Holooly Plus Logo

Question 10.4: How many component plane-waves does the TM21 mode have in a ......

How many component plane-waves does the TM_{21} mode have in a rectangular waveguide with dimensions a and b?

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

From Eq. (10-41) we write

\boxed{E_{z}(x, y) = E_{o}\sin \left\lgroup\frac{m \pi }{a} x \right\rgroup \sin \left\lgroup\frac{ n \pi }{b} y \right\rgroup }       (m n , 1,2,3,… = )         (10-41)

E_{z} = E_{o}\sin \left\lgroup\frac{2 \pi }{a} x \right\rgroup \sin \left\lgroup\frac{ \pi }{b} y \right\rgroup \\ \quad =E_{o} \frac{1}{2j}(e^{j 2 \pi x / a} – e^{-j 2 \pi x /a}) \frac{1}{2 j }(e^{j \pi y / b} – e^{-j \pi y /b}) \\ \quad = -\frac{E_{o}}{4}\left[e^{j 2 \pi x / a + j \pi y /b} – e^{j 2 \pi x /a – j \pi y /b} – e^{- j 2 \pi x / a + j \pi y /b} + e^{-j 2 \pi x /a – j \pi y / b}\right]

From Eq. (10-56a) write

\boxed{E_{x} = – \frac{\gamma }{h^{2}} \left\lgroup\frac{m\pi }{a} \right\rgroup E_{o}\cos \left\lgroup\frac{m \pi }{a} x \right\rgroup \sin \left\lgroup\frac{ n \pi }{b} y \right\rgroup }

(m n , 1,2,3,… = )       (10-56a)

E_{x} = – \frac{\gamma }{h^{2}} \left\lgroup\frac{2 \pi }{a} \right\rgroup E_{o}\cos \left\lgroup\frac{2 \pi }{a} x \right\rgroup \sin \left\lgroup\frac{ \pi }{b} y \right\rgroup \\ \quad = – \frac{\gamma }{h^{2}} \left\lgroup\frac{2 \pi }{a} \right\rgroup E_{o} \frac{1}{2}(e^{j 2 \pi x / a} + e^{-j 2 \pi x /a}) \frac{1}{2 j }(e^{j \pi y / b} – e^{-j \pi y /b}) \\ \quad = – \frac{\gamma }{h^{2}} \left\lgroup\frac{2 \pi }{a} \right\rgroup \frac{E_{o}}{4j} \left[e^{j 2 \pi x / a + j \pi y /b} – e^{j 2 \pi x /a – j \pi y /b} + e^{- j 2 \pi x / a + j \pi y /b} – e^{-j 2 \pi x /a – j \pi y / b}\right]

From Eq. (10-56b) write

\boxed{E_{y} = – \frac{\gamma }{h^{2}} \left\lgroup\frac{n\pi }{b} \right\rgroup E_{o}\sin \left\lgroup\frac{m \pi }{a} x \right\rgroup \cos \left\lgroup\frac{ n \pi }{b} y \right\rgroup }                                   (10-56b)
E_{y} = – \frac{\gamma }{h^{2}} \left\lgroup\frac{ \pi }{b} \right\rgroup E_{o} \sin \left\lgroup\frac{2 \pi }{a} x \right\rgroup \cos \left\lgroup\frac{ \pi }{b} y \right\rgroup \\ \quad = – \frac{\gamma }{h^{2}} \left\lgroup\frac{ \pi }{b} \right\rgroup E_{o} \frac{1}{2}(e^{j 2 \pi x / a} – e^{-j 2 \pi x /a}) \frac{1}{2 j }(e^{j \pi y / b} + e^{-j \pi y /b}) \\ \quad = – \frac{\gamma }{h^{2}} \left\lgroup\frac{ \pi }{b} \right\rgroup \frac{E_{o}}{4j} \left[e^{j 2 \pi x / a + j \pi y /b} + e^{j 2 \pi x /a – j \pi y /b} – e^{- j 2 \pi x / a + j \pi y /b} – e^{-j 2 \pi x /a – j \pi y / b}\right]

The three components of the electric field can be combined into the electric field phasor, in a lossless rectangular waveguide( \gamma = j \beta _{21}), as

\pmb{E} = \left\{E_{x} (x, y) \pmb{a}_{x} + E_{y} (x, y) \pmb{a}_{y} + E_{z}(x, y) \pmb{a}_{z}\right\} e^{- j \beta _{21} z}

In view of these, we note that the TM_{21} mode comprises four plane waves,

e^{j 2 \pi x / a + j \pi y /b – j \beta _{21}z} , e^{j 2 \pi x /a – j \pi y /b – j \beta _{21}z} , e^{- j 2 \pi x / a + j \pi y /b – j \beta _{21}z} , and e^{-j 2 \pi x /a – j \pi y / b – j \beta _{21}z}

It is important to note that the four component plane-waves have the same wavenumber k as well as the same phase constant \beta _{21}.

Related Answered Questions