//chapter-4 page 147 example 4.9
//==============================================================================
clc;
clear;

f=8.6*10^9;//frequency in Hz
c=3*10^10;//Velocity of Light in cm/sec
a=2.5;//Length of a Waveguide in cm
b=1;//Width of a Waveguide in cm

//CALCULATION
disp('The condition for the wave to propagate along a guide is that wc>w0.');
w0=c/f;//free space wavelength in cm
disp('Free space wavelength w0 in cm is');
disp(w0);
disp('For TE waves, wc=(2ab/sqrt((mb)^2+(na)^2))');
disp('For TE01 waves');
m1=0;
n1=1;
wc1=((2*a*b)/(sqrt((m1*b)^2+(n1*a)^2)));//Cutoff wavelength for TE01 mode in cm
disp('Cutoff wavelength for TE01 mode in cm is');
disp(wc1);
disp('Since wc for TE01=2cm is not greater than w0 TE01,will not propagate for TE01 mode.');
disp('For TE10 waves');
m2=1;
n2=0;
wc2=((2*a*b)/(sqrt((m2*b)^2+(n2*a)^2)));//Cutoff wavelength for TE10 mode in cm
disp('Cutoff wavelength for TE10 mode in cm is');
disp(wc2);
disp('Since wc TE10 > w0 TE10 is a possible mode.');
fc=(c/wc2)/10^9;//Cutoff frequency in GHz
disp('For TE11 and TM11 waves');
m3=1;
n3=1;
wc3=((2*a*b)/(sqrt((m3*b)^2+(n3*a)^2)));//Cutoff wavelength for TE11 mode in cm
disp('Cutoff wavelength for TE11 and TM11 modes in cm is');
disp(wc3);
disp('As wc for TE11 and TM11 is < w0 both TE11 and TM11 do not propagate as higher modes.');
wg=(w0/sqrt(1-(w0/wc2)^2));//Guide wavelength in cm
disp('From the above analysis we conclude that only TE10 mode is possible');

//OUTPUT
mprintf('\nCutoff frequency is fc=%1.0f GHz \nGuide wavelength is wg=%1.3f cm',fc,wg);

//=========================END OF PROGRAM===============================