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clc
disp("Problem 13.2")
printf("\n")
printf("Given")
disp("|Hv|=1/sqrt(2) (1)")
disp("Resistance R1=5kohm")
R1=5000;
disp("Hv(w)=1/1+%i*(w/wx) (2)")
//wx=1/(R1*C2)
//On solving we get
disp("wx=2*10^-4/C2 (3)")
disp("a)")
C2=10*10^-9;
//Taking modulus of (2)
disp("|Hv(w)|=1/sqrt(1+(w/wx)^2)")
//Equating (1) and (2)
wx=2*10^-4/C2;
fx=(wx/(2*%pi))*10^-3
printf("Frequency(a) is %3.2fkHz\n",fx)
disp("b)")
C2b=1*10^-9;
//As frequency is inversely proportional to C2 (from (3))
fx1=(C2/C2b)*fx
printf("Frequency(b) is %3.2fkHz\n",fx1)
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