# Gaussian, Rayleigh, and Exponential RVs

I am doing some simulations on the performance of OFDM systems in multi-path channels and I am confused about something. It is explained below:

Assume that we have a multi-path channel with a number of taps equal to $L$. Each tap has a power of $\sigma$, hence, the total power in the multi-path channel in time domain is $\sigma\:L$. If the generated channel in time domain of length $L$ is denoted as $h$, then the channel in frequency domain of length $N$ is $H = \rm{FFT}(h,N)$, where $N$ is the number of OFDM subcarriers.

• If we have $X \sim N(0,\frac{\sigma^2}{2})$ and $Y \sim N(0,\frac{\sigma^2}{2})$, then $h = \sqrt{X^2 + Y^2} \sim \rm{Rayleigh}(\frac{\sigma}{\sqrt(2)})$.
• Additionally, if channel in frequency domain $H\sim \rm{Rayleigh}(\frac{1}{\sqrt{2\lambda}})$, then $|H|^2 \sim e^{\lambda}$? In this line correct? Is $\lambda = \frac{1}{\sigma^2}$?

I am confused about the transition from $h$ to $|H|^2$ and the relation between their means and variances.

• Are you considering a random channel or a deterministic channel with a random signal? What exactly is $X$ and $Y$? Can you give a reference on which you base your notation etc.? The basic fact that the magnitude of a complex Gaussian RV has a Rayleigh distribution, and that the squared magnitude has an exponential distribution is clear, but I think there's some confusion in your question as to what $X$, $Y$, and $h$ exactly mean. Please clarify to make your question easier to answer. – Matt L. Mar 18 '15 at 13:25
• Thanks for your response. I consider a random channel $h$. The $X$ and $Y$ are normally distributed RVs used to generate the channel $h$ with a certain variance. My question is given $h$ with a certain variance (based on variance of $X$ and $Y$), what will be the variance of $|H|^2$ (given that $H$ is the frequency response of $h$)? – Noor Mar 18 '15 at 16:34

So you have a complex impulse response, and the real and imaginary parts of each tap are modeled as i.i.d. zero mean Gaussian variables with variance $\sigma^2/2$ (correct me if I'm wrong). The DFT of the impulse response is a weighted sum of the filter taps:

$$H[k]=\sum_{n=0}^{N-1}h[n]e^{-j2\pi nk/N}\tag{1}$$

where I assume $N\ge L$ (where $L$ is the number of taps), and in Eq. (1) $h[n]=0$ for $n\ge L$. From (1) it follows that $H[k]$ is also a complex Gaussian RV with zero mean and with variance $L\sigma^2$ (the variance of the real and imaginary part is $L\sigma^2/2$, respectively). From this it follows that the magnitude $|H[k]|$ has a Rayleigh distribution:

$$f_{|H|}(x)=\frac{2x}{L\sigma^2}e^{-x^2/(L\sigma^2)}\tag{2}$$

And, accordingly, the squared magnitude $|H[k]|^2$ has an exponential distribution:

$$f_{|H|^2}(x)=\lambda e^{-\lambda x},\quad x\ge 0\tag{3}$$

With $\lambda=1/(L\sigma^2)$, i.e. with mean $1/\lambda=L\sigma^2$, and variance $1/\lambda^2=L^2\sigma^4$.

• Your assumptions are correct and the answer makes sense. Thanks for your help. – Noor Mar 18 '15 at 21:06