Question

6. The circuit shown below is common in radio receivers. The dependent source is created using a device known as a transistor. R2 VX VOUT - RS 9m Vx R = 5002 R2 = 20000 9m = 0.0385 A/V C = 2.3374nF L = 1nH Ry = 100k 22 a) Determine the transfer function H(s) = OUT. It should have the form: VIN Vout aus? + ans + do H(S) = 1 VIN b2s2 + bis + 1 Provide the values of az, aj, ao, b2, and by b) What are the pole and zero locations for this circuit? If you have complex poles, make sure to enter the pole value with the positive imaginary part as p1. c) What is the impedance of the parallel RLC circuit formed by L, C, and Rz when w = 1/VLC? d) Use MATLAB to create the exact Bode Plot for this circuit. The voltage source vin(t) is the voltage provided by the antenna in the radio. The source vi(t) contains a sine wave at 104.1 MHz and a sine wave at 104.9MHz. Each sine wave has an amplitude of

Answer 1

a) from the (RIAR2) Ritra given circuits . Va= Vin( R ) Im Vx = Vin (R2 M) =) 9m Ve(s) = VEN(S) Regu Apply KCl at nede ® , we get r y you can vary best way to u avalioso. * To leto -Vovies [* *&+c8) -- Peine web. — ® substituting all the given parameter values in (A) 3 Vout Vour(s) VouT(S). R3 Vour (s) .. 2+ m Ve(s)=0. > Vour(s) - Rz9m Ritr we haves al's . -3.756*10's . ro Vour (s) VING) - 18 v -14 2.337x10 st 10 stil 5.756 x, az= ao=1; a = -3.756 *10 by = 2.337x10'8; a = 1014 b. Zeros from ® , are szo. poles from ® are p = -10 Vize 4×2-337x103 -18 2x 2.337x10 >> P = -2114x10 +6.54%10 g & P2 = -2.14 x103-6.54X108 when w= , the crant will be in resonance, (ie. Inductive reactance will cancel out Capacitive reactance) Then the Impedance will be pure resistance R3, i z = 100ks.

MATLAB code is given below in bold letter.

clc;
close all;
clear all;

% define R L C
R1 = 50;
R2 = 2000;
R3 = 100e3;
gm = 0.0385;
C = 2.3374e-9;
L = 1e-9;

% define s
s = tf('s');

% define H(s)
H = -R2*gm/(R1+R2)/(1/R3+1/(L*s)+C*s)

% plot the bode figure
figure;
bode(H);grid on;

BODE PLOT IS GIVEN BELOW.

Bode Diagram Magnitude (dB) -50 -- Phase (deg) 10 Frequency (rad/s)

from the above figure it is observed that the magnitude gain at 104.1MHz is 71.5 dB (3758) and that at 104.9MHz is 2.23 dB (1.29).

Therefore the value of sinusoids amplitudes will respectively be 37.58 V and 12.9 mV.

Question e

from the above it is clearly understood that the radio is currently tuned to 104.1MHz.