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+// Chapter 6 example 1
+//------------------------------------------------------------------------------
+clc;
+clear;
+// Given data
+gs      = 0.0025;       // output conductance in mho
+gl      = 0.0025;       // load conductance
+r       = -250;         // negative resistance of microwave device
+
+// calculations
+
+//  P1  = Vl^2 *gl          // power that is transferred to load
+//  P   = Vl^2 *gs          // source is matched to load
+//  P   = [Is/(gl+gs)]^2 *gs
+//      = ((Is^2)/(4*gs^2))*gs 
+//      = (Is^2)/(4*gl)
+//  P2  = Vl^2 *gl          // Load power
+//      = [Is/(gs+gl-g)]^2 *gl
+//      = (Is^2 *gl)/(2gl - g)^2
+// P2/P1 = ((Is^2 *gl)/(2gl - g)^2)*(4*gl)/(Is^2)
+//       = (4*gl^2)/(2gl - g)^2;
+//       = (4*gl^2)/(4gl^2 + g(g-4gl))
+// For P2/P1 > 1 , 4gl > g so that denominator is less than numerator 
+g       = 1/r
+// let k = P2/P1
+k       = (4*gl*gl)/((2*gs)+ g)^2
+
+// output
+    mprintf('Power gain = %d',k);
+