Ferkans — Interactive Telecom Tutor

Fading in the Time Domain

Section 6.3 characterised the distribution of the fading envelope — but not how fast it changes. A pedestrian at 3 km/h and a vehicle at 120 km/h experience the same Rayleigh distribution, yet the temporal behaviour is completely different. The Doppler effect governs how rapidly the channel changes, which determines whether the channel is "fast fading" or "slow fading" relative to the symbol duration.

Definition:
Doppler Shift

A mobile moving at velocity $v$ receives a signal from a scatterer at angle $\theta$ relative to the direction of motion. The Doppler shift on that path is

$f_d = \frac{v}{\lambda} \cos\theta = f_D \cos\theta$

where $f_D = v/\lambda = vf_0/c$ is the maximum Doppler shift (or maximum Doppler frequency).

Velocity	$f_0$	$f_D$
3 km/h (pedestrian)	2 GHz	5.6 Hz
60 km/h (urban)	2 GHz	111 Hz
120 km/h (highway)	2 GHz	222 Hz
300 km/h (train)	3.5 GHz	972 Hz

Theorem: Clarke/Jakes Doppler Spectrum

Under the Clarke model (2D isotropic scattering, uniform angle of arrival $\theta \sim \text{Uniform}[0, 2\pi)$ , many scatterers), the Doppler power spectrum of the fading process is

$S_H(f) = \begin{cases} \dfrac{1}{\pi f_D \sqrt{1 - (f/f_D)^2}}, & |f| < f_D \\ 0, & |f| \geq f_D \end{cases}$

This is the characteristic U-shaped (bathtub) spectrum: the power density diverges at $f = \pm f_D$ and has a minimum at $f = 0$ .

Scatterers at $\theta = 0$ and $\theta = \pi$ produce the extreme Doppler shifts $\pm f_D$ . Since $f_d = f_D \cos\theta$ and $\cos\theta$ changes slowly near $\theta = 0, \pi$ , many angles map to Doppler shifts near $\pm f_D$ , creating the peaks. Near $\theta = \pi/2$ ( $f_d \approx 0$ ), $\cos\theta$ changes rapidly, so fewer angles contribute, creating the dip.

Proof

Change of variables

The scattered power arriving from angle $\theta$ is $p(\theta) = 1/(2\pi)$ (uniform).

The Doppler shift is $f = f_D\cos\theta$ , so $\theta = \arccos(f/f_D)$ and $|d\theta/df| = 1/\sqrt{f_D^2 - f^2}$ .

There are two solutions ( $\theta$ and $2\pi - \theta$ ), so:

$S_H(f) = \frac{2}{2\pi} \cdot \frac{1}{\sqrt{f_D^2 - f^2}} = \frac{1}{\pi f_D \sqrt{1 - (f/f_D)^2}}$ . $\blacksquare$

,

Clarke/Jakes Doppler Spectrum

Adjust the mobile velocity to see how the maximum Doppler shift $f_D$ changes the width of the U-shaped spectrum and the corresponding autocorrelation function.

Parameters

Mobile velocity (km/h)60

Carrier frequency (GHz)2

Definition:
Coherence Time

The coherence time $T_c$ is the time duration over which the channel's fading coefficient remains approximately constant (correlation $\geq 0.5$ ).

Under the Clarke model, the autocorrelation of the fading process is

$R_h(\Delta t) = J_0(2\pi f_D \Delta t)$

where $J_0$ is the zeroth-order Bessel function of the first kind. The coherence time is approximately

$T_c \approx \frac{9}{16\pi f_D} \approx \frac{0.423}{f_D}$

(defined as the time where $|R_h| = 0.5$ ).

A common engineering approximation is

$T_c \approx \frac{1}{4 f_D} = \frac{\lambda}{4v}$

Classification:

$T_s \ll T_c$ : slow fading — channel constant over one symbol
$T_s \gg T_c$ : fast fading — channel changes within one symbol

,

Definition:
Level Crossing Rate and Average Fade Duration

The level crossing rate (LCR) is the expected number of times per second the fading envelope crosses a threshold $R$ in the positive direction:

$N_R = \sqrt{2\pi}\, f_D\, \rho\, e^{-\rho^2}$

where $\rho = R / R_{\text{rms}}$ is the normalised threshold.

The average fade duration (AFD) is the average time the envelope stays below $R$ :

$\bar{\tau}_R = \frac{P(r \leq R)}{N_R} = \frac{1 - e^{-\rho^2}}{\sqrt{2\pi}\, f_D\, \rho\, e^{-\rho^2}} \approx \frac{e^{\rho^2} - 1}{\sqrt{2\pi}\, f_D\, \rho}$

These quantities are critical for designing error-correction coding and interleaving depth.

,

Example: Coherence Time at Highway Speed

A vehicle at $v = 120$ km/h communicates at $f_0 = 3.5$ GHz.

(a) Compute the maximum Doppler shift $f_D$ .

(b) Compute the coherence time $T_c$ .

(c) If the OFDM symbol duration is $T_s = 71.4\,\mu$ s (5G NR), is this slow or fast fading?

Solution

Maximum Doppler shift

$v = 120/3.6 = 33.33$ m/s.

$f_D = v f_0 / c = 33.33 \times 3.5 \times 10^9 / 3 \times 10^8 = 389$ Hz.

Coherence time

$T_c \approx 0.423 / f_D = 0.423 / 389 = 1.09$ ms.

Classification

$T_s = 71.4\,\mu\text{s} \ll T_c = 1.09$ ms.

The channel is slow fading — it remains approximately constant over one OFDM symbol. However, over a slot of 14 symbols ( $\approx 1$ ms), the channel changes appreciably, requiring channel estimation updates. $\blacksquare$

Common Mistake: Fast Fading Does Not Mean Deep Fading

Mistake:

Assuming "fast fading" means the fading is more severe or deeper.

Correction:

"Fast" and "slow" refer to how quickly the channel changes relative to the symbol duration, not to the depth of fades. A slow-fading channel can have deep Rayleigh fades; a fast-fading channel averages them out within a symbol (which can actually improve performance through time diversity).

Quick Check

A pedestrian at 5 km/h uses a 900 MHz phone. What is the maximum Doppler shift?

4.2 Hz

15 Hz

42 Hz

1.5 Hz

Correction:

4.2 Hz

$v = 5/3.6 = 1.389$ m/s. $f_D = 1.389 \times 900 \times 10^6 / 3 \times 10^8 = 4.17$ Hz $\approx 4.2$ Hz.

⚠️Engineering Note

Doppler Estimation in Practice

In real systems, the maximum Doppler shift $f_D$ is not known a priori and must be estimated from the received signal. Practical considerations:

Pilot-based estimation: 5G NR uses DMRS (Demodulation Reference Signals) spaced in time. The pilot spacing in the time domain must satisfy $\Delta t_{\text{pilot}} < 1/(2f_D)$ (Nyquist criterion on the fading process). For 120 km/h at 3.5 GHz ( $f_D \approx 389$ Hz), this requires pilots every $\sim 1.3$ ms — hence NR places DMRS in every slot.
Doppler ambiguity: If pilot spacing violates Nyquist, the estimated Doppler is aliased, causing catastrophic channel estimation errors. This limits the maximum supportable velocity for each numerology: 15 kHz spacing supports up to $\sim 500$ km/h at 2 GHz, but 120 kHz spacing (used at mmWave) supports only $\sim 63$ km/h at 28 GHz.
Numerical issues: The Clarke/Jakes spectrum $S_H(f)$ has integrable singularities at $f = \pm f_D$ . In simulation, clip the spectrum at $|f| < 0.999 f_D$ or use the sum-of-sinusoids approach to avoid numerical overflow.

Practical Constraints

•
Pilot spacing in time: $\Delta t < 1/(2 f_D)$
•
Maximum velocity per numerology is limited by pilot density

📋 Ref: 3GPP TS 38.211, §7.4.1 (DMRS configuration)

Doppler Shift

Frequency shift $f_d = (v/\lambda)\cos\theta$ experienced by a signal arriving at angle $\theta$ relative to the mobile's direction of motion.

Coherence Time

$T_c \approx 0.423/f_D$ : the time duration over which the channel remains approximately constant. Inversely proportional to maximum Doppler shift.

Doppler Effect and Temporal Fading Statistics