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

The speed of a point of constant phase of $e^{j (ω t - τ z)}$ is the phase velocity

v_{p} = \frac{ω}{τ} = \frac{v}{1 - ( f _{c} / f ) ^{2}} \geq v . (7.1.16)

The envelope of a signal, and with it the energy, travels at the group velocity $v_{g} = d ω / d τ$ (Section 6.2). Differentiate $τ^{2} = ω^{2} μ ε - h^{2}$ , in which $h$ is fixed by the geometry and does not depend on $ω$ :

2 τ d τ = 2 ω μ ε d ω ⟹ v_{g} = \frac{d ω}{d τ} = \frac{τ}{ω μ ε} = \frac{1}{μ ε} \frac{τ}{ω} = \frac{v ^{2}}{v _{p}} . (7.1.17)

Phase and group velocity

v_{p} v_{g} v_{p} v_{g} = \frac{v}{1 - ( f _{c} / f ) ^{2}} \geq v, = v 1 - (f_{c} / f)^{2} \leq v, = v^{2} . (7.1.18)

In plain words

The crests run along the guide faster than light in the filling, the energy slower, and the product of the two speeds is always $v^{2}$ . At cutoff the energy stands still ( $v_{g} = 0$ ) while the crests race ( $v_{p} \to \infty$ ).

Phase and group velocity of a TM or TE mode against f/f_c. Both approach v far above cutoff. The TEM wave (f_c=0) has v_p=v_g=v at every frequency. — **Figure 7.1.5.** Phase and group velocity of a TM or TE mode against $f / f_{c}$ . Both approach $v$ far above cutoff. The TEM wave ( $f_{c} = 0$ ) has $v_{p} = v_{g} = v$ at every frequency.

A guide is dispersive

$v_{g}$ depends on frequency, so the spectral components of a pulse arrive at different times and the pulse spreads (Application 7.1.1). This waveguide dispersion is caused by the geometry alone. It appears even with a perfectly non-dispersive filling, in air. The TEM wave, with $f_{c} = 0$ , is not dispersive: $v_{p} = v_{g} = v$ .

Common mistake · Reading

v_{p} > c

as a signal faster than light

In an air-filled guide $v_{p} > c$ . This carries no information and no energy, and it does not conflict with relativity. Information and energy travel at $v_{g} < c$ , and $v_{g} \to 0$ at cutoff. Section 7.2 shows with the zig-zag picture where the two speeds come from.

Quick check

In an air-filled guide a mode has $v_{p} = 1.25 c$ . What are $v_{g}$ and $f / f_{c}$ ?

$v_{g} = c^{2} / v_{p} = 0.8 c = 2.4 \times 1 0^{8} m/s$ , and $1 - (f_{c} / f)^{2} = 0.8$ gives $f_{c} / f = 0.6$ , so $f = 1.667 f_{c}$ .

7.1.5 Phase and group velocity in a guide