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

The boundary conditions on the walls allow only discrete values of $h$ , one for each mode, and these values depend on the shape and size of the cross-section only (Section 7.2 finds them for two plates, $h_{n} = nπ / b$ ). For a given mode, Eq. (7.1.10) gives

τ = k^{2} - h^{2} = ω^{2} μ ε - h^{2}, (7.1.13)

which is real only if $k > h$ . The frequency at which $k = h$ , that is $2 π f_{c} μ ε = h$ , is the cutoff frequency of the mode. Writing $k^{2} - h^{2} = k^{2} (1 - h^{2} / k^{2}) = k^{2} (1 - f_{c}^{2} / f^{2})$ ,

Cutoff frequency

f_{c} τ α = \frac{h}{2 π μ ε}, = k 1 - (\frac{f _{c}}{f})^{2} (f > f_{c}), = h 1 - (\frac{f}{f _{c}})^{2} (f < f_{c}) . (7.1.14)

$f > f_{c}$ : $τ$ is real and the mode propagates as $e^{- j τ z}$ .
$f < f_{c}$ : $τ^{2} < 0$ . Choosing the root that decays in the direction of travel, $τ = - j α$ with $α = h^{2} - k^{2}$ , the field varies as $e^{- j τ z} = e^{- α z}$ . It does not oscillate along $z$ , it only decays: the mode is evanescent, and no power travels (Section 7.1.7).

Figure 7.1.4 plots both. The phase constant rises from zero at cutoff towards $k$ , and the attenuation constant falls from $h$ at low frequency to zero at cutoff.

Phase constant above cutoff, \tau/k=\sqrt{1-(f_c/f)^2}, and attenuation constant below cutoff, \alpha/h=\sqrt{1-(f/f_c)^2}. Far above cutoff \tau\to k, the mode behaves like a plane wave. Far below cutoff \alpha\to h, independent of frequency. — **Figure 7.1.4.** Phase constant above cutoff, $τ / k = 1 - (f_{c} / f)^{2}$ , and attenuation constant below cutoff, $α / h = 1 - (f / f_{c})^{2}$ . Far above cutoff $τ \to k$ , the mode behaves like a plane wave. Far below cutoff $α \to h$ , independent of frequency.

The period of the field along $z$ is the guide wavelength,

Guide wavelength

λ_{g} λ_{c} = \frac{2 π}{τ} = \frac{λ}{1 - ( f _{c} / f ) ^{2}} > λ, = \frac{v}{f _{c}} = \frac{2 π}{h} . (7.1.15)

In plain words

A mode exists only above a frequency set by the size of the guide. Just above cutoff $τ \to 0$ and $λ_{g} \to \infty$ : the field hardly changes along the guide. Far above cutoff $τ \to k$ and $λ_{g} \to λ$ . Below cutoff the field does not travel at all, it dies away exponentially. The cutoff wavelength $λ_{c}$ is the wavelength in the filling at $f = f_{c}$ , and the mode propagates when $λ < λ_{c}$ .

Common mistake · Measuring along the guide with

λ

Using the wavelength $λ = v / f$ of the filling for distances along the guide, for instance between two maxima of a standing wave or for the length of a quarter-wave section. Along the guide the field repeats after $λ_{g}$ , which is always longer than $λ$ (Home exercise 7.2.1 in Section 7.2 shows how $λ_{g}$ is measured).

Quick check

An air-filled guide carries a mode with $f_{c} = 5 GHz$ at $f = 13 GHz$ . Find $λ$ and $λ_{g}$ .

$λ = c / f = 2.308 cm$ and $1 - (5/13)^{2} = 12/13$ , so $λ_{g} = 2.308 \times 13/12 = 2.50 cm$ .

Quick check

A mode is driven at half its cutoff frequency. By how many decibels does its field fall over one cutoff wavelength $λ_{c}$ ?

$α = h 1 - 1/4 = 0.866 h$ with $h = 2 π / λ_{c}$ , so $α λ_{c} = 2 π \times 0.866 = 5.44 Np$ , that is $8.686 \times 5.44 = 47.3 dB$ . Evanescent fields die within a fraction of a cutoff wavelength.

7.1.4 Cutoff frequency and guide wavelength