Chapter 2 - Discrete Time Signals and Systems

7/29/2019 Chapter 2 - Discrete Time Signals and Systems

1/812-1

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 signal

carries both amplitude and phase information.

Throughout this course, we use square brackets [ ] to denote a DT signaland round brackets ( )g 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

[ ] ( )a

x n x nT=


2/812-2

Some common DT signals are1.Unit sample

1 0[ ]

0 otherwise

nn ==

2.Unit step1 0

[ ]

0 0

nu n

n

=


3/812-3

[ ] nx n A=

where both andA 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+ =


4/812-4

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-impulseAnswers:

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

(b)0

[ ] [ ]k

u n n k

=

=

Example: Express1 2,3,...,10

[ ]0 otherwise

nx n

==

in terms of the unit step function.

Answer:

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


5/812-5

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 2oje

= =

It is also possible to express the signal as

( ){ }1 1[ ] 2 Ren

x n A =

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

[ ][ ] [ ]k

x n x k n k

=

=


6/812-6

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 theprinciple 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


7/812-7

when the input is

[ ]

[ ]

[ ] [ ]

[ ]

k

k

k

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-invariantif[ ] [ ]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 the

mobile radio environment.


8/812-8

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

[ ][ ] [ ]ky n h k x n k

== ,

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


9/812-9

Example: A moving averager computes the signal2

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 the

impulse 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 impulse

response 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


10/812-10

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 the

input/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 of

the 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.


11/812-11

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 finite

number of non-zero samples.

Finite Impulse Response (FIR) andInfinite Impulse Response (IIR):FIR => An impulse response of finite duration, hence a finite number of

non-zero samples. Always stable.

IIR => The impulse response is infinitely long. Can be unstable (for

example 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

= = =

= =


12/812-12

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

1[ ]

1kS h k

a

=

= =

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

As far as the input/output relationship is concerned, it really does not

matter 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=


13/812-13

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 objective

of equalizer design is to find a digital filter with an impulse response 2 [ ]h n

so 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 j

k 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.


14/812-14

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 xy 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


15/812-15

2[ ] [ ]nh n a u n=

It is clear that the ZIR corresponds to the bias term in [ ]y n . Since it is

independent 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

1

1 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

+

=

+

=

= +

= +


16/812-16

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 n

j 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 the

concept 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 consistent

with that in matrix theory where the eigenvector v and the eigenvalue b ofa matrix A is defined as

b=Av v .


17/812-17

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 n

x 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

=

= .


18/812-18

It is clear from the above analysis thatj n

e

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 from

the 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 n

x n A e A e = +


19/812-19

where*

1 2/ 2j

A Ae A= = are the weighting coefficients.

- Since the system is linear, the sinusoidal response is1 1 2 2

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

( )1[ ]j j

y n H e e =

and

( )2[ ]j j

y n H e e = ,

are the outputs of the system when the inputs are 1[ ]j n

x n e

= and2[ ]

j nx n e = respectively.

- Since the impulse response is real,

( ) ( )*

*[ ] [ ]j j k j k j

k 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


20/812-20

Note that ( )jH e and ( ) are respectively the magnitude and phase of

the frequency response, i.e.

( ) ( ) ( )j j jH e H e e =

In conclusion, when the input is a sinusoid, the output is also a sinusoid

at the same frequency but with the amplitude scaled by ( )jH e and with

the 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 d

n nH 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


21/812-21

[ ] [ ]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

where

sin( )( )

1 cos( )

a

a

=

,

this means the magnitude of the Fourier transform is

( ) ( )2 2 21

1 cos( ) sin ( )

jX e

a a

= +

and the phase is simply

( ) ( ) = .


22/812-22

Existence of the Fourier Transform:- If we set the parameter a in the above example to unity, then the signal

becomes a unit-step. The Fourier Transform in this case, however, does

not 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 LTI

system.

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

Example: Determine the Fourier Transform of the signal1/( 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

+

= =

+ + +

+

= = =+ +

=

+

=+


23/812-23

The magnitude of the transform is

( )( )( )( )

12

2

sin1

1 sin

M

jX eM

+=

+

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

additional 180 degree phase shift.

Plots of the magnitude and phase of the transform for the case of 4M =are shown below.


24/812-24

The Inverse Fourier Transform is defined mathematically as( )

1[ ]

2

j j nx n X e e d

=

Proof:

( )

( )

( )

1 1[ ]

2 2

1[ ]

2

1[ ]

2

j j n j k j n

k

j n k

k

j n k

k

X e e d x k e e d

x k e d

x k e d

=

=

=

=

=

=

Since

( ) ( )

( )

( ) ( )

( ) ( )

1 1

2 2

1

2

1

2

sin sin

sinc

c c

c c

c

c

c c

j mj m

j m

j m j m

c cc

c

c c

e d e d j mj m

ej m

e e

m j

m m

m m

m

=

=

=

= = =

this means


25/812-25

( )

( )

( )1 1[ ]2 2

[ ] sinc

[ ] [ ]

[ ]

j j n j n k

k

k

k

X e e d x k e d

x k n k

x k n k

x n

=

=

=

=

=

=

=

Physically, the inverse Fourier transform states that the time domain signalis the sum of infinitesimally small complex sinuoids of the form( )j j nX e e d

where ( )jX e denotes the relative amount of each complex sinusoidalcomponent. Consequently, it is a synthesizingformula.

Example: Determine the impulse response of a LTI system whosefrequency response is given by

( ) 412

1

1

j

jH e

e

=

Solution:

We first rewrite the frequency response as


26/812-26

( )

( )

( )

412

412

0

412

0

1

1

j

j

kj

k

k j k

k

H ee

e

e

=

=

=

=

=

Comparing the above infinite series with the definition of a Fourier

transform, we come to the conclusion that

( )/ 4

12

0,4,8,12,...[ ]

0 otherwise

nn

h n ==

Example: The frequency response of an ideal low pass (i.e. brickwall)filter is

( )1

0 otherwise

j c

lpH e =

What is the corresponding impulse response?

c

c

1

( )j

lpH e


27/812-27

Solution:

( )

( )

1[ ]

2

1

2

sin

c

c

j j n

lp lp

j n

c

h n X e e d

e d

n

n

=

=

=

Observation: the impulse response of an ideal low pass filter is NOT

absolute summable.

Question: But then why does the frequency response exist?

Answer: The absolute-summability of a signal is a sufficient condition for

( )jH e < , not a necessary condition.

If we impose the constraint that the magnitude of a valid Fourier Transformmust be finite, is there any reason why we shouldnt impose the constraint

that the derivative(s) of a Fourier Transform should also be finite? After all,

any paremeter associated with a real world signal should be finite, right?

If we impose this additional constriant on the derivative, then the ideal lowpass filter is not a valid frequency response because of the discontinuities

in the spectrum.


28/812-28

Exercise: Show that a sufficient requirement for( )j

dH e

d

<

is

( )k

k h k

=

<

Verify the result using the ideal low pass example.

It appears that if we were to be able to deal with a wide variety of signalsin our analysis, we should relax on the requirement that magnitude of avalid transform or its derivative(s) must be finite. This leads us to the

impulse function in the frequency domain ( ) . Some important

properties of this function are:

1. ( )0 is undefinied by infinitely large,2. ( ) 0 = for 0 , and

3. ( ) ( ) (0)f d f

=

With the introduction of this function, we can now have proper definitions

for the Fourier transforms of signals such as a DC signal, a complexsinusoidal, and the unit-step signal. The fact that these signals exist at

discrete frequencies is consistent with the above properties of the impulse

function.


29/812-29

The Fourier Transform of the DC signal[ ] 1x n =

is

( )

( )

[ ]

2 2

j j n

n

j n

n

n

X e x n e

e

n

=

=

=

=

=

= +

Proof:

The function ( )jX e can be treated as an analog signal in . Since this

analog signal has a period of 2P = , it can be represented by thecomplex Fourier series

( )2

expj

k

k

X e X j kP

=

=

(1)

where

( )/ 2

/ 2

1 2exp

P

j

k

P

X X e j k dP P

= (2)

is the k-th complex Fourier coefficient. Substituting

( ) ( )2 2jn

X e n

=

= +


30/812-30

into (2) yields

( )

( )

/ 2

/ 2

1 2exp

1 22 2 exp

2 2

1

P

j

k

P

n

X X e j k dP P

n j k d

=

=

= +

=

Substituting 1kX = into (1) yields

( )

( )

2exp

2exp

2

exp

j

kk

k

k

X e X j kP

j k

jk

=

=

=

=

=

=

The Fourier Transform of the complex sinuoid[ ] oj nx n e =

is

( )( )0

oj nj j n

n

j n

n

X e e e

e

=

=

=

=


31/812-31

This is simply the Fourier Transform of the DC signal shifted to the

frequency o = . Consequently,

( ) ( )2 2j on

X e n

=

= +

It can be shown that the Fourier Transform of the unit-step function is( ) ( )

12

1

j

jn

U e ne

=

= + +

2.4 Properties of Fourier Transforms

Linearity:If

( )

( )

1 1

2 2

[ ] ,

[ ]

j

j

x n X e

x n X e

then

( ) ( )1 2 1 2[ ] [ ]j j

ax n bx n aX e bX e

+ +


32/812-32

Time Shi f ting and F requency Shi f ting

[ ] [ ]

[ ] ( )

[ ]

( )

j n

n

j n d j d

n

j m j d

m

j d j

x n d x n d e

x n d e e

x m e e

e X e

=

=

=

=

=

=

[ ] [ ]

[ ]( )

( )

( )

0

0

o oj n j n j n

n

j n

n

j

e x n x n e e

x n e

X e

=

=

=

=

Time reversal

[ ] [ ]

[ ]

[ ]

( )

( )( )

( )

j n

n

j n

n

j m

m

j

x n x n e

x n e

x m e

X e

=

=

=

=

=

=


33/812-33

We showed earlier that if [ ]x n is real, then

( ) ( )*j jX e X e = .

So if [ ]x n is real, then

[ ] ( )* jx n X e

Di ff erentiation in F requency

( ) [ ]

[ ]

[ ]( )

[ ]

j j k

k

j k

k

j k

k

j k

k

d dX e x k ed d

dx k e

d

x k jk e

j kx k e

=

=

=

=

=

=

=

=

The above implies

( )[ ] jd

nx n j X ed

Convolution:If 3 1 2[ ] [ ] [ ]x n x n x n= , then( ) ( ) ( )3 1 2

j j jX e X e X e =


34/812-34

Proof: Since

3 1 2

1 2

[ ] [ ] [ ]

[ ] [ ]k

x n x n x n

x k x n k

=

=

= ,then

( )

( )

( )

3 3

1 2

1 2

1 2

1 2

[ ]

[ ] [ ]

[ ] [ ]

[ ] [ ]

[ ] [ ]

j j n

n

j n

n k

j n k j k

n k

j n k j k

k n

j m

X e x n e

x k x n k e

x k x n k e e

x k x n k e e

x k x m e

=

= =

= =

= =

=

=

=

=

=

( )

( )

( ) ( )

1 2

2 1

1 2

[ ]

[ ]

j k

k m

j j k

k

j j k

k

j j

e

x k X e e

X e x k e

X e X e

= =

=

=

=

=

=

The result is known as the Convolution Theorem.


35/812-35

Energy of a signal and the Parseval s TheoremThe energy of a discrete-time signal is defined as

( )22 1

[ ] 2j

nE x n X e d

= = =

The result is known as the Parsevals Theorem.

Proof:

Let [ ]

*[ ]h n x n=

This means

( ) ( )*j jH e X e = .

If

*

[ ] [ ] [ ]

[ ] [ ]

[ ] [ ]

k

k

y n x n h n

x k h n k

x k x k n

=

=

= =

=

,

then

( ) ( ) ( ) ( )2j j j j

Y e X e H e X e = =

Taking the inverse Fourier Transform of ( )jY e at 0n = yields


36/812-36

( )

( )

2

2

1[0] [ ]

2

1

2

j

k

j

y x k Y e d

X e d

=

= =

=

The Windowing Theorem

( ) ( )

( ) ( )

( ) ( ) ( )

( )

[ ] [ ] [ ]

1 1 2 2

1 1

2 2

1 1

2 2

1 1

2 2

j jn j jn

j jn j jn

jnj j

j

y n x n w n

X e e d W e e d

X e e d W e e d

X e W e e d d

X e W

+

= =

= =

=

=

=

=

=

( )( )

( ) ( )( )

( )

1 1

2 2

1

2

j jn

jj jn

j jn

e e d d

X e W e d e d

Y e e d

= =

=

=

=

where

( ) ( ) ( )( )1

2

jj jY e X e W e d

=

=


37/812-37

It is important to realize that the above is a circular convolution in thefrequency domain.

Example: Use the Windowing Theorem to determine the FourierTransform of the signal

( )sin

[ ]0 otherwise

cn

nM n M

y n

=

Solution

This signal can be considered as the product of the ideal low pass signal

( )sin[ ]

c

lp

nh n

n

=

and the rectangular window

1[ ]

0 otherwise

M n Mw n =

As shown earlier,

( )1

0 otherwise

j c

lpH e =

.

It can also be shown that (please verify)

( )( 1) ( 1)1 1

11 1

j M j Mj

j j

e eW e

e e

+ + +

+

= +


38/812-38

Results for different values ofM are shown below

Note that ( )jMH e is this figure is equivalent to ( )jY e

Observations:

- the oscillation (also known as the Gibbs phenomenon) is more rapid forlargerM,

- the amount of ripples does not decrease though

Exercise: Plot ( )jW e and find out what causes the ripples in thesediagrams.


39/812-39

2.5 Sampling of an Analog Signal

Reference: Sections 4.0-4.3, 4.6 of Text

Now that we know what the Fourier Transform of a discrete time signal[ ]x n is, we want to relate it to the Fourier transform of its continuous-time

counterpart ( )cx t , under the condition that

[ ] ( )c

x n x nT= ,

where T is the sampling period and

1sf

T=

is thesampling rate.

Let( ) ( ) j tc cX j x t e dt

=

be the continuous-time (CT) Fourier transform of ( )cx t , where is theanalog frequency. This means

( )1

( ) (inverse Fourier Transform)2

j t

c cx t X j e d

= .

Since [ ] ( )cx n x nT= , this means


40/812-40

( )

( )( )

( )

( )( )( )

( )

( )( )

212

212

2/

2 2

/

/

2

/

1[ ] ( )

2

12

1

2

1

2

T

T

T

j nT

c c

k

j nTc

k k

Tj k nT

c T T

k T

T

j nT

c T

T

x n x nT X j e d

X j e d

X j k e d k

X j k e d

+

=

+

=

= =

=

= + +

= +

( )( )

( )( )

( )( )

/

2

/

/

2

/

2

1

2

1

2

1 1

2

k

T

j nT

c T

kT

T

j nT

c T

kT

j n

c T T

k

X j k e d

X j k e d

X j k e d

T

=

=

=

=

= +

= +

= +

In comparing the above with the inverse Fourier transform of a DT signal,

we come to the conclusion that

Exercise: Show that the ( )jX e shown above is periodic in with aperiod of 2 .

( ) ( )( )21j

c T T

k

X e X j kT

=

= +


41/812-41

To construct ( )jX e from ( )cX j , we can adopt the following procedure1.Divide ( )cX j into intervals of width 2 / T according to the formula

k-th interval:2 2

k kT T T T

< +

2.Define the spectral segment in the k-th interval mathematically as

( )( ) ( ) ( )1 12 22 / 2 /

0 otherwise

c

k

X j k T k TX j

< + =

3.Shift the different spectral segments to the center of the spectrum andadd them together according to

( ) ( )( )2k Tk

Y j X j k

=

= +

4.The Fourier transform of the DT signal is obtain by mapping the analogfrequency in ( )Y j to T = , i.e.

( )1

,j

X e Y jT T

=


42/812-42

Example: Sampling of a triangular analog spectrum with a one-sidedbandwidth of 480W= rad/s. The sampling frequency, in rad/s, is

2640s

T

= =

Fig: (a) ( )cX j , (b) 1( )X j , (c) 0( )X j , and (d) 1( )X j


43/812-43

Fig: (a) the various shifted spectra ( )kX j , (b) the summed spectrum ( )Y j ,and (c) the Fourier transform ( )jX e . Note that the x-variable in diagram (c)

is the digital frequency divided by .

It is observed that the summed spectrum ( )Y j is no longer triangular.Consequently the Fourier transform ( )jX e in the interval [ , ] is also not

triangular. This is known as the aliasing effectand is caused by using toosmall a sampling frequency.

The aliasing effect can be eliminated if the sampling frequency satisfies theNyquist criterion :


44/812-44

22

sW

T

=

where W is the (one-sided) bandwidth of the analog signal ( )cx t in rad/s.

Proof:

Consider the term ( )12 2 /k T for any 0k> . If the Nyquist samplingcriterion is satisfied, this means

1 2 1 12

2 2 2s

k k k W W

T

=

This means the spectral segments ( )kX j , k=1,2,, are all zero.Similarly, it can be shown that all the spectral segments with negative kare

zero. Consequently,

( )1j

cX e X j

T T

=

,

i.e. the shape of ( )cX j is preserved after sampling. In this case, theFourier transform of the sampled signal is simply a (frequency and

amplitude) scaled version of ( )cX j .

For the previous example involving the triangular ( )cX j , a samplingfrequency of 640 rad/s was used. This is less than the Nyquist frequency of

2 960W= rad/s. Consequently there is aliasing in the sampled signal.When the sampling frequency is at or above the Nyquist frequency, say

960, 2 960, 3 960s = rad/s, then the aliasing disappears and ( )jX e

becomes


45/812-45

Fig: Effect of varying the sampling frequency on ( )jX e : (a) the sampling frequency

is at the Nyquist rate of 960 rad/s, (b) the sampling frequency is twice the

Nyquist rate or 1920 rad/s, and (c) the sampling frequency is three times theNyquist rate or 2880 rad/s.

It is evident from the diagram that the spectral width of ( )jX e in theinterval [ , ] is proportional to the bandwidth to sampling-rate ratio

2

s

W =

The larger s is, the narrower the spectrum of the digital signal.


46/812-46

2.5.1 Down Sampling by an I nteger Factor

Suppose ( )cx t was sampled at a rate of1/ T Hz to obtain

[ ] ( )cx n x nT=A down sampler will convert [ ]x n to another DT-signal [ ]dx n according to

[ ][ ]dx n x nM=

Example:M=2

[ ]dx n[ ]x n down sampler

M

0

n

[4]x

[2]x

[0]x

0 1 2 3 54

n

[2]d

x

[1]dx

[0]d

x

1 2


47/812-47

The samples [1], [3], [5],...x x x are not used in forming the signal [ ]dx n ,i.e. they are decimated. Consequently a down sampler is also known as a

decimator.

What is the relationship between ( )cX j , ( )jX e , and ( )jdX e ?From our earlier discussion we know

( ) ( )( )21j

c T Tk

X e X j kT

=

= +

Since the down-sampled signal [ ]dx n can be viewed as a sampled version

of ( )cx t with a sampling period of

'T MT= ,

so ( )jdX e must be

( ) ( )( )

( )( )

( )( )

( )( )

2' '

2

12

0

12 2

0

1

'

1

1 1

[ ]

1 1

j

d c T Tr

c MT MT r

M

c MT MT i k

M

c MT T MT i k

X e X j rT

X j rMT

X j kM iM T

X j k iM T

=

=

= =

= =

= +

= +

= + +

= + +


48/812-48

( )( )

( )

12 21

0

1[ 2 ]/

0

1 1

1

Mi

c M T Ti k

Mj i M

i

X j kM T

X eM

+

= =

+

=

= +

=

This equation says that the FT of the down-sampled signal is the

superposition ofM frequency shifted and scaled copies of ( )jX e .

It should be emphasized that the above expression should only be used todetermine ( )jdX e

in the range < . For outside this range,

the value of ( )jdX e

can be determined by making use of the fact that it is

periodic with a period of 2 .

On the other hand if you blindly apply the formula above, you will end up

with a ( )

j

dX e

with a period of 2 .M

Example: for 2M = , we have( ) ( ) ( ){ }/ 2 ( 2 ) / 2

1

2

j j jdX e X e X e

= +

There will be no aliasing effect in [ ]dx n if the sampling frequency of [ ]x n

is at least twice the Nyquist frequency, i.e.

( )2

2 2s WT

= .


49/812-49

If this is the case, then the effective sampling frequency of [ ]dx n is

( )' 22

ss W

=

which satisfies the Nyquist criterion for zero-aliasing. Alternatively you

can prove this using the diagram below.

Fig: (a) ( )jX e when the sampling frequency is twice the Nyquist frequency, (b)

( )/ 212 jX e , (c) ( )( 2 ) / 212 jX e , and (d) ( )j

dX e .

In general, the sampling frequency of [ ]x n must be at least M times theNyquist rate in order to avoid aliasing in [ ]dx n .


50/812-50

2.5.2 Up Sampli ng by an I nteger Factor

When there is no aliasing in the sampled signal, then theoretically it ispossible to compute ( )cx t exactly at any value of t from [ ]x n . This

process is known as interpolation.

Argument:

( ) ( )[ ] ( )j c cx n X e X j x t=> => =>

Derivation:

( )

( )

( )

( )

/

/

/

/

/

/

1( )2

1

2

1

2

1

2

1[ ]

2

1[ ]2

j t

c c

T

j t

c

T

j t T

c

j j t T

j n j t T

n

j t T n

x t X j e d

X j e d

X j e d

T T

X e e d

x n e e d

x n e d

=

=

=

=

=

=

=

[ ]sinc

n

n

tx n n

T

=

=

=


51/812-51

In practice, it is not possible to calculate ( )cx t exactly from its samplesbecause the sinc function is, straightly speaking, infinitely long.

Interpolators that are commonly used in practice are:

- Lagrange,- Cubic spline

Note that sinc( / )t T is the inverse Fourier transform of the following ideallow pass spectrum:

Thus ideal interpolation is equivalent to feeding the discrete time signal

[ ]x n as a sequence of (continuous-time) impulses to an ideal low passfilter, i.e.

( ) [ ] ( ) sinccn

tx t x n t nT

T

=

=

Again, since the ideal low pass filter can never be implemented exactly

because of the abrupt transitions, we can never interpolate the analog signal

exactly from its samples.

/T /T

T

( )lpH j


52/812-52

Let[ ] ( ')i cx n x nT=

be the DT signal obtained by sampling ( )cx t at a rate of

( )' 1 1 Hz

' /s

Lf

T T L T = = =

Since this signal can be obtained from the signal

[ ] ( )c

x n x nT=

according

'[ ] [ ]sinc

[ ]sinc

[ ]sinc ,

i

k

k

k

nTx n x k k

T

nTx k k

LT

nx k k

L

=

=

=

= =

=

it is called an up-sampled signal.

[ ]i

x n[ ]x n

up sampler

L


53/812-53

What is the Fourier Transform of this up-sampled signal?- Since the sampling rate 1/sf T= used in generating [ ]x n satisfies the

Nyquist criterion ,

( )1j

cX e X j

T T

=

- Since 1/sf T= satisfies the Nyquist criterion, the new sampling rate' /sf L T= used in generating [ ]ix n will also satisfy the Nyquist

criterion. This means

( )

( )

1

' '

/

0 /

j

i c c

j L

L LX e X j X j

T T T T

LX e L

L

= =

= <

In otherword, ( )jiX e is simply a compressed version of ( )jX e .

The existence of the two distinct frequency bands in ( )jiX e suggests(again) a low pass filtering effect in interpolation. This can be traced back

to the sinc function

sinc n kL

in [ ]ix n .


54/812-54


55/812-55

Exercise: What is the Fourier Transform of the signal

[ ] [ ]sincik

nx n x k k

L

=

=

in general? Do not assume that [ ]x k is obtained from sampling an analog

signal.

2.5.3 Changing the Sampli ng Rate by a Non-I nteger Factor

Suppose we want to change the sampling rate of the system by a factor off . How?

Approximate f as a rational numberL

fM

and then up-sample the signal by a factor of L , followed by down-sampling (the up-sampled signal) by a factor ofM .

[ ]y n[ ]x n

up sampler

L down sampler

M


56/812-56

2.6 Discrete Time Random Processes

A random variable is a parameter whose value can not be predicted exactly.

Associated with a random variable is its probability density function (pdf).

Example 1: Uniform variable x :

1/( )

( ) 0 otherwisexb a a v b

p v

=

Fig: pdf of a uniform random variable.


57/812-57

Example 2: Gaussian random variable x :

( )( )

2

22

1exp

2

x

x

xx

v mp v

=

where

[ ]xm E= xand

( )22

x xE m = x

are respectively the mean and variance of x , with [ ]E beingthe average operator.

Fig:pdf of Gaussian random variables.


58/812-58

Exercise: Determine the mean and variance of a uniform random variable.

Note that[ ]Pr ( )

b

x

a

a b p v dv = xwith

( ) 1x

p v dv

= .

A discrete time signal [ ]x n is a random process when every [ ]x n is arandom variable.

Notations:

1. [ ]n x n=x2.The pdf of nx is ( )nxp v3.The mean of nx is nxm4.The variance of nx is 2nx

If the random process [ ]x n is (wide-sense) stationary, then1. the pdf ( )nxp v is independent of the time index n , and2. the autocorrelation function, defined as


59/812-59

[ ] *, [ ] [ ]xx n m E x n x m = ,

depends only on the time difference n m , i.e.

[ ]

[ ]

*, [ ] [ ]

xx

xx

n m E x n x m

n m

= =

Basically what we are saying is that the first and second other statistics of a

(wide-sense) stationary random process do not depend on absolute time. In

this course, we will focus only on these random processes.

Exercise: Show that[ ] [ ]*xx xxn n =

Subsequently show that for a real random process

( ) ( )j je e =

The autocorrelation function at a time difference of 0n = is

[ ] 2*0 [ ] [ ] [ ]xx E x n x n E x n = =

and is called the average powerof the random process [ ]x n . If the random

process has zero mean, i.e. 0nxm = , then


60/812-60

[ ]22

2

0 [ ] [ ]

,

nxx x

x

E x n E x n m

= = =

where2

x is the variance of [ ]x n .

We will only focus on zero-mean processes in this course.

The autocorrelation function provides information as to how fast a randomprocess varies with time. A fast process will have a relatively narrow

autocorrelation function. The exact frequency contents of a random process

can be obtained from its power spectral density (psd) function, defined as

( ) [ ]j j nxx xxn

e n e

=

=

Since this equation is simply the Fourier transform of [ ]xx n , consequently

1[ ] ( )

2

j j n

xx xxn e e d

=

Note that

1[0] ( )

2

area under psd function

= average power

j

xx xxe d

=

=


61/812-61

White noise:A random process is white if there is no correlation between the values of

the process at different time instants. Mathematically this means

2[ ] [ ]xx x

n n =and

( ) 2

2

[ ] [ ]

j j n j n

xx xx x

n n

x

e n e n e

= =

= =

=

Exercise: Convince yourself that a white noise must have zero mean.

A commonly encountered random process is the white Gaussian noise.Remember, white refers to the spectral shape, and Gaussian refers to

the pdf.

Let the (stationary) random process [ ]x n be the input to a LTI system withan impulse response [ ]h n . Then the output of the system is

[ ] [ ] [ ]k

y n x k h n k

=

= .

We want to examine the statistical properties of the output process.

The mean value of [ ]y n is


62/812-62

[ ]

[ ]

( ) 0

[ ] [ ] [ ]

[ ] [ ]

[ ]

[ ]

ny

k

k

x

k

x

k

j

x

m E y n E x k h n k

E x k h n k

m h n k

m h n k

m H e

=

=

=

=

=

= =

=

=

=

=

,ym=where

( ) [ ]j j mm

H e h m e

=

=

is the frequency response of the system.

Basically, what the result is saying is that if the mean of the input isstationary, then the mean of the output is also stationary.

How about the autocorrelation function?By definition

[ ] *

*

, [ ] [ ]

[ ] [ ] [ ] [ ]

yy

k r

n m E y n y m

E h k x n k h r x m r

= =

=

=


63/812-63

* *

* *

* *

*

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [

k r

k r

k r

E h k x n k h r x m r

E h k h r x n k x m r

h k h r E x n k x m r

h k h

= =

= =

= =

=

=

=

=

[ ]]

[ ]

xx

k r

yy

r n m k r

n m

= =

+

=

Thus the autocorrelation function of the output process is also independent

of absolute time.

Since both the first and second order statistics of the output process areindependent of absolute time, the process is (wide-sense) stationary.

Let d n m= , then the autocorrelation function of the output process canbe written as

[ ] [ ]*

[ ] [ ]yy xxk r

d h k h r d k r

= == + Taking Fourier transform of both sides yields


64/812-64

( ) [ ]

[ ]

[ ]

[ ]

*

* ( )

*

[ ] [ ]

[ ] [ ]

[ ] [ ]

j j d

yy yy

d

j d

xx

d k r

j d k r j r j k

xx

k r d

j m

xx

m

e d e

h k h r d k r e

h k h r d k r e e e

h k h r m e

=

= = =

+

= = =

=

=

= +

= +

=

( )

( ) ( )

( ) ( ) ( )

( ) ( )

*

*

*

2

[ ] [ ]

[ ]

j r j k

k r

j j r j k

xx

k r

j j j k

xx

d

j j j

xx

j j

xx

e e

h k e h r e e

e H e h k e

e H e H e

e H e

= =

= =

=

=

=

=

=

According to an earlier result, if the input process [ ]x n has zero mean,then the output process [ ]y n will also have zero mean. This means thevariance of the output process equals the output power:

( )

( ) ( )

22

2

1[ ] [0]2

1

2

j

y yy yy

j j

xx

E y n e d

H e e d

= = =

=


65/812-65

If ( )jH e is a very narrow band filter centered at c = , then

( ) ( ) ( ){ }2

2 c c cj j j

y xx xxH e e e

+ ,

where 0 is the bandwidth of the filter.

For simiplicity assume a real [ ]x n 1. Then [ ] [ ]xx xxn n = (see an earlierexercise) and ( ) ( )j je e = . Consequently,

( ) ( )22 2 0c cj jy xxH e e

or simply

( ) 0cjxx e .

This property of ( )cjxx e is consistent with the fact that it represents a

power density.

Example: The input/output relationship of a LTI system is given by[ ] [ 1] [ ]y n ay n x n= +

where a is a positive constant less than 1. Determine the psd and the pdf of[ ]y n if [ ]x n is a white Gaussian process with zero mean and unit

variance.

1The proof is a bit lengthy for complex [ ]x n but the end result is the same.


66/812-66

Solution 1

- Since the input has zero mean, the output must also have zero mean.- Since the input is Gaussian, the output must also be Gaussian.- Taking the expectation of the square of both sides of [ ]y n yields

[ ]

2 2 2

2 2 2

2 2 2

2 2 2

[ ] ( [ 1] [ ])

[ 1] [ ] 2 [ 1] [ ]

[ 1] [ ] 2 [ 1] [ ]

2 [ 1

y

y x

E y n E ay n x n

E a y n x n ay n x n

a E y n E x n aE y n x n

a aE y n

= = + = + +

= + + = + + [ ] [ ]

2 2 2

] [ ]

y x

E x n

a = +or

22

2 2

1

1 1

xy

a a

= =

- The pdf of [ ]y n is simply

( )2

22

1exp

2y

yy

vp v

=

- The output [ ]y n can be expressed recursively as

( )

[ ] [ 1] [ ]

[ 2] [ 1] [ ]

y n ay n x n

a ay n x n x n

= +

= + +


67/812-67

2

3 2

1

0

[ 2] [ 1] [ ]

[ 3] [ 2] [ 1] [ ]

[ ] [ ]m

m k

k

a y n ax n x n

a y n a x n ax n x n

a y n m a x n k

=

= + +

= + + +

= + Multiplying both sides by [ ]y n m , 0m > , and taking expectationyields

[ ]

[ ]

1

0

1

0

12

0

[ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [ ] [ ]

mm k

k

mm k

k

mm k

k

E y n y n m E a y n m a x n k y n m

E a y n m y n m a x n k y n m

a E y n m a E x n k y n m

=

=

=

= +

= +

= +

2

[ ]

m

y

yy

a

m

=

=

Since the process [ ]y n is real, [ ] [ ]yy yym m = . So in general

2[ ]m

yy ym a =

- The psd is

( )

2

[ ]

j j myy yy

m

m j m

y

m

e m e

a e

=

=

=

=


68/812-68

( ) ( ) ( )( )( )

02 2 2

0

2 2 2

0 0

2 2

2

2 2 2

1 1

1 1 1 1

1 1

m mj m j m

y y y

m m

n j n m j m

y y y

n m

y y

yj j

j j j j

y y y

j

a e a e

a e a e

ae ae

ae ae ae ae

ae a

= =

= =

= +

= +

= +

+ =

( )

( )

( )

( )

2 2

2

2

1

1 2 cos

1

1 2 cos

j

y

e

a

a a

a a

=

+ =

+

Solution 2

-

The LTI has an impulse response of

[ ] [ ]nh n a u n=

where [ ]u n is the unit-step function.

- The corresponding frequency response is

( )0

1[ ]

1

j j n n j n

jn n

H e h n e a eae

= =

= = =

This means


69/812-69

( )

( ) ( )

2

2

2 2

2

1

1

1

1 cos( ) sin( )

1

1 2 cos( )

j

jH e

ae

a a

a a

=

=

+

=+

- Since [ ]x n is white and has a variance of unity, its psd is simply

( ) 2 1jxx xe = =

- The psd of [ ]y n is

( ) ( ) ( )2

2

1

1 2 cos( )

j j j

yy xxe e H e

a a

= =

+

2.7 Linear Predictiive Coding and the Autocorrelation Function

Consider the last example in Section 2.6 where we have a random process[ ]y n governed by the equation

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


70/812-70

Here a is a positive constant less than 1 and [ ]x n is a white process withzero mean and unit variance. As shown in Section 2.6, the variance of

[ ]y n is related to the variance of [ ]x n according to the equation

2

22 2

11 1

xy

a a = =

Suppose [ ]y n is a sampled-speech signal and we want to send this speechsignal digitally through a communication channel. A simple way is to

quantize each sample into yB bits and feed the resultant bit stream to a

digital modulator to generate the transmitted signal

This encoding method, which is referred to as Pulse Code Modulation(PCM) in the literature, is not very efficient because [ ]y n has a relatively

large dynamic range, at least compared to [ ]x n .

Suppose the parameter parameter a (which will vary from speaker tospeaker) is known to both the encoder and the decoder. Then a more

efficient encoding scheme can be obtained by first subtracting

[ ] [ 1]py n ay n= from [ ]y n to obtain

[ ]y n bit

Quantizer

yB Digital

Modulator


71/812-71

[ ] [ ] [ ] [ ] [ 1]p

x n y n y n y n ay n= =

and then quantize [ ]x n using a xB -bit quantizer. The resultant bits,

together with the parameter a (also in quantized form), are then sent to the

receiver. Upon receiving these information, the decoder can approximate[ ]y n according to

[ ] [ 1] [ ]y n ay n x n= +% % ,

where [ ]x n% and a% are respectively the quantized versions of [ ]x n and a .

This encoding scheme, known as linear predictive coding(LPC), is more

efficient than PCM because [ ]x n has a smaller dynamic range than [ ]y n .

In the speech coding literature, [ ] [ 1]py n ay n= is referred to as the

predicted value of [ ]y n and [ ] [ ] [ ]px n y n y n= is called the residual orexcitation.

[ 1]ay n

+ [ ]x n[ ]x n

( )

Speech Model

1

1

j

jH e

ae

=

( )

Prediction Filter

j jF e ae =

[ ]y n

Eqv. response = 1j

ae


72/812-72

As shown earlier, the autocorrelation function of [ ]y n is 2[ ] myy xm a = .This means

[1] [0]yy yya =or

[1]=

[0]

yy

yy

a

So as long as [0] and [1]yy yy are known, then the parameter a will be

known to both the encoder and decoder. In practice, [0] and [1]yy yy can be

estimated according to

1

1 [ ] [ ] [ ].N d

yy

k

d y n y n d N

=

= +

This suggests that in order to implement the LPC encoder, the signal [ ]y n

must first be analyzed to obtain its autocorrelation function. Once the

autocorrelation function is estimated, then the information will be used to

obtain the prediction filter at the encoder and the synthesizing filter at the

decoder.

[ 1]ay n

[ ]y n[ ]x n%[ ]x n

[ ]Q

Quantizer1az

Synthesizing filter


73/812-73

As the parameter 1a , the variance of [ ]y n is going to much greater thanthat of [ ]x n . Consequently LPC will be much more efficient than PCM.

As 0a , the variance of [ ]y n approaches that of [ ]x n and LPC provideslittle improvement to the encoding efficiency.

Since 0a implies little correlation between samples in [ ]y n while1a corresponds to high correlation, we can conclude that LPC achieves

a higher encoding efficiency because it removes the redundancy in [ ]y n

through linear prediction.

The expression [ ] [ 1] [ ]y n ay n x n= + represents a first-order model. Ingeneral, speech signals can be modeled accurately using aN-th order model

(typical value ofNis 10):

1

[ ] [ ] [ ]N

k

k

y n a y n k x n=

= +

where [ ]x n is a white noise. The model also represents a system described

by a Linear Constant Coefficient Difference Equation with [ ]x n being the

excitation and [ ]y n being the output.

For theN-th order model, the prediction filter in the encoder computes

1

[ ] [ ]N

p k

k

y n a y n k=

=

and subtract it from [ ]y n to obtain the residual (excitation)


74/812-74

1

[ ] [ ] [ ] [ ] [ ]N

p k

k

x n y n y n y n a y n k=

= =

The residual, as well as the filter coefficients,

1

2

N

a

a

a

=

AM

are then quantized and transmitted to the receiver. Upon receiving these

information, the decoder synthesizes the original speech signal according to

1

[ ] [ ] [ ]N

k

k

y n a y n k x n=

= + % % ,

where [ ]x n% and ka% are respectively the quantized versions of [ ]x n and

ka .

How to determine linear predictorA? If we multiply both sides of [ ]y n by[ 1]y n and taking average, we obtain

[ ]

[ ] [ ]

1

1

1

[ ] [ 1] [ ] [ 1] [ ] [ 1]

[ ] [ 1] [ ] [ 1]

[ 1]

N

kk

N

k

k

N

k yy

k

E y n y n E a y n k y n x n y n

a E y n k y n E x n y n

a k

=

=

=

= +

= +

=


75/812-75

or simply

1

[1] [ 1]N

yy k yy

k

a k =

=

Similarly if we multiply both sides of [ ]y n by [ 1]y n and takingaverage, we obtain

[ ]

[ ] [ ]

1

1

1

[ ] [ 2] [ ] [ 2] [ ] [ 2]

[ ] [ 2] [ ] [ 2]

[ 2]

N

k

k

N

kk

N

k yy

k

E y n y n E a y n k y n x n y n

a E y n k y n E x n y n

a k

=

=

=

= +

= +

=

or

1

[2] [ 2]N

yy k yy

k

a k =

=

It is evident that in general

1

1

[ ] [ ];

; 1,2,...,

N

yy k yy

k

N

k yy

k

m a k m

a k m m N

=

=

=

= =


76/812-76

These equations can be written in matrix form as2

1

2

3

1

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

[1] [0] [1] [ 3] [ 2] [2]

[2] [1] [0] [ 4] [ 3] [3

[ 2] [ 3] [ 4] [0] [1]

[ 1] [ 2] [ 3] [1] [0]

N

N

aN N

aN N

aN N

aN N N

aN N N

=

L

L

LMM M M M M M

L

L

]

[ 1]

[ ]

N

N

M

or

N N N=U A V ,

where

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

[1] [0] [1] [ 3] [ 2]

[2] [1] [0] [ 4] [ 3]

[ 2] [ 3] [ 4] [0] [1]

[ 1] [ 2] [ 3] [1] [0]

N

N N

N N

N N

N N N

N N N

=

U

L

L

L

M M M M M M

L

L

,11

,22

,33

, 11

,

N

N

N

N

N NN

N NN

aa

aa

aa

aa

aa

=

=

AMM and

[1]

[2]

[3]

[ 1]

[ ]

N

N

N

=

VM

2For convenience, we drop the subscript yy in [ ]yy m .


77/812-77

So the coefficients for the prediction filter can be obtained by solving

1

N N N

=A U V ,

provided that the autocorrelation function [ ]m is known.

A brute-force computation of NA results in a complexity in the order of( )3O N . However, since the matrix is Toeplitz, i.e. all elements along any

diagonal are identical, a more efficient algorithm called the Levison and

Durbin (LD) algorithm can be used.

The LD algorithm only has a complexity of only ( )2O N .

The LD algorithm is a order-recursive algorithm, i.e. the m -th orderpredictor mA can be obtained from the ( 1)m -th predictor 1mA .

Specifically, we rewrite mA

as

,1

,2 11

,

0

m

m mm

m

m

m m

a

a

k

a

= = +

dAA

M (1)

where the vector 1md and the scaler mk are quantities to be determined.

The covariance matrix mU itself can be written in terms of 1mU and 1mVas:


78/812-78

( )

1 1

1[0]

r

m m

tm r

m

=

U VU

V(2)

where

1

[ 1]

[ 2]

[ 3]

[2]

[1]

r

m

m

m

m

=

VM (3)

is the correlation vector 1mV arranged in reverse order, and [ ]t

stands for

the transpose of a matrix.

Substituting (1)-(3) into the equation m m m=U A V implies

( )

1 1 11 1

10 [ ][0]

r

m m mm m

tr

mmk m

+ =

U V dA V

V(4)

One of the equation we can obtain from (4) is1 1 1 1 1 1

1 1 1 1

r

m m m m m m m

r

m m m m m

k

k

= + +

= + +

V U A U d V

V U d V

or


79/812-79

1

1 1 1

r

m m m mk

= d U V (5)

Since 1 1 1m m m =V U A ,

1 1

1 1

1 1

1 1

rm m

m m

r

m m

r

m m

==

=

=

V PV

PU A

PU PA

W A

, (6)

where

0 0 1

0 1 0

1 0 0

=

P

L

L

M M M M

L

is the permutation matrix representing vector reversal, and

1 1m m =W PU P

Because of the property of 1mU ,

1 1m m =W U (7)

Combining (5)-(7) implies

1

1 1 1

1

1 1 1

1

1 1 1

1

,

r

m m m m

r

m m m m

r

m m m m

r

m m

k

k

k

k

= =

=

=

d U V

U W A

U U A

A

(8)


80/812-80

i.e. the vector 1md is the vector containing the coefficients of the ( 1)m -th

predictor arranged in reverse order and scaled by the term mk .

The second equation we can derive from (4) is( ) ( )

( ) ( )

1 1 1 1

1 1 1 1

[ ] [0]

[0]

t tr r

m m m m m

t tr r r

m m m m m m

m k

k k

= + +

= +

V A V d

V A V A

or

( )

( )

( )1 1 1 1

11 1

[ ] [ ]

[0]

t tr r

m m m m

m tr r

mm m

m mk

= =

V A V A

V A(9)

where

( )

( )

1 1 1

1 1

[0]

[0]

tr r

m m m

t

m m

=

=

V A

V A (10)

In summary,1.At the end of the (m-1)-th iteration, the available information to the LPC

encoder are 1m and the (m-1)-th order predictor 1mA .

2.In the m-th iteration, the encoder computes mk according to (9) and setthe highest order term in the m-th order predictor to (see Eqn. (1))

,m m ma k=


81/81

3.The other predictor coefficients are computed according to (1) and (8) as, 1, 1,

1, 1, ; 1,2,..., 1

m k m k m k

m k m m m k

a a d

a k a k m

= +

= = , (11)

where 1,m kd is the k-th component of the vector 1md in (8)

4.Update the term m and increase m by 1.

Exercise: Show that the term m can also be calculated recursively.

Equations (9)-(11) each has a computational complexity of 1m multiplication. Summing over all m from 1 toNyields a complexity of

( ) 21

3 1 3 ( 1) / 2N

m

m N N N =

=

Note that LPC-based speech codec can produce communication qualityspeech at a rate as low as 2.4 kbps (though 4.8-9.6 kbps are more typical).

Thi i h l th th 64 kb i d i PCM b d d

Chapter 2 - Discrete Time Signals and Systems

Documents