14.3 Solutions
701
The general solution is
H z = (A cos k x x + B sin k x x)(C cos k y y + D sin k y y)e
−jkz
(2)
∴
∂ H z
∂ x
= (−k 2 A sin k x x + Bk 2 cos k x x)
(C cos k y y + D sin k y y)e
−jkz
For
∂ H z
∂ x
= 0 at x = 0, it is necessary that B = 0.
Similarly
∂ H z
∂ y
= (A cos k x x + B sin k x x)(−Ck y sin k y y + D cos k y y)e
−jkz
For
∂ H z
∂ y
= 0, D = 0. Therefore (2) becomes
H z = K cos(k x x) cos(k y y) e
−jkz
(3)
where K is the product of A and B is another constant. Imposing
boundary conditions at x = a and y = b, we have
∂ H z
∂ x
= −K k x sin k x x cos k y y = 0
yielding
k x a = mπ
(4)
and
∂ H
∂ y
= −K k y cos k x x sin k y y = 0
yielding
k y b = nπ
(5)
∴ H z = K cos
mπ x
a
cos
nπ y
b
(6)
Substituting (2) in (1) we obtain
k
2
x + k
2
y = ω
2
με − k
2
(7)
which is identical with (3) of prob. (14.104).
(b) Substituting the values of k x and k y from (4) and (5) in (7) and setting
k = 0 gives the cut-off frequency.