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+// Chapter 3 example 18

+//------------------------------------------------------------------------------

+clc;

+clear;

+// let 'a' and 'b' be the broad and narrow dimensions of the rectangular guide and 'r' be internal radius of circular guide

+// Dominant mode in rectangular guide =TE10

+// cutoff wavelength = 2a

+// dominant mode in circular guide = TE11

+// cut-off wavelength = 2*pi*r/1.841  = 3.41r

+// for the two cut-off wavelengths to equal

+// 2a  = 3.41r

+// a = 1.705r

+// now area of cross section of rectangular guide = a*b

+//assuming a= 2b,which is very reasonable assumption ,we get 

+// area of cross section of rectangular waveguide = a*a/2 = ((1.705^2)*r*r)/2 = 1.453r^2

+// area of cross-section of circular guide = pi*r*r = 3.14r^2

+// ratio of two cross sectional areas = (3.14r^2)/(1.453r^2) = 2.16

+mprintf('Circular guide is 2.16 times larger in cross section as compared to rectangular guide');

+//------------------------------------------------------------------------------