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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)
+
+
+
+