelx(t)
Find the FScoefficients {ck}
Time-domain model Frequency-domain model
Converting time-domain signal model into
a frequency-domain singal model
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Q: How do we find the (Exp. Form) Fourier Series Coefficients?
A: Use this formula (it can be proved but we wont do that!)
+ =
Tt
t
tjk
k dtetxT
c0
0
0)(1 Slightly different
than book
It uses t0 = 0.
Integrate over
any complete
period
where: T= fundamental period ofx(t) (in seconds)
0= fundamental frequency ofx(t) (in rad/second)
= 2/T
t0 = any time point (you pickt0 to ease calculations)
k
all integers
Comment: Note that for k= 0 this gives
+
=Tt
tdttx
Tc
0
0
)(10
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Summarizing rules for converting between the
Time-Domain Model & the Exponential Form FS Model
Analysis Synthesis
+ =
Tt
t
tjk
k etxT
c0
0
0)(1
=
=k
tjk
kectx0)(
Use signal to figure out
the FS Coefficients
Eat food and figure out recipe
Use FS Coefficients to Build
the Signal
Read recipe and cook food
Time-Domain Model: The Periodic Signal Itself
Frequency-Domain Model: The FS Coefficients
There are similar equations for finding the FS coefficients for
the other equivalent forms But we wont worry about them
because once you have the ckyou can get all the others easily
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K
,3,2,121
21
00
=
=
=
=
keAc
eAc
Ac
k
k
j
kk
j
kk
( )
( )
K,3,2,1
2
1
21
00
=
+=
=
=
k
jbac
jbac
ac
kkk
kkk
Exponential Form
=
=k
tjk
kectx0)(
Trig Form: Amplitude & Phase
=
++=1
00 )cos()(k
kk tkAAtx
Trig Form: Sine-Cosine
[ ]
=
++=1
000 )sin()cos()(k
kk tkbtkaatx { }
{ }K
K
,3,2,1,Im2
,3,2,1,Re2
00
==
==
=
kcb
kca
ca
kk
kk
K,3,2,1
2
00
=
=
=
=
k
c
cA
cA
kk
kk
=
+=
=
k
kk
kkk
a
b
baA
aA
1
22
00
tan)sin(
)cos(
00
kkk
kkk
Ab
Aa
ca
=
=
=
Three (Equivalent) Forms of FS and Their Relationships
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Example of Using FS Analysis
In electronics you have seen (or will see) how to use diodes and an RC filter circuit
to create a DC power supply:
60Hz Sine
wave with
around110V RMS
msT 67.1660
11 ==
x(t) R C
60 Hz
t
ms67.16
x(t)
T = T/2 = 8.33ms
t
1. This signal goes into the RC
filter what comes out?
(Assume zero ICs)2. How do we choose the desired
RCvalues?
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t )(th)(tx ?)(
x(t)=tyA
Later (Ch. 5) we will use it to analyze whaty(t) looks like
The equation for the FS coefficients is: =
Ttjk
k dtetxT
c0
0)(1
T
20 =
Choose t0 = 0 here to make things easier:
t
x(t)A
t0 t0+T
This kind of choice would
make things harder: t
x(t)
A
t0t0+T
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Now what is the equation forx(t) over t [0,T]?
TttT
Atx
= 0sin)(
So using this we get:
=
=
T tT
jk
Ttjk
k
dtetT
AT
dtetxT
c
0
2
0
sin1
)(1
0
T
20 =
Determined bylooking at the plot
t
x(t)A
t0 t0+T
So now we just have to evaluate this integral as a function ofk
jk2
1
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we do a Change of Variables. There are three steps:
1. Identify the new variable and sub it into the integrand
2. Determine its impact on the differential
3. Determine its impact on the limits of integration
=
T tT
jk
k dtetT
AT
c0
2
sin1
To evaluate the integral:
dtT
d
=
dT
dt=Step 2:
when t = 0
when t= T
Step 3: 00 ==T
== TT
tT
=Step 1: 2)sin( jke
( )
( )
=
=
=
0
2
0
2
0
2
sin
sin
1
sin1
deA
d
T
eAT
dtetTATc
jk
jk
T t
T
jk
k
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use your favorite Table of Integrals (a short one is available on the
course web site):
+
=22
)]cos()sin([
)sin( ba
bxbbxaedxbxe
axax
We get our case with: a = -j2k b = 1
So
0
2
2
41
)]cos()sin(2[
=
k
kjeAc
kj
k
So [ ]
0
2
2 )cos()41(
kj
k ek
A
c
=
Recall: sin(0) = sin() = 0 So the sin term above goes away
(Finesse the problem dont use brute force!)
( )=
0
2sin deA
cjk
kSo to evaluate the integral given by:
S
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So[ ])0cos()cos(
)41(
022
2
kjkj
k eek
Ac
=
=1 =1=-1 =1
=-2So
)41(
22
k
Ack
=
Notes: 1. This does not depend on T
2. ckis proportional toA
So: 1. If you change the input sine wave's
frequency the ckdoes not change
2. If you, say, doubleA youll double ck
Things you would
never know if youcant work
arbitrary cases
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Re
Im
ckk0
0
2
0
0
=
=
c
A
c
0
0)14(
22
=
=
kc
kk
A
c
k
k
Because c0 is real and > 0 Because ckis real and < 0
Now find the magnitude and phase of the FS coefficients:
)41(
22
k
A
ck =
Re
Im
c0
+
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So the two-sided spectrum after the rectifier:
A2
3
2A
0 02 03 04 05002030405
0 02 03 04 05002030405
kc
kc
The spectrum before the rectifier (input is a single sinusoid):
2
0
2
0
2
0
2
0
kc~
kc~
Note: Therectifier
creates
frequencies
because it
isnon-
linear
Now you can find the Trigonometric form of FS
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The rectified sine wave has Trig. Form FS:
=
+
+=1
02)cos(
)14(
42)(
k
tkk
AAtx
Ak kA0
Now you can find the Trigonometric form of FS
Once you have the ckfor the Exp. Form, Eulers formula gives the Trig Form as:
=
++=1
00 )cos(2)(k
kk ctkcctx
So the input to the RC circuit
consists of a superposition of
sinusoids and you know
from circuits class how to:1. Handle a superposition of
inputs to an RC circuit
2. Determine how a single
sinusoid of a given frequencygoes through an RC circuit
A2
3
4A
0 02 03 04 05
The one-sided spectrum is:
0 02 03 04 05
MagnitudeSpectrum
Phase
Spectrum
Ak
k
Preliminary to Parsevals Theorem (Not in book)
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Preliminary to Parseval s Theorem (Not in book)
Imagine that signalx(t) is a voltage.
Ifx(t) drops across resistanceR, the instantaneous power is R
tx
tp
)()(
2
=
Sometimes we dont know whatR is there so we normalize this by ignoring theR
value: )()( 2 txtpN =
Energy in one period ++ ==
0
0
0
0
)()( 2tTt
tT
tdttxtdE
Recall: power = energy per unit time
(1 W = 1 J/s)dt
tdEtp
)()( = dttxtdE )()( 2=
differential increment
of energy
Once we have a specificR we can always un-normalize viaR
tpN )(
(In Signals & Systems we will drop theNsubscript)
The Total Energy
= dttx )(2
= for a periodic signal
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+=
0
0 )(
1 2tT
tdttx
TP
Average power over one period =T
PeriodOneinEnergy
Recall: power = energy per unit time
For periodic signals we use the average power as measure of the sizeof a signal.
The Average Power of practical periodic signals is finite and non-zero.
(Recall that the total energy of a periodic signal is infinite.)
P l Th
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Parsevals Theorem
We just saw how to compute the average power of a periodic signal if we are
given its time-domain model:
+
=0
0
)(1 2tT
tdttx
TP
Q: Can we compute the average power from the frequency domain modelA: Parsevals Theorem says Yes!
{ } ,...2,1,0, =kck
=
=k
kcP2
Another way to view Parsevals theorem is this equality:
=
+=
k
k
Tt
tcdttx
T
220
0
)(1
Parsevals theorem gives this equation
as an alternate way to compute the average power of a periodic signal whose
complex exponential FS coefficients are given by ck
I t ti P l Th
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=
+
=k
k
Tt
t cdttxT
220
0 )(
1
sum of squares in
time-domain model
sum of squares in
freq.-domain model
=
)(2 tx = power at time t(includeseffects of all frequencies)
=2
kcpower at frequency k0
(includes effects of all times)
Interpreting Parsevals Theorem