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+// Exa 7.7
+
+clc;
+clear;
+
+// Given data
+fH= 400; // Higher cutoff frequency(Hz)
+fL=2000; // lower cutoff frequency(Hz)
+Ao=2;  // Pass band gain
+
+// Solution
+
+disp("For HPF, fL=2 kHz.");
+disp("Assume C2=0.1 μF. ");
+C2=0.1*10^-6; // Farads
+// Since fL= 1/(2*%pi*R*C2);
+// Therefore
+RL= 1/(2*%pi*C2*fL);
+printf(' The calculated value of R = %d Ω.',int(RL));
+printf('\n Let R = 800 Ω.');
+// Since Ao=Ao2 = 1+ (Rf/Ri);  
+disp("Let Rf = Ri =10 kΩ(say) to give A02 of 2.");
+disp("");
+disp("");
+disp("For LPF, fL=400 Hz.");
+disp("Assume C1=0.1 μF. ");
+C1=0.1*10^-6; // Farads
+// Since fH= 1/(2*%pi*R*C1);
+// Therefore
+RF= 1/(2*%pi*C1*fH);
+printf(' The calculated value of R = %d Ω.',int(RF));
+printf('\n Let R = 4 kΩ.');..
+// Since Ao=Ao1 = 1+ (Rf/Ri);  
+disp("Let Rf = Ri =10 kΩ(say) to give A01 of 2.");
+
+disp("");
+disp("");
+
+disp("The schematic arrangement and the frequency response is shown in figs. 7.16(a,b) on page no. 280.")