Frequency dispersion and Doppler spread derivation

Significance of frequency dispersion

Frequency dispersion is the spreading of the signal spectrum in the frequency domain, caused by the relative motion between the mobile and the base station (or moving objects). Each multipath wave arrives from a different angle, so it gets a different Doppler shift, and a single tone is received as a band of frequencies of width up to 2fm.

• It shows that the channel is time-variant; it is measured by Doppler spread BD and its dual, coherence time Tc ≈ 1/BD.
• If BD is comparable to the signal bandwidth (Ts > Tc), the channel causes fast fading and distortion.
• It limits how slowly data can be sent and how often the receiver must track the channel; it decides interleaver depth and time diversity design.

Derivation of Doppler shift and Doppler spread

A mobile moves at speed v from point X to Y (distance l) in time Δt, while receiving from a distant source S. The wave arrives at angle θ to the direction of motion.

                    S (far source)
                   /
                  /  waves nearly parallel
                 /
         X ----θ------------ Y    → v
           \___ Δl = l cosθ
           l = vΔt

Since the source is far, rays at X and Y are parallel. The difference in path length is:

Δl = l cosθ = vΔt cosθ

The phase change of the received signal:

Δφ = 2πΔl/λ = 2πvΔt cosθ / λ

Frequency change (Doppler shift) is the rate of change of phase:

fd = (1/2π)·(Δφ/Δt) = (v/λ)·cosθ

• θ = 0° (moving towards source): fd = +v/λ (maximum, frequency rises)
• θ = 180° (moving away): fd = −v/λ
• θ = 90°: fd = 0

The maximum Doppler shift is fm = v/λ. Since the multipath waves arrive from all angles θ (0 to 2π), the received spectrum spreads over fc − fm to fc + fm. The Doppler spread BD is the measure of this width, taken as

BD = fm = v/λ   (spectrum occupies 2fm in total)

Example: fc = 900 MHz, v = 72 km/h = 20 m/s → λ = 0.333 m, fm = 20/0.333 = 60 Hz.