Applicability

Densely built Manhattan has been shown to approach a Rayleigh-fading environment.

One second of Rayleigh fading with a maximum Doppler shift of 10 Hz.

One second of Rayleigh fading with a maximum Doppler shift of 100 Hz.

The requirement that there be many scatterers present means that Rayleigh fading can be a useful model in heavily built-up city centres where there is no line of sight between the transmitter and receiver and many buildings and other objects attenuate, reflect, refract, and diffract the signal. Experimental work in Manhattan has found near-Rayleigh fading there.^[3] In tropospheric and ionospheric signal propagation the many particles in the atmospheric layers act as scatterers and this kind of environment may also approximate Rayleigh fading. If the environment is such that, in addition to the scattering, there is a strongly dominant signal seen at the receiver, usually caused by a line of sight, then the mean of the random process will no longer be zero, varying instead around the power-level of the dominant path. Such a situation may be better modelled as Rician fading.

Note that Rayleigh fading is a small-scale effect. There will be bulk properties of the environment such as path loss and shadowing upon which the fading is superimposed.

How rapidly the channel fades will be affected by how fast the receiver and/or transmitter are moving. Motion causes Doppler shift in the received signal components. The figures show the power variation over 1 second of a constant signal after passing through a single-path Rayleigh fading channel with a maximum Doppler shift of 10 Hz and 100 Hz. These Doppler shifts correspond to velocities of about 6 km/h (4 mph) and 60 km/h (40 mph) respectively at 1800 MHz, one of the operating frequencies for GSM mobile phones. This is the classic shape of Rayleigh fading. Note in particular the 'deep fades' where signal strength can drop by a factor of several thousand, or 30–40 dB.

Level crossing rate

The level crossing rate is a measure of the rapidity of the fading. It quantifies how often the fading crosses some threshold, usually in the positive-going direction. For Rayleigh fading, the level crossing rate is:^[4]

${\displaystyle \mathrm {LCR} ={\sqrt {2\pi }}f_{d}\rho e^{-\rho ^{2}}}$

where ${\displaystyle f_{d}}$ is the maximum Doppler shift and ${\displaystyle \,\!\rho }$ is the threshold level normalised to the root mean square (RMS) signal level:

${\displaystyle \rho ={\frac {R_{\mathrm {threshold} }}{R_{\mathrm {rms} }}}.}$

Average fade duration

The average fade duration quantifies how long the signal spends below the threshold ${\displaystyle \,\!\rho }$. For Rayleigh fading, the average fade duration is:^[4]

${\displaystyle \mathrm {AFD} ={\frac {e^{\rho ^{2}}-1}{\rho f_{d}{\sqrt {2\pi }}}}.}$

The level crossing rate and average fade duration taken together give a useful means of characterizing the severity of the fading over time.

For a particular normalized threshold value ${\displaystyle \rho }$, the product of the average fade duration and the level crossing rate is a constant and is given by

${\displaystyle \mathrm {AFD} \times \mathrm {LCR} =1-e^{-\rho ^{2}}.}$

Doppler power spectral density

The normalized Doppler power spectrum of Rayleigh fading with a maximum Doppler shift of 10 Hz.

The Doppler power spectral density of a fading channel describes how much spectral broadening it causes. This shows how a pure frequency, e.g., a pure sinusoid, which is an impulse in the frequency domain, is spread out across frequency when it passes through the channel. It is the Fourier transform of the time-autocorrelation function. For Rayleigh fading with a vertical receive antenna with equal sensitivity in all directions, this has been shown to be:^[5]

${\displaystyle S(\nu )={\frac {1}{\pi f_{d}{\sqrt {1-\left({\frac {\nu }{f_{d}}}\right)^{2}}}}},}$

where ${\displaystyle \,\!\nu }$ is the frequency shift relative to the carrier frequency. This equation is valid only for values of ${\displaystyle \,\!\nu }$ between ${\displaystyle \pm f_{d}}$; the spectrum is zero outside this range. This spectrum is shown in the figure for a maximum Doppler shift of 10 Hz. The 'bowl shape' or 'bathtub shape' is the classic form of this Doppler spectrum.

Jakes's model

In his book,^[6] Jakes popularised a model for Rayleigh fading based on summing sinusoids. Let the scatterers be uniformly distributed around a circle at angles ${\displaystyle \alpha _{n}}$ with ${\displaystyle k}$ rays emerging from each scatterer. The Doppler shift on ray ${\displaystyle n}$ is

${\displaystyle \,\!f_{n}=f_{d}\cos \alpha _{n}}$

and, with ${\displaystyle M}$ such scatterers, the Rayleigh fading of the ${\displaystyle k^{\text{th}}}$ waveform over time ${\displaystyle t}$ can be modelled as:

${\displaystyle {\begin{aligned}R(t,k)=2{\sqrt {2}}\left[\sum _{n=1}^{M}\right.&\left(\cos \beta _{n}+j\sin \beta _{n}\right)\cos \left(2\pi f_{n}t+\theta _{n,k}\right)\\[4pt]&\left.{}+{\frac {1}{\sqrt {2}}}\left(\cos \alpha +j\sin \alpha \right)\cos(2\pi f_{d}t)\right].\end{aligned}}}$

Here, ${\displaystyle \,\!\alpha }$ and the ${\displaystyle \,\!\beta _{n}}$ and ${\displaystyle \,\!\theta _{n,k}}$ are model parameters with ${\displaystyle \,\!\alpha }$ usually set to zero, ${\displaystyle \,\!\beta _{n}}$ chosen so that there is no cross-correlation between the real and imaginary parts of ${\displaystyle R(t)}$:

${\displaystyle \,\!\beta _{n}={\frac {\pi n}{M+1}}}$

and ${\displaystyle \,\!\theta _{n,k}}$ used to generate multiple waveforms. If a single-path channel is being modelled, so that there is only one waveform then ${\displaystyle \,\!\theta _{n}}$ can be zero. If a multipath, frequency-selective channel is being modelled so that multiple waveforms are needed, Jakes suggests that uncorrelated waveforms are given by

${\displaystyle \theta _{n,k}=\beta _{n}+{\frac {2\pi (k-1)}{M+1}}.}$

In fact, it has been shown that the waveforms are correlated among themselves — they have non-zero cross-correlation — except in special circumstances.^[7] The model is also deterministic (it has no random element to it once the parameters are chosen). A modified Jakes's model^[8] chooses slightly different spacings for the scatterers and scales their waveforms using Walsh–Hadamard sequences to ensure zero cross-correlation. Setting

${\displaystyle \alpha _{n}={\frac {\pi (n-0.5)}{2M}}{\text{ and }}\beta _{n}={\frac {\pi n}{M}},}$

results in the following model, usually termed the Dent model or the modified Jakes model:

${\displaystyle R(t,k)={\sqrt {\frac {2}{M}}}\sum _{n=1}^{M}A_{k}(n)\left(\cos \beta _{n}+j\sin \beta _{n}\right)\cos \left(2\pi f_{d}t\cos \alpha _{n}+\theta _{n}\right).}$

The weighting functions ${\displaystyle A_{k}(n)}$ are the ${\displaystyle k}$^th Walsh–Hadamard sequence in ${\displaystyle n}$. Since these have zero cross-correlation by design, this model results in uncorrelated waveforms. The phases ${\displaystyle \,\!\theta _{n}}$ can be initialised randomly and have no effect on the correlation properties. The fast Walsh transform can be used to efficiently generate samples using this model.

The Jakes's model also popularised the Doppler spectrum associated with Rayleigh fading, and, as a result, this Doppler spectrum is often termed Jakes's spectrum.

Filtered white noise

Another way to generate a signal with the required Doppler power spectrum is to pass a white Gaussian noise signal through a Gaussian filter with a frequency response equal to the square-root of the Doppler spectrum required. Although simpler than the models above, and non-deterministic, it presents some implementation questions related to needing high-order filters to approximate the irrational square-root function in the response and sampling the Gaussian waveform at an appropriate rate.

Butterworth filter as Doppler power spectral density

Filtered by Butterworth filter Rayleigh time series (sampling frequency is 120 Hz)

According to ^[9][10][11] Doppler PSD can also be modeled via Butterworth filter as:

${\displaystyle S_{Doppler}(f)=|H_{Butterworth}|^{2}={\frac {B}{\sqrt {1-(f/f_{0})^{2k}}}}}$

where f is a frequency, ${\displaystyle H_{Butterworth}}$ is the Butterworth filter response, B is the normalization constant, k is the filter order and ${\displaystyle f_{0}}$ is the Cutoff frequency which should be selected with respect to maximum Doppler shift.