INC 212 Signals and systems Lecture#5: Frequency response (cont.) - PowerPoint PPT Presentation

INC 212 Signals and systems Lecture#5: Frequency response (cont.) Assoc. Prof. Benjamas Panomruttanarug benjamas.pan@kmutt.ac.th

Fr Frequency equency re response of of an an LT LTI sy system We can find y(t) from the frequency response of the system.           ( ) cos( ) ( ) cos( ) x t M t y t M M t LTI System i i i i s i i s y ( t ) x ( t )         ( ) ( ) j H j h e d        ( ) i H j M e M s s s s amplitude of output = amplitude of input*magnitude response phase of output = phase of input+phase response 2 BP INC212

8.3 Ideal Low‐Pass Filters 8.3  The frequency response of a filter is characterized by a passband and a stopband, which are separated by a transition band. (   (  ) ) H j H j   c       c c c 3 BP INC212

Practical low ‐ pass filter: the passband, transition band, and stopband are shown for positive frequencies. (  ) H j 1 2   c Cut ‐ off frequency 4 BP INC212

Practical low ‐ pass filter: more general form (  ) H j Maximum passband attenuation or ripple in passband   1 Maximum stopband magnitude or stopband attenuation     p s Passband edge frequency Stopband edge frequency 5 BP INC212

Linear phase lowpass filter 1 / RC   ( ) H j   1 / RC j Cutoff frequency =  c = 1/( RC )       j  arg H j  H  2  4    4    2 6 BP INC212

Linear phase lowpass filter 1 / RC   ( ) H j   1 / RC j  K    Cutoff frequency =  c = 1/( RC ) 1 10 For R C F       j  arg H j  H  1    100 / rad s 2    3 6 c 10 10 10  4  16 f c Hz  Any input with frequency   4  lower than 100 rad/s  can pass through the filter  2 7 BP INC212

Matlab code clear;clc; close all R = 1e3; C = 10e-6; for i = 0:100 H(i+1) = (1/(R*C))/(1/(R*C)+j*(2*pi*i)); M(i+1) = abs(H(i+1)); P(i+1) = angle(H(i+1)); end figure; subplot 211; plot((0:100),M); xlabel('Frequency (Hz)'); ylabel('Magnitude'); subplot 212; plot((0:100),P); xlabel('Frequency (Hz)'); ylabel('Phase'); 8 BP INC212

Example  K    1 10 Calculate the output y(t), giving that R C F      ( ) 5 * 0 . 95 cos( 10 0 . 3 ) ( ) 5 cos( 10 ) y t t 1. x t t    ( ) 0 . 174 sin( 90 1 . 396 )   ( ) sin( 90 ) y t t 2. x t t     ( ) 5 cos( 10 ) sin( 90 ) 3. x t t t       ( ) 5 * 0 . 95 cos( 10 0 . 3 ) 0 . 174 sin( 90 1 . 396 ) y t t t 9 BP INC212

Matlab code • t = 0:0.001:1; • x = 5*cos(2*pi*5*t); 5 • y = 5*0.95*cos(2*pi*5*t ‐ 0.3); 0 • figure; subplot 311; plot(t,x,'b',t,y,'r ‐‐ '); -5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 • x = sin(90*pi*t); 0.5 • y = 0.174*sin(90*pi*t ‐ 1.396); 0 -0.5 • subplot 312; plot(t,x,'b',t,y,'r ‐‐ '); -1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 10 5 • x = 5*cos(2*pi*5*t)+sin(90*pi*t); 0 • y = 5*0.95*cos(2*pi*5*t ‐ -5 0.3)+0.174*sin(90*pi*t ‐ 1.396); -10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 • subplot 313; plot(t,x,'b',t,y,'r ‐‐ '); 10 BP INC212