LTI systems3studs

7/28/2019 LTI systems3studs

1/23

LTI systems: Convolution and

difference equations


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8/23

In Matlab

>> x = [ 3 2 1]x =

3 2 1

>> h = [ 2 1 3]

h =

2 1 3

>> conv(x,h)ans =

6 7 13 7 3

length = length(x) + length(h) - 1


9/23

Compute by (1) hand

(2) with the use of Matlab


10/23

>> x = [ 0 0 1 2 1 0];

>> h = [ 0 0 1 -1 0 0];>> y = conv(x,h)

y =

Columns 1 through 7

0 0 0 0 1 1 -1

Columns 8 through 11

-1 0 0 0

A

>> x = [ 0 1 1 1 1 0];

>> h = [0 0 1 -1 1 0];>> y = conv(x,h)

y =

Columns 1 through 7

0 0 0 1 0 1 1


0 1 0 0

B


11/23

)(*)()()()( nhnxknhkxnyk

)()()()()( nxnhknxkhnyk

)(*)()(*)( nxnhnhnx Commutative Law

Convolution

Equation

Therefore

impulsenh

inputnx

outputny

)(

)(

)(


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)(*)](*)([)(*)()(

)](*)([*)()(*)](*)([)(

2121

2121

nhnhnxnhnyny

nhnhnxnhnhnxny

Associative Law

)()(*)( 21 nhnhnh Cascade Form

impulsenh

inputnx

outputny

)(

)(

)(


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)(*)()(*)()]()([*)()( 2121 nhnxnhnxnhnhnxny

Distributive Law

)()()( 21 nhnhnh Parallel Form

impulsenh

inputnx

outputny

)(

)(

)(


14/23

Plot from 0 n 9,

Convolution is commutative!!!


15/23

>> x = [ 1 1 1 1 1 0 0 0 0 0];

>> h1 = [ 1 -1 3 0 1 0 0 0 0 0];>> h2 = [ 0 2 5 4 -1 0 0 0 0 0];

>> y = conv(x,h1)

y =

Columns 1 through 6

1 0 3 3 4 3Columns 7 through 12

4 1 1 0 0 0


0 0 0 0 0 0

Column 19

0

>> y = conv(h1,x)

y =

Columns 1 through 71 0 3 3 4 3 4


1 1 0 0 0 0 0


0 0 0 0 0

Therefore convolution is commutative!!!



16/23

Convolution is distributive!!!!

Plot from 0 n 9,


17/23

>> y= conv(x,h1+h2)

y =

Columns 1 through 7

1 2 10 14 14 13 12

Columns 8 through 144 0 0 0 0 0 0


0 0 0 0 0

>> x = [ 1 1 1 1 1 0 0 0 0 0];

>> h1 = [ 1 -1 3 0 1 0 0 0 0 0];

>> h2 = [ 0 2 5 4 -1 0 0 0 0 0];

>> y2 = conv(x,h1) + conv (x,h2)

y2 =


1 2 10 14 14 13 12 4 0 0 0Columns 12 through 19

0 0 0 0 0 0 0 0

Therefore convolution is distributive!!!!



18/23

Convolution is associative!!!

>> x = [ 1 1 1 1 1 0 0 0 0 0];

>> h1 = [ 1 -1 3 0 1 0 0 0 0 0];

>> h2 = [ 0 2 5 4 -1 0 0 0 0 0];

>> y = conv(conv(x,h1),h2)

y =

Columns 1 through 7

0 2 5 10 20 35 35


36 30 20 5 3 -1 0


0 0 0 0 0 0 0


0 0 0 0 0 0 0

>> y2 = conv(x,conv(h1,h2))

y2 =

Columns 1 through 7

0 2 5 10 20 35 35


36 30 20 5 3 -1 0


0 0 0 0 0 0 0


0 0 0 0 0 0 0



19/23

Ex2: Convolution

Determine the unit step response of the system

described by the impulse response. Plot using

stem command evaluate from n = 0 to 9.

h(n) = 0.75nsin(2*pi*n) [u(n) u(n-10)]

ones(1,10)


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n=0:9;

hna = 0.75.^n.*sin(2*pi*n).*ones(1,10);

x = ones(1,10);

y = conv(x,hna);

ma = length(y)-1;

stem(0:ma,y);


21/23

Difference Equations

Filteringfilter(b,a,x) filters the vectorx using a difference equation where vectors a and b

represent the equation coefficients. This command implements the equation

Ifa = 1, then the command

performs the convolution of vectors x and b with the length of the output equal to the

length of the input data.

If a = [ 1 2 3 4], b = [5 6], x = [ 1 2 3] then:conv(a,b) = 5 16 27 38 24,

conv(b,a) = 5 16 27 38 24,

filter (b,a,x) = 5 6 0,

filter(b,1,x) = 5 16 27,

filter(x,1,b) = 5 16.


22/23

Able to properly represent the input/output relationship to a given LTI

system

A linear constant-coefficient difference equation (LCCDE) serves as a

way to express just this relationship

y[n]+7y[n1]+2y[n2]=x[n]4x[n1]


23/23

Write a MATLAB program to simulate thefollowing difference equation 8y[n] - 2y[n-

1] - y[n-2] = x[n] + x[n-1] initial conditions:

x(n) = 5u(n)

Example

>> b = [1 1 0];

>> a = [ 8 -2 -1];

>> x = 5*ones(1,5);

>> y = filter(b,a,x)

y =

0.6250 1.4063 1.6797 1.8457 1.9214

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