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
power_tx = 1

path_loss = K * ((d0/d)**gamma)

beta = np.random.normal(0,
(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

$$A_\beta = J_0(2\pi f_D \Delta t) \pi f_D P_r$$

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

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

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 $$\frac{f}{f_D}$$