In the RC circuit, as frequency increases, the value of the resistance and the capacitance remained...

In the RC circuit, as frequency increases, the value of the resistance and the capacitance remained constant and the output gain began to decay exponentially. From a theoretical point of view the impedance of a capacitor is (-j)/wc. By applying the mathematical representation of voltage gain [H(w0)= v0/vi] , we end up with the following formula: 1/(1+jwRC), where V0 is 1/jwC, and Vi is the summation of the total input impedance as seen by the voltage source which is R + (-j)/wC.
Therefore, to determine the graphical representation of the magnitude of the transfer function |H(w)|= 1/(√(1+(WRC)^2))=1/(√(1+(W/W0)^2)) , as the value of w/w0 increases the graph should begin to decay.
Based on these facts, the behavior of our circuit make sense. (using mathlab)

Homework Answers

Answer #1

Frequency response of RC circuit can be determined by making a transfer function for given values of R and C in MATLAB.

Plotting frequency response by Bode Plot.

Following code will analyze the frequency response-

s = tf('s');

disp('Enter the value of R')

R = input('R')

disp ('Enter the value of C')

C= input ('C')

% calculate cutoff frequency

disp('cutoff frequency')

w0 = 1/(R*C)

G = 1/(1+s*R*C);

bode (G)


Enter the value of R


R =


Enter the value of C


C =


cutoff frequency

w0 =


Magnitude and Phase plot-

It can be seen that after 100 rad/sec which is its cutoff frequency, the magnitude starts decaying.

So conclusion is this RC circuit behaves as a low pass filter of cut off frequency 100 rad/sec.

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