I have a confusion regarding the distribution of noise and channel fading. As we know channel fading is usually modeled with Rayleigh fading since it has Gaussian distributed real and imaginary components. On the other hand, we express a complex noise as a Gaussian distributed signal (at least for AWGN channels) while it also has the same real and imaginary components.

Why do we express complex noise as Gaussian distributed while the channel is called Rayleigh distributed? Am I missing something here?

Thank you.


2 Answers 2


There are too many things named after Rayleigh floating around in this problem.

Consider first additive white Gaussian noise with two-sided power spectral density $\frac{N_0}{2}$. This is generally taken as the model for channel noise, though in fact the source of the noise actually is thermal noise in the front-end of the receiver. NASA once even experimented with cooling the receiver front end in a liquid helium bath to reduce the noise temperature to about $5^\circ$K. Be that as is may, the receiver has filters (and amplifiers) so that the noise at the receiver output is definitely band-limited. Because the front-end noise has been amplified greatly, the contribution of the impedances of later receiver stages to the thermal noise is negligible. Indeed, the noise at later stages is a (stationary) band-pass Gaussian process $\left\{N(t)\colon -\infty < t < \infty\right\}$ where $$N(t) = X(t)\cos(2\pi f_ct) - Y(t)\sin(2\pi f_ct)\tag{1}$$ with $\left\{X(t)\colon -\infty < t < \infty\right\}$ and $\left\{Y(t)\colon -\infty < t < \infty\right\}$ being independent zero-mean low-pass Gaussian processes with common autocorrelation function $R(\tau)$. And Yes, I do want a $-$ sign in $(1)$.

The autocorrelation function of the $\{N(t)\}$ process is \begin{align}\require{cancel}R_N( \tau) &= E[N(t)N(t+\tau)]\\ &= E\big[\big(X(t)\cos(2\pi f_ct) - Y(t)\sin(2\pi f_ct)\big)\\ &= \hspace{0.2in}\cdot \big(X(t+\tau)\cos(2\pi f_c(t+\tau)) - Y(t+\tau)\sin(2\pi f_c(t+\tau))\big)\big]\\ &= R(\tau)\big[\cos(2\pi f_c t)\cos(2\pi f_c(t+\tau))+\sin(2\pi f_c t)\sin(2\pi f_c(t+\tau))\big]\\ &= \hspace{0.2in}~ - \cancelto{0}{R_{X,Y}(t)}\big[\cos(2\pi f_c t)\sin(2\pi f_c(t+\tau))-\sin(2\pi f_c t)\cos(2\pi f_c(t+\tau))\big]\\ \\&= R(\tau)\cos(2\pi f_c \tau) \end{align} which is a band-pass function. Note that we can also express the $\{N(t)\}$ process in terms of its complex baseband equivalent process $\{\mathcal Z(t)\colon -\infty < t < \infty\}$ where $\mathcal Z(t) = X(t) + jY(t)$ as $$N(t) = \mathrm{Re}\left( \mathcal Z(t)e^{j2\pi f_c t}\right)\tag{2}$$ But, if we choose to represent the complex number $\mathcal Z(t)$ in polar form as $B(t)e^{j\Theta(t)}$ where $B(t) \geq 0$ and $\Theta(t) \in [0,2\pi)$ are real numbers, then $B(t) = \sqrt{X^2(t)+Y^2(t)}$ is a Rayleigh random variable and $\Theta(t)$ is uniformly distributed on $[0,2\pi)$. (This idea is the basis of the Box-Muller method of simulating Gaussian random variables. The usual probability transform method does not work well for Gaussian random variables because the Gaussian CDF does not have an inverse that is an elementary function whereas a Rayleigh CDF $1-\exp(-t^2/2\sigma^2)$ is readily invertible. So, the Box-Muller method calls a random number generator twice to simulate $B$ and $\Theta$ and then sets $X = B\cos(\Theta)$ and $Y = B\sin(\Theta)$ to get realizations of two independent Gaussian random variables).

The other place where Rayleigh's name comes in is in the transmission of signals over fading channels. Real-life signals are real-valued, not complex-valued. When a sinusoid (say $\cos(2\pi f_ct), -\infty < t < \infty$) is transmitted over a (slowly fading) channel, the received signal is of form \begin{align}r(t) &= b(t)\cos(2\pi f_c t + \theta(t)) \tag{3}\\ &= b(t)\cos(\theta(t))\cos(2\pi f_c t) - b(t)\sin(\theta(t))\sin(2\pi f_c t) \tag{4}\end{align} where $b(t)$ and $\theta(t)$ are slowly varying signals. For the purposes of digital communications, it is reasonable to treat $b(t)$ and $\theta(t)$ as constants over the duration of transmission of one bit/symbol, so that $(4)$ simplifies to $$b\cos(\theta)\cos(2\pi f_c t) - b\sin(\theta)\sin(2\pi f_c t).\tag{5}$$ Slow fading occurs in ionospheric channels because the transmitted signal is reflected back to earth from the ions in a ion cloud and the received signal is a sum of a very large number of very weak reflections with nearly equal delays. As a consequence, with a lot of handwaving and genuflections in the direction of the Central Limit Theorem, it is possible to regard $b\cos(\theta)$ and $b\sin(\theta)$ in $(5)$ as realizations of two independent Gaussian random variables which allows us to treat $b$ and $\theta$ as realizations of Rayleigh and uniform random variables respectively. Hence, the complex baseband equivalent of a Rayleigh fading channel is that complex input $\mathrm{Re}(\mathcal Ae^{j2\pi f_c t})$ is received as output $\mathrm{Re}(\mathcal{AB}e^{j2\pi f_c t})$ where $\mathcal B = Be^{j\Theta}$ with $B$ and $\Theta$ being independent Rayleigh and uniform random variables respectively. Fading is sometimes referred to as multiplicative noise since the fading multiplies the input amplitude instead of adding to it as additive noise does.

"But, but, but," you say, "In a Rayleigh channel model, the fading parameters $B_{n+1}$ and $\Theta_{n+1}$ for the next bit/symbol interval are independent of the parameters $B_n$ and $\Theta_n$ for the current interval. How does that fit into your model of $b(t)$ and $\theta(t)$ being slowly varying functions? If $b(t)$ and $\theta(t)$ really vary so slowly that you can treat them as constant over the interval, then $B_{n+1}$ and $B_{n}$ should be very highly correlated, as should with $\Theta_{n+1}$ and $\Theta_n$." Well, the dirty little secret that no one will ever tell you about is that communication systems intended for use over Rayleigh fading channels are very heavily interleaved so that what appears to be the next symbol in the mathematical model is actually a symbol that occurs $N\gg 1$ bit durations later where $N$ is chosen to be large enough that heavily-faded consecutive symbols appear in different frames rather than wiping out four bars of Bethoven's Fifth Symphony or a whole stripe from an image of Old Glory if interleaving were not used.


Because fading and noise affect the received signal in different ways.

The received matched-filter output is $$r = hs + n,$$ where $s$ is the transmitted symbol, $h$ is the fading coefficient, and $n$ is Gaussian noise. All numbers are complex and both $h$ and $n$ are complex Gaussian random variables.

To estimate $s$, the receiver calculates \begin{align} y &= \frac{h^*}{|h|} r \\ & = |h| s + \frac{h^*}{|h|} n, \end{align} where $h^* $ is the complex conjugate of $h$. You can see that the symbol $s$ is scaled by $|h|$, which is a Rayleigh random variable. However, the noise $h^*n/|h|$ is still Gaussian.

  • $\begingroup$ Thanks for your reply MBaz. However, my question is about the random characteristics of these two random variables. Both are represented as complex numbers and both has a Gaussian distributed real and a Gaussian distributed imaginary parts (as randn(..)+1i*randn(...) in MATLAB). Why channel is Rayleigh while the noise is still Gaussian? $\endgroup$
    – abdVef
    Aug 30, 2017 at 23:20
  • $\begingroup$ @abdVef That is what I tried to convey in my answer... the channel is called Rayleigh because you can model it as having a random gain that is Rayleigh distributed. Note that the received symbol can be written as $y = \alpha s + n$, where $\alpha$ is a Rayleigh RV, and $n$ is Gaussian noise. Please let me know if it's still not clear. $\endgroup$
    – MBaz
    Aug 30, 2017 at 23:22
  • $\begingroup$ @abdVef In case it helps; The pure physical channel, by itself, is Gaussian. However, the channel toghether with the coherent detector in the receiver produce a Rayleigh gain. $\endgroup$
    – MBaz
    Aug 30, 2017 at 23:24
  • $\begingroup$ Wow, I really appreciate your quick responses. Unfortunately, it still looks confusing to me. Obviously, the way they affect the signal is different; one is additive while the other one is multiplicative. However, the channel comes as a Gaussian gain for real and imaginary parts of the signal and when we look at the combined effect it becomes a Rayleigh gain. I am clear on that. However, noise does exactly the same thing by shifting real and imaginary parts of the signal by Gaussian RVs. Why do we still call the noise Gaussian then? Sorry if I'm missing something. $\endgroup$
    – abdVef
    Aug 30, 2017 at 23:34
  • $\begingroup$ It's the transformation of Gaussian random variables. If you square a sequence of Gaussian random variables and sum them you get a Chi squared random variable. Further, taking the square root gives another distribution. The signal portion transforms that way. For the noise portion - a linear combination of independent Gaussian random variables is still a Gaussian random variable - so the noise component still remains Gaussian. Look up transformations of Gaussian random variable and their relationship to Chi squared, Rician, and Rayleigh random variables. $\endgroup$
    – David
    Aug 31, 2017 at 13:35

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