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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


  1. INC 212 Signals and systems Lecture#4: Frequency response of LTI systems Assoc. Prof. Benjamas Panomruttanarug benjamas.pan@kmutt.ac.th

  2. Fourier Representations of f Sig ignals & LTI I Systems A signal can be represented as a weighted superposition of complex sinusoids. y ( t ) x ( t ) LTI System Output = A weighted superposition of the system response to each complex sinusoid.  Frequency response  The response of an LTI system to a sinusoidal input.

  3. Continuous-time LTI system • Impulse response of continuous-time LTI system = h ( t ), input = x ( t ) = e j  t   • Output:               j ( t ) j t j y t ( ) h ( ) e d e h ( ) e d      j t H j ( ) e         • Frequency response: j H j ( ) h ( ) e d 

  4.  Polar form complex number c = a + jb :        j arg{ } c 2 2 1 b c c e c a b and arg{ } c tan where a  Polar form for H ( j  ):     j arg{ H j ( )} H j ( ) H j ( ) e       H j ( ) Magnitude response and arg H j ( ) Phase respnse where                j t arg H j y t H j e amplitude of output = amplitude of input*magnitude response phase of output = phase of input+phase response

  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 1   t RC / h t ( ) e u t ( ) RC Find an expression for the frequency response, and plot the magnitude and phase response.

  6. Frequency response: Magnitude response: 1         1         j RC H j e RC u e d   H j   RC 2   1      2   RC    1     1 1   j    RC e   1 RC     j   RC 0  1 1     0 1   Phase response: RC 1     j             RC arg H j arctan RC 1 RC  1   j RC

  7. Matlab code 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);

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