Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

We need only a few results of the guide analysis (Sections 7.1 and 7.3). A mode of a uniform guide filled with a homogeneous medium has a transverse pattern $ψ (x, y)$ and travels as $e^{\mp j τ z}$ , with $τ$ fixed by Eq. (7.8.1). In the rectangular guide the cutoff wavenumbers are

h_{mn}^{2} f_{co, mn} = (\frac{mπ}{a})^{2} + (\frac{nπ}{b})^{2}, = \frac{v}{2} (\frac{m}{a})^{2} + (\frac{n}{b})^{2}, (7.8.2)

and the dominant TE $_{10}$ mode has $E_{y} \propto sin (π x / a)$ and $f_{co, 10} = v /2 a$ . For a TM mode the transverse electric field follows from the longitudinal one. For a wave travelling along $+ z$ it is $E_{T} = - (j τ / h^{2}) \nabla_{T} E_{z}$ , and for a wave travelling along $- z$ the sign of $τ$ changes, $E_{T} = + (j τ / h^{2}) \nabla_{T} E_{z}$ .

Now close the guide by conducting walls at $z = 0$ and $z = d$ . A mode travelling along $+ z$ is reflected by the wall at $z = d$ , the reflected mode is reflected again at $z = 0$ , and the field inside is the sum of the two travelling waves. For a TM mode,

E_{z} = (C^{+} e^{- j τ z} + C^{-} e^{j τ z}) ψ (x, y), (7.8.3)

E_{T} = - \frac{j τ}{h ^{2}} (C^{+} e^{- j τ z} - C^{-} e^{j τ z}) \nabla_{T} ψ . (7.8.4)

The minus sign inside the second bracket is the change of sign of $E_{T}$ with the direction of travel. The end walls are perfect conductors, so the tangential electric field, which is $E_{T}$ , must vanish on them. At $z = 0$ this requires $C^{-} = C^{+}$ , and then

E_{z} E_{T} = 2 C^{+} cos τ z ψ (x, y), = - \frac{2 τ}{h ^{2}} C^{+} sin τ z \nabla_{T} ψ, (7.8.5)

since $e^{- j τ z} - e^{j τ z} = - 2 j sin τ z$ . The wave has become a standing wave: $E_{T}$ has a node on the wall $z = 0$ and $E_{z}$ , normal to that wall, a maximum. At $z = d$ we need $E_{T} = 0$ as well, that is $sin τ d = 0$ .

Resonance condition of a shorted guide

The two end walls allow only the phase constants

⟺ τ = \frac{p π}{d}, p = 0, 1, 2, \dots d = p \frac{λ _{g}}{2}, (7.8.6)

where $λ_{g} = 2 π / τ$ is the guide wavelength. The end walls sit at nodes of the transverse electric field, a whole number of half guide wavelengths apart.

In plain words

A cavity is a piece of guide whose length fits the standing wave exactly. Lengthen it by half a guide wavelength and another standing wave fits, at the same frequency, with one more half-period along $z$ .

The standing wave between the end walls for p=2 (d=\lambda_g). The transverse electric field has nodes on both walls and in the middle. The transverse magnetic field, which is linked to it by one derivative along z, has its maxima there. — **Figure 7.8.2.** The standing wave between the end walls for $p = 2$ ( $d = λ_{g}$ ). The transverse electric field has nodes on both walls and in the middle. The transverse magnetic field, which is linked to it by one derivative along $z$ , has its maxima there.

The same reasoning holds for every component and for TE modes as well, and it gives a rule that we shall use all the time.

The rule for every component

The components that must vanish on the end walls, the transverse $E$ and the normal $H$ ( $H_{z}$ ), vary as $sin (p π z / d)$ . The components linked to them by one derivative $\partial / \partial z$ vary as $cos (p π z / d)$ . These are $E_{z}$ and $H_{T}$ for TM modes and $H_{T}$ for TE modes. Figure 7.8.2 shows the pattern for $p = 2$ . Electric and magnetic maxima are a quarter of a guide wavelength apart, as in the standing wave in front of a conducting plane (Section 6.3).

7.8.2 From a shorted waveguide to a cavity