Question 10.7: In a hollow rectangular waveguide with an aspect ratio a/b =......
In a hollow rectangular waveguide with an aspect ratio a/b = 2 (a < 10[cm] ), the TE mode of propagation gives \mathscr{E} = 4(\pmb{a}_{x} – \pmb{a}_{y})\sin (20\pi z – 1.8\pi \times 10^{10}t) [V/m] at a point ( x = a/8 and y = b / 4 ) on the cross section of the waveguide. Find
(a) phase velocity,
(b) cutoff frequency, and
(c) complete expression for \mathscr{E} in the waveguide.
(a) From the given \mathscr{E}, we obtain \beta _{mn} = 20\pi , and \omega = 1.8\pi \times 10^{10}.
The phase velocity is, from Eq. (10-52a),
v_{p(mn)} = \frac{\omega }{\beta _{mn}} = f \lambda _{mn} (10-52a)
v_{p(mn)} = \frac{\omega }{\beta _{mn}} = \frac{1.8 \pi \times 10^{10}}{20\pi } = 9 \times 10^{8}[m/s](b) Rewriting Eq. (10-48) we have
\boxed{\beta_{mn} = k \sqrt{1 – \left\lgroup \frac{f_{c(mn)}}{f} \right\rgroup ^{2} } } [rad/m] (10-48)
f_{c(mn)} = f \sqrt{ 1 – \left\lgroup\frac{\beta _{mn} c}{\omega }\right\rgroup ^{2} } = 9 \times 10^{9} \sqrt{1 – \left\lgroup\frac{20\pi \times 3\times 10^{8}}{1.8\pi \times 10^{10}} \right\rgroup ^{2}} = 8.485[GHz](c) Inserting x = a/8 and y = b/4 into the general expression for the TE_{mn} mode given in Eqs. (10-75a) and (10-75b), we have
If the electric fields in Eq. (10-84) are combined, it should be of the form 4(\pmb{a}_{x} – \pmb{a}_{y}) as given in the problem. Thus, using a = 2b , from Eq. (10- 84) we obtain
n\pi \cos \left\lgroup\frac{m\pi }{8} \right\rgroup \sin \left\lgroup\frac{n\pi }{4} \right\rgroup = (m\pi /2) \sin \left\lgroup \frac{m\pi }{8} \right\rgroup \cos \left\lgroup\frac{n\pi }{4} \right\rgroupRewriting it we obtain
\left\lgroup\frac{n\pi }{4} \right\rgroup \tan \left\lgroup\frac{n\pi }{4} \right\rgroup = \left\lgroup \frac{m \pi }{8} \right\rgroup \tan \left\lgroup\frac{m \pi }{8} \right\rgroupThis is of the form X tanX = Y tan Y, and thus the solution must be
X = Y , or
m = 2n (10-85)
Next, from Eq. (10-46), the cutoff frequency is
f_{c(mn)} = \frac{1}{2\sqrt{\mu \epsilon } } \sqrt{ \left\lgroup \frac{m}{a} \right\rgroup^{2} + \left\lgroup\frac{n}{b} \right\rgroup^{2} } [Hz] (10-46)
f_{c(mn)} = \frac{1}{2 \sqrt{\mu _{o} \epsilon _{o}} }\sqrt{\left\lgroup\frac{m}{a} \right\rgroup^{2} + \left\lgroup\frac{n}{b} \right\rgroup ^{2}} = 8.485 \times 10^{9} (10-86)
Solving Eqs. (10-85) and (10-86) for the mode numbers, using a = 2b, we get
n /b = 40 (10-87a)
m /a = 40 (10-87b)
Inserting Eq. (10-87) into the general expressions for E given in Eq. (10-75), we get
\mathscr{E} = A \left[ \pmb{a}_{x} \cos (40 \pi x) \sin (40\pi y) – \pmb{a}_{y} \sin (40 \pi x) \cos (40\pi y)\right] \\ \quad \times \sin (20\pi z – 1.8\pi \times 10^{10}t) [V/m] (10-88)
where A is a constant.
Inserting x = a/8 and y = b/4 into Eq. (10-88), and comparing Eq. (10-88) with the given \mathscr{E} in the problem, we obtain
a = 1/20 , b = 1/40 , and A = 8
Finally, the real instantaneous electric field of the mode of operation is
\mathscr{E} = 8 \left[ \pmb{a}_{x} \cos (40 \pi x) \sin (40\pi y) – \pmb{a}_{y} \sin (40 \pi x) \cos (40\pi y)\right] \\ \quad \times \sin (20\pi z – 1.8\pi \times 10^{10}t) [V/m].