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Doppler spread and Doppler shift derivation

Short answer

Doppler spread is the spectral broadening of a received tone over fc-fm to fc+fm, with fm=v/lambda. For a mobile moving at speed v, path difference is v*t*cos(theta), giving phase change 2*pi*v*t*cos(theta)/lambda, so Doppler shift fd=(v/lambda)cos(theta). Maximum is v/lambda.

What is Doppler spread? Derive the expression for Doppler shift.

Doppler spread (BD) is the measure of spectral broadening of a received signal caused by the time rate of change of the mobile radio channel. It is the range of frequencies over which the received Doppler spectrum is non-zero. When a pure tone fc is sent, the received spectrum spreads between fc − fm and fc + fm, where fm = v/λ is the maximum Doppler shift; BD ≈ fm.

Derivation of Doppler shift

A mobile moves with constant speed v from point X to point Y in time Δt, receiving a signal from a remote source S. The direction of arrival makes angle θ with the direction of motion.

                 S (remote source)
                /|
               / |
              /  |  rays from S are
             /   |  almost parallel
            / θ  |
    X ----------- Y ------>  v
    |<---- l --->|
    path difference Δl = l cosθ

Distance travelled:

l = v·Δt

Since S is very far, the two rays (to X and to Y) are nearly parallel, so the difference in path length is:

Δl = l cosθ = v·Δt·cosθ

Phase change due to this path difference:

Δφ = 2π·Δl/λ = (2π·v·Δt·cosθ)/λ

The apparent change in frequency is the rate of change of phase divided by 2π:

fd = (1/2π)·(Δφ/Δt)
fd = (v/λ)·cosθ

Results:

  • Moving towards the source (θ = 0): fd = +v/λ, received frequency increases.
  • Moving away (θ = 180°): fd = −v/λ, frequency decreases.
  • Moving perpendicular (θ = 90°): fd = 0.
  • Maximum Doppler shift fm = v/λ.

Example: fc = 1850 MHz, v = 60 mph = 26.82 m/s, λ = 0.162 m → fm = 26.82/0.1622 = 165.4 Hz; moving towards the source, received frequency = 1850.000165 MHz.

✓ Answer checked by Claude Opus 5.5 (high effort)

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