Question

Solve for H(s) and plot the magnitude and phase of that function in MATLAB. Make sure to use the freqs() command when doing so.

1Ω 1 F 1 H 1Ω 18

Please include all the steps and add comments to the code to help me understand the problem. Thank you

Answer 1

Hello,
       Please find the answer attached below. If the answer has helped you please give a thumbs up rating. Thank you and have a nice day!

********** Matlab Code************

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% transfer function derivation

s = tf([1 0],1);        % transfer function s
z1 = (1/s) + 1;         % impedance of R and C in series
z2 = 1/s;               % impedance of C
z = (z1*z2)/(z1+z2);    % z1 and z2 in parallel
clc;
h1 = z/(z+1)           % voltage division between R and z
num = [1 3 2 0 0 0 0 0];
den = [1 5 7 2 0 0 0 0 0];
w = logspace(-2,4);
freqs(num,den,w)
title('Frequency response for system 1');

%%%%%%%%%%%% sys 2
figure;
z1 = 1 + s;     % R + L in series
z2 = 1;         % R=1
z = (z1*z2)/(z1+z2);    % z1 and z2 in parallel
h2 = z/(s+z)           % voltage division between L and z
num = [1 3 2];
den = [1 5 7 2];
freqs(num,den,w)
title('Frequency response for system 2');

************** End of Code ************

Output:

h1 =

      s^7 + 3 s^6 + 2 s^5
---------------------------
s^8 + 5 s^7 + 7 s^6 + 2 s^5

Continuous-time transfer function.

h2 =

      s^2 + 3 s + 2
---------------------
s^3 + 5 s^2 + 7 s + 2

Continuous-time transfer function.

NOTE: You can see that both the transfer functions are the same (cancellation). Thus, we get two similar frequency responses:

Frequency response for system 1 10 10 10 10 10 10 -2 10 10 10 Frequency (rad/s) 2 -50 -100 10 10 -2 10 10 10 10 Frequency (rad/s)

Frequency response for system 2 10 10 10 10 10 10 -2 10 10 10 Frequency (rad/s) 2 -50 -100 10 10 -2 10 10 10 10 Frequency (rad/s)

**************************************************************

PS: Please do not forget the thumbs up!!