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 olution In Integral al Input Output LTI system x ( t ) y ( t ) H                 y(t) = x t h t x h t d  where h ( t ) = H {  ( t )}  Impulse response of LTI system H = the output from the system given that the input is the delta function 2 BP INC212

Exam Ex ample ple 2.6 2.6 Reflect‐and‐shift Convolution Evaluation                    1 3 2 Given and , evaluate the convolution x t u t u t h t u t u t integral y ( t ) = x ( t )  h ( t ). 3 BP INC212

  0, 1 t     1, 1 3  t t      y t   5 , 3 5 t t    0, 5  t (a) The input x (  ) depicted above the reflected and time ‐ shifted impulse response. (b) The product signal w t (  ) for 1  t < 3. (c) The product signal w t (  ) for 3  t < 5. (d) The system output y ( t ). 4 BP INC212

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

Ex Exam ample ple 2.14 2.14 RC Cir Circuit: uit: St Step ep Respo sponse se 1 t   ( ) ( ) The impulse response of the RC circuit h t e RC u t RC Find the step response of the circuit. 6 BP INC212

Solving by convolution    1  t   ( ) ( ) . s t e RC u d  RC   0, 0 t    ( )  s t 1   t    ( ) 0  RC e u d t   RC   0, 0 t    ( )  s t 1   t   0  RC e d t  0 RC   0, 0 t    t     1 , 0  RC e t 7 BP INC212

1 t Matlab: conv   x ( t ) = u ( t ) ( ) ( ) RC h t e u t RC • 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 ‐‐ ');

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). 9 BP INC212

(  (  ) ) 1. Fix and flip h x Shrinking Growing       ( ) ( ) ( 2 ) x u u t>2 0<t<2  ( ) ( ) * ( )           ( ) ( ) * ( ) y t x t h t ( ) ( ) ( 2 ) y t x t h t x u u           ( ) ( )       ( ) ( )  u h x t d      h x t d ( ) ( ) h e       t t             1 1 e d e d    2   0 t t   t       e e  2 0 t      2  1    t 1 t e e e

(  (  ) ) x h 2. Fix and flip (Harder!!!) Shrinking Growing       t>2 0<t<2 ( ) ( ) ( 2 ) x u u  ( ) ( ) * ( )  ( ) ( ) * ( ) y t x t h t y t x t h t  u      ( ) ( ) h e           ( ) ( )       ( ) ( ) x h t d x h t d  u        ( ) ( )    h e 2 t             1 t 1 t e d e d   0   0 2   t      t t e e e 0 0      2  1   t 1 t e e e

  0, 0 t          1 , 0 2 t  y t e t       2 1 , 2 t  e e t  (a) The input x (  ) superimposed over the reflected and time ‐ shifted impulse response h ( t –  ), depicted as a function of  . (b) The product signal w t (  ) for 0  t < 2. (c) The product signal w t (  ) for t  2. (d) The system output y ( t ). 12 BP INC212

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