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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.