Problem 6.2.3: Group delay through an ionospheric layer

Problem 6.2.3

Group delay through an ionospheric layer

An HF signal consists of two tones of equal amplitude at $f_{0} \pm Δ f$ , with $f_{0} = 10 MHz$ and $Δ f = 5 kHz$ . It crosses vertically an ionospheric layer of thickness $d = 200 km$ and uniform electron density $N = 4.46 \times 1 0^{11} m^{- 3}$ (collisions neglected).

Find the plasma frequency of the layer.
Find $v_{p}$ and $v_{g}$ at $f_{0}$ and check that $v_{p} v_{g} = c^{2}$ .
How much later than in vacuum does the envelope leave the layer?
Find the carrier wavelength in the layer and the distance between successive maxima of the envelope.

Try it yourself first.

Strategy

The carrier crests travel with $v_{p} = ω_{0} / β_{0}$ , the envelope of the two-tone packet with $v_{g} = d ω / d β$ . In a plasma both follow from the one factor $1 - f_{p}^{2} / f^{2}$ , Eq. (6.2.51).

$f_{p} = 8.9836 4.46 \times 1 0^{11} = 6.00 MHz$ .

$f_{p} / f_{0} = 0.6$ , so $1 - f_{p}^{2} / f_{0}^{2} = 1 - 0.36 = 0.8$ and

v_{p} v_{g} = \frac{c}{0.8} = 3.75 \times 1 0^{8} m/s, = 0.8 c = 2.40 \times 1 0^{8} m/s,

v_{p} v_{g} = 3.75 \times 2.40 \times 1 0^{16} = 9 \times 1 0^{16} = c^{2} .

The envelope crosses the layer with $v_{g}$ , in

t_{g} = \frac{d}{v _{g}} = \frac{2 \times 1 0 ^{5}}{2.40 \times 1 0 ^{8}} = 0.833 ms, against t_{0} = \frac{d}{c} = 0.667 ms in vacuum,

so the excess group delay is $Δ t = t_{g} - t_{0} = 0.167 ms$ .

The carrier wavelength is $λ = v_{p} / f_{0} = 37.5 m$ , against $30 m$ in vacuum: longer in the plasma. The envelope $∣2 E_{0} cos (Δ ω t - Δ β z) ∣$ of Eq. (6.2.48) repeats when $Δ β z$ grows by $π$ , and $Δ β \approx Δ ω / v_{g}$ :

Δ z = \frac{π}{Δ β} = \frac{π v _{g}}{2 π Δ f} = \frac{v _{g}}{2Δ f} = \frac{2.40 \times 1 0 ^{8}}{1 0 ^{4}} = 24 km .

Answer

f_{p} = 6.00 MHz, v_{p} = 3.75 \times 1 0^{8} m/s, v_{g} = 2.40 \times 1 0^{8} m/s, Δ t = 0.167 ms, λ = 37.5 m, Δ z = 24 km

Check without the derivative

$β (f) = \frac{2 π}{c} f^{2} - f_{p}^{2}$ gives $β (f_{0} + Δ f) = 0.16769 rad/m$ and $β (f_{0} - Δ f) = 0.16743 rad/m$ , so $2Δ β = 2.618 \times 1 0^{- 4} rad/m$ and $π /Δ β = 24.0 km$ .

Problem 6.2.3. A short packet with a Gaussian envelope at two instants in the layer, not to scale (\lambda is enlarged). The solid blue line marks the envelope peak, the green dashed line one particular crest of the carrier. In the time t_1 the envelope advances v_gt_1, the crest v_pt_1, which is v_p/v_g=1.25/0.8=1.56 times farther. — **Figure 6.2.19.** Problem 6.2.3. A short packet with a Gaussian envelope at two instants in the layer, not to scale ( $λ$ is enlarged). The solid blue line marks the envelope peak, the green dashed line one particular crest of the carrier. In the time $t_{1}$ the envelope advances $v_{g} t_{1}$ , the crest $v_{p} t_{1}$ , which is $v_{p} / v_{g} = 1.25/0.8 = 1.56$ times farther.

What the numbers say

The information arrives late ( $v_{g} < c$ ) while the carrier phase runs ahead ( $v_{p} > c$ ), by the same factor $0.8$ . Figure 6.2.19 shows what this means for a packet: crests are born at its back, run through it and die at its front. Energy sits where the envelope is, so it moves with $v_{g} = 2.40 \times 1 0^{8} m/s < c$ . Nothing physical travels with a crest, so $v_{p} = 3.75 \times 1 0^{8} m/s > c$ breaks no law. An ionosonde measures exactly this group delay, and GPS corrects it: at $1.6 GHz$ the effect is tiny, but it is still metres of range error (Exercise 6.2.8).