Waveguide Cutoff Frequency

Equivalent forms

f_{c,mn} = \frac{c}{2}\sqrt{(m/a)^2+(n/b)^2}

\lambda_g = \frac{\lambda}{\sqrt{1-(f_c/f)^2}}

A pipe becomes a high-pass filter purely from geometry — its width alone sets the frequency below which the universe refuses to let a wave through it.

Symbol	Name	SI unit	Dimension	Range
$f_c$	Cutoff frequencyoutput Lowest frequency that propagates (TE₁₀)	GHz	T⁻¹	0 – 30
$a$	Guide width Broad inner dimension of the guide	mm	L	10 – 100
$f$	Operating frequency Frequency of the signal sent in	GHz	T⁻¹	1 – 30

Unit systems

Symbol	SI	CGS	Imperial
$f_c$	GHz	GHz	GHz
$a$	mm	mm	in
$f$	GHz	GHz	GHz

Where it holds

Formula

f_c = c/2

a is the

TE_{10}

mode of an ideal lossless rectangular guide filled with vacuum/air. Dielectric filling replaces c by

c/\surd \varepsilon _r

; lossy walls and higher modes need the general dispersion relation.

Dimensional analysis

c/(2a $) = (m/s)/m = s^{-1} = Hz \checkmark$

Discovery

Lord Rayleigh · 1897

Rayleigh proved theoretically that electromagnetic waves can travel down hollow conducting tubes, but only above a cutoff frequency set by the cross-section. Decades later, WWII radar drove waveguides into practical microwave engineering.

Try this

A hollow metal pipe carries radar and satellite signals with almost no loss — but only above a sharp frequency. Below it, nothing gets through. Why?

For a rectangular waveguide of inner width a = 22.86 mm (WR-90), find the TE₁₀ cutoff frequency and explain what happens to a 7 GHz signal.

Research status: stable

Real-world applications

Radar and satellite microwave links
Cavity filters and microwave plumbing
Why a microwave-oven mesh (sub-mm holes) blocks 2.45 GHz but passes light
Below-cutoff attenuators and waveguide-beyond-cutoff seals

Common misconceptions

Any frequency travels down a pipe — only those above cutoff propagate
The signal speeds up above c inside — phase velocity exceeds c but energy (group velocity) stays below c
Cutoff depends on the material inside — for an air guide it's pure geometry, c/2a

Experimental verification

Sweeping a microwave source through a guide shows transmission switch on abruptly at

f_c = c/2

a, and the measured guide wavelength

\lambda _g = \lambda /\surd (1-(f_c/f)^{2})

matches slotted-line measurements. Standard WR-bands are defined by these cutoffs.

Derivation

Solving Maxwell's equations with conducting-wall boundary conditions gives modes with transverse wavenumber

k_c = \surd ((m\pi

/a

)^{2}+(n\pi /b)^{2})

.

Propagation needs

k_z^{2} = (\omega /c)^{2} - k_c^{2}

> 0, i.e.

\omega

> ck_c.

For

TE_{10} (m=1

,

n=0)

,

k_c = \pi

/a, giving

f_c = c/(2

a).

Limiting cases

f < f_c⟶ evanescent (blocked)No real propagation constant — the field decays exponentially down the guide.

f = f_c

⟶

\lambda _g \to \infty

At cutoff the guide wavelength diverges and group velocity

\to 0

; the wave stands still.

f ≫ f_c⟶

\lambda _g \to \lambda

,

v_g \to c

Far above cutoff the wave behaves almost like free space.

What if…

What if you send a 5 GHz signal into WR-90?

Below the 6.56 GHz cutoff it can't propagate — it decays exponentially; the guide acts as a high-pass filter.

What if you fill the guide with a dielectric

\varepsilon _r = 4

?

Cutoff halves to $c/(2a\surd \varepsilon _r)$ , letting lower frequencies through — a way to shrink guides.

What if you widen the guide?

f_c drops as 1/a, but very wide guides start supporting unwanted higher-order modes.

1

WR-90 X-band guide

Given ·

a:: 0.02286
f:: 7000000000

Find · f_c and behavior

Steps

$f_c = c/(2$ a $) = 2.998\times 10^{8} / 0.04572 \approx 6.56\,\mathrm{GHz}$
7 GHz > 6.56 GHz, so $TE_{10}$ propagates
$\lambda _g = \lambda /\surd (1-(6.56/7)^{2}) \approx \lambda /0.35$ — stretched well beyond free-space $\lambda$

Result ·

f_c = c/2

a = 3\times 10^{8}/(2\times 0.02286) \approx 6.56\,\mathrm{GHz}

; a 7 GHz signal (> f_c) propagates

Speed of Electromagnetic Waves→Maxwell's Equations (Unified)→Poynting Vector→