Signal and Systems Chapter 3: Fourier Series Representation...

Signal and SystemsChapter 3: Fourier Series Representation of Periodic Signals

• 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• CT Fourier series reprise, properties, and examples• DT Fourier series • DT Fourier series examples and differences with CTFS• Fourier Series and LTI Systems• Frequency Response and Filtering• Examples and Demos

Portrait of Jean Baptiste Joseph Fourier

Image removed due to copyright considerations.

Signals & Systems, 2nd ed. Upper Saddle River, N.J.: Prentice Hall, 1997, p. 179.

Book Chapter 3 : Section 1

Computer Engineering Department, Signal and Systems 2

Desirable Characteristics of a Set of “Basic” Signals

a. We can represent large and useful classes of signals using these building blocks

b. The response of LTI systems to these basic signals is particularly simple, useful, and insightful

Previous focus: Unit samples and impulses

Focus now: Eigenfunctions of all LTI systems



The eigenfunctions and their properties

Eigenfunction in →same function out with a “gain”


(Focus on CT systems now, but results apply to DT systems as well.)

( )k t

From the superposition property of LTI systems:

Now the task of finding response of LTI systems is to determine λk.


Complex Exponentials as the Eigenfunctions of any LTI Systems


stetx )(

dehty ts )()()(

sts edeh

)(

eigenvalue eigenfunction

)(sHste

eigenvalue eigenfunction

m

mnzmhny ][][

n

m

m zzmh

][

)(zHnz


DT:

dtethsH st)()(

k k

ts

kk

ts

kkk easHtyeatx )()()(

nznhzH ][)(

k

n

kkk

k

n

kk zazHnyzanx )(][][


What kinds of signals can we represent as “sums” of complex exponentials?

For Now: Focus on restricted sets of complex exponentials

CT:

DT:

s = jω – purely imaginaly,i.e., signals of the form ejωt

i.e., signals of the form ejωn

Magnitude 1

CT & DT Fourier Series and Transforms

Periodic Signals

jZ e


for all t


Fourier Series Representation of CT Periodic Signals

- smallest such T is the fundamental period- is the fundamental frequency

Periodic with period T

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

0 2 /( )jk t jk t T

k kx t a e a e

( ) ( )x t x t T



Question #1: How do we find the Fourier coefficients?First, for simple periodic signals consisting of a few sinusoidal terms

tttx 8sin24cos)(

][2

2][

2

1 8844 tjtjtjtj eej

ee

0 4 0

2 2 1

4 2T


For real periodic signals, there are two other commonly used forms for CT Fourier series:

Because of the eigenfunction property of e jωt , we will usually use the complex exponential form.



or

- A consequence of this is that we need to include terms for both positive and negative frequencies:

0 0 0

1

( ) [ cos sin ]k k

k

x t a k t k t

0 0

1

( ) [ cos( )]k k

k

x t a k t


Now, the complete answer to Question #1

denotes integral over any interval of lengthHere

Next, note that

Orthogonality

0( )jk t

k

k

x t a e

0 0 0( )jn t jk t jn t

k

kT T

x t e dt a e e dt

0( )j k n t

k

k T

a e dt

0( )j k n t

T

e dt


Given x(t), how to find 𝑎𝑘


CT Fourier Series Pair

(Synthesis equation)

(Analysis equation)

0 0( )( ) . [ ]

jn t j k n t

k k

T T

x t e dt a e dt a T k n

0( )jn t

n

T

x t e dt a T

0

2( )

T

0( )jk t

kx t a e

01

( )jk t

k

T

a x t e dtT



Ex: Periodic Square Wave

DC component is just the average

2

2

10

2)(

1T

TT

Tdttx

Ta

10 0

1

2

2

1 1( )

TT

jk t jk t

TkT

a x t e dt e dtT T

0 1

1

0 1

0

sin1|

jk t T

T

k Te

jk T k

0

2( )

T


How can the Fourier series for the square wave possibly make sense?

The key is: What do we mean by

One useful notion for engineers: there is no energy in the difference



Convergence of CT Fourier Series

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

0( )jk t

kx t a e

0( ) ( )jk t

ke t x t a e

T

dtte 0|)(| 2

?

T

dttx 2|)(|


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

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

AndCondition 2. In a finite time interval,

x(t) has a finite number of maxima and minima.

Ex. An example that violates Condition 2.

AndCondition 3. In a finite time interval, x(t) has only a

finite number of discontinuities.Ex. An example that violates

Condition 3.

T

dttx |)(|

10)2

sin()( tt

tx


Dirichlet conditions are met for the signals we will encounter in the real world. Then

Still, convergence has some interesting characteristics:



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

- The Fourier series = “midpoint” at points of discontinuity

- As N→ ∞, xN(t) exhibits Gibbs’ phenomenon at points of discontinuity

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

0( )N

jk t

N k

k N

x t a e

CT Fourier Series Pairs

Book Chapter3: Section2

Computer Engineering Department, Signals and Systems 17

𝜔0 =2𝜋

𝑇

Review:

𝑥(𝑡) =

𝑘=−∞

∞

𝑎𝑘𝑒𝑗𝑘𝜔0𝑡 =

𝑘=−∞

∞

𝑎𝑘𝑒𝑗2𝜋𝑘 Τ𝑡 𝑇

𝑎𝑘 =1

𝑇න

𝑇

𝑥(𝑡)𝑒−𝑗𝑘𝜔0𝑡𝑑𝑡

Skip it in future

for shorthand

𝑥(𝑡)𝐹𝑆𝑎𝑘

Another (important!) example: Periodic Impulse Train

𝑥 𝑡 = 𝑛=−∞

∞)𝛿(𝑡 − 𝑛𝑇 sampling function

important for sampling

𝑎𝑘 =1

𝑇න−𝑇2

𝑇2𝑥(𝑡)𝑒−𝑗𝑘𝜔0𝑡 𝑑𝑡 =

1

𝑇න−𝑇2

𝑇2𝛿(𝑡)𝑒−𝑗𝑘𝜔0𝑡𝑑𝑡 =

1

𝑇𝑓𝑜𝑟 𝑎𝑙𝑙 𝑘!

𝑥(𝑡) = 𝑘=−∞

∞𝑎𝑘𝑒

𝑗𝑘𝜔0𝑡



─ All components have:(1) the same amplitude,

&(2) the same phase.

(A few of the) Properties of CT Fourier Series

Linearity

Conjugate Symmetry

𝑥(𝑡)𝑖𝑠 𝑟𝑒𝑎𝑙 ⇒ 𝑎−𝑘 = 𝑎𝑘∗

⇓𝑎𝑘 = Re{𝑎𝑘} + 𝑗Im{𝑎𝑘} = |𝑎𝑘|𝑒

𝑗∠𝑎𝑘

Re{𝑎𝑘} 𝑖𝑠 𝑒𝑣𝑒𝑛, Im{𝑎𝑘} 𝑖𝑠 𝑜𝑑𝑑𝑜𝑟

|𝑎𝑘| 𝑖𝑠 𝑒𝑣𝑒𝑛, ∠𝑎𝑘 𝑖𝑠 𝑜𝑑𝑑

Time shift 𝑥 𝑡 → 𝑎𝑘𝑥(𝑡 − 𝑡0) → 𝑎𝑘𝑒

−𝑗𝑘𝜔0𝑡0 = 𝑎𝑘𝑒−𝑗𝑘2𝜋 Τ𝑡0 𝑇



𝑥(𝑡) ↔ 𝑎𝑘 , 𝑦(𝑡) ↔ 𝑏𝑘 ⇒ 𝛼𝑥(𝑡) + 𝛽𝑦(𝑡) ↔ 𝛼𝑎𝑘 + 𝛽𝑏𝑘

Introduces a linear phase shift of to

Example: Shift by half period



using

𝑦(𝑡) = 𝑥(𝑡 − Τ𝑇 2) ↔ 𝑎𝑘𝑒−𝑗𝑘𝜋 = −1 𝑘𝑎𝑘

𝑒−𝑗𝑘𝜔0 Τ𝑇 2 = 𝑒−𝑗𝑘𝜋

𝑦(𝑡) ↔ −1 𝑘𝑎𝑘 𝑎𝑘 =1

𝑇= 𝐹. 𝐶. 𝑜𝑓

−∞

∞

)𝛿(𝑡 − 𝑛𝑇

||

−1 𝑘

𝑇

Parseval’s Relation

Multiplication Property



1

𝑇

𝑇

|𝑥(𝑡)|2𝑑𝑡 =

−∞

∞

|𝑎𝑘|2

Power in the kth harmonic Average signal power

kk btyatx )(,)( (Both x(t) and y(t) are periodic with the same period T)

l

kklklk babactytx *)().(

Energy is the same whether measured in the time-domain or the frequency-domain

Proof:

0 0 0 0( )

,

.jl t jm t j l m t jk tl m k

l m l m l k l

l m l m k l

a e b e a b e a b e

X(t) y(t) ck

Periodic Convolutionx(t), y(t) periodic with period T



dtyxtytx )()()(*)( - Not very meaningful

E.g. If both x(t) and y(t) are positive, then

)(*)( tytx

Periodic Convolution (continued)Periodic convolution : Integrate over any one period (e.g. -T/2 to T/2)

z(t) is periodic with period T

𝑥𝑇 𝑡 = ቐ𝑥 𝑡 −𝑇

2< 𝑡 <

𝑇

20 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒



/ 2

/ 2

( ) ( ) ( ) ( ) ( )

T

T

T

z t x y t d x y t d

where

Periodic Convolution (continued) Facts



1) z(t) is periodic with period T (why?)From Lecture #2, x(t) = x(t + T) → y(t) = y(t + T) for LTI systems.In the convolution, treat y(t) as the input and xT(t) as h(t)

2) Doesn’t matter which period we choose to integrate over:

Periodic3) Convolution

in time

)()()()()( tytxdtyxtzT

kkk ctzbtyatx )(,)(,)(

0 01 1

( ) ( ) ( )jk t jk t

k

T T T

c z t e dt x y t d e dtT T

0 0( )1( ) ( )

jk t jk

T T

y t e dt x e dT

0( )jk

k k k

T

b x e d Ta b

MultiplicationIn frequency!

Fourier Series Representation of DT Periodic Signals

x[n] -periodic with fundamental period N, fundamental frequency

Only e jωn which are periodic with period N will appear in the FS

There are only N distinct values of signals of this form

So we could just use

However, it is often useful to allow the choice of N

consecutive values of k to be arbitrary.



][][ nxNnx and0

2

N

02 , 0, 1, 2,...N k k k

0 0 0 0( )j k N n jk n jN n jk ne e e e

2πn

DT Fourier Series Representation



Nk

nNjk

keanx )/2(][

Sum over any N consecutive values of k

— This is a finite series

𝑘= 𝑁

=

𝑎𝑘 - Fourier (series) coefficients

Questions:

1) What DT periodic signals have such a representation?

2) How do we find ak?

Answer to Question #1:Any DT periodic signal has a Fourier series representation



0

0

0

0

2

( 1)

[ ]

[0]

[1]

[2]

[ 1]

jk n

k

k N

k

k N

jk

k

k N

j k

k

k N

j N k

k

k N

x n a e

x a

x a e

x a e

x N a e

N equations for N unknowns, a0, a1, …, a N-1

A More Direct Way to Solve for akFinite geometric series



1,

1,1

1

1

0

NN

n

nN

0jkwe

0 0

1 12 /

0 0

( ) ( )N N

jk n jk n jk N n

n N n n

e e e

( 2 / )

0

, 0, , 2 ,...

10 ,

1

jk N N

jk

N k N N

eotherwise

e



So, from0[ ]

jk n

k

k N

x n a e

multiply both sides by

and then

0jm ne

Nn

0 0 0[ ]jm n jk n jm n

k

n N n N k N

x n e a e e

0( )

[ ]

j k m n

k

k N n N

N k m

a e

orthogonality

mNa

DT Fourier Series Pair



𝜔0 =2𝜋

𝑁

0

0

[ ]

1[ ]

jk n

k

k N

jk n

k

n N

x n a e

a x n eN

(Synthesis equation)

(Analysis equation)

Note : It is convenient to think of ak as being defined for all integers k. So:

1) ak+N = ak—Special property of DT Fourier Coefficients.

2) We only use N consecutive values of ak in the synthesis equation. (Since x[n] is periodic, it is specified by N numbers, either in the time or frequency domain)

Example #1: Sum of a pair of sinusoids



)4/4/cos()8/cos(][ nnnx

- periodic with period N = 16 → ω0 = π/8

0 0 0 02 2/ 4 / 41 1[ ] [ ] [ ]

2 2

j n j n j n j nj jx n e e e e e e

0

0

2/

2/

2/1

2/1

0

3

3

4/

2

4/

2

1

1

0

a

a

ea

ea

a

a

a

j

j

2/

2

1

4/

2164266

116115

jeaaa

aaa

Example #2: DT Square Wave

using n = m - 𝑁1



1

1

10 6

2 11[ ]

N

N N N

n N

Na x n a a a

N N

𝐹𝑜𝑟 𝑘 ≠ 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓𝑁:

1 1

0 0 1

1

2( )

0

1 1N N

jk n jk m N

k

n N m

a e eN N

0 11

0 1 0 0 1

0

(2 1)2

0

1 1 1( )

1

jk NNjk N jk jk Nm

jkm

ee e e

N N e

1 0 1

0

sin[ ( 1/ 2) sin[2 ( 1/ 2) / ]1 1

sin( / 2) sin( / )

k N k N N

N k N k N

Example #2: DT Square Wave (continued)



)/sin(

]/)2/1(2sin[1 1

Nk

NNk

Nak



Convergence Issues for DT Fourier Series:Not an issue, since all series are finite sums.

Properties of DT Fourier Series: Lots, just as with CT Fourier Series

Example: [ ] kx n a

0 [ ] ?jM n

ke x n b

0 0 0

0

[ ]jM n jr n jM n

r

r N

jk n

k M

k N

x n e a e e

k r m a e

Frequency shift

Mkk ab 0 0( )jk j k M

The Eigenfunction Property of Complex Exponentials

CT

CT System Function:

DT

DT System function:

Book Chapter#: Section#


dtethsH st)()(

n

nznhzH ][)(

Fourier Series: Periodic Signals and LTI Systems

,

So , or powers of signals get modified through filter/ system.

,



0( )jk t

k

k

x t a e

0

0( ) ( )jk t

k

k

y t H jk a e

0

" "

( )k k

gain

a H jk a 0( )

0 0( ) | ( ) |j H jk

H jk H jk e

Includes both amplitude & phase

0[ ]jk n

k

k N

x n a e

0 0[ ] ( )jk jk n

k

k N

y n H e a e

0

" "

( )jk

k k

gain

a H e a

00 0 ( )( ) | ( ) |

jkjk jk j H eH e H e e

Includes both amplitude & phase

0| | | ( ) || |k ka H jk a

The Frequency Response of an LTI System

CT Frequency Response:

DT Frequency Response:



( ) ( ) j tH j h t e dt

( ) [ ]j j n

n

H e h n e

Frequency Shaping and Filtering

By choice of (or ) as a function of , we can shape the frequency composition of the output

Preferential amplification

Selective filtering of some frequencies

Example #1:

For audio signals, the amplitude is much more important than the phase



( )H j ( )jH e

Frequency Shaping and Filtering



Example #2: Frequency Selective Filters

Filter out signals outside of the frequency range of interest

( 2 )( ) ( )j jH e H e

Lowpass Filters:

Only show

amplitude here.

Note for DT:

Highpass Filters



Remember:

njn e)1(

Bandpass Filters

Demo: Filtering effects on audio signals



Idealized Filters

CT

DT

Note: |H| = 1 and ∠H = 0 for the ideal filters in the passbands, no need for the phase plot.



ωc : cutoff frequency

Highpass

CT

DT



Bandpass

CT

DT



lower cut-off upper cut-off



]}1[][]1[{3

1][ nxnxnxny

]}1[][]1[{3

1][ nnnnh

Example #3: DT Averager/Smoother

FIR (Finite Impulse

Response) filters

1 1 2( ) [ ] [ 1 ] cos

3 3 3

j j n j j

n

H e h n e e e

Example #4: Nonrecursive DT (FIR) filters



[( ) / 2]

1 1[ ] [ ] [ ] [ ]

1 1

1 1 sin[ ( 1) / 2]( )

1 1 sin( / 2)

M M

k N k N

Mj jk j N M

k N

y n x n k h n n kN M N M

M NH e e e

N M N M

Rolls off at lower

ω as M+N+1

increases

Example #5: Simple DT “Edge” Detector

DT 2-point “differentiator”



]}1[][{2

1][

]}1[][{2

1][

nnnh

nxnxny

/ 2 / 2 / 2 / 2[ ( ) sin( / 2)2

j j j jje e e je

j

/ 21( ) (1 ) sin( / 2)

2

j j jH e e je

| ( ) | | sin( / 2) |jH e

Demo: DT filters, LP, HP, and BP applied to DJ

Industrial average



Example #6: Edge enhancement using DT

differentiator



Courtesy of Jason Oppenheim.

Used with permission.

Courtesy of Jason Oppenheim.

Used with permission.

Example #7: A Filter Bank



Demo: Apply different filters to two-dimensional image signals.

Note: To really understand these examples, we need to understand frequency contents of aperiodic signals ⇒ the Fourier Transform



Face of a monkey.

Image removed do to copyright

considerations

Signal and Systems Chapter 3: Fourier Series Representation...

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