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+// Exa 4.5
+
+clc;
+clear;
+
+// Given data
+
+App = 20; // peak gain in dB
+A = 17; // Actual gain in dB
+w = 10000; // Angular Frequancy in rad/sec
+C = 0.01*10^-6; // Farads
+
+// Solution
+
+printf('From Eq.(4.84) , we see  that gain is at its peak when w = 0.\n Therefore, 20*log(Rf/R1) = 20.\n');
+
+// Therefore,
+// Rf = 10 R1;   ............Eq. (1)
+z = 10;  // z is ratio of Rf/R1
+printf(' i.e Rf/R1 = %d. \n',z );
+printf(' At w = 10^4 rad/sec, gain in dB is down from its peak of 20 dB. \n Therefore, convering gain to dB in Eq.(4.84) and sub    situting for w,C, and Rf/R1 we can get value of Rf.\n\n');
+//  20*log10               10                = 17 dB
+ //          --------------------------
+ //          sqrt(1 + [10^4*10^-8*Rf]^2)
+ 
+deff('y=f(x)','y = 20*log10( 10 / sqrt(1+[10^-4*x]^2))-17');// x is Rf(Ω)
+    [x,v,info] = fsolve(10,f);
+    printf(' The calculated value of Rf is %d Ω. Rounding off to nearest possible value i.e 10 kΩ. \n',x); 
+    Rf = 10000; // Ω
+    printf(' Since we have ratio of Rf  by R1 so, \n The value of R1 can be given as R1 = %d kΩ. \n',0.1*(Rf/1000)); //as R1/Rf = 0.1
+