25-Frequency Response

# 25-Frequency Response - Frequency Response Frequency...

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requency esponse Frequency Response © Fred Terry all, 008 Fall, 2008 1

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Transfer Function H( ω ) inear Circuit + + Linear Circuit (Linear System) - V g (t) V o (t) - () cos gg Assume Vt V t ω = j i j jt jt H og g VV e VH V V H e ωω +∠ = == 2 ( ) ( ) Vt V H t H =+
ourier Series & Transfer Functions Fourier Series & Transfer Functions () I ffti sa ny periodic function with period T Any Periodic Signal f(t) Can Be Represented in A Fourier Series Expansion: 0 0 2 jn t n n ft ce w h e r e T if ω π =−∞ == 0 0 1 T jn t n cf t e d t T = b S iti if h i di i t f(t) ( lt So by Superposition, if we have a periodic input f(t) (voltage or current) into our linear circuit with transfer function H( ω ), we will have an output g(t) given by: 0 00 2 jn t n n gt cH n e whe re T ωω 3

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ourier Series & Transfer Functions Fourier Series & Transfer Functions So by Superposition, if we have a periodic input f(t) (voltage or rrent) into our linear circuit with transfer function H( we will current) into our linear circuit with transfer function H( ω ), we will have an output g(t) given by: ( ) 0 00 2 () jn t n n gt cH n e whe re T ω π ωω =−∞ == hi i f l i it th t b d i AC St d St t This is powerful, since it means that by doing AC Steady State Analysis on a Circuit, We Can Easily Calculate the Response to ANY Periodic Signal!! 4
Sawtooth Wave Example 0 0 1 () ) 0 for 1st period to T jn t n cf t e d t T t t t T ω = = 00 2 11 os t p e o d t o TT jn t jn t n ft T t c e dt te dt T ωω −− ⎛⎞ == ⎜⎟ ⎝⎠ ∫∫ [] 2 1 10 /2 0 using constant if we get: ax ax xe dx ax e a a i f n =− + + = 2 2 1/2 1 2 0 (2 ) n jn n ci cj n e f o r n n π ⎡⎤ + + ⎣⎦ 5

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Sawtooth Wave 6
Sawtooth Wave 7

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Sawtooth Wave 8
xample: 1 st rder Low Pass Example: 1 Order Low Pass Filter + R i j j i 1 1 1 1 og g jC VV V jR C R ω == + + +- V g (t) V o (t) C () j ( ) /2 1 1 1 o g V H C V H + = - ( ) () ( ) 1/2 2 1 1 tan RC HR C ωω ⎡⎤ + ⎣⎦ ∠= 9

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Example: Sawtooth Wave into 1 st Order Low Pass Filter 1/2 0 n ci f n = = ( ) 2 2 1 112 0 (2 ) jn n cj n e f o r n n π ⎡⎤ = −+ + ⎣⎦ 10
Matlab Code For Example %sawtooth wave into RC filter clear; tau=1; % assume 1 WLOG T=5*tau; % vary this for fun, the period of the sawtooth vs. tau t=0:1e-3:5*max(T,tau); N=1000; v0=0*t; vg=0*t; for jj=-N:N, if abs(jj)>0, cn=-1/(4*(jj*pi)^2)+(1+2*j*jj*pi)*1/(4*(jj*pi)^2)*exp(-2*j*jj*pi); else cn=0.5; nd; end; w=2*pi/T*jj; H=1/(1+j*tau*w); % transfer function v0=v0+cn*H*exp(j*w*t); vg=vg+cn*exp(j*w*t); end; igure(1); figure(1); h=plot(t,vg,t,v0, '--'); grid on; set(h,'linewidth',2); title(['T= ' num2str(T) ' \tau=RC= ' num2str(tau) ' n= -' num2str(N) ' to +' num2str(N)]); xlabel('Time'); l b l('V lt ') 11 ylabel('Volts'); % As check, rounding produces imaginary part figure(2); plot(t,imag(vg),t,imag(v0));

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## This note was uploaded on 01/31/2011 for the course EECS 215 taught by Professor Phillips during the Spring '08 term at University of Michigan.

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25-Frequency Response - Frequency Response Frequency...

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