Question

i About Exercise 9.5.8 For the circuit shown in the figure above: Obtain an expression for H (u) = Ve /V, in standard form. G

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Answer #1

Part-a

The given circuit can be redrawn in the frequency domain as shown below.

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the parallel combination of a capacitor and resistor is given by

Z_1=\frac{R\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}}=\frac{R}{j\omega RC+1}

Now the above circuit can be redrawn as shown below.

According to the Voltage Division principle, the output voltage is given by

V(u) Z1

juRC+ jwRC+1

V4(w)

Rearranging the terms in the above equation

H(\omega)=\frac{V_o(\omega)}{V_i(\omega)} =\frac{R}{R(1-\omega^2 LC)+j\omega L}

Part-b

The spectral plots of magnitude and phase are generated using MATLAB.

Below is the MATLAB code.

%values of circuit elements
R=100;
L=0.1*10^(-3);
C=10^(-6);
%vector of angular frequency (10^4<=w<=10^6)
w=10000:10:10^6;
%calculating the gain H(w)
H=R./(R*(1-w.^2*L*C)+j*w*L);
figure(1)
%plotting magnitude spectrum
subplot(2,1,1)
semilogx(w,20*log10(abs(H)));
grid on
xlabel('\omega(rad/sec)')
ylabel('20log_{10}|H(\omega)|(dB)')
title('Magnitude Spectrum')
%plotting phase spectrum
subplot(2,1,2)
semilogx(w,angle(H)*180/pi);
grid on
xlabel('\omega(rad/sec)')
ylabel('\angle H(\omega)(deg)')
title('Phase Spectrum')

The spectral plot generated is as shown below.

Magnitude Spectrum 30 20 10 3 0 -10 -20 -30 40 104 106 105 (rad/sec) Phase Spectrum -50 E -100 150 104 105 u(radsec) 106

Part-c

From the above plot, we can clearly see that the upper cut-off frequency(frequency at which gain is 3dB less than the peak gain) is 10°rad/sec 1,046 .

Magnitude Spectrum 30 X: 1.046e+05 Y: 17 20 10 3 0 -10 -20 -30 40 104 106 105 (rad/sec) Phase Spectrum -50 E -100 150 104 105

From the above figure, we can clearly see that the slope of the magnitude plot is close to 40dB /decade

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