I have a circuit that I want to gain its transfer function.I've attached a JPG file.

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Plz look at my JPG file. How can I find its transfer function???
Thank you so much.
  2 件のコメント
Azzi Abdelmalek
Azzi Abdelmalek 2014 年 12 月 20 日
Do you need Matlab to do it ?
vahid torabi
vahid torabi 2014 年 12 月 20 日
Yes I want to do that by using matlab

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Shoaibur Rahman
Shoaibur Rahman 2014 年 12 月 20 日
Input voltage (across left C) = Vi
Output voltage (across right C) = Vo
Apply voltage divider rule in RL and C (at right) branch:
Vo = ((1/jwC)/(R+jwL+1/jwC))*Vi, that gives the transfer function H:
H(jw) = Vo/Vi = (1/jwC)/(R+jWL+1/jwC) = |Ho(jw)|e^(j*theta0)
where, Ho(jw) = abs(H(jw)) and theta0 = angle(H(jw))
Matlab code:
R = 1; L = 1; C = 1; % use values what you want
w = 0:0.1:10*pi; % define the frequency range
H = 1./(1+w.^2*L*C+1j*w*R*C);
H0 = abs(H);
theta0 = angle(H);
subplot(211), plot(w,H0)
subplot(212), plot(w,theta0)
  3 件のコメント
vahid torabi
vahid torabi 2015 年 1 月 6 日
Thank you so much sir; but : H = 1./(1-w.^2*L*C+1j*w*R*C) is correct. Am I right??? you wrote 1./(1+w.^2*L*C+1j*w*R*C)!
Shoaibur Rahman
Shoaibur Rahman 2015 年 1 月 14 日
Yes, you are correct! j^2 = -1.

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