# Is it safe to call this WSSUS channel a Gaussian process?

BACKGROUND:

Equation (3.6) of Wireless Communications by Goldsmith gives the baseband impulse response of a time-varying channel as: $$c(\tau,t) = \sum_{n=0}^{N(t)}\alpha_n(t)e^{-j\phi_n(t)}\delta(\tau-\tau_n(t)),$$ where $$N(t)$$ is the number of multipath components ($$n=0$$ is the line of sight path), $$\alpha_n$$, $$\phi_n$$, and $$\tau_n$$ are the amplitude, phase, and lag of the $$n^\text{th}$$ path, respectively. If $$u(t)$$ is the baseband transmitted signal, then the received signal $$r(t)$$ (at baseband) would be: $$r(t) = \int_{-\infty}^\infty c(\tau, t) u(t-\tau) d\tau$$

Frequently the impulse response $$c(\tau,t)$$ is modeled as a zero-mean wide-sense stationary uncorrelated scattering (WSSUS) stochastic process. Coupled with this is frequently the assumption that $$c(\tau,t)$$ is a complex Gaussian random process, which gives a Rayleigh fading channel as the result.

Normally when a stochastic process $$x(t)$$ is called a Gaussian process it is because the marginal distribution of $$x$$ at any time $$t$$ follows a Gaussian distribution.

In the case of $$c(\tau,t)$$, the "value" of the random process is "infinite", due to the multiplication with the Dirac delta above (which is not actually a function, but a distribution).

QUESTION:

If a Gaussian random variable is multiplied by a Dirac delta distribution, is it still fair to call it a Gaussian random process? Does it still have all the "nice" properties that a Gaussian random process has, such as:

• Being completely specified by its first and second-order statistics
• Any linear transformation being done on it still resulting in a Gaussian distribution at the output