INC 212 Signals and systems Lecture#3: Convolution integral Assoc. - - PowerPoint PPT Presentation

INC 212 Signals and systems

Lecture#3: Convolution integral

• Assoc. Prof. Benjamas Panomruttanarug

benjamas.pan@kmutt.ac.th

The The Con Convolution

• lution In

Integral al

where h(t) = H{ (t)}  Impulse response of LTI system H = the output from the system given that the input is the delta function

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LTI system H

Input x(t) Output y(t)

       

x t h t x h t d   

 

  



y(t) =

Ex Exam ample ple 2.6 2.6 Reflect‐and‐shift Convolution Evaluation

Given and , evaluate the convolution integral y(t) = x(t)  h(t).

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     

1 3 x t u t u t    

     

2 h t u t u t   

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(a) The input x() depicted above the reflected and time‐shifted impulse

• response. (b) The product signal wt() for 1  t < 3. (c) The product

signal wt() for 3  t < 5. (d) The system output y(t).

 

0, 1 1, 1 3 5 , 3 5 0, 5 t t t y t t t t                

Matlab: conv

• clear;clc; close all;
• ts = 0.01;
• t = 1:ts:3;
• x = square(t);
• figure; plot(t,x);
• m = length(x);
• t = 0:ts:2;
• h = square(t);
• hold on; plot(t,h,'r‐‐');
• n = length(h);
• t = 1:ts:5;
• y = ts*conv(h,x);
• for i = 1:401
• t(i) = 1+ts*(i‐1);
• if t(i) < 3
• ya(i) = t(i)‐1;
• else
• ya(i) = 5‐t(i);
• end
• end
• figure; plot(t,y,'b',t,ya,'r‐‐');

     

1 3 x t u t u t    

     

2 h t u t u t   

Ex Exam ample ple 2.14 2.14 RC Cir Circuit: uit: St Step ep Respo sponse se

The impulse response of the RC circuit Find the step response of the circuit.

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1 ( ) ( )

t RC

h t e u t RC





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1 ( ) ( ) .

t RC

s t e u d RC



 

 

  0, ( ) 1 ( )

t RC

t s t e u d t RC



 

 

       



0, ( ) 1 0, 1 ,

t RC t RC

t s t e d t RC t e t





 

                



Solving by convolution

Matlab: conv

• clear;clc; close all;
• ts = 0.01;
• t = 0:ts:50;
• x = square(t,100);
• figure;plot(t,x);
• h = exp(‐t);
• y = ts*conv(x,h);
• ya = 1‐exp(‐t);
• figure; plot(t,y(1:5001),'b',t,ya,'r‐‐');

x(t) = u(t)

1 ( ) ( )

t RC

h t e u t RC





Ex Exam ample ple 2.7 2.7 RC Cir Circuit uit Output Output

Assume that the circuit’s time constant is RC = 1 sec.

• Find the impulse response of the circuit.
• Use convolution to determine the capacitor voltage, y(t), resulting

from an input voltage x(t) = u(t)  u(t  2).

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• 1. Fix and flip

) ( x ) ( h

) 2 ( ) ( ) (          u u x ) ( ) (  



 

 u

e h 1 1 ) ( ) ( ) ( * ) ( ) (           

      

 

t t t

e e d e d t x h t h t x t y

   

   

Growing Shrinking  

1 1 ) ( ) ( ) ( * ) ( ) (

2 2 2

         

        

 

e e e d e d t x h t h t x t y

t t t t t    

   

0<t<2 t>2

) 2 ( ) ( ) (       u u x

• 2. Fix and flip (Harder!!!)

) ( x ) ( h

) 2 ( ) ( ) (       u u x ) ( ) (  



 

 u

e h ) ( ) (  

u

e h  

t t t t t

e e d e d t h x t h t x t y

      

        

 

1 1 ) ( ) ( ) ( * ) ( ) (

   

   

Growing Shrinking  

1 1 ) ( ) ( ) ( * ) ( ) (

2 2 2

         

      

 

e e e e d e d t h x t h t x t y

t t t    

   

0<t<2 t>2

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(a) The input x() superimposed over the reflected and time‐shifted impulse response h(t – ), depicted as a function of . (b) The product signal wt() for 0 t < 2. (c) The product signal wt() for t  2. (d) The system output y(t).

 

 

2

0, 1 , 2 1 , 2

t t

t y t e t e e t

 

             

Matlab: conv

• clear;clc; close all;
• ts = 0.01;
• t = 0:ts:2;
• x = square(t,100);
• t = 0:ts:50;
• h = exp(‐t);
• y = ts*conv(x,h);
• for i = 1:5001
• t(i) = (i‐1)*ts;
• if t(i) < 2
• ya(i) = 1‐exp(‐t(i));
• else
• ya(i) = (exp(2)‐1)*exp(‐t(i));
• end
• end
• figure; plot(t,y(1:5001),'b',t,ya,'r‐‐');

x(t) = u(t)  u(t  2)

1 ( ) ( )

t RC

h t e u t RC



