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diff --git a/3830/CH5/EX5.3/Ex5_3.sce b/3830/CH5/EX5.3/Ex5_3.sce
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+// Exa 5.3
+
+clc;
+clear;
+
+// Given
+
+fs = 10000; // frequency of modulated signal(Hz)
+fm = 200*10^3; // modulation frequency(Hz)
+Ri = 10; // Input resistance(ohms)
+e2_by_e1 = 1.3; // limit for lowest frequency(in %)
+
+// Solution
+
+F_lower = fm - fs ; 
+
+printf(' For a double-section filter, \n  e2/e1 =  1/sqrt(1+(w*Rf*Cf)^2) \n');
+// Therefore,
+
+function y=f(x) 
+    y =(1/(sqrt(2*%pi*F_lower*x)^2+1))-e2_by_e1/100;
+endfunction
+[x,v,info] = fsolve(0,f);
+printf('  The product of Rf*Cf = %.4f  sec \n ',x);
+printf(' Let Rf = 10^5 Ohms, so that attenuation is 10:1. Therefore, Cf = ');
+Cf = x*10^-7;
+printf(' %.3f pf \n ',Cf*10^12);
+
+// The answer provided in the textbook is wrong