# Understanding Communication Theory Jargon for Statistical Linear Models

Coming from a statistical background, I understand the following formula:

$$$$y = X\beta + \epsilon$$$$

as a linear regression model, with data input-output pair: $$(X,y)$$, noise term, $$\epsilon$$, and regression coefficient vector $$\beta$$.

Now I am confused in how this system is described in communication theory. I see terms such as Rayleigh Fading, MIMO, Multi-access, Multi-antenna, Band-limited etc ... in relation to equations that look exactly like statistical linear regression.

So far I suppose I can work out that for communication theory,

$$$$y = Hx + \epsilon$$$$

$$x$$ is some input signal at a transmitter antenna, $$\epsilon$$ is noise again, $$y$$ is the signal at the receiver antenna ... but what is $$H$$? I don't understand how this term should be described, and how it is modelled with respect to all those previous terms I have listed? Reading journal papers I see the simple system as above described with lots of jargon I can't follow, and I'm unaware of the implications this has on the modeling.

If someone could clarify all this jargon and connections for me that would be suuuper helpful!

The equation $$y = hs + n$$ is a discrete-time model of a certain kind of wireless communications system.

• $$s$$ is a complex number, drawn from a finite set called a "constellation", that represents the information being transmitted.
• $$n$$ is a complex, circularly-symmetric Gaussian random variable with zero mean and variance usually denoted by $$N_0$$.
• $$h$$ represents the effect of the communications channel. In this simple model of a wireless channel, $$h$$ is a complex Gaussian random variable with zero mean and variance one. This model is called Rayleigh-fading because $$|h|^2$$ is a Rayleigh-distributed random variable.
• $$y$$ is the received complex number. From it, the receiver must make an estimate of $$s$$. This estimate is usually calculated by first estimating $$h$$, and then finding $$y' = \frac{h^* y}{|h|^2} = s + \frac{h^*n}{|h|^2}$$ Here you can see the importance of $$|h|^2$$ to enhance the noise, which is why the channel is called Rayleigh. The estimate of $$s$$ is calculated by finding the constellation element that is closest (in Euclidean-distance sense) to $$y'$$.
• Thanks! Could you also clarify the mutli access/multi antenna/bandlimited (and also cdma i think?) terms in your answer if possible? What effect would these notions have on modeling – tisPrimeTime Apr 8 '20 at 18:45
• @tisPrimeTime That broadens your question quite a bit! Each of those merits their own, separate question. I suggest repeating the format of this question: show the model you're interested in, how you understand it as a statistician, and ask how the model is interpreted in DSP/communications theory. – MBaz Apr 8 '20 at 19:39
• @tisPrimeTime: I come from statistics background and have the same problem that you describe. I think some of these "Signal Processing" books give a slightly more "statistical-time series style" than others. Let me look at my book shelves ( I have way more of these DSP books than I've actually read ) and see which I would recommend in terms of not being totally EE language based. Off the top of my head, I think Monson's "statistical signal processing" does a pretty decent job. I'm only at the DSP level and not the "communications systems" level so it may be too basic for you ? – mark leeds Apr 9 '20 at 7:04
• Thanks mate! Ill give it a look – tisPrimeTime Apr 9 '20 at 7:44
• Also, check this one out. Sometimes it's hard to know what's helpful because, when the notation is unfamiliar, how can you know !!!!! amazon.com/Random-Processes-Engineers-BruceHajek/dp/1107100127/… – mark leeds Apr 9 '20 at 8:14

The equation you wrote down describes the case for a narrowband (flat fading) MIMO system.

Narrowband/flat fading describes the scenario when your signal bandwidth is small relative to the channel bandwidth (inverse delay spread). Whatever system you are modeling, you'd have an idea of the delay spread and signal bandwidth, and could use that as a justification for using the model: $$\mathbf{y}=\mathbf{Hx}+\mathbf{\epsilon}$$.

In the frequency domain, use this picture to understand why it is also called flat flading. All frequencies in the signal will experience the same channel, that is, the channel response is "flat" over all frequencies (blue signal). This is opposed to wideband or frequency selective fading, which would correspond to the orange signal in the picture. The term OFDM is often times used when talking about this equation because the main idea of OFDM is to split up the signal into smaller subchannels so that each subchannel is flat (imagine splitting the orange signal into a bunch of smaller bandwidth slots like the blue one).

You can also look at this same thing in the time domain. Remember what a constant in frequency domain means for time domain, an impulse. For the narrowband case, even if there are multiple taps in the channel (never going to be exactly a flat channel response), if they all lie sufficiently within a symbol period then those energies get combined in the end. This is why the model equation does not contain any dependence on previous transmitted symbols. But in the wideband case, the multiple channel taps are sufficiently dispersed so that even after sampling at the symbol instants, there is energy from different symbol interfering with each other. For this case, using the model: $$\mathbf{y}=\mathbf{H}\mathbf{x}+\mathbf{\epsilon}$$ is not appropriate.

MIMO stands for multiple input multiple output, and in a communication system context this means multiple transmit antennas and multiple receive antennas. The picture to have in your mind is that the transmitter uses multiple antennas to input signals into the channel, and the receiver uses multiple antennas to get the output signals from the channel.

You also mention Rayleigh as an unknown. The matrix $$\mathbf{H}$$ is a $$N_{RX} \times N_{TX}$$ matrix with entries that are zero mean complex Gaussian. The term Rayleigh fading comes up because the distribution of the magnitude squared elements, $$|h_{ij}|^2$$, follows the Rayleigh distribution.