Question

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)

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)

Response-

Enter the value of R

R10

R =

10

Enter the value of C

C0.001

C =

1.0000e-03

cutoff frequency

w0 =

100

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.