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authorpriyanka2015-06-24 15:03:17 +0530
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+// Chapter 3 example 15
+//------------------------------------------------------------------------------
+clc;
+clear;
+// Given data
+a = 4*10^-2; // width of waveguide
+b = 2*10^-2; // narrow dimension of waveguide
+c = 3*10^8; // velocity of EM wave in m/s
+m1 = 1; // for TE10
+m2 = 2; // for TE20
+n = 0; // for TE10
+// Calculations
+lamda_c = 2*a // cutoff wavelength for TE10 mode
+f1 = c/lamda_c // frequency in Hz
+// the frequency range for single mode operation is the range of frequencies corresponding to the dominant mode and the second highest cutoff wavelength
+lamda_c_2 = 2/sqrt((m2/a)^2 + (n/b)^2)
+f2 = c/lamda_c_2; // freq at second largest cutoff wavelength
+
+// Output
+mprintf('Therefore,single mode operating range = %3.2f Ghz to %3.1f Ghz\n',f1/10^9,f2/10^9 );
+mprintf(' Note: instead of 3.75,3.5 is printed in textbook');
+//------------------------------------------------------------------------------
+