rayleigh fading model

Table of Contents

From narrowband fading, the signal can be shown to be modified by one sum of multipaths, \(\beta(t)\).

Signal arrives without strong LOS path, and instead comes from a scatterer. There are \(N\) scatterers distributed around the reciever.

With large \(N\), the channel has complex gaussian coefficients, where the variance comes from shadowing and path loss.

The complex envelope now has a rayleigh distribution.

and variance is \(2\sigma^2\).

1. Simulation procedure

  1. Decide on \(K\) \(\gamma\) \(P_t\) \(d0\) and \(d\) for path loss
  2. Generate path loss \(K\frac{d0}{d}^\gamma\)
  3. Decide on \(sigma^2\) for shadowing variance
  4. Generate shadowing \(\mathcal{N}(0, \sigma^2)\)
  5. Generate shadowing power -> \(10^{0.1S_h}\)
  6. fast multipath fading is exponential random variable
  7. Power is product of all 4
  8. Channel gain \(\beta = \mathcal{CN}(0, P_l * 10^{0.1S_h})\)

In summary

n = 10
K = 8e-4
d0 = 1
d = 150
gamma = 2
shadowing_variance = 4
power_tx = 1

path_loss = K * ((d0/d)**gamma)
shadowing_power = 10**(0.1 * (np.random.normal(0, np.sqrt(shadowing_variance))))
fading = np.random.exponential(1, n)

power_rx = power_tx * path_loss * shadowing_power * fading
beta = np.random.normal(0,
                        path_loss * shadowing_power,
                        (n, 2)).view(np.complex128)

2. Correlation

For derivation of the below, see Wireless Communications 3.2.1, pg 71.

The autocorrelation function is given by

\begin{equation} A_\beta = J_0(2\pi f_D \Delta t) \pi f_D P_r \end{equation}

Where \(J_0\) is the 0th order Bessel function. Note this is a function of both delay and the doppler frequency shift.

It can be shown that because the autocorrelation function plotted against \(\Delta t f_D\) has a root at 0.4, when the reciever moves 0.4 of a wavelength, the channel becomes uncorrelated.

In other words, the coherence time is \(\frac{0.4}{f_D}\)

The corresponding power spectral density is given by

\begin{equation} S_\beta(f) = \frac{P_r}{\sqrt{1-(\frac{f}{f_D})^2}} \end{equation}

Because it's a function of doppler frequency shift, it can be interpreted as the pdf of the random frequency due to Doppler.

This has a U-shape over the ratio \(\frac{f}{f_D}\)

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