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Signals and SystemsSpring 2008

Lecture #5(2/20/2008)

• Covers O & W pp. 177-201• Complex Exponentials as Eigenfunctions of LTI Systems• Fourier Series representation of CT periodic signals• How do we calculate the Fourier coefficients?• Convergence and Gibbs’ Phenomenon

“Figures and images used in these lecture notes by permission,copyright 1997 by Alan V. Oppenheim and Alan S. Willsky”

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The eigenfunctions φk(t) and their properties(Focus on CT systems now, but results apply to DT systems as well.)

System

!k(t) "

k{!k(t)

{

eigenvalue eigenfunction

Eigenfunction in → same function out with a “gain”

From the superposition property of LTI systems:

LTIx(t) = ak!k (t)k

" y(t ) = #kak!k (t)k

"

Now the task of finding response of LTI systems is to determine λk.The solution is simple, general, and insightful.

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Two Key Questions

1. What are the eigenfunctions of a general LTI system?

2. What kinds of signals can be expressed assuperpositions of these eigenfunctions?

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A specific LTI system can have more than one type of eigenfunction

Ex. #1: Identity system

x(t) !(t) x(t) "!(t) = x(t)

Any function is an eigenfunction for this LTI system.

x(t) !(t " T) x(t " T )

Any periodic function x(t) = x(t+T) is an eigenfunction.

Ex. #2: A delay

cos!t h(t) = h("t) y(t)

Ex. #3: h(t) even

y(t) = h(!) cos["(t #! )]d! =#$

$

% h(! )[cos"t & cos"! + sin"t & sin"!)]d!#$$

%

= cos"t h(! )cos"!d!#$

$

%H( j" )

1 2 4 4 4 3 4 4 4 — cos"t is an eigenfunction

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Complex Exponentials are the only Eigenfunctions of any LTI Systems

y(t) = h(! )"#

+#

$ es( t"! )

d!

= h(! )"#

+#

$ e"s!d![ ]est

= H (s){est

{

x(t) = est

h( t)

eigenvalue eigenfunction

x[n] = zn

h[n]

y[n]= h[m]zn!mm=!"

"

#

= h[m]z!m

m= !"

"

#[ ]zn

= H(z){ zn

{

eigenvalue eigenfunction

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System Functions H(s) or H(z) – Eigenvalues fornow, and more about them later

H (s) = h( t)e! stdt

!"

"

#x(t) = ake

sk t$ % & % y(t ) = akH(sk )esk t$

x(t) h(t) y(t)est H(s)est

H(z) = h[n]z! n

n=!"

"

#x[n] = akzk

n# $ % $ y[n] = akH (zk )zkn#

x[n] h[n] y[n]zn H(z)zn

CT:

DT:

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Question 2. What kinds of signals can werepresent as “sums” of complex exponentials?

s = j! - purely imaginary ,

i.e. signals of the form e j!t

z = e j! , i.e. signals of the form e j!n

"

For Now: Focus on restricted sets of complex exponentials

CT & DT Fourier Series and Transforms

CT:

DT:

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Joseph Fourier (1768-1830)

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Fourier Series Representation of CT Periodic Signals

x(t) = x(t + T ) for all t

ej!t

periodic with period T " ! = k!o

#

x(t) = akejk! 0t

k=$%

+%

& = akejk2't /Tk= $%+%

& = x(t + T )

-periodic with period T- {ak} are the Fourier (series) coefficients- k = 0 DC- k = ±1 first harmonic- k = ±2 second harmonic

- smallest such T is the fundamental period

- is the fundamental frequency!o=2"

T

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• For real periodic signals, there are two other commonly usedforms for CT Fourier series:

x(t) = ao+ !

kcosk"

ot + #

ksink"

ot[ ]

k =1

$

%

or

x(t) = ao+ &

kcos(k"

ot +'

k)[ ]

k =1

$

%

ejk! 0t , e

" jk! 0t

• Because of the eigenfunction property of ejωt, we will only usethe complex exponential form in 6.003.

–– A nontrivial consequence of this is that we need to includeterms for both positive and negative frequencies:

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Let’s first take a detour by studying a three-dimensional vector:

How do we find the coefficients Ax, Ay, and Az?

Easy,

Project the vector onto the x-, y-, and z-axis.

Why does it work this way?

Orthogonality:

Question: How do we find the Fourier coefficients ak?

!

r A = Ax ˆ x + Ay ˆ y + Az ˆ z ,

ˆ x , ˆ y , and ˆ z are unit vectors.

!

Ax =r A • ˆ x , Ay =

r A • ˆ y , and Az =

r A • ˆ z .

!

ˆ y • ˆ x = ˆ z • ˆ y = ˆ x • ˆ z " 0Ax

Ay

Az

!

r A

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Now let’s imagine that x(t) is a vector in a space of ∞-dimensions and is a unit vector in thisspace. (Such a space does exist, it is called Hilbert space, justnobody lives there.) Then,

Now all we need to do is to “project” x(t) onto the ,then we will obtain the coefficients ak. How to do that? Again,need orthogonality.

!

ejk"ot (k = 0, ±1, ± 2, ...)

!

ejk"ot # axis

Back to Fourier series

!

x(t) = akejk"ot

k = #$

+$

% , with the analog y

x(t)&r A

ejk"ot & ˆ x , ˆ y , ˆ z

ak & Ax,Ay,Aza0

a1

a2

!

x(t)

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Obtaining the Fourier series coefficientsOrthogonality in the Hilbert space:

!

1

Tejk"ot # e$ jn"ot

T% dt =

1

Tej(k$n )"ot

T% dt =

1, k = n

0, k & n

' ( )

= *[k $ n] .

(T% = integral over any interval of length T, and the operation of

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Te$ jn" ot

T% dt # is to take an "inner product" with e jn"ot )

!

Now if we "project" x(t) onto e jn" ot by taking the operation :

1

Tx(t) # e$ jn" ot

T% dt =

1

Tak

k = $&

+&

' e j(k$n )"otT% dt,

then only one term (an ) will be nonzero. That is :

1

Tx(t) # e$ jn" ot

T% dt = an .

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Finally

!

1

Tx(t) " e# jn$ot

T% dt = an .

&

CT Fourier Series Pair ($o = 2' /T)

x(t) = akejk$ot

k=#(

+(

) (Synthesis equation)

ak =1

Tx(t)e

# jk$otdt (Analysis equation)T%

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15!

ak =1

Tx(t)e

" jk#otdt"T / 2

T / 2

$ =1

Te" jk#otdt

"T1 / 2

T1 / 2

$

=1

" jk#oT[e

" jk#ot ]"T1

T1(#o =

2%

T)

=sin(k#oT1)

k%

Example: Periodic Square Wave

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Convergence of CT Fourier Series• How can the Fourier series (composed of continuous sine and

cos functions) for the square wave (with many discontinuities)possibly make sense?

• The key is: What do we mean by

• One useful notion for engineers: there is no energy in thedifference

(just need x(t) to have finite energy per period)

x(t) = akejk! 0t

k ="#

+#

$ ?

e(t) = x(t) ! akejk" 0t

k= !#

+#

$

| e(t) |2 dt = 0T%

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Under a different, but reasonable set of conditions(the Dirichlet conditions)

Condition 1. x(t) is absolutely integrable over one period, i. e.

| x(t) | dt < !T"

Condition 2. In a finite time interval,x(t) has a finite numberof maxima and minima.

Ex. An example that violatesCondition 2.

Condition 3. In a finite time interval, x(t) has only a finite number of discontinuities.

Ex. An example that violatesCondition 3.!

x(t) = sin(2" / t) 0 < t #1

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The Dirichlet Conditions (cont.)

• Still, convergence has some interesting characteristics:

– As N → ∞, xN(t) exhibits Gibbs’ phenomenon at pointsof discontinuity

xN (t) = akejk!0 t

k= "N

N

#

Dirichlet conditions are met for most of the signals we willencounter in the real world. Then

– The Fourier series = x(t) at points where x(t) is continuous

– The Fourier series = “midpoint” at points of discontinuity!

akejk"ot

k=#$

$

% =1

2[x(t + 0) + x(t # 0)]

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•Demo: Fourier Series for CT square wave (Gibbs phenomenon).

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Next lecture covers:O & W pp. 202-221