# Chapter 2 - Discrete Time Signals and 2 - Discrete...Discrete Time Signals and Systems ... • The complex signal ejnw is an important signal in discrete time signal processing

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2-1

2. Discrete Time Signals and Systems

We will review in this chapter the basic theories of discrete time signalsand systems. The relevant sections from our text are 2.0-2.5 and 2.7-2.10.

The only material that may be new to you in this chapter is the section onrandom signals (Section 2.10 of Text)

2.1 Discrete Time Signals

A discrete-time (DT) signal is signal that exists at specific time instants.The amplitude of a discrete-time signal can be continuous though.

When the amplitude of a DT signal is also discrete, then the signal is adigital signal.

A DT signal can be either real or complex. While a real signal carries onlyamplitude information about a physical phenomenon, a complex signalcarries both amplitude and phase information.

Throughout this course, we use square brackets [ ] to denote a DT signaland round brackets ( )i to denote a continuous time signal.

Example : If the n-th sample of the DT signal [ ]x n is the value of the

analog signal ( )ax t at t nT= , then

[ ] ( )ax n x nT=

2-2

Some common DT signals are

1. Unit sample

1 0[ ]

0 otherwise

nn

==

2. Unit step

1 0[ ]

0 0

nu n

n

=

2-3

[ ] nx n A=

where both and A are real. If 0A > and 0 1< < , then [ ]x ndecreases as n increases; see figure below.

4. Sinusoidal

( )[ ] cos ox n A n = +

where A is the amplitude, o is the frequency, and is the phase.

5. Periodic

[ ] [ ]x n N x n+ =

2-4

for any time index n . Here N denotes the period.

Exercise: Is the sinusoidal signal defined above periodic in general?

Example : Express the unit step function in terms of the unit-impulse

Answers:

(a) [ ] [ 1] [ ]u n u n n =

(b) 0

[ ] [ ]k

u n n k

=

=

Example : Express

1 2,3,...,10[ ]

0 otherwise

nx n

==

in terms of the unit step function.

Answer:

[ ] [ 2] [ 11]x n u n u n=

2-5

Example : Express the sinusoidal signal in terms of complex exponentialsignals

Answer:

( )( ) ( )

( ) ( )1 1 2 2

[ ] cos

2

o o

o

j n j n

n n

x n A n

e eA

A A

+ +

= +

+=

= +

where

*1 22

jAeA A

= =

*1 2

oje = =

It is also possible to express the signal as

( ){ }1 1[ ] 2 Re nx n A =

Example : Express an arbitrary DT signal in terms of the unit impulse

Answer

[ ][ ] [ ]k

x n x k n k

=

=

2-6

2.2 Discrete Time Systems

Let [ ]T represents the transformation a discrete time system performedon its input [ ]x n . The corresponding output signal of the system is

[ ][ ] [ ]y n T x n= .

The system is linear if

[ ]1 2 1 2[ ] [ ] [ ] [ ]T ax n bx n ay n by n+ = +

where 1 2[ ] and [ ]y n y n are the responses of the system to inputs of

1 2[ ] and [ ]x n x n respectively.

The above equation illustrates the principle of superposition.

Assume the system is linear and let [ ]hh n be the output of the system whenthe input is

[ ] [ ]kp n n k=

(i.e. a unit-impulse at time n k= ). Then according to the linearity property,

[ ]y n[ ]T [ ]x n

2-7

when the input is

[ ]

[ ]

[ ] [ ]

[ ]

k

kk

x n x k n k

x k p n

=

=

=

=

,

the output will be

[ ][ ] [ ] kk

y n x k h n

=

=

The system is time-invariant if

[ ] [ ]kh n h n k= ,

i.e. the output is delayed if the input is delayed. In this case

[ ][ ] [ ]

[ ] [ ]k

y n x k h n k

x n h n

=

=

=

The signal [ ]h n is called the impulse response of the time-invariantsystem.

While the focus of this course is on linear, time-invariant (LTI) system,there are many real-life applications where the system is non-linear andtime-variant. A good example is a digital FM demodulator operating in themobile radio environment.

2-8

Example : Provide a physical interpretation of a LTI system whose impulseresponse is

[ ] [ ]h n u n=

Answer: The output of the system is

[ ]

[ ]

1

[ ] [ ]

[ ]

[ ]

[ ] [ ]

[ 1] [ ]

k

k

n

k

n

k

y n x k h n k

x k u n k

x k

x k x n

y n x k

=

=

=

=

=

=

=

= +

= +

Thus the system is an integrator.

A system is casual if and only if the output at time n depends only on theinput up to time n . According to the equation

[ ][ ] [ ]k

y n h k x n k

=

= ,

this means the impulse response [ ]h n is zero when 0n < .

2-9

Example : A moving averager computes the signal

2

11 2

1[ ] [ ]

1

M

k M

y n x n kM M =

= + +

from its input [ ]x n . Here 1M and 2M are positive integers. What is theimpulse response of the system? Is the system casual?

Answer:

The output can be rewritten as

1

2

1 2

1 2

1

1 2

1[ ] [ ]

1

1 [ ] [ ]

1

n M

m n M

n M n M

m m

y n x mM M

x m x mM M

+

=

+

= =

=+ +

= + +

Compared to the output of the integrator, we can deduce that the impulseresponse of the system is

[ ] [ ]( )1 21 2

1[ ] 1

1h n u n M u n M

M M= +

+ +

Since the impulse response is non-zero when 0n < , so the system is notcasual.

A real physical system can not be non-casual, i.e. it can not generate anoutput before there is an input. So in practice what a non-casual system

2-10

means is that there is a processing delay. For example, you can view themoving averager as a device that computes the local mean of the signal

[ ]x n at time n after it observes the sample 1[ ]x n M+ . So 1M is the delay.

A system is stable if a bounded input results in a bounded output. Therequirement for having a stable system can be derived from theinput/output relationship of a LTI system, which states that

[ ][ ] [ ]k

y n x k h n k

=

=

This means

[ ]

[ ]

[ ] [ ]max max

[ ] [ ]

[ ]

k

k

k m

y n x k h n k

x k h n k

x h n k x h m

=

=

= =

=

=

where is the absolute value operator and maxx is the largest magnitude ofthe input signal.

So if the impulse response of the system is absolute-summable, i.e. when

[ ]k

S h k

=

= <

then the system is stable.

2-11

Example : Is the integrator a stable system?

Answer: Since [ ] 1h n = for 0n and zero otherwise, the impulse responseis not absolute-summable. Consequently the system is not stable.

Example : Is the moving averager a stable system?

Answer: Yes, because the impulse response consists of only a finitenumber of non-zero samples.

Finite Impulse Response (FIR) and Infinite Impulse Response (IIR):

FIR => An impulse response of finite duration, hence a finite number ofnon-zero samples. Always stable.

IIR => The impulse response is infinitely long. Can be unstable (forexample the integrator).

Example : Comment on the stability of a LTI system with the exponentialimpulse response

0[ ]

0 otherwise

na nh n

=

Solution:

0 0

[ ]kk

k k k

h k a a

= = =

= =

2-12

This is summable if 1a < . In this case,

1[ ]

1kS h k

a

=

= =

Cascading of LTI systems serial connection of two or more systems; seethe example below.

As far as the input/output relationship is concerned, it really does notmatter what the order of the concatenation is. For the example above, bothpossibilities yields the same combined impulse response of

1 2[ ] [ ] [ ]h n h n h n=

2-13

In many applications, we have to concatenate a system to an existing oneso that the combined system yields the desired response. A good example isthe equalizer used in a digital communication system.

Many communication channels introduces intersymbol interference (ISI)ef. This means the received signal [ ]r n depends not only on the data bit

[ ]b n , but also on some adjacent bits. For example,

1 1[ ] [0] [ ] [1] [ 1]r n h b n h b n= +

where 1[ ]h n represents the impulse response of the channel. The objectiveof equalizer design is to find a digital filter with an impulse response 2 [ ]h nso that the combined response of the channel and the equalizer,

1 2[ ] [ ] [ ]h n h n h n= , is the unit-impulse function. This means afterequalization, we have [ ] [ ]y n b n= , i.e. the ISI is removed.

Exercise: Consider an ISI channel with 1[ ] [ ]nh n a u n= , where 0 1a< < ,

and [ ]u n is the unit step function. Determine the equalizer that completelyremoves the ISI.

Systems governed by the Linear Constant Coefficient Difference Equation(LCCDE):

1 0

[ ] [ ] [ ]N M

k jk j

y n a y n k b x n j= =

= +

The above equation suggests that current output of the system depends onthe previous output as well as the current and previous input.

2-14

In analyzing the above system, we assume the input is applied at time0n = (i.e. [ ] 0x n = for negative n ) and the initial state of the system is

defined as

( )[0] [ 1], [ 2],..., [ ]y y y N= Y

(a) Zero State Response (ZSR) response of the system to an unitimpulse applied at time 0n = , under the condition that [0]Y is theall-zero vector.

(b) Zero Input Response (ZIR) response of the system due to a non-zero initial state but no input.

Example : [ ] [ 1] [ ]y n ay n x n= +

Let the initial state be [0] [ 1]y b= =Y , then

2

3 2

[0] [0]

[1] [0] [1]

[2] [0] [1] [2]

y ab x

y a b ax x

y a b a x ax x

= +

= + +

= + + +

or in general

1

0

[ ] [ ]n

n n k

k

y n a b a x k+

=

= +

The ZIR is1

1[ ] [ ]nh n a bu n+=

and the ZSR is

2-15

2[ ] [ ]nh n a u n=

It is clear that the ZIR corresponds to the bias term in [ ]y n . Since it isindependent of the input, the system can NOT be classified as a linearsystem. Note that the response of the system to

3 1 1 2 2[ ] [ ] [ ]x n w x n w x n= +

is

{ }

13

0

11 1 2 2

0

[ ] [ ]

[ ] [ ]

nn n k

k

nn n k

k

y n a b a x k

a b a w x k w x k

+

=

+

=

= +

= + +

,

which is different from

3 1 1 2 2[ ] [ ] [ ]y n w y n w y n= + ,

where

11 1

0

12 2

0

[ ] [ ],

[ ] [ ]

nn n k

k

nn n k

k

y n a b a x k

y n a b a x k

+

=

+

=

= +

= +

2-16

2.3 Fourier Transform of Discrete Time Signals

Consider the sinusoidal signal

( )[ ] cosx n A n = +

It can be written in terms of two complex exponential functions as

( ) ( )

1 2[ ] 2

j n j nj n j ne ex n A A e A e

+ ++= = +

where

*1 22

jAA e A= =

The complex signal j ne is an important signal in discrete time signalprocessing it is an eigenfunction of a linear system and it leads us to theconcept of Fourier Transform of a discrete-time signal.

Again let us use [ ]T to represent the operation a discrete time systemperforms on its input. A signal [ ]f n is an eigenfunction of the system if

[ ][ ] [ ]T f n a f n= ,

where the constant a is called an eigenvalue. This definition is consistentwith that in matrix theory where the eigenvector v and the eigenvalue b ofa matrix A is defined as

b=Av v .

2-17

Here the matrix A is analogous to our linear system.

As shown in Section 2.2, the transformation performed by a LTI on itsinput [ ]x n is described by the convolution formula:

[ ] [ ] [ ]k

y n h k x n k

=

= ,

where [ ]h n is the impulse response of the system and [ ]y n is thetransformed signal or output of the system. If

[ ] j nx n e = ,

then the output signal becomes

( )

( )

[ ] [ ]

[ ]

[ ]

j n k

k

j n j k

k

j n j k

k

j n j

y n h k e

h k e e

e h k e

e H e

=

=

=

=

=

=

=

,

where

( ) [ ]j j kk

H e h k e

== .

2-18

It is clear from the above analysis that j ne is indeed an eigenfunction of a

discrete-time LTI system with ( )jH e being the corresponding eigenvalue.

In the linear system literature, ( )jH e is called the frequency response of adiscrete-time LTI system.

In general, the expression

( ) [ ]j j kk

X e x k e

==

is called the Fourier Transform of the discrete-time signal [ ]x n .

One important property of the Fourier Transform of a discrete time signalis that it is periodic in with a period of 2 . This is quite different fromthe Fourier Transform of a continuous time signal, which in general is notperiodic.

Example: Express the output of a LTI system in terms of its frequencyresponse when the input is the sinusoid ( )[ ] cosx n A n = + . Assume theimpulse response of the sytem is a real signal.

Solution:

- The sinusoidal input can be written as a weighted sum of two complexexponential functions as

1 2[ ]j n j nx n A e A e = +

2-19

where *

1 2/ 2jA Ae A= = are the weighting coefficients.

- Since the system is linear, the sinusoidal response is

1 1 2 2[ ] [ ] [ ]y n A y n A y n= +where

( )1[ ] j jy n H e e =and

( )2[ ] j jy n H e e = ,

are the outputs of the system when the inputs are 1[ ]j nx n e = and

2[ ]j nx n e = respectively.

- Since the impulse response is real,

( ) ( )*

*[ ] [ ]j j k j k jk k

H e h k e h k e H e

= =

= = =

.

This means we can write the output of the system as

( ) ( )

( ) ( )( ){ }

( ) ( )

1 2

1 2

[ ] ]

* *1 1

1

[ ]

2Re

cos ( )

j j n j j n

y n y n

j j n j j n

j j n

j

y n A H e e A H e e

A H e e A H e e

A H e e

A H e n

= +

= +

=

= + +

14243 1442443

2-20

Note that ( )jH e and ( ) are respectively the magnitude and phase ofthe frequency response, i.e.

( ) ( ) ( )j j jH e H e e =In conclusion, when the input is a sinusoid, the output is also a sinusoidat the same frequency but with the amplitude scaled by ( )jH e and withthe phase shifted by an amount ( ) .

Example : Determine the frequency response of a delay element describedby the impulse response

[ ] [ ]h n n d=

Solution

( ) [ ] [ ]j j n j n j dn n

H e h n e n d e e

= =

= = =

This means

( ) 1jH e = (constant magnitude response)and

( ) d = (linear phase)

Example : Determine the Fourier Transform of the one-sided exponentialsignal

2-21

[ ] [ ]nx n a u n=

where 0 1a< < and [ ]u n is the unit-step function.

Solution:

( ) ( )0 0

1[ ]

1

nj j n n j n j

jn n n

X e x n e a e aeae

= = =

= = = =

Since

( )

( )( ) ( )

( ) ( ) ( ){ }( )

2 2 2

2 22 2 2 2

2 2 2

2

1 1 cos( ) sin( )

1 cos( ) sin( ) 1 cos( ) sin ( )

1 cos( ) sin ( ) 1 cos( ) sin ( )

1 cos( ) sin ( ) cos ( ) sin ( )

1 cos( )

jae a ja

a aa a j

a a a a

a a j

a a

= +

= + + + +

= + +

= + ( )2 2sin ( )exp ( )

j

wheresin( )

( )1 cos( )

a

a

=

,

this means the magnitude of the Fourier transform is

( )( )2 2 2

1

1 cos( ) sin ( )

jX ea a

=

+

and the phase is simply

( ) ( ) = .

2-22

Existence of the Fourier Transform:

- If we set the parameter a in the above example to unity, then the signalbecomes a unit-step. The Fourier Transform in this case, however, doesnot exist in the finite magnitude sense.

- A sufficient condition for the existence of the Fourier transform (in thefinite-magnitude sense) is that the signal is absolute-summable, i.e.

[ ]k

S x k

=

= <

The proof is the same as that we used to proof the stability of a LTIsystem.

- We can deduce from the above that the Fourier Transform always existsfor signals with finite duration.

Example : Determine the Fourier Transform of the signal

1/( 1) 0[ ]

0 otherwise

M n Mx n

+ =

Solution

( )( )

( ) ( ) ( ){ }{ }

( )( )( )

1

0

1 / 2 1 /2 1 / 2

/2 /2 /2

12/2

2

1 1 1[ ]1 1 1

1

1

sin1

1 sin

j MMj j n j n

jn n

j M j M j M

j j j

Mj M

eX e x n e e

M M e

e e e

M e e e

eM

+

= =

+ + +

+

= = =+ +

=

+

=+

2-23

The magnitude of the transform is

( ) ( )( )( )1

2

2

sin11 sin

MjX e

M

+=

+

At a first glance, the phase of the Fourier Transform is ( ) / 2M = .However, the sin( ) / sin( )i function can take on either + or ve value.When there is a sign change in this function, that corresponds to anadditional 180 degree phase shift.

Plots of the magnitude and phase o...

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