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+// Exa 7.5
+
+clc;
+clear;
+
+// Given data
+
+// A wide-band pass filter
+fL=400; // Lower cutoff frequency(Hz)
+fH=2000; // Higher cutoff frequency(Hz)
+A0=4; // passband gain
+
+// Solution
+
+printf('Since, the pass band gain is 4. so each of LPF and HPF section may be designed to give gain of 2,\n that is Ao=1+ (Rf/Ri) = 2.\n So, Rf and Ri should be equal. \n Let Rf=Ri=10 kΩ for each of LPF and HPF sections.');
+
+disp("");
+disp("");
+disp("For HPF, fL=400 Hz.");
+printf(' Assume C2=0.01 μF. ');
+C2=0.01*10^-6; // Farads
+// Since fL= 1/(2*%pi*R2*C2);
+// Therefore
+R2= 1/(2*%pi*C2*fL);
+printf(' \n The calculated value of R = %.1f kΩ.',int(R2)/1000);
+
+disp("");
+disp("");
+disp("For LPF, fH=2000 Hz.");
+printf(' Assume C1=0.01 μF.');
+C1=0.01*10^-6; // Farads
+// Since fH= 1/(2*%pi*R1*C1);
+// Therefore
+R1= 1/(2*%pi*C1*fH);
+printf(' \n The calculated value of R = %.2f kΩ.',R1/1000);
+
+disp("");
+disp("");
+
+fo=sqrt(fL*fH);
+Q=fo/(fH-fL);
+
+printf(' The value of cutoff frequency = %.1f Hz.\n ',fo);
+printf('\n The quality factor = %.2f (<10) since wide passband filter.',Q);