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

For the TM case $H_{z} = 0$ but $E_{z} \neq = 0$ :

Δ E_{z} (x, y, z) + k^{2} E_{z} (x, y, z) = 0, (7.2.15)

and Eq. (7.1.11) with $H_{z} = 0$ , $\partial / \partial x = 0$ gives

H_{x} E_{x} H_{y} E_{y} = \frac{1}{h ^{2}} j ω ε \frac{\partial E _{z}}{\partial y}, = - \frac{1}{h ^{2}} j τ \frac{\partial E _{z}}{\partial x} = 0, = - \frac{1}{h ^{2}} j ω ε \frac{\partial E _{z}}{\partial x} = 0, = - \frac{1}{h ^{2}} j τ \frac{\partial E _{z}}{\partial y} . (7.2.16)

The boundary conditions are now on $E_{z}$ itself:

E_{z} (x, 0, z) = 0, E_{z} (x, b, z) = 0. (7.2.17)

Only $E_{z}$ needs a condition: $E_{x}$ is proportional to $\partial E_{z} / \partial x = 0$ , so the other tangential component vanishes by itself. $E_{y}$ is normal to the plates and may be nonzero there, where it ends on a surface charge.

Exactly as for TE, $\partial^{2} E_{z} / \partial y^{2} + h^{2} E_{z} = 0$ , and with $E_{z} (y, z) = Y (y) e^{- j τ z}$

Y (y) E_{z} (y, z) = C_{1} cos (h y) + C_{2} sin (h y), = [C_{1} cos (h y) + C_{2} sin (h y)] e^{- j τ z} .

It is the same equation as in the TE case. Only the boundary condition differs: now the function itself, not its slope, must vanish on the plates. At $y = 0$ ,

E_{z} (x, 0, z) = C_{1} e^{- j τ z} = 0 ⟹ C_{1} = 0, E_{z} (y, z) = C_{2} sin (h y) e^{- j τ z},

and at $y = b$ , $C_{2} sin (hb) = 0$ , so again $h_{n} b = nπ$ with $n = 1, 2, \dots$ and

E_{z n} (y, z) = C_{2 n} sin (\frac{nπ}{b} y) e^{- j τ_{n} z} . (7.2.18)

With $\partial E_{z n} / \partial y = (nπ / b) C_{2 n} cos (nπ y / b) e^{- j τ_{n} z}$ in Eq. (7.2.16),

E_{z} (y, z) = n = 1 \sum \infty C_{2 n} sin (\frac{nπ}{b} y) e^{- j τ_{n} z}, (7.2.19)

H_{y} (y, z) = 0, E_{x} (y, z) = 0, (7.2.20)

E_{y} (y, z) = - j n = 1 \sum \infty (\frac{nπ τ _{n}}{b h _{n}^{2}}) C_{2 n} \times cos (\frac{nπ}{b} y) e^{- j τ_{n} z}, (7.2.21)

H_{x} (y, z) = j ω ε n = 1 \sum \infty (\frac{nπ}{b h _{n}^{2}}) C_{2 n} \times cos (\frac{nπ}{b} y) e^{- j τ_{n} z} . (7.2.22)

The $n$ th term is the TM $_{n}$ mode.

TM$_n$ mode of the parallel-plate guide

E_{z} E_{y} H_{x} = C sin (h_{n} y) e^{- j τ_{n} z}, h_{n} = \frac{nπ}{b}, = - \frac{j τ _{n}}{h _{n}} C cos (h_{n} y) e^{- j τ_{n} z}, = \frac{j ω ε}{h _{n}} C cos (h_{n} y) e^{- j τ_{n} z}, (7.2.23)

with $- \frac{E _{y}}{H _{x}} = \frac{τ _{n}}{ω ε} = Z_{T M}$ and $E_{x} = H_{y} = H_{z} = 0$ .

In plain words

The magnetic field of a TM $_{n}$ mode points along the plates ( $x$ ), and it is strongest on the plates. The electric field lies in the $y z$ plane: its lines start on positive surface charge on a plate, arch along $z$ , and end on negative charge on the same plate (Figure 7.2.3).

The TM_1 mode, side view, real fields at t=0 (f=1.5f_{co,1}). The electric field lines (orange) are the contours of \cos(\pi y/b)\cos\tau z. They leave the positive surface charges of a plate and return to the negative charges of the same plate, half a guide wavelength away. The magnetic field is along \pm x and is strongest at the plates, where it is tangential and carries the surface current. — **Figure 7.2.3.** The TM $_{1}$ mode, side view, real fields at $t = 0$ ( $f = 1.5 f_{co, 1}$ ). The electric field lines (orange) are the contours of $cos (π y / b) cos τ z$ . They leave the positive surface charges of a plate and return to the negative charges of the same plate, half a guide wavelength away. The magnetic field is along $\pm x$ and is strongest at the plates, where it is tangential and carries the surface current.

Cutoff of the TM modes

The phase constant is again $τ_{n} = k^{2} - (nπ / b)^{2}$ , and in order to have a propagating TM $_{n}$ mode $τ_{n}$ must be real: $f > f_{co, n} = n v /2 b$ , Eq. (7.2.14). The cutoff frequency is the same as that of the TE waves, and the lowest TM mode with $n \geq 1$ is TM $_{1}$ .

Definition · Degenerate modes

TE $_{n}$ and TM $_{n}$ have the same $h_{n} = nπ / b$ , hence the same $f_{co, n}$ , $τ_{n}$ , $λ_{g}$ , $v_{p}$ and $v_{g}$ . Two modes with the same cutoff are called degenerate. They differ in the field pattern and in the wave impedance, $Z_{T M} < η < Z_{T E}$ .

Quick check

Show from Eq. (7.2.23) that the TM $_{n}$ field satisfies both boundary conditions on the plates.

The tangential components are $E_{z} \propto sin (nπ y / b)$ , which is zero at $y = 0$ and $y = b$ , and $E_{x} = 0$ . The normal $H_{y}$ is zero everywhere. $E_{y} \propto cos (nπ y / b) = \pm 1$ on the plates is normal to them and ends on a surface charge $ρ_{s} = \pm ε E_{y}$ .

7.2.3 TM modes

Cutoff of the TM modes