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// Exa 4.5
clc;
clear;
// Given data
App = 20; // peak gain in dB
A = 17; // Actual gain in dB
w = 10000; // Angular Frequancy in rad/sec
C = 0.01*10^-6; // Farads
// Solution
printf('From Eq.(4.84) , we see that gain is at its peak when w = 0.\n Therefore, 20*log(Rf/R1) = 20.\n');
// Therefore,
// Rf = 10 R1; ............Eq. (1)
z = 10; // z is ratio of Rf/R1
printf(' i.e Rf/R1 = %d. \n',z );
printf(' At w = 10^4 rad/sec, gain in dB is down from its peak of 20 dB. \n Therefore, convering gain to dB in Eq.(4.84) and sub situting for w,C, and Rf/R1 we can get value of Rf.\n\n');
// 20*log10 10 = 17 dB
// --------------------------
// sqrt(1 + [10^4*10^-8*Rf]^2)
deff('y=f(x)','y = 20*log10( 10 / sqrt(1+[10^-4*x]^2))-17');// x is Rf(Ω)
[x,v,info] = fsolve(10,f);
printf(' The calculated value of Rf is %d Ω. Rounding off to nearest possible value i.e 10 kΩ. \n',x);
Rf = 10000; // Ω
printf(' Since we have ratio of Rf by R1 so, \n The value of R1 can be given as R1 = %d kΩ. \n',0.1*(Rf/1000)); //as R1/Rf = 0.1
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