Show that the TMmn and TEmn modes are independent of each other in a rectangular waveguide, even though they have the same phase constant and the same group velocity.

Question

Show that the [latex]TM_{mn}[/latex] and [latex]TE_{mn}[/latex] modes are independent of each other in a rectangular waveguide, even though they have the same phase constant and the same group velocity.

Accepted Answer

The electric and magnetic field phasors of the [latex]TM_{mn}[/latex] mode, [latex]\pmb{E}_{TM}[/latex] and [latex]\pmb{H}_{TM}[/latex], are expressed by Eqs. (10-41) and (10-56), whereas the phasors [latex]\pmb{E}_{TE}[/latex] and [latex]\pmb{H}_{TE}[/latex] of the [latex]TE_{mn}[/latex] mode are expressed by Eqs. (10-74) and (10-75).

[latex]\boxed{E_{z} (x, y) = E_{o} \sin \left\lgroup\frac{m\pi }{a} x ight group \sin \left\lgroup\frac{n\pi }{b} y ight group } (m, n = 1,2,3,...) (10-41) \ \boxed{E_{x} = - \frac{\gamma }{h^{2}} \left\lgroup\frac{m\pi }{a} ight group E_{o}\cos \left\lgroup\frac{m \pi }{a} x ight group \sin \left\lgroup\frac{ n \pi }{b} y ight group } (m, n= 1,2,3,... ) (10-56a) \ \boxed{E_{y} = - \frac{\gamma }{h^{2}} \left\lgroup\frac{n\pi }{b} ight group E_{o}\sin \left\lgroup\frac{m \pi }{a} x ight group \cos \left\lgroup\frac{ n \pi }{b} y ight group } (10-56b) \ \boxed{H_{x} = \frac{j \omega \epsilon }{h^{2}} \left\lgroup\frac{n \pi }{b} ight group E_{o}\sin \left\lgroup\frac{m \pi }{a} x ight group \cos \left\lgroup\frac{ n \pi }{b} y ight group } (10-56c) \ \boxed{H_{y} = - \frac{j \omega \epsilon }{h^{2}} \left\lgroup\frac{m\pi }{a} ight group E_{o}\cos \left\lgroup\frac{m \pi }{a} x ight group \sin \left\lgroup\frac{ n \pi }{b} y ight group } (10-56d) \ \boxed{H_{z}(x,y) = H_{o} \cos \left\lgroup\frac{m\pi }{a} x ight group \cos \left\lgroup\frac{n\pi }{b}y ight group } ( m,n = 0,1,2,...) (10-74) \ \boxed{E_{x} =\frac{j\omega \mu }{h^{2}} \left\lgroup\frac{n\pi }{b} ight group H_{o} \cos \left\lgroup \frac{m\pi }{a}x ight group \sin \left\lgroup\frac{n\pi }{b}y ight group} (m,n = 0,1,2,...) (10-75a) \ \boxed{E_{y} = - \frac{j\omega \mu }{h^{2}} \left\lgroup \frac{m\pi }{a} ight group H_{o} \sin \left\lgroup\frac{m\pi }{a}x ight group \cos \left\lgroup\frac{n\pi }{b}y ight group } (10-75b) \\boxed{H_{x} = \frac{\gamma }{h^{2}} \left\lgroup\frac{m \pi }{a} ight group H_{o} \sin \left\lgroup \frac{m\pi }{a}x ight group \cos \left\lgroup \frac{n\pi }{b} y ight group} (10-75c)\ \boxed{H_{y} = \frac{\gamma }{h^{2}} \left\lgroup\frac{n \pi }{b} ight group H_{o} \cos \left\lgroup \frac{m\pi }{a}x ight group \sin \left\lgroup \frac{n\pi }{b} y ight group} (10-75d)[/latex]

The total electric and magnetic fields in the waveguide are therefore

[latex]\pmb{E} = \pmb{E}_{TM} + \pmb{E}_{TE} (10-106a) \ \pmb{H} = \pmb{H}_{TM} + \pmb{H}_{TE} (10-106b) [/latex]

The time-average power density in the waveguide is

[latex]\left\langle P ight angle =\int_{S}{\left\langle \pmb{S} ight angle } \pmb{\cdot } d \pmb{s} = \frac{1}{2} Re\int_{S}{ \left\lgroup \pmb{E} imes \pmb{H}^{*} ight group \pmb{\cdot } \pmb{a}_{z} dx dy} \ \quad \quad = \frac{1}{2} Re \left[\int_{S}{\left\lgroup\pmb{E}_{TM} imes \pmb{H}^{*}_{TM} + \pmb{E}_{TE} imes \pmb{H}^{*}_{TE} ight group \pmb{\cdot } \pmb{a}_{z} dx dy} ight] \ \quad \quad + \frac{1}{2} Re \left[\int_{S}{\left\lgroup\pmb{E}_{TM} imes \pmb{H}^{*}_{TE} + \pmb{E}_{TE} imes \pmb{H}^{*}_{TM} ight group \pmb{\cdot } \pmb{a}_{z} dx dy} ight] [/latex] (10-107)

The integrand of the second integral on the right-hand side of Eq. (10-107) is written as

[latex] \pmb{E}_{TM} imes \pmb{H}^{*}_{TE} + \pmb{E}_{TE} imes \pmb{H}^{*}_{TM} = (E_{x , TM} H^{*}_{y , TE} + E_{y , TM} H^{*}_{x , TE}) \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + (E_{x , TE} H^{*}_{y , TM} + E_{y , TE} H^{*}_{x , TM})[/latex] (10-108)

Substitution of Eqs. (10-56) and (10-75) into Eq. (10-108), in conjunction with [latex]\gamma = j \beta _{mn}[/latex], shows that each parenthesis on the right-hand side of Eq. (10-108) vanishes during the integration. Thus, Eq. (10-107) becomes

[latex]\left\langle P ight angle = \frac{1}{2} Re \left[\int_{S}{\left\lgroup\pmb{E}_{TM} imes \pmb{H}^{*}_{TM} ight group \pmb{\cdot } \pmb{a}_{z} dx dy} ight] + \frac{1}{2} Re \left[\int_{S}{\left\lgroup\pmb{E}_{TE} imes \pmb{H}^{*}_{TE} ight group \pmb{\cdot } \pmb{a}_{z} dx dy} ight] [/latex] (10-109)

From (10-109), we see that there is no coupling between the two waves: the first term on the right-hand side of Eq.(10-109) is the total power of the [latex]TM_{mn}[/latex] mode, while the second term is that of the [latex]TE_{mn}[/latex] mode. Thus, the [latex]TM_{mn}[/latex] and [latex]TE_{mn}[/latex] waves are independent of each other in the waveguide.

Question 10.10: Show that the TMmn and TEmn modes are independent of each ot......

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