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+clc
+// Fundamental of Electric Circuit 
+// Charles K. Alexander and  Matthew N.O Sadiku  
+// Mc Graw Hill of New York 
+// 5th Edition 
+
+// Part 2     :  AC Circuits 
+// Chapter 14 :  Frequency Response
+// Example 14 - 17
+
+clear; clc; close; 
+//
+// Given data
+f0_upper  =  1600.0000 * 10^3;                                                     
+f0_lower  =   540.0000 * 10^3;
+L         =     10^-6
+//
+// Calculations C1 The High End of The AM Band 
+C1  = 1/(4*(%pi)^2*(f0_upper)^2*L);
+// Calculations C1 The Low End of The AM Band 
+C2  = 1/(4*(%pi)^2*(f0_lower)^2*L);
+// 
+disp("Example 14-18 Solution : ");
+printf(" \n C1  = Capacitance of Capacitor The High End of The AM Band   = %.3f nanoFarad",C1*10^9)
+printf(" \n C2  = Capacitance of Capacitor The Low End of The AM Band    = %.3f nanoFarad",C2*10^9)
+
+
+