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Waveguides and Cavities

9.1 Rectangular Waveguides

A rectangular waveguide with dimensions aa (width) and bb (height) supports electromagnetic waves propagating in the zz-direction. Two families of modes exist: TE (transverse electric, Ez=0E_z = 0) and TM (transverse magnetic, Bz=0B_z = 0).

TEmn_{mn} modes. The longitudinal field is Bz=B0cos(mπx/a)cos(nπy/b)ei(kzωt)B_z = B_0\cos(m\pi x/a)\cos(n\pi y/b)\,e^{i(kz-\omega t)}.

The transverse fields are determined from BzB_z via:

Ex=iωkc2Bzy,Ey=iωkc2BzxE_x = \frac{i\omega}{k_c^2}\frac{\partial B_z}{\partial y}, \quad E_y = -\frac{i\omega}{k_c^2}\frac{\partial B_z}{\partial x}

Bx=ikkc2Bzx,By=ikkc2BzyB_x = \frac{-ik}{k_c^2}\frac{\partial B_z}{\partial x}, \quad B_y = \frac{-ik}{k_c^2}\frac{\partial B_z}{\partial y}

Where kc2=(mπ/a)2+(nπ/b)2k_c^2 = (m\pi/a)^2 + (n\pi/b)^2 is the cutoff wavenumber.

Cutoff frequency: Waves propagate only when ω>ωc,mn\omega > \omega_{c,mn} where:

fc,mn=c2(ma)2+(nb)2f_{c,mn} = \frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}

The dominant (lowest frequency) mode is TE10_{10} with fc,10=c/(2a)f_{c,10} = c/(2a) (for a>ba > b).

Dispersion relation:

k=ω2c2kc2,vphase=ωk=c1(ωc/ω)2>ck = \sqrt{\frac{\omega^2}{c^2} - k_c^2}, \quad v_{\text{phase} = \frac{\omega}{k} = \frac{c}{\sqrt{1 - (\omega_c/\omega)^2}} > c}

vgroup=dωdk=c1(ωcω)2<cv_{\text{group} = \frac{d\omega}{dk} = c\sqrt{1 - \left(\frac{\omega_c}{\omega}\right)^2} < c}

The product vpvg=c2v_p \cdot v_g = c^2.

9.2 Waveguide Impedance and Power Flow

The wave impedance for TE modes:

ZTE=ExHy=ωμ0k=Z01(fc/f)2Z_{\text{TE} = \frac{E_x}{H_y} = \frac{\omega\mu_0}{k} = \frac{Z_0}{\sqrt{1 - (f_c/f)^2}}}

Where Z0=μ0/ε0377ΩZ_0 = \sqrt{\mu_0/\varepsilon_0} \approx 377\,\Omega is the impedance of free space.

The time-averaged power carried by TE10_{10} mode:

P=ab4E02βωμ0=ab4ZTEE02\langle P \rangle = \frac{ab}{4}E_0^2\frac{\beta}{\omega\mu_0} = \frac{ab}{4Z_{\text{TE}}E_0^2}

Where β=k\beta = k is the propagation constant and E0E_0 is the peak electric field.

9.3 Resonant Cavities

A rectangular cavity of dimensions a×b×da \times b \times d supports standing waves at resonant frequencies:

fmnp=c2(ma)2+(nb)2+(pd)2f_{mnp} = \frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 + \left(\frac{p}{d}\right)^2}

Where m,n,pm, n, p are non-negative integers (not all zero). For TM modes, p1p \geq 1; for TE modes, mm and nn cannot both be zero.

Quality factor:

Q=ω×energystoredpowerdissipated=2π×energystoredenergylostpercycleQ = \frac{\omega \times \text{energy} stored}{\text{power} dissipated} = \frac{2\pi \times \text{energy} stored}{\text{energy} lost per cycle}

For a cavity with conducting walls of conductivity σ\sigma:

QVSδ32Q \approx \frac{V}{S\,\delta} \cdot \frac{3}{2}

Where VV is the cavity volume, SS is the surface area, and δ\delta is the skin depth.

Worked Example 9.1: X-Band Waveguide

Standard X-band waveguide (WR-90) has a=22.86a = 22.86 mm, b=10.16b = 10.16 mm.

(a) Cutoff frequency of TE10_{10} mode:

fc,10=c2a=3×1082×22.86×103=3×1084.572×102=6.56 GHzf_{c,10} = \frac{c}{2a} = \frac{3 \times 10^8}{2 \times 22.86 \times 10^{-3}} = \frac{3 \times 10^8}{4.572 \times 10^{-2}} = 6.56\ \text{GHz}

(b) At f=10f = 10 GHz (within X-band), is TE10_{10} the only propagating mode?

Cutoff of TE01_{01}: fc,01=c/(2b)=3×108/(2×10.16×103)=14.76f_{c,01} = c/(2b) = 3 \times 10^8/(2 \times 10.16 \times 10^{-3}) = 14.76 GHz.

Cutoff of TE20_{20}: fc,20=c/a=13.12f_{c,20} = c/a = 13.12 GHz.

Since 6.56<10<13.126.56 < 10 < 13.12 GHz, only TE10_{10} propagates. This single-mode operation is essential for low-loss, distortion-free signal transmission.

(c) Guide wavelength at 10 GHz:

\lambda_g = \frac{\lambda}{\sqrt{1 - (f_c/f)^2}} = \frac{30\ \text{mm}{\sqrt{1 - (6.56/10)^2}} = \frac{30}{\sqrt{1 - 0.430}} = \frac{30}{0.755} = 39.7\ \text{mm}}

(d) Phase and group velocities:

vp=c1(fc/f)2=3×1080.755=3.97×108 m/s=1.32cv_p = \frac{c}{\sqrt{1 - (f_c/f)^2}} = \frac{3 \times 10^8}{0.755} = 3.97 \times 10^8\ \text{m}/s = 1.32\,c

vg=c1(fc/f)2=3×108×0.755=2.27×108 m/s=0.756cv_g = c\sqrt{1 - (f_c/f)^2} = 3 \times 10^8 \times 0.755 = 2.27 \times 10^8\ \text{m}/s = 0.756\,c

Check: vp×vg=1.32c×0.756c=c2v_p \times v_g = 1.32c \times 0.756c = c^2. \checkmark