3
$\begingroup$

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
$\endgroup$
0

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.