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+// Exa 10.12
+clc;
+clear;
+close;
+format('v',5)
+// Given data
+Ao = 100;
+f_L = 20;// in Hz
+f_H = 40;// in kHz
+f_H = f_H*10^3;// in Hz
+Beta = 0.1;
+Af = Ao/(1 + (Beta*Ao));
+disp(Af,"The overall gain at mid frequency is");
+f_Hf = f_H*(1+(Ao*Beta));// in Hz
+f_Hf = f_Hf * 10^-3;// in kHz
+disp(f_Hf,"The upper cutoff frequency with negative feedback in kHz is");
+f_Lf = f_L/(1+(Ao*Beta));// in Hz
+disp(f_Lf,"The lower cutoff frequency with negative feedback in Hz is");
+
+// Note: The calculated value of lower cutoff frequency with negative feedback i.e f_Lf is wrong. So the answer in the book is wrong.