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DIGITAL SIGNAL PROCESSING

Unit 1

UNIT I OVERVIEW OF SIGNALS AND SYSTEM S 9

--Basic elements of digital signal Processing

Concept of frequency in continuous time and discrete time signals

Discrete time signals. Discrete time systems

Analysis of Linear t ime invariant systems

Z transform

Convolution and correlation.

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Introduction

W hat is signal processing ?

Signal processing is the action of changing one or more features (parameters)

of a signal according to a predetermined requirement. The parameters that is to

be changed may be , amplitude, frequency, phase etc... of the signal . The signal

which undergoes such a process is know n as input signal. And the entity which

performs this processing is known as "Signal processing system" or simply

system. ie; the input signal is fed to the system and the appropriately processed

signal is coming out of the system. This signal is known as the "Output signal".

This process is illustrated below .

Some of the signal processing tasks are Amplification, Attenuation, Filtering,

Modulation, etc... . To accomplish the required task, the "system" is designed with

appropriate behavior and features. ie; for an amplifier, the "system" is designed

such that any input signal going to the "system" will be amplified. In actual circuit,

such a behavior can be realized using transistors , FET or Op-Amps. This sort of

signal processing is know n as analog signal processing. Well ! , then what is Digital

Signal Processing ....?

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Analog signal processing Vs Digital Signal Processing

In analog processing system the information is represented

as an analog quantity. ie; something that varies continuously with time. This

"something" can be voltage , current etc.. . for example, the signal that comes out

of a microphone is analog in nature. The varying voltage from the microphone is

processed using an analog signal processing system known as "amplifier" , and

reproduced through a speaker.

Where in Digital Signal Processing system , the information is represented in

digital format. The signals that are to be processed is converted into numerical

form before any processing. This conversion is known as sampling. The

information contained in an analog signal is f irst converted to digital samples which

are equally spaced in time. The figure below shows an analog signal and its

sampled version.

Each of these samples are converted in to a numerical value and stored in the

computer' s memory. Processing is then done on these samples. The sampling in a

Digital Signal Processing system is governed by a theorem know n as sampling

theorem.

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Classification of Signals: Continuous-Time verses Discrete-Time Signals

Continuous time or analog signals are signals that are defined for every value of a

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1

DIGITAL SIGNAL PROCESSING

Lecture 1 - Chapter 1

Classification of Signals: Continuous-Time verses Discrete-Time Signals

Continuous time or analog signals are signals that are defined for every value of a < t N

Infinite duration

2. Right-Left sided

( ) 10 Nnnx = left-sided

Some Elementary Discrete-Time Signals

- unit sample sequence ( )

==

00

01

n

nn

- unit step sequence ( )

+

=

+

=

zzz

z

zz

z

nun

Zn

From Table on page 174

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18

( )

+

=

21

11

25.05.01

5.01

zz

z

dz

dzzX

( )4321

543

0625.025.075.01

0625.05.025.0

++

+=

zzzz

zzzzX ; |z| > 0.5

Now, lets use MATLAB to see if weve computed correctly.

b = [0, 0, 0, 0.25, - 0.5, 0.0625];

a = [1, -1, 0.75, -0.25, 0.1625];

n = 0: 20 % checking the fist 21 samples ofx(n)

delta = [n =0]; % creating (n)

x = filter (b, a, delta),

plot (n, x), hold

x = [zeros (1, 2) n.* (0.5.n) * cos(pi * n/3)]; % creating the original signal

n1 = 0:22;

plot (n1, x, r)

RationalZ-Transforms

( )

=

=

=N

k

k

k

M

k

k

k

za

zb

zX

0

0 ifao and bo0, then we can rewrite it as: ( )( )

( )kN

k

k

M

kNM

pz

zzz

a

bzX

=

=

=+

1

1

0

0

It has Mfinite zeros atz1, z2, , zM andNfinite poles atp1, p2, ., pN as well asN Mzeros or

M Npoles at origin and a possible zero/pole at . Depending on the location of the poles, the

signal has different behaviors. Read Section 3.3.2.

The System Function of a LTI System

Y(z) = H(z) X(z) H(z) is called the system-function. A system in general can be presented

by a difference equation:

( ) ( ) ( )

( ) ( ) ( )zXzbzYzazY

knxbknyany

kN

kk

kN

kk

M

kk

N

kk

=

=

==

+=

+=

01

01

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19

( )( )( )

=

=

=

=

=+

==N

k

k

k

M

k

k

k

N

k

k

k

M

ok

k

k

za

zb

za

zb

zX

zYzH

0

0

1

1

, where a0 = 1

Special Cases:

Ifak= 0 for1 < k < N, then ( ) kMM

kkM

kM

kk zb

zzbzH

=

= ==

00

1, which is an all-zeros system.

The system has Mtrivial poles at the origin. Such a system has a finite duration impulse response

and therefore is called FIR system.

On the other hand, ifbk= 0 for Mk1 then ( ) 1,1

0

0

0

1

0 ==+

=

=

=

a

za

zb

za

bzH

N

K

kN

k

N

N

k

k

k

.

This system is an all-pole system (has N trivial zeros at origin) and therefore, has an infinite

duration impulse response and thus is called IIR system. A pole-zero system is still IIR because

of the poles.

The Inverse of Z Transform

( ) ( ) kk

zkxzX +

==

By multiplying both sides of the above formula byzn-1

and integrating both sides over a closed

contour within ROC ofX(z), which encloses the origin, we have:

( ) ( ) dzznxdzzzX kn

c k

n

c

+

=

=11

Since the series converges on this contour, we can interchange and . Then

( )43421

==

=

=

kn knj

c

kn

kc

n dzzkxdzzzX

0 2

11)(

Cauchy Integral Theorem

( ) ( ) dzzzxj

nx n

c

1

2

1 =

One of Caushy Theorems states ( )

=

= 1

1

2 n

n

j

odzzz

n

c

o

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20

Let zo = o. Then, f(z) = zn. Ifn is positive the antiderivitive

1

1

+

+

n

znis analytic every where and

therefore, its contour integral is zero. But only forf(z) = z-1

it doesnt have an antiderivitive even

in a punctured plane. For 2n , it is analytic in a punctured plane with origin deleted.

Remember that iff is analytic in a simply connected domain, D, and is any loop (close

contour) in D, then 0)( =

dzzf because in a simply connected domain any loop can be shrunk

to a point. Therefore, the integral of a continuous function over a shrinking loop converges to

zero.

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21

Lecture 5

Calculating the Inverse Z-Transform

( ) ( ) dzzzXj2

1nx 1n=

Three methods to calculate it:

1) Direct Method by contour integration

2) Expansion into a series of termsz/z-1

3) Partial Fraction expansion and look-up table.

Cauchy-Residue Theorem

Let f(z) be a function of the complex variable zand Cbe a closed path in the z-plane. If the

derivative ( )zfdzd

exists on and inside the contour Cand iff(z) has no poles at z = z0, then

( )( )

( )

= else

Cinsideiszifzfdz

zz

zf

j c

00

0 02

1

More generally:

( )( )

( )( )

=

=

else

Cinsidezifdz

zfd

kdzzz

zf

jzzk

k

ck

01

1

0

0

0!1

1

2

1

RHS of the above equation is called residue of the pole zo.

Now suppose the function can be written as( )( )zgzf

wheref(z) has no poles inside Candg(z) is a

polynomial with simple rootsz1, z2, , zninside C. Then,

( ) ( ) ( )( )i

n

i

i

i

in

ic

n

i i

i

c

zAdzzz

zA

jdz

zz

zA

jdz

zg

zf

j ===

=

=

=

111 2

1

2

1

)(2

1

,

where ( ) ( )

( )

( )zgzf

zzzA ii = .

( ) ( ) ( ) ( )izz

nN

i

i

n zzXzzdzzzxj

nx =

=

==1

1

1

2

1

= sum at residue ofx(z)zn-1

at z = zi and N=

number of poles.

Example: Problem 3.56 (C)

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22

( )a

zaz

azzX

1

1>

=

( ) dzzaz

az

ajdzz

az

az

jnx n

c

n

c

11

1

1

2

1

12

1

=

=

arz

1>= This is the contour C.

Three cases:

1) For 1n then ( ) ( ) 11

= nzaz

azf and the only pole inside c is

a

1. Therefore, for 1n

( ) ( )

11

1

1

1

11

1111

+

=

=

=

=

nn

n

az

n

aa

aa

aazaz

anx

2) For ( )( )

dzazz

az

jnxon

==

12

1

then, ( ) ( )aza

zf

=1

and two poles (0 anda

1) inside C. Therefore,

( )( ) ( )

az

az

az

az

anx

az

oz

a

1111

1

=

+

=

==

3) Forn < 0, ( ) ( ) dzzazaz

ajnx

n

c

1

1

1

1

2

1

=

Therefore, it has a pole at zero with order of (n1) and a pole at 1/a. Since

ROC ( )nxa

1z > is right-sided and therefore, it is enough to find where it reaches the zero

on the left side.

( ) ( ) 011

1

1

2

11 122 =

+

=

= == ozaz az

az

dz

d

z

az

adz

azz

az

ajx

With the same method, we can prove thatx(-2)=x(-3)==0 .

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23

The Inverse z-Transform by Power Series Expansion

The method is to use long division. How to divide.

Example:

( ) ( )( ) 232311

211

12

2

2111+=+==

zz

z

zzzzzX

It has two poles: z = 1 andz = 2. Therefore, if the signal is casual, then ROC: |z| > 2 and if the

signal is non-casual, ROC: |z| < 1.

Case 1:

ROC: |z| > 2 signal is casual and therefore ( ) ( ) non

znxzX

=

= has terms with negative power

forz. Therefore, we divide as:

...31715731

3032304515

14150

14217

670

693

230

231

1

231

4321

54

543

43

432

32

321

21

2121

+++++

++

+

+

+

+

++

zzz

zzzzz

zz

zzz

zz

zzz

zz

zzzz

x(n) = {1, 3, 7, 15, 31,}

Case 2: ROC: |z| < 1 signal is non-casual ( ) ( ) nznxzX

=0

has terms with positive

powers ofz. Therefore, we divide in a way to getzwith positive powers.

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24

...8

7

4

3

2

1

4

3

4

70

4

3

4

9

2

3

2

1

2

30

2

1

2

31

1

132

432

32

32

2

21

12

+++

+

+

+

+

+

zzz

zz

zzz

zz

zz

zz

( )

= 0,0,2

1,

4

3,

8

7...,nx

Question: How would you use this method for a case like ROC: 1 < |z| < 2?

The Inverse z-Transform by Partial- Fraction Expansion

( )( )( )zDzN

zX = The goal is to write it in the form

( )( ) ( ) ( ) ( )113

3

1

2

2

1

1

1

1...

111 ++

+

+

=

zp

A

zp

A

zp

A

zp

AzX

N

N

Let ( )N

N

MM

zazaza

zbzbbzX

+++++++=...1

....2

2

1

1

110 . If it is an improper rational function (M > N) then the

first step is to write it as:

( ) ( )( )( )zDzN

zczcczX NMNM11

10 ..... ++++=

, where now the degree ofN1(z) is less thanN.

The inverse z-transformation ofc0 + c1z-1

+ . is very straight-forward. Next step is to write

( )( )zDzN1 or in general a proper function

( )( )zDzN

, where M < N, in terms of positive powers ofz,

factorize denominator and then write( )

z

zXin terms of partial fractions. In general, lets assume

anX(z) hasNsimple poles and L multiple poles at k = j. Then:

( )( ) ( ) ( )N

N

j

j

j

j

j

j

P

A

p

A

p

A

p

A

pz

A

pz

A

z

zX

+

++

+

++

+

=

1...

1....

11....

2

21

2

2

1

1

l

l

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25

Then( ) ( )

kpzk

kz

zXpzA =

= k = 1, 2, ., n excluding k = j and

( ) ( )

=

z

zXpz

zd

dA

k

i

k

k

jk l

l

, k = 1, 2, ..,l

Then

( ) ( )( ) ( )

=

k

N

k

k

n

k

k pzROCnup

pzROCnup

zpZ

:1

:

1

11

1

Example

( )( )( )

2

115.05.11

1 2

21

=+

=

zz

z

zzzX

( )

( )( )2

112

11

21

+

=

=

z

A

z

A

zz

z

z

zX

( )( )

2

21

11 11 === =z

z

zXzA

( ) ( ) 12

12

1

21

212 =

==

=zz

zXzA

( )11

211

1

1

2

21

1

1

2

=

=

zzzzzzX

Now depending on ROC, we get differentx(n).

1) ROC: |z| > 1 casual

( ) ( ) ( ) ( ) ( ) ( ) ( )nunununx nnn 5.022

112 ==

2) ROC: |z| < non-casual and left-sided

( ) ( ) ( ) ( )

( ) ( )15.0215.0112

+=

+=

nu

nununx

n

nn

3) 12

1

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26

( ) ( ) ( ) ( ) ( )nununx nn 5.0112 =

One-Sided Z Transform

One-sided Z Transform is defined as ( ) ( ) ( ) ( ) nnon

znunxznxzX +

=

+

== . It does not contain

information aboutx(n) forn < 0. It is unique only for the causal signals that are zero forn < 0. It

is useful to solve difference equations of the systems that are not relaxed initially, but the input is

not necessarily zero before applying to the system.

Properties

If ( ) ( )zXnx Z ++

then ( ) ( ) ( ) okznxzXzknxk

n

nkZ >

+

=

++ ,1

Proof: ( ){ } ( ) non

zknxknxZ

=

+ = . Let n k = lthen n = 0 l = -kand also -n = -l k.

( )

{( )

( )

( ) ( )

+=

+

=

=

=

+

=

=

=

+

K

1n

nK1

K

zx

on

K

K

K

zXznxzzxzxz

zzx

l l

ll

l

l

43421

ll

l

Time Advance: ( ) ( ) ( ) 0,1

>

+

=

++ kznxzXzknxk

on

nkZ

Final Value Theorem: ( ) ( )zXzimnximzn

+

=1

)1(ll

This limit exists if the ROC of(z-1)X+(z) includes the unit circle.

This can be proved by the following analogy. If the limit )(lim nxn exists, then the function

x(n) can be written as 0)(limand)(limwhere),()( ==+=

nfcnxnfcnxnn

. Thentake the one

sided Z-Transform from both sides and prove the theorem.

Analysis of the Systems in Z-Domain

Without too much restriction, lets assume

( )( )( )zQzN

zx = and ( )( )( )zAzB

zH =

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27

( )( ) ( )( ) ( )zAzQ

zNzBzY

= . If the system is relaxed, the initial condition = o

If the system is not relaxed, then

( ) ( ) ( )( )

( )( )( )

321

43421

zY

o

zY

zi

zs

zA

zN

zxzHzY

+

+= where ( ) ( )n

k

n

kN

kk znyzazN

=

= = 110

H(z) hasp1, ., pNpoles andX(z) has q1,., qL poles.

First lets assume that the poles are distinct and not common. Then

( )

( ) ( ) ( ) ( ) ( )44 344 2144 344 21

responseforce

L

k

n

kk

responsenatural

N

k

n

kk

L

kk

k

kN

kk

k

k

nuqQnuPAny

zq

Q

zP

AzY

==

=

=

+=

+

=

11

11 11

Note that natural response zero-input response. It is in fact the no-input response.

Example

( ) ( ) ( )nxnyny += 12

1Find the output when ( ) ( )nu

nnx

4cos10

=

Solution

( ) ( ) ( )zXzYzzY += 12

1

( )2

1

211

111=

=

P

zzH

However, ( )21

1

2

21

1110

+

=zz

z

zX 4/*124/

1

jj eqqeq ===

( ) ( ) ( )44444 344444 2143421

responseforce

14j

728j

14j

728j

responsenatrual

1 ze1

e786

ze1

e786

z211

36zxzHzY

oo

+

+

==

/

.

/

. ..

/

.

( ) ( )nunyn

nr

=

2

13.6

( ) ( )nunny ofr

= 7.28

4cos5.13

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28

Testing it with MATLAB:

( )

( ) 321

1

2

1

2

21

2

121

21110

++

=

zzz

z

zY

[ ] ( )

++==

2

1,

2

21,

212,1;2/10,10 ab ;

[R, P, K] = residuez (b, a)

+

=

9074.1

2562.39537.5

2562.39537.5

i

i

R ][

5.0

7071.07071.0

7071.07071.0

=

+

= Ki

i

polesP

norm (R (1)) = 6.78

angle (R (1)) = -0.5005 rad = - 28.7o

zplane (b, a) % plots the zero-poles

Example of a Non-Relaxed System

( ) ( ) ( ) ( ) 022

11

2

3=+ nnxnynyny

( ) ( )nunxn

= 4

1 find y(n) ify (-1) = 4 andy (-2) = 10

Solution

TakingZ+

from both sides:

( ) ( ) ( )[ ] ( ) ( ) ( )[ ]1

211

411

1)(12

2

11

2

3

+++

==++++

zzXzYzyzyzYzyzY

( ) ( )11

21 21

4

11

1

2

1

2

31

+ +

=

+ z

zzzzY

Finally: ( )21

21

21

231

21

492

+

+

+=

zz

zzzY

or ( )111

411

31

1

32

211

1

+

+

+

=

zzzzY

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29

( ) ( )nu4

1

3

1

3

2

2

1ny

responseforce

n

responsenatural

n

++

=

32143421

Note that natural response is due to system poles and force response is due to the input poles.

Transient response is due to the poles inside the unit circle and steady-state response is due to

poles on the unit circle. In this case,

Transient Response )(2

1

4

1

3

1nu

nn

+

and Steady-State Response )(3

2nu .

Zero-Input (or Initial condition) Response is: )().()( zXzHzY ICZI = andZero-State Response is:

)().()( zXzHzYZS = . In this case,

)(3

8

2

12

4

1

3

1)( nuny

nn

zs

+

=

and )(22

13)( nuny

n

zi

= .

Note that complete response is eitherTransient Response+Steady-State Response, orNatural Response + Force Response, or Zero-Input Response + Zero-State Response. Each response

emphasizes a different aspect of system analysis.

Checking with MatLab:

n = [0: 7]; % just checking the first 8 samples

; n4

1x

= xic = [1, -2] % terms due to initial conditions (1 2z-1)

b = [1, 0];

21

231=a ;

y1 = filter (b, a, x, xic);

( ) ( ) ( )8,1*3

22

141*

31 onesnny ++= ;

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30

Since y andy1are the same, then our solution is correct!

However, for large order difference equations, it is tedious to determine xic(n) analytically.

MatLab command filtic does findxic as well.

Xic = filtic (b, a, Y, x) where Yandx are initial conditions.

Y = [y(-1), y(-2), , y(-N)]

x = [x(-1), x(-2), , x(-M)]

Ifx(n) = 0 forn < -1, then x need not to be defined. In our example:

Y = [4, 10];

Xic = 1, -2

Causality and Stability

Ifh(n) = 0, forn < 0, then the system is causal. Then its ROC is the exterior of a circle. The

stability of a system is quarantined by the condition that the ROC includes the unit circle.

Because the necessary and sufficient condition for a BIBO system is that

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31

Lecture 6- Chapter 4

Frequency Analysis of Signals and Systems

Continuous Signals and Discrete-Time Signals

Periodic Aperiodic

Starting with periodic CT signals:

Recall that a linear combination of harmonically related complex exponentials of the form

( ) tkFjk

keCtx02

+

=

= is a periodic signal with fundamental periodo

pF

T1

= . In order to find Ck,

multiply both sides byFot2je l and integrate over one period:

dteCdteCedtetx

lkTlk

T

tFlkj

k

k

tkFj

k

k

T

tlFj

T

tlFj

P

PPP

43421

=

+

=

+

=

==

0

0

)(22

0

2

0

2 0000)(

( ) plFotj

Tp

oTCetx = l

2

( ) FotjT

p

etxT

Cp

l

l

21 = Fourier Series

An important issue is that whether kFotj

k

keC2

0+

=

representation is equal to x(t) for every moment

oft. The Dirichlet conditions guarantee that this series is equal to x(t) except at the values oftfor

which x(t) is discontinuous. At those values oft, the series converges to the midpoint (average

value) of the discontinuity.

Dirichlet conditions are:

1) x(t) has a finite number of discontinuity in any period.

2) x(t) has a finite number of maxima and minima during each period sufficient but not

3) x(t) is absolutely integrable in any period: necessary

( )

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32

Power Density Spectrum of Periodic Signals

A periodic signal has infinite energy but finite average power.

Parsevals Theorem:

( ) ( ) ( )

( ) ( )

2

K

CT

TpKFot2j

Kp

KFot2jKTp

p

Tpp

2

Tpp

x

C

dtetxCT

1dteCtx

T

1

dttxtxT

1

dttxT

1

P

Kp

+

+

+

=

==

==

444 3444 21

**

*

Ifx(t) is real then PSDCCCC kkkk == 2

*2* is an even function in frequency and the

phase is an odd function.Example:

( )

( )

0

0

0

0

2

20

22/

2/

2

sin

sin

2

2

1

00

0

kFcT

A

kF

kF

T

Aee

kTF

A

kFj

e

T

AdtAe

TC

p

p

kFjtkFj

p

kFotj

p

tkFj

p

k

=

=

=

==

x(t)

t

-Tp Tp-/2 /2

po

T

1F =

pTA

F

kC

A

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33

Now ifpT

decreases ,pT then Ck 0, which means the signal becomes aperiodic

average power becomes zero.

CT Aperiodic Signals

We can say ( ) ( ) = txtx pTplim

( ) ( ) ( )

( ) ( ) ( ) dtetxdtetxFX

anddeXdFeFxtx

tjtFj

tjFtj

+

+

+

+

==

==

2

2

2

1

Aperiodic signals are energy signals.

( )

( ) dFFX

dttxEx

2

2

+

+

=

=

Energy Density Spectrum: Sxx(F) = |X(F)|2

A couple of points:

1) Remember that from only ESD or PSD we cannot reconstruct x(t)because phase information

is lost.

2) Ckforxp(t) is just samples ofX(F)

( )op

k kFXT

C1

=

DT Frequency Analysis

First consider a periodic DT signalx(n) = x(n + N)

( )n

N

kjN

ok

keCnx

=

=21

Multiply both sides byn

N2j

e

l

and sum over one period.

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34

( )( )

=

=

=

=

=

=

elseo

N2NoKifN

eCenx1N

on

1N

oK

nN

K2j

K

nN

2j1N

on

,,l

ll

( )l

l

CNenxn

NjN

on

.21

=

=

( )

( )

( )2121

1/2

/21

1

1,....,1

=

=

=

=

==

=

==

N

on

N

ok

kx

N

oK

Nknj

k

NknjN

onk

nxN

CP

eCnx

NokenxN

C

*DTFS is periodic like Periodic DT*

Ck+N= Ck. Therefore, the spectrum of a periodic DT,x(n), is also periodic with periodN.

( ) ( ) ( ) kN

on

NknjN

on

NnNkj

Nk CenxN

enxN

C ===

=

=

++

1/2

1/2 11

Example

Find DTFS of the following signals:

( )

( ) ( )

155

2sin

3

2cos

53 22

1

1

=+=

==

Nnnnx

Nnx

Nnx

4342143421

For this case, we can directly write it is a sum of exponentials.

( )( )

( )

+=

+=

43421

3

n22j

e

3

n232j

3

n2j

3

n2j

3

n2j

1 ee2

1ee

2

1nx

2

1C

2

1C 21 == , forx1(n)

Interchange the

sums

=

=

= 1if,1

1

1if,1

aa

a

aN

a NN

on

n

Power

The smallest common denominator

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35

( )( )

=

==

5

n254j

5

n2j

5

n2j

5

n2j

2 eej2

1ee

j2

1n

5

2nx

sin

j2

1C

j2

1C

41

== , forx2(n) and 0 else where.

Ckforx(n) is like 21 35x

k

x

k CC +

=

=

=

=

else

122

1

1052

1

32

1

o

kj

,k

kj

Ck

Fourier Transform for Aperiodic D.T. Signals

( ) ( ) ( )

( ) ( )

=

=

+

=

deXnx

enxXeX

nj

n

njj

22

1

(recall that for C.T. signals it was over+

and here is over 2 which means that X() is

periodic).

Two Basic Differences Between CTFT and DTFT:

1) X() X(ej) is periodic with period 2( ) ( ) ( ) ( ) ( ) XenxenxkX njnkj

n

===+ +

++

= 22

2) SinceX() is periodic, (in fact a it is a periodic C.T. signal), then it has a Fourier Seriesand in factx(n) are the coefficients of that Fourier Series.

Before visiting a famous example, lets review the concept of convergence.

If we have a limited observation, we will have the truncation effect, and the famous theory of the

Gibbs. Let ( ) ( ) njN

Nn

N enxX

== if ( ) ( ) 0lim

44 344 21

N

N XX , then XN() converges uniformly

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36

toX() as N . This convergence is guaranteed ifx(n) is absolutely summable (3rd Dirichlet

condition).

( )

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37

( ) njN

N

cN e

n

nX

=sin

Matlab definition: ( )x

xxc

sinsin =

Section 4.2.12

There are two time-domain characteristics that determine the type of signal spectrum and they

are: Periodicity and Continuity

Signals

Continuous Discrete-Time

** Periodicity with period in one domain automatically implies discretion with spacing

1 in

the other domain.

Properties of the Fourier Transform for Discrete-Time Signals

1. Real signals if x(n) is real, theX*() = X(-)

Spectrum magnitude: ( ) ( ) = XX even function

Spectrum phase ( ) ( )= XX pp odd function.

2. Real and evenx(n) Real and EvenX()

3. Real and oddx(n) Imaginary and oddX()

4. Imaginary and oddx(n) Real and oddX()

5. Imaginary and evenx(n) Imaginary and evenX()

6. Linearity a1x1(n) + bx2(n) F aX1() +bX2()

7. Time-Shiftingx(n) F X()

x(n-k) F e-jk

X()

Aperiodic

&

Continuous

Spectrum

Periodic PeriodicAperiodic Aperiodic

Aperiodic

&

Discrete

Spectrum

Periodic

&

Continuous

Spectrum

Periodic

&

Discrete

Spectrum

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38

8. Time-Reversalx(n) F X()

x(-n) F X(-) Therefore, FT of an even function is an even function too.

9. Convolution: x1(n) * x2(n)F

X1(

).X2(

)Proof:

( ) ( ) ( )knxkxnxk

= +

=

21

( ) ( ) ( ){

( ) ( )

)().( 21

)(

21

.

21)(

XX

eknxekx

eknxkxX

n

knj

k

kj

ee

nj

n k kjknj

=

=

=

+

=

+

=

+

=

+

=

10. Correlation Theorem: ( )nr21xx

FX1()X2(-)

Proof:

( ) ( ) ( )nkxkxnrk

xx = +

=

2121

FT

( ) ( ) ( ) ( )( )

321KjKnj

2121

ee

nj2

n K

1nj

n

xxxx enKxKxenrS

+

=

+

=

+

=

==

( )

( )

( )[ ] ( )

( )44444 344444 21444 3444 21

+

=

+

=

=

21 x

n

Knj2

x

K

Kj1 eKnxeKxRHS

Now ifx(n) is real, thenX*() =X(-)

( ) ( ) ( ) ( )( )

( ) 2

x

xxxx xxxSr

==321

l*

Energy Spectral Density

11. Frequency Shifting

ejon

F X( -o)

12 Modulation Theorem

Cross-Energy

Density Spectrum

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39

( ) ( ) ( ) ( )[ ]

[ ]njnj

oo

een

XXnnx

00

2

1)cos(

2

1cos

0

0

+=

++

13. Parseval Theorem

( ) ( ) ( ) ( )

dXXnxnxn

*

21

*

212

1

+

=

=

Proof:

( )

( ) )()(2

1)(

)(2

1

*

21

)(

*

21

*

21

*2

nxnxdeXnx

dXenxRHS

n

nx

nj

n

nj

n

=

=

=

==

=

444 3444 21

Special case: ( ) ( ) ( ) ( )

dx2

1nxnxnx

22

2

12 ==

( ) ( ) ( )( )

===+

=

2

dS

22

nxxx

xx

dx2

1nx0rE

43421

14. Windowing

( ) ( )nxnx 21 F ( ) ( )[ ]

21 *2

1XX

( ) ( ) ( ) ( ) nj21nj enxnxenxx

+

+

==

( ) ( )

( ) ( )

( ) ( ) ( ) ( )[ ]

2122

1

22

1

22

1

*

2

1

2

1

2

1

2

1

XXdXX

eenxdX

enxdeXRHS

njnj

n

njnj

=

=

=

+

=

+

15. Differentiation in Frequency Domain

n(x)n F j( )d

dx

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40

Skipping to Section 4.4.8

Correlation Function and Power spectra for Random Input Signals

When the input signal is random, then we have to consider statistical moments of input and

output. So here is a bit of introduction about Stationary Random Process. Starting with the

Definition of Stationary Signals:

If X(t) is a random process with a point Probability Density Function (PDF),

( ) nxxxxPxP tttt ,...,,, 31 2= forn random variables. ( ) ( ) nIii itxtX ,...2,1,

If the joint probability of ( ) ( ) += 11 ,.........,,, 111t

n

t

n xxPxxxP for all t1 and then the random

process X(t) is stationary in strict sense. In other words, statistical properties of a stationary

random process is time-invariant, meaning that its mean and variance and other moments are

time invariant.

One Observation

Set 1

Set 2

Set 3

t

t

t

t1

t1

t1

t1+

t1+

t1+

x(t,S1)

x(t,S2)

x(t,S3)

x1(t1)

x2(t

1)

x3(t1)

x3(t+)

x1(t1+)

x2(t+)

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41

Statistical (ensemble) Average

( ) ( )iiii tttt

dxxPxxE +

=

If we dont haveP(xti) but have many observations, then ( ) ( )1111 ...

121

t

N

ttt

xxxNxE +++= , which of

a stationary process it is equal toE(xti) for any ti.

Also autocorrelation function:

( ) ( )

[ ]21

21212121 ,,

tt

tttttttt

xx

xxE

dxdxxxPxxxx

=

= +

If this xx depends only on the time difference t1 t2 = , then ( ) [ ],, 11 += ttxx xxE which is the

case for stationary process. If a process has two features:

1) Its xx depends only on time difference, , and

2) E(xt1) = E(x

t2) = E(x

ti)

then the process is said to bestationary in wide sense.

Now if all statistical averages can be obtained by one single realization (or one sample set), then

the process is also Ergodic meaningEnsemble average time average.

( )nx xE= and ( )

=

=1

2

1 N

on

nx

N

)

x)

is an estimate ofx. It will be said it is an unbiased estimate if ( ) xxE =)

. Also, it is a

good estimator if

( ) ( ) 022 = xxx EVar ))

as N

Therefore, time average ensemble average.

Autocorrelation: ( ) ( ) ( )mnxnxN

mN

on

xx +=

=

1*1

( )[ ] ( )mrmE xxxx = the true autocorrelation.

Now back to systems:

x(n) y(n)h(n)

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42

( ){ } ( ) ( )

( ) ( ){ } ( ) ( )0HkhknxEkh

knxkhEnyE

xx

n

y

===

==

+

+

+

=

The autocorrelation sequence:

( ) ( ) ( ){ } ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ){ }

( ) ( ) ( )kmhkh

mnxknxEhkh

mnxhknxkhEmnynyEm

xx

k

k

k

yy

+=

+=

+=+=

+

=

+

=

ll

ll

ll

l

l

l

.*

**

Special Form: when x(n) is a white noise, then ( ) ( )mm xxx 2= and ( )02 xxx = . Then

( ) ( )mm hhxyy 2=

( ) ( ) ( )

dHxhhxyy2

2

22

2

100 ==

by getting the Fourier transform in general:

2

)()(realissignaltheIf

)().().(

)()()()(

..let

)()()(

)()(

H

HH

euelhekh

eeeeklmu

eklmlhkh

em

xx

xx

u

uj

xx

l

ljkj

k

yy

kjljujmj

mj

m k l

xx

m

mj

yyyy

=

=

=

=+=

+=

=

=

=

=

=

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1. The Concept and Representation of Periodic

Sampling of a CT Signal

2. Analysis of Sampling in the Frequency Domain

3. The Sampling Theorem the Nyquist Rate

4. In the Time Domain: Interpolation

5. Undersampling and Aliasing

Signals and SystemFall 2003

Lecture #13

21 October 2003

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We live in a continuous-time world: most of the signa

encounter are CT signals, e.g.x(t). How do we conversignalsx[n]?

SAMPLING

How do we perform sampling

Sampling, taking snap shots ofx(t) every Tse

T sampling period

x[n] x(nT), n = ..., -1, 0, 1, 2, ... regularly spaced

Applications and Examples

Digital Processing of Signals

Strobe

Images in Newspapers

Sampling Oscilloscope

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Why/When Would a Set of Samples Be A

Observation: Lots of signals have the same sample

By sampling we throw out lots of information

all values ofx(t) between sampling points are l

Key Question for Sampling:Under what conditions can we reconstruct the orix(t) from its samples?

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Impulse Sampling Multiplying x(t) by the sam

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Analysis of Sampling in the Frequency D

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Illustration of sampling in the frequency-do

band-limited (X(j)=0 for | |> M) s

No overlap between shifted spectra

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Reconstruction ofx(t) from sampled s

If there is no overlap between

shifted spectra, a LPF canreproduce x(t) from xp(t)

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The Sampling Theorem

Suppose x(t) is bandlimited, so t

Then x(t) is uniquely determined by its samp

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Observations on Sampling

(1) In practice, we obviously

dont sample with impulses

or implement ideal lowpass

filters.

One practical example:

The Zero-Order Hold

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Observations (Continued)

(2) Sampling is fundamentally a time-varyingoperati

multiply x(t) with a time-varying function p(t). H

is the identity system (which is TI) for bandlimi

the sampling theorem (s > 2M).

(3) What ifs 2M? Something different: more la

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Time-Domain Interpretation of Reconst

Sampled Signals Band-Limited Inter

The lowpass filter interpolates the samples assuming

no energy at frequencies c

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Graphic Illustration of Time-Domain Inte

Original

CT signal

After sampling

After passing the LPF

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Interpolation Methods

Bandlimited Interpolation

Zero-Order Hold First-Order Hold Linear interpolation

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Undersampling and Aliasing

When s 2 M Undersampling

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Undersampling and Aliasing (contin

Higher frequencies ofx(t) are folded back and

aliases of lower frequencies

Note that at the sample times, xr(nT) = x(nT)

Xr(j

Dis

ofa

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Demo: Sampling and reconstruction of cosot

A Simple Example

Picture would beModified

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M ultiple Choice

1. The system y(t) = x(t) + 2x(t+3) is(a) causal system (b) non-causal system

(c) part ly (a) and (b) (d) none of these

2. The systemdt

tdy )(+3y(t) = x( t) is a

(a) time invariant system (b) time variant system

(c) partly (a) and (b) (d) none of these

3. The system y(t) = 3x(t) + 4 is(a) linear system (b) non linear system

(c) partly (a) and (b) (d) none of these

4. A power signal has inf ini te energy whereas an energy signal has zeroaverage power.

(a) true (b) false

5. Find dynamic system of the follow ing(a) y(n) = 3x(n) (b) y(n) = x(n) + x(n-1)

6. z transform of the signal {0,1,3,2,0,0,0.} is(a) 1+

z

2+

2

3

z

(b) 1+

z

3

(c) 1+z

3+

2

2

z(d) 1+

2

2

z

7. The ROC of Question no.6 is(a) whole of z plane (b) whole of z plane except z =

(c) whole of z plane except z = 0 (d) none of these

8.

If x(n) and y(n) are two finite sequences, then x(n) * y(n) is(a) X(z)/ Y(z) (b) Y(z)/ X(z)

(c) X 2 (z) Y(z) (d) X(z) Y(z)

9. If X(z) =)45.095.0)(1(

5.02

2

zzz

z, the initial value of x(n) is

(a) 0.5 (b) 1

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(c) 0 (d) 0.45

10.z transform of x(nk) is(a) kz X(z) (b) nz X(z)

(c) 1z X(z) (d) none of these

11. Find correct representation for the following(a) x1(n) * x2(n) = x2(n) * x1(n)

(b) x1(n) * [x2(n) * x3(n)] = [x1(n)* x2(n)] * x3(n)

(c) x1(n)* [x2(n)+x3(n) ] = x1(n) * x2(n) + x1(n) * x3(n)

(d) all are valid

12.The ROC of the z transform of the discrete time sequencex(n) = n)

3

1( u(n) n)

2

1( u(-n-1) is

(a)3

1

2

1

(c) |z|