INC 212 Signals and systems Lecture#4: Frequency response of LTI - - PowerPoint PPT Presentation

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INC 212 Signals and systems Lecture#4: Frequency response of LTI - - PowerPoint PPT Presentation

INC 212 Signals and systems Lecture#4: Frequency response of LTI systems Assoc. Prof. Benjamas Panomruttanarug benjamas.pan@kmutt.ac.th Fourier Representations of f Sig ignals & LTI I Systems A signal can be represented as a weighted


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SLIDE 1

INC 212 Signals and systems

Lecture#4: Frequency response of LTI systems

  • Assoc. Prof. Benjamas Panomruttanarug

benjamas.pan@kmutt.ac.th

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SLIDE 2

Fourier Representations of f Sig ignals & LTI I Systems

A signal can be represented as a weighted superposition of complex sinusoids.

LTI System x(t) Output = A weighted superposition of the system response to each complex sinusoid. y(t)

  • Frequency response  The response of an LTI system to a sinusoidal input.
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SLIDE 3
  • Impulse response of continuous-time LTI system = h(t), input = x(t) =

e j  t

  • Output:
  • Frequency response:

Continuous-time LTI system

( )

( ) ( ) ( ) ( )

j t j t j j t

y t h e d e h e d H j e

    

    

     

  

 

( ) ( )

j

H j h e d



  

  



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SLIDE 4
  • Polar form complex number c = a + jb:

arg{ } j c

c c e 

where

 

2 2 1

and arg{ } tan

b a

c a b c

  

  • Polar form for H (j):

arg{ ( )}

( ) ( )

j H j

H j H j e

  

where

 

( ) Magnitude response and arg ( ) Phase respnse H j H j    

   

 

 

 

arg j t H j

y t H j e

 

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

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SLIDE 5

Example 3.1 .1 RC Circuit: Frequency response

The impulse response of the system relating to the input voltage to the voltage across the capacitor Find an expression for the frequency response, and plot the magnitude and phase response.

/

1 ( ) ( )

t RC

h t e u t RC

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SLIDE 6

   

  

 

d e u e RC j H

j RC     

 1

1

1 1 1

j RC

e RC j RC

 

        

        

 

1 1 1 1           RC j RC  RC j RC 1 1   

Frequency response: Magnitude response:

 

2 2

1 1         RC RC j H  

Phase response:

     

RC j H   arctan arg  

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SLIDE 7

clear;clc; R = 1; C = 1; for i = 0:100 H(i+1) = 1/(1+j*(2*pi*i)*R*C); M(i+1) = abs(H(i+1)); P(i+1) = angle(H(i+1)); end figure; subplot 211; plot((0:100),M); subplot 212; plot((0:100),P);

Matlab code