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

For the TE case $E_{z} = 0$ but $H_{z} \neq = 0$ , and $H_{z}$ satisfies the same Helmholtz equation,

\nabla^{2} H_{z} (x, y, z) + k^{2} H_{z} (x, y, z) = 0, \frac{\partial ^{2} H _{z}}{\partial x ^{2}} + \frac{\partial ^{2} H _{z}}{\partial y ^{2}} + h^{2} H_{z} = 0. (7.3.20)

With $E_{z} = 0$ in Eq. (7.3.3) the transverse fields follow from $H_{z}$ :

H_{x} E_{x} = - \frac{j τ}{h ^{2}} \frac{\partial H _{z}}{\partial x}, = - \frac{j ω μ}{h ^{2}} \frac{\partial H _{z}}{\partial y}, H_{y} E_{y} = - \frac{j τ}{h ^{2}} \frac{\partial H _{z}}{\partial y}, = \frac{j ω μ}{h ^{2}} \frac{\partial H _{z}}{\partial x} . (7.3.21)

The boundary conditions are different. $H_{z}$ is a magnetic field and is not required to vanish on a conductor. What must vanish is the tangential $E$ , which is now transverse: $E_{x}$ on the walls $y = 0$ , $b$ and $E_{y}$ on the walls $x = 0$ , $a$ . By Eq. (7.3.21)

E_{x} (x, 0, z) = 0, E_{x} (x, b, z) = 0 \Rightarrow \frac{\partial H _{z}}{\partial y} (x, 0, z) = 0, \frac{\partial H _{z}}{\partial y} (x, b, z) = 0, E_{y} (0, y, z) = 0, E_{y} (a, y, z) = 0 \Rightarrow \frac{\partial H _{z}}{\partial x} (0, y, z) = 0, \frac{\partial H _{z}}{\partial x} (a, y, z) = 0. (7.3.22)

The normal derivative of $H_{z}$ vanishes on every wall, $\partial H_{z} / \partial n = 0$ , because the tangential transverse $E$ is proportional to it.

The equation is the same as in the TM case, so the separation $H_{z} = X (x) Y (y) e^{- j τ z}$ goes through word for word: Eqs. (7.3.12), (7.3.13) and (7.3.14) hold with $H_{z}$ in place of $E_{z}$ . Only the boundary conditions change: now the slope of $X$ and $Y$ must vanish on the walls, not the functions themselves, and sines become cosines:

X^{'} (0) X^{'} (a) Y^{'} (0) Y^{'} (b) = k_{x} A_{2} = 0 \Rightarrow X (x) = A_{1} cos (k_{x} x), = - k_{x} A_{1} sin (k_{x} a) = 0 \Rightarrow k_{x m} = \frac{mπ}{a}, = k_{y} B_{2} = 0 \Rightarrow Y (y) = B_{1} cos (k_{y} y), = - k_{y} B_{1} sin (k_{y} b) = 0 \Rightarrow k_{y n} = \frac{nπ}{b},

now with $m, n = 0, 1, 2, \dots$ The same $h_{mn}$ and $τ_{mn}$ as in Eq. (7.3.15) result, and the TE $_{mn}$ mode is

H_{z, mn} (x, y, z) = C_{mn} cos (\frac{mπ}{a} x) cos (\frac{nπ}{b} y) \times e^{- j τ_{mn} z} . (7.3.23)

Either $m$ or $n$ can be zero, but not both at the same time. A cosine with index zero is the constant $1$ , a perfectly good function, so TE $_{10}$ , TE $_{20}$ , TE $_{01}$ , … exist. With $m = n = 0$ , however, $H_{z}$ is a constant, $h = 0$ , and every transverse component in Eq. (7.3.21) is a derivative of a constant: there is no transverse field and no wave.

Differentiating each term of the sum $H_{z} = \sum_{m = 0}^{\infty} \sum_{n = 0}^{\infty} H_{z, mn}$ according to Eq. (7.3.21) gives the complete TE field ( $m$ and $n$ not both zero):

H_{z} (x, y, z) = m = 0 \sum \infty n = 0 \sum \infty C_{mn} cos (\frac{mπ}{a} x) \times cos (\frac{nπ}{b} y) e^{- j τ_{mn} z}, E_{x} (x, y, z) = j ω μ m = 0 \sum \infty n = 0 \sum \infty \frac{nπ}{b h _{mn}^{2}} C_{mn} \times cos (\frac{mπ}{a} x) sin (\frac{nπ}{b} y) e^{- j τ_{mn} z}, E_{y} (x, y, z) = - j ω μ m = 0 \sum \infty n = 0 \sum \infty \frac{mπ}{a h _{mn}^{2}} C_{mn} \times sin (\frac{mπ}{a} x) cos (\frac{nπ}{b} y) e^{- j τ_{mn} z}, H_{x} (x, y, z) = j m = 0 \sum \infty n = 0 \sum \infty \frac{mπ τ _{mn}}{a h _{mn}^{2}} C_{mn} \times sin (\frac{mπ}{a} x) cos (\frac{nπ}{b} y) e^{- j τ_{mn} z}, H_{y} (x, y, z) = j m = 0 \sum \infty n = 0 \sum \infty \frac{nπ τ _{mn}}{b h _{mn}^{2}} C_{mn} \times cos (\frac{mπ}{a} x) sin (\frac{nπ}{b} y) e^{- j τ_{mn} z} . (7.3.24)

The checks are as before: $E_{x} / H_{y} = - E_{y} / H_{x} = ω μ / τ_{mn} = Z_{T E}$ , and $E_{x} \propto sin (nπ y / b)$ vanishes on $y = 0, b$ , $E_{y} \propto sin (mπ x / a)$ on $x = 0, a$ .

Cutoff and the dominant mode

Since $h_{mn}$ is the same as for TM, so is the cutoff,

f_{co, mn} = \frac{1}{2 μ ε} (\frac{m}{a})^{2} + (\frac{n}{b})^{2}, (7.3.25)

but now with one index allowed to be zero. The lowest cutoff is that of TE $_{10}$ or of TE $_{01}$ , whichever of $a$ and $b$ is larger. With the convention $a > b$ the dominant mode is TE $_{10}$ ,

f_{co, 10} = \frac{v}{2 a}, that is a = \frac{λ}{2} at cutoff . (7.3.26)

TE$_{mn}$ modes of the rectangular guide

H_{z} = C_{mn} cos \frac{mπ x}{a} cos \frac{nπ y}{b} e^{- j τ_{mn} z}, h_{mn}^{2} = (\frac{mπ}{a})^{2} + (\frac{nπ}{b})^{2}, m, n = 0, 1, 2, \dots not both 0. (7.3.27)

The transverse fields follow from Eq. (7.3.21), the cutoff is Eq. (7.3.25) and $Z_{T E} = ω μ / τ_{mn}$ . For $a > b$ the dominant mode of the guide is TE $_{10}$ , $f_{co, 10} = v /2 a$ .

In plain words

For TE modes the walls fix the slope of $H_{z}$ , not its value, so the patterns are half cosines, and a pattern that does not vary at all in one direction is allowed. The simplest of them, one half cosine across the width and nothing along the height, is TE $_{10}$ , the mode with the lowest cutoff of all.

Definition · Degenerate modes

Modes with different field patterns but the same cutoff frequency are degenerate. TE $_{mn}$ and TM $_{mn}$ with $m, n \geq 1$ have the same $h_{mn}$ , hence the same cutoff, $τ$ , $λ_{g}$ , $v_{p}$ and $v_{g}$ . They differ in field pattern and in wave impedance. A square guide ( $a = b$ ) adds the degeneracy of TE $_{10}$ with TE $_{01}$ , and a guide with $a = 2 b$ that of TE $_{20}$ with TE $_{01}$ .

Common mistake · Imposing

H_{z} = 0

on the walls

Copying the TM boundary condition into the TE problem gives $sin$ instead of $cos$ and loses the modes with a zero index, among them the dominant TE $_{10}$ . The boundary condition is on the electric field: in a TE mode the tangential $E$ is proportional to the normal derivative of $H_{z}$ , so $\partial H_{z} / \partial n = 0$ .

7.3.3 TE modes

Cutoff and the dominant mode