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

• f an

an LT LTI sy system

We can find y(t) from the frequency response of the system.

LTI System x(t) y(t)

s s i s

M e M j H

s

 



   ) (

amplitude of output = amplitude of input*magnitude response phase of output = phase of input+phase response

) cos( ) (

i i i

t M t x    

( ) ( )

j

H j h e d



  

  

 

) cos( ) (

s i i s i

t M M t y      

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8.3 8.3 Ideal Low‐Pass Filters

• The frequency response of a filter is characterized by a passband

and a stopband, which are separated by a transition band.

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) (  j H

c

 

c

 ) (  j H 

c



c

   

Practical low‐pass filter: the passband, transition band, and stopband are shown for positive frequencies.

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Cut‐off frequency

c



2 1

 ) (  j H

Practical low‐pass filter: more general form

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Maximum passband attenuation

• r ripple in passband

Maximum stopband magnitude or stopband attenuation Passband edge frequency Stopband edge frequency

  1 

p



s

  ) (  j H

Linear phase lowpass filter

2  4  2   4   

 

 

arg H j



 

H j

Cutoff frequency = c = 1/(RC)

  j RC RC j H   / 1 / 1 ) (

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Linear phase lowpass filter

2  4  2   4   

 

 

arg H j



 

H j

Cutoff frequency = c = 1/(RC)

  j RC RC j H   / 1 / 1 ) (

Any input with frequency lower than 100 rad/s can pass through the filter For

  K R 1 F C  10  s rad

c

/ 100 10 10 10 1

6 3

   



 Hz fc 16 

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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');

Matlab code

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Example

Calculate the output y(t), giving that 1. 2. 3.

  K R 1 F C  10  ) 90 sin( ) ( t t x   ) 10 cos( 5 ) ( t t x   ) 90 sin( ) 10 cos( 5 ) ( t t t x     ) 3 . 10 cos( 95 . * 5 ) (   t t y  ) 396 . 1 90 sin( 174 . ) (   t t y  ) 396 . 1 90 sin( 174 . ) 3 . 10 cos( 95 . * 5 ) (     t t t y  

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Matlab code

• t = 0:0.001:1;
• x = 5*cos(2*pi*5*t);
• y = 5*0.95*cos(2*pi*5*t‐0.3);
• figure; subplot 311; plot(t,x,'b',t,y,'r‐‐');
• x = sin(90*pi*t);
• y = 0.174*sin(90*pi*t‐1.396);
• subplot 312; plot(t,x,'b',t,y,'r‐‐');
• x = 5*cos(2*pi*5*t)+sin(90*pi*t);
• y = 5*0.95*cos(2*pi*5*t‐

0.3)+0.174*sin(90*pi*t‐1.396);

• subplot 313; plot(t,x,'b',t,y,'r‐‐');

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

• 5

5 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

• 1
• 0.5

0.5 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

• 10
• 5

5 10

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