Waveguides & Cavities

Step 1: Separation of variables

Write $E_z(x,y) = X(x)Y(y)$. The Helmholtz equation separates into $X'' + k_x^2 X = 0$ and $Y'' + k_y^2 Y = 0$ with $k_x^2 + k_y^2 = \gamma^2$.

Step 2: Apply boundary conditions

For TM modes, $E_z = 0$ on all walls. This requires $X(0) = X(a) = Y(0) = Y(b) = 0$:

$$E_z = E_0\sin\!\left(\frac{m\pi x}{a}\right)\sin\!\left(\frac{n\pi y}{b}\right), \quad m,n = 1,2,3,\ldots$$

Step 3: Eigenvalues and cutoff frequencies

$$\gamma_{mn}^2 = \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2$$

The cutoff frequency is $f_{c,mn} = c\gamma_{mn}/(2\pi)$. Below this frequency, the mode is evanescent.

Step 4: Propagation constant

$$k_z = \sqrt{\frac{\omega^2}{c^2} - \gamma_{mn}^2} = \frac{\omega}{c}\sqrt{1 - \left(\frac{f_c}{f}\right)^2}$$

Step 5: Transverse fields

The transverse E and H fields are determined from $E_z$ via: $$\mathbf{E}_T = \frac{ik_z}{\gamma^2}\nabla_T E_z, \qquad \mathbf{H}_T = \frac{-i\omega\epsilon_0}{\gamma^2}\hat{z}\times\nabla_T E_z$$

The lowest TM mode is TM$_{11}$, since both $m$ and $n$ must be at least 1.

Step 1: TE$_{10}$ longitudinal field

With $m=1$, $n=0$: $B_z = B_0\cos(\pi x/a)$. The cutoff frequency is $f_c = c/(2a)$, the lowest of all modes when $a > b$.

Step 2: Transverse fields

$E_y = E_0\sin(\pi x/a)e^{i(k_z z - \omega t)}$, $H_x = -(k_z/\omega\mu_0)E_y$. The electric field is polarized in the $y$-direction and varies as a half-sine across the width.

Step 3: Wave impedance

$$Z_{\text{TE}} = \frac{E_y}{H_x} = \frac{\omega\mu_0}{k_z} = \frac{Z_0}{\sqrt{1-(f_c/f)^2}}$$

The TE wave impedance exceeds the free-space impedance $Z_0 = 377$ $\Omega$ and diverges at cutoff.

Step 4: Time-averaged power flow

$$\langle P\rangle = \frac{1}{2}\text{Re}\int_S(\mathbf{E}\times\mathbf{H}^*)\cdot\hat{z}\,da = \frac{E_0^2 ab}{4Z_{\text{TE}}}$$

Step 5: Single-mode bandwidth

For a standard rectangular waveguide with $a = 2b$, only TE$_{10}$ propagates in the range $f_c < f < 2f_c$. This 2:1 bandwidth is the usable single-mode range, free from higher-order mode excitation.

Step 1: Helmholtz equation in cylindrical coordinates

$$\frac{1}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial B_z}{\partial\rho}\right) + \frac{1}{\rho^2}\frac{\partial^2 B_z}{\partial\phi^2} + \gamma^2 B_z = 0$$

Step 2: Separation of variables

$B_z(\rho,\phi) = R(\rho)\Phi(\phi)$. The azimuthal part gives $\Phi = e^{im\phi}$ with integer $m$. The radial part is Bessel's equation.

Step 3: Solution and boundary condition

$B_z = J_m(\gamma\rho)e^{im\phi}$. The TE boundary condition $\partial B_z/\partial\rho|_{\rho=R} = 0$ requires $J_m'(\gamma R) = 0$.

Step 4: Cutoff frequencies

If $x'_{mn}$ is the $n$-th zero of $J_m'$: $\gamma_{mn} = x'_{mn}/R$ and $f_{c,mn} = cx'_{mn}/(2\pi R)$.

Step 5: Dominant mode

The lowest TE mode is TE$_{11}$ ($x'_{11} = 1.841$), which is the dominant mode in a circular waveguide. The TE$_{01}$ mode ($x'_{01} = 3.832$) has the special property that its attenuation decreases with frequency, making it attractive for long-distance transmission.

Step 1: Definition of Q

$$Q = \omega_0\frac{W_{\text{stored}}}{P_{\text{loss}}} = \frac{2\pi \times \text{energy stored}}{\text{energy lost per cycle}}$$

Step 2: Stored energy in TE$_{101}$ mode

The volume integral of $\epsilon_0|E|^2/2 + |B|^2/(2\mu_0)$ over the cavity gives $W = \epsilon_0 E_0^2 abd/8$ at resonance.

Step 3: Wall losses from surface currents

The surface resistance is $R_s = 1/(\sigma\delta) = \sqrt{\omega\mu_0/(2\sigma)}$. The power loss per unit area is $R_s|H_{\text{tan}}|^2/2$. Integrating over all six walls:

$$P_{\text{loss}} = \frac{R_s}{2}\oint_S |\mathbf{H}_{\text{tan}}|^2\,da$$

Step 4: Evaluate the surface integral

Computing $|\mathbf{H}_{\text{tan}}|^2$ on each wall face and integrating yields a result proportional to $R_s$ times a geometric factor involving $a, b, d$.

Step 5: Final Q expression

$$\boxed{Q = \frac{\omega_0\mu_0}{2R_s}\cdot\frac{V}{S_{\text{eff}}}}$$

where $V$ is the cavity volume and $S_{\text{eff}}$ is an effective surface area. For copper at 10 GHz, $Q \sim 10^4$. Superconducting cavities achieve $Q \sim 10^{10}$.

Step 1: Unperturbed fields

Assume the fields are well-approximated by the perfect-conductor solutions. The tangential magnetic field at the wall determines the surface current: $\mathbf{K} = \hat{n}\times\mathbf{H}$.

Step 2: Power dissipated per unit length

$$\frac{dP_{\text{loss}}}{dz} = \frac{R_s}{2}\oint_C |\mathbf{H}_{\text{tan}}|^2\,dl$$

where the integral is around the waveguide perimeter.

Step 3: Transmitted power

$P = P_0 e^{-2\alpha z}$, so $dP/dz = -2\alpha P$.

Step 4: Solve for the attenuation constant

$$\boxed{\alpha = \frac{R_s\oint_C|\mathbf{H}_{\text{tan}}|^2\,dl}{4\text{Re}\int_S(\mathbf{E}\times\mathbf{H}^*)\cdot\hat{z}\,da}}$$

Step 5: Frequency dependence

For TE modes, $\alpha$ diverges near cutoff (vanishing group velocity) and at high frequency (increasing $R_s \propto \sqrt{f}$), with a minimum in between. The TE$_{01}$ mode in circular waveguide is exceptional: its attenuation decreases monotonically with frequency.