00:01
Hello students, for rectangular waveguide minimum mode the equation is given by hx of x comma y is equal to cos of am x into cos of bmy into e power minus j beta square z.
00:29
Here am is equal to m pi divided by a, bm is equal to n pi divided by a and beta is equal to root of k square minus am square minus bm square where k is equal to root of am square plus bm square.
01:03
In the question given that a is equal to 1 .5 centimeter which is equal to 0 .015 meter, b is equal to 0 .8 centimeter which is equal to 0 .08 meter, then m and n will be taken as 1.
01:29
Now substituting all these values in the above condition to find am, bm and k, we get the equation for minimum mode will be equal to of x comma y is equal to cos of 209 .4 x cos of 39 .26 y into e power plus j 20 z.
02:07
For the next question b, here all the field components are 0 as given conductivity sigma is equal to 0.
02:18
Next we need to find cutoff frequency which is given by cutoff frequency is equal to c divided by 2 into a, where c is the speed of light which is 3 into 10 power 8 meter per second and a is the largest internal dimension of waveguide which is given in the question.
02:43
In meter it is taken as 0 .015 meter given 1 .5 centimeter.
02:52
From this calculation we obtain cutoff frequency is equal to 1 into 10 power 10 hertz.
02:59
Next we need to find phase constant which is already taken in the above equation value of beta that is root of k square minus am square minus bm square.
03:17
Now k is equal to root of am square plus bm square.
03:25
We know the value of am and bm is given by m divided by pi a and for bm also n pi divided by a.
03:38
By substituting for m n from the above calculations we obtain for beta is equal to k is 213.
03:47
Now 213 and again for am pi divided by a value 0 .015 square and bm pi divided by 0 .08 whole square.
04:05
From this calculation we obtain phase constant value beta is equal to minus 38 .03 and hence it is equal to 6 .16 j...