Oct'04 CS3291 1

CS3291 : Digital Signal Processing '04-05

Section 3 : Discrete-time LTI systems

3.1 Introduction:

A discrete time system takes discrete time input signal { x[ n ] }, & produces an output signal { y[ n ] }.

SYSTEM input {x[n]}

output {y[n]}

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• Implemented by general purpose computer, microcomputer or

dedicated digital hardware capable of carrying out arithmetic

operations on samples of {x[n]} and {y[n]}.

• {x[n]} is sequence whose value at t=nT is x[n].

• Similarly for {y[n]}.

• T is sampling interval in seconds.

• 1/T is sampling frequency in Hz.

• {x[n-N]} is sequence whose value at t=nT is x[n-N].

{x[n-N]} is {x[n]} with every sample delayed

by N sampling intervals.

• Many types of discrete time system can be considered, e.g.

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(i) Discrete time "amplifier" with output:

y[n] = A . x[n].

• Described by "difference equation": y[n] = A x[n].

• Represented in diagram form by a "signal flow graph” .

x[n] y[n] A

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(ii) Processing system whose output at t=nT is obtained by weighting & summing present & previous input samples:

y[n] = A0 x[n] + A1 x[n-1] + A2 x[n-2] + A3 x[n-3] + A4x[n-4]

‘Non-recursive difference equatn’ with signal flow graph below. Boxes marked " z -1 " produce a delay of one sampling interval.

z-1 z-1 z-1 z-1 x[n]

A0 A1 A2 A3 A4

y[n]

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(iii) System whose output y[n] at t = nT is calculated according to the following recursive difference equation: y[n] = A 0 x[n] - B 1 y[n-1]

whose signal flow graph is given below. • Recursive means that previous values of y[n] as well as present & previous values of x[n] are used to calculate y[n].

z-1

x[n] A0

B1

y[n]

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(iv) A system whose output at t=nT is: y[n] = (x[n]) 2

as represented below.

x[n]

y[n]

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3.2. LTI Systems:

To be LTI, a discrete-time system must satisfy:

(I) Linearity ( Superposition ):

•Given any two discrete time signals {x 1 [n]} & {x 2 [n]}, if {x 1 [n]} {y 1 [n]} & {x 2 [n]} {y 2 [n]} then for any values of k 1 and k 2 , response to k 1{x 1[n]} + k 2{x 2[n]} must be k 1{y1[n]} + k 2 {y 2 [n]} .

•To multiply a sequence by k, multiply each element by k, k{x[n]} = {k x[n]}. •To add two sequences together, add corresponding samples, {x[n]} + {y[n]} = {x[n] + y[n]}.)

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(II) Time-invariance :

• Given any discrete time signal {x[n]}, if response to {x[n]} is {y[n]}, response to {x[n-N]} must be {y[n-N]} for any N.

• Delaying input by N samples only delays output by N samples.

• Examples (i), (ii) and (iii) are LTI whereas (iv) is not LTI .

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3.3. Impulse-response:

Useful to consider response of LTI systems to a discrete time unit impulse , or in short an impulse denoted by {d [n] } with:

0 n : 0

0 n : 1d[n]

... ...

1

-3 -2 -1 1 2 3 4 5

n

0

d[n-4]

... ...

1

-3 -2 -1 1 2 3 4 5

d[n]

n

0

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• {d[n-N]} is delayed impulse where the only non-zero sample occurs at n=N rather than at n=0. • When input is {d[n]}, output is impulse response {h[n]}.

• If impulse-response of an LTI system is known, its response to any other input signal may be predicted.

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3.4. Computing an impulse response:

• Consider a discrete time system with signal-flow graph below with A1, A2, A3, A4, A5 set to specific constants. • Notice the labels X1, X2, etc.

• Realised by microcomputer running a program with flow-diagram & high-level language implementation as listed next.

z-1 z-1 z-1 z-1 x[n]

A1 A2 A3 A4 A5

y[n]

X1

Y

X5X4X2 X3

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Zero X1, X2, X3, X4,X5

Set A1,A2,….

INPUT X

Y = A1*X + A2*X2 + A3*X3 +...

OUTPUT Y

X5 = X4; X4 = X3; X3 = X2; X2 =X1

clear all;

A1=1; A2=2; A3=3; A4=-4; A5=5;

x1=0; x2=0; x3=0; x4=0; x5=0;

while ~isempty(X)

X = input( 'X = '); x1 = X;

Y= A1*x1 + A2*x2 + A3*x3

+ A4*x4 + A5*x5 ;

disp([' Y = ' num2str(Y)]);

x5 = x4 ; x4 = x3;

x3 = x2 ; x2 =x1;

end;

% MATLAB program

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Here is another version using MATLAB arrays:

A = [1 2 3 -4 5 ] ;x = [0 0 0 0 0 ] ; while 1 X = input( 'X = '); x(1)=X; Y=A(1) * X; for k = 2 : 5 Y = Y + A(k)*x(k); end; disp([' Y = ' num2str(Y)]); for k=5:-1:2 x(k) = x(k-1); end;end;

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And an even more efficient version

A = [1 2 3 -4 5 ] ;x = [0 0 0 0 0 ] ; while 1

x(1) = input( 'X = ');Y = A(1)*x(1);for k = 5 : -1: 2

Y = Y + A(k)*x(k);x(k) = x(k-1);

end;disp([' Y = ' num2str(Y)]);

end;

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• These can be run in MATLAB

• The ‘while 1’ statement initiates an infinite loop.

• Program will run for ever

or until interrupted by ‘CONTROL+C’

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• Impulse-response for this example may be deduced by running

program & entering values for X on the keyboard :

0, 0, 0, 1, 0, 0, 0, 0,.... • Sequence of output samples printed out will be :

0, 0, 0, A1, A2, A3, A4, A5, 0, 0, .... • Impulse-response can also be obtained by tabulation (later). • Output must be zero until input becomes equal to 1 at n=0• Impulse response is:

{..., 0, ..., 0, A1, A2, A3, A4, A5, 0, 0, ... ,0,...}

where the sample at n=0 is underlined. • Only five non-zero output samples are observed.• This is a " finite impulse-response ” (FIR).

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Exercise 3.1: Calculate impulse-responses for difference equations:

(i) y[n] = x[n] + 2 x[n-1] + 3 x[n-2] - 4 x[n-3] + 5 x[n-4]

(ii) y[n] = 4 x[n] - 0.5 y[n-1]

Solutions: • If computer not available we may use a tabulation. • Difference eqn (i) will produce a finite impulse-response.• Difference eqn (ii) produces infinite response whose samples gradually reduce in amplitude but never quite become zero.

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(i) y[n] = x[n] + 2 x[n-1] + 3 x[n-2] - 4 x[n-3] + 5 x[n-4]

n x[n] x[n-1] x[n-2] x[n-3] x[n-4] y[n]-1 0 0 0 0 0 00 1 0 0 0 0 11 0 1 0 0 0 22 0 0 1 0 0 33 0 0 0 1 0 -44 0 0 0 0 1 55 0 0 0 0 0 0: : : : : : :

Solution is:- {.. 0, .., 0, 1, 2, 3, -4, 5, 0, .., 0, ...}

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Difference equation (ii): y[n] = 4 x[n] - 0.5 y[n-1]

n x[n] y[n-1] y[n]0 1 0 41 0 4 -22 0 -2 13 0 1 -0.54 0 -0.5 0.255 0 0.25 -0.125: : : : Solution is: {.., 0, .., 0, 4, -2, 1, -0.5, 0.25, -0.125, ...)

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3.5. Digital filters:

•A "digital filter” is a digitally implemented LTI discrete time system governed by a difference equation of finite order; e.g. :

(i) y[n] = x[n] + 2 x[n-1] + 3 x[n-2] - 4 x[n-3] + 5 x[n-4]

(ii) y[n] = 4 x[n] - 0.5 y[n-1]

• Difference equation (i) is "non-recursive" & produces a finite impulse response (FIR). • Difference equation (ii) is " recursive " . • Impulse-response of a recursive difference equation can have an infinite number of non-zero terms. In this case it is an infinite impulse-response (IIR).

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• Next slide has yet another way of programming difference equations (i.e. digital filters) in MATLAB.

• But this time, we use a MATLAB function and operate on a whole array of music at once.

• You can find the music file ‘cap4th.wav’ in

•www.cs.man.ac.uk/~barry/mydocs/CS3291/MATLAB

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MATLAB function ‘filter’ used for non-recursive diffce eqn

y = filter(A, 1, x) filters signal in array x to create array y.

For each sample x(n) of array x:

y(n) = A(1)*x(n) + A(2)*x(n-1) + ... + A(N+1)*x(n-N)

Example (i): y[n] = x[n] + 2 x[n-1] + 3 x[n-2] - 4 x[n-3] + 5 x[n-4]

[x, fs, nbits] = wavread ('cap4th.wav');sound(x,fs,nbits);pause;y = filter ( [1 2 3 -4 5], 1, x);sound(y,fs,nbits);

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Use of ‘filter’ for recursive difference equatn

y = filter(A, B, x) filters signal in array x to create array y

using a recursive difference equation.

Example (ii): y[n] = 4 x[n] - 0.5 y[n-1]

[x, fs, nbits] = wavread ('cap4th.wav');sound(x,fs,nbits);pause;y = filter ( [4], [1 0.5], x);sound(y,fs,nbits);

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3.6. Discrete time convolution:

If impulse-response of an LTI system is {h[n]} its response to any input {x[n]} is an output {y[n] } whose samples are given by the following "convolution" formulae:

m

m] x[nh[m] y[n]

k

k]h[n x[k]y[n]or

Consider the response of a system with impulse response: {h[n]} = { ..., 0,..., 0, 1, 2, 3, -4, 5, 0, .....0, .... }

to {x[n]} = { ... 0, ... , 0 , 1, 2, 3, 0, ..., 0, ....}

4]5x[n3]4x[n2]3x[n1]2x[n x[n] m] x[nh[m] y[n]4

0m

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• Result is not too surprising.

• It is the difference equation for an LTI system whose impulse

response is {.., 0, .., 0, 1, 2, 3, -4, 5, 0, .., 0, ..}.

• Program discussed earlier implements this difference equation,

•We could complete the example by running it, entering samples of

{ x[n] }, and noting the output sequence.

• Alternatively, we could use tabulation as follows:

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y[n] = x[n] + 2x[n-1] + 3x[n-2] - 4x[n-3] + 5x[n-4]

n x[n] x[n-1] x[n-2] x[n-3] x[n-4] y[n]: : : : : : :-1 0 0 0 0 0 00 1 0 0 0 0 11 2 1 0 0 0 42 3 2 1 0 0 103 0 3 2 1 0 84 0 0 3 2 1 65 0 0 0 3 2 -26 0 0 0 0 3 157 0 0 0 0 0 0: : : : : : :{y[n]} = { .... 0, ....., 0, 1, 4, 10, 8, 6, -2, 15, 0, ...., 0, ....}

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3.7. Proof of discrete time convolution formulae:

Only use defns of linearity, time-invariance & impulse-response. Since d[n-m] is non-zero only at n = m, given any sequence {x[n]},

m

m]x[m]d[nx[n]

m]}[m]{d[nx{x[n]}

m

{x[n]} is sum of infinite number of delayed impulses{d[n-m]} each multiplied by a single element, x[m]. Response to {d[n-m]} is {h[n-m]} for any value of m. response to {x[n]} is :

n allfor m][m]h[nxy[n] i.e. m]}[m]{h[nx{y[n]}

mm

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• Replacing n-m by k gives the alternative formula. • Study the graphical explanation of this proof in Section 3.17.

n

2

3

x[n]

2

h[n]

n

{h[n]} = { ..., 0, ..., 0, 1, 2, 1, 2, 0, ..., 0...}

{x[n]} = { ..., 0, ..., 0, 1, 2, 3, 0, 0, ..., 0, ...}

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3.17 Discrete time convolution graphically

n

23

x[n]

d[n]

n

2

2d[n-1]

n

3

3d[n-2]

n

2

h[n]

n

2

42h[n-1]

n

3

63h[n-2]

n

{h[n]} = { ..., 0, ..., 0, 1, 2, 1, 2, 0, ..., 0,...}

{x[n]} = { ..., 0, ..., 0, 1, 2, 3, 0, 0, ..., 0, ...}

Oct'04 CS3291 30

2

4

6

8

10

y[n]

n

Express:

{x[n]} = {d[n]} + 2 {d[n-1]} + 3 {d[n-2]}

Response is:

{y[n]} = {h[n]} + 2 {h[n-1} + 3 {h[n-2]}

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2

4

6

8

10y[n]

n

n

2

3

x[n]

{ d[n] } {h[n]}

+ 2{ d[n-1 ]} + 2 { h[n-1]}

+ 3 { d[n-2] } + 3 { h[n-2] }

{h[n]} = { ..., 0, ..., 0, 1, 2, 1, 2, 0, ..., 0,...}

{x[n]} = { ..., 0, ..., 0, 1, 2, 3, 0, 0, ..., 0, ...}

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d[n]

n

2

2d[n-1]

n

3

3d[n-2]

n

2

4

2h[n-1]

n

3

63h[n-2]

n

2

h[n]

n

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This means that {h[n]} must be either an FIR or an IIR whose samples decay towards zero as n .

3.9. Causality:

An LTI system operating in real time must be "causal " which

means that its impulse-response {h[n]} must satisfy:

h[n] = 0 for n < 0.

Non-causal system would need “crystal ball ” to predict future.

3.8. Stability:

An LTI system is stable if its impulse-response {h[n]} satisfies:

finite is ][

n

nh

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• = T is “ relative" frequency of sampled sine-wave. • Units of are 'radians per sample'. • To convert back to true frequency (radians/s ) multiply by 1/T.

• Note: 1/T = sampling frequency, f S, in samples/second (Hz).

‘radians / sample’ times ‘samples / second’ = ‘radians / second’

•Normally process analogue signals in range 0 to f S /2 ( = 1/(2T) ),

• Restricts to the range 0 to .

3.10. Relative Frequency

• Study effect of digital filters on sine-waves of different frequencies.• Discrete time sine-wave obtained by sampling A cos(t + ). • If sampling period is T seconds, resulting discrete time signal is :

)} cos( A{ )} cos( A{ n Tn

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T

3T-T

-3T 4T-4T

cos( t )

1

3-1

-3 4-4

cos( T n )

t

n

= 2 / 8

=2 / (8T)

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Table gives values of & corresponding true frequencies :-

Relative frequency True frequency(radians/sample) (radians/s) (Hz)

0 0 0/6 fS/6 fS/12/4 fS/4 fs/8/3 fS/3 fs/6/2 fS/2 fs/42/3 2fS/3 fs/3 fS fS/2

‘radians / sample’ ‘samples / second’ = ‘radians / second’

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3.11. Relative frequency response

Response to sampled sine-waves calculated by defining:

)}sin(){cos(}{ njne nj

If this is applied to a system with impulse response {h[n]}, output would be {y[n]} with :

(DTFT) ][ ][)( where

)( ][

][][][ )(

n

nj

m

mjj

njj

m

njnj

m m

mjnjmnj

enhemheH

eeHemhe

eemhemhny

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n

n

1

-1-2

3

1

3-1

-3 4-4

cos( (2/8) n )

sin( (2/8) n)

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• H( e j ) is the " relative frequency response " of LTI system.

• Complex number for any value of .

• It is DTFT of { h[n] }.

• Note similarity to the analogue Fourier transform.

(FT) )()(

(DTFT) ][)(

atj

a

n

njj

ethjH

enheH

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• If input is {e j n }, output is same sequence with each

element multiplied by H( e j ).

•When { h[n] } is real, H( e -j ) = H*( e j ).

•Exercise: Prove this.

{e j n }

{h[n]}

{ e j n H(e j n ) }

n

njj enheH ][)(

LTI

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• Remember the analogue case:-

• When h(t) is real, Ha(j ) = Ha*(j )

e j t

ha(t)

e j t Ha(j )LTI

)()(atj

a ethjH

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3.12. Gain and phase responses

• Expressing H( e j ) = G() e j ()

• G() = |H( e j )| is "gain ” • () = arg ( H( e j ) ) is "phase lead”. • Both vary with .

• If input is {A cos(n)}, • output is: { G()A cos(n + ()) } Exercise: Prove this

•When input is sampled sine-wave of rel frequency , output is sine-wave of same frequency , but with amplitude scaled by G() & phase increased by ().

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Gain & phase response graphs again

/4 /2 3/40

20log10[G()] dB-()

-()

G() in dB

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•Common to plot graphs of G() & () against . •G() often converted to dBs by calculating 20 log10( G() ). •Restrict to lie in range 0 to & adopt a linear horizontal scale.

•Example 3.2: Calculate gain & phase response of FIR digital filter in Figure 3.3 where A1 = 1, A2 = 2, A3 = 3, A4 = -4, A5 = 5 .•Solution:Impulse response is: {.., 0, .., 0, 1, 2, 3, -4, 5 ,0, .., 0,..}. By the formula established above, H( e j ) = 1 + 2 e - j + 3 e -2 j - 4 e -3 j + 5 e - 4 j

To obtain the gain and phase at any take modulus and phase. Best done by computer, either by writing & running a simple program, or by running a standard package. A simple program is given at the end of these notes.

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• Can use a spread-sheet such as excel (see lecture notes)

• Another way of calculating gain & phase responses for H( e j ) = 1 + 2 e - j + 3 e -2 j - 4 e -3 j + 5 e - 4 j

is to start MATLAB and type:

A = [ 1 2 3 -4 5 ]; freqz( A );

• See later section 3.18 for graphs produced.

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• Produces graphs of gain & phase responses of H( ej ).

• Frequency scale normalised to fS/2 & labelled 0 to 1 instead of 0 to . • See Section 3.18

• In next section we calculate coeffs of an FIR filter to achieve a particular type of gain & phase response.

• To calculate gain & phase responses for an IIR filter would be difficult by method used above because impulse-response has infinite number of non-zero terms.

•There is an easier method in a later section.

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Phase delay

• Assume input is {A cos(n)},

• Output is: { G() A cos ( n + () ) }

= { G() A cos ( [ n + () / ] ) }

= { G() A cos ( [ n - (-() / ) ] ) }

•Phase lead () delays sine-wave by -()/ sampling intervals

• This is ‘phase-delay’.

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3.13. Linear phase response

• If -()/ remains constant for all values of ,

i.e. if -() = k for constant k,

system has ‘linear phase’ response.

• Graph of -() against on linear scale would then be

straight line with slope k where k is "phase delay"

i.e. the delay measured in sampling intervals.

• This need not be an integer.

Oct'04 CS3291 49

-()/

/4 /2 3/40

-()

G() in dB

-()/

Gain, phase & ‘phase-delay’ response graphs

Not ‘linear-phase’.

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-()/

/4 /2 3/40

-()

G() in dB

-()/

Gain, phase & ‘phase-delay’ response graphs again

This is ‘linear-phase’ as -()/ is constant.

k

Oct'04 CS3291 51

• Linear phase systems of special interest since all sine-waves are

delayed by same number of sampling intervals.

• Input signal expressed as Fourier series will have all its

sinusoidal components delayed by same amount of time.

• In absence of amplitude changes, output signal will be exact

replica of input.

•LTI systems are not necessarily linear phase.

• A certain class of digital filter can be made linear phase.

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3.14. Inverse DTFT

• Frequency -response H( e j ) is DTFT of {h[n]}.

• Inverse DTFT formula allows {h[n]} to be deduced from H(e j ) :

<<-: )(2

1][ ndeeHnh njj

• Observe similarities with analogue Fourier transform:

(i) (1/2) factor, (ii) sign of jn (n replaces t). (iii) variable of integration (d)

<<-: )(

2

1)( tdejHth tj

aa

Oct'04 CS3291 53

Also note that with DTFT:

• Negative frequencies are again included.

• Range of integration is now - to rather than - to .

• Forward transform is a summation

& inverse transform is an integral.

• This is because {h[n]} is a sequence

whereas H(e j ) is function of the continuous variable .

Oct'04 CS3291 54

3.15 Problems:1. Show that y[n]=(x[n]) 2 is not LTI.2. If {h[n]}= { .. 0, .. , 0, 1, -1, 0, .. 0, .. }, calculate response to {x[n]} = { .... 0, .., 0, 1, 2, 3, 4, 0, .., 0, ... }.3.. Produce a signal flow graph for each of the following difference equations: (i) y[n] = x[n] + x[n-1] (ii) y[n] = x[n] - y[n-1]4. For each difference equation in question 3, determine the impulse response, & deduce whether the system it represents is stable and/or causal..5. Calculate, by tabulation, the output from difference equation (ii) in question 3 when the input is the impulse response of difference equation (i).6. If fS =8000 Hz, what true frequency corresponds to /5 radians/sample?7. Sketch the gain & phase responses of the system referred to in question 2. (You don't need a computer program.) Is it a linear phase system?8. Calculate the impulse response for y[n] = 4x[n] - 2y[n-1]. Is it stable?9. Show that input {cos(n)} produces an output {G()cos(n+())}.10. For y[n]=x[n]+2x[n-1]+3x[n-2]+2x[n-3]+x[n-4], sketch phase response and comment on its significance. Show that () = -2.

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3.16. Progam for gain & phase responses of FIR digital filter. (using complex arithmetic)

fs = input('enter fs:'); M = input('Enter order:'); disp('Enter coefficients:');for N = 0:M C(N+1)=input('coeff = '); end;disp(' FREQ(Hz) GAIN(dB.) PHASE LEAD(deg)');K = 0; F =0.0; % F is frequency in Hz}while F < fs/2 H=0; W =2*pi*F/fs; for N=0:M H=H+C(N+1)*exp(-j*W*N); end; MD=abs(H); PH = angle(H); GAIN(K+1)=-999; if MD>0 GAIN(K+1)=20*log10(MD);end; PHAS(K+1)=PH*180.0/pi; disp([num2str(F) ' 'num2str(GAIN(K+1)) ' 'num2str(PHAS(K+1))]); F=F+fs/200; K=K+1;end; %while loop

Oct'04 CS3291 56

3.16. Progam for gain & phase responses of FIR digital filter. (without complex arithmetic)fs = input('enter fs:'); M = input('Enter order:'); disp('Enter coefficients:');for N = 0:M H(N+1)=input('coeff = '); end;disp(' FREQ(Hz) GAIN(dB.) PHASE LEAD(deg)');K = 0; F =0.0; % F is frequency in Hz}while F < fs/2 Hre =0.0; Him =0.0; W =2*pi*F/fs; for N=0:M Hre=Hre+H(N+1)*cos(W*N); Him=Him-H(N+1)*sin(W*N); end; MD=sqrt(Hre*Hre+Him*Him); PH = atan2(Him,Hre); GAIN(K+1)=-999; if MD>0 GAIN(K+1)=20*log10(MD);end; PHAS(K+1)=PH*180.0/pi; disp([num2str(F) ' 'num2str(GAIN(K+1)) ' 'num2str(PHAS(K+1))]); F=F+fs/200; K=K+1;end; %while loop

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Sample run:- FIR FREQUENCY RESPONSE Enter sampling frequency(Hz): 1000 Enter order of FIR filter: 3 Enter coefficients:- H[0] = 1 H[1] = 2 H[2] = 1 H[3] = 2

Output obtained:-

Freq(Hz) Gain(DB) Phase(deg) 0 15.56 0.0 5 15.56 -3.0 10 15.56 -6.0 : : : 500 6.01 -180.0 }

Oct'04 CS3291 58

3.18. Frequency response of FIR digital filter by MATLAB

• Given discrete time LTI system with difference equation:y[n] = x[n] + 2 x[n-1] + 3 x[n-2] - 4 x[n-3] + 5 x[n-4] • Its impulse response is: { ... , 0, ... , 0, 1 , 2, 3, -4, 5, 0, ..., 0, ... }• Its frequency response is:H( e j ) = 1 + 2 e - j + 3 e -2 j - 4 e -3 j + 5 e - 4 j

• To calculate its gain and phase responses, type:

freqz ( [ 1 2 3 -4 5 ]) ;

• Resulting graphs have a normalised frequency scale with “1” corresponding to = radians/sample; i.e. one half of the sampling frequency.

Oct'04 CS3291 59

Oct'04 CS3291 60