I have been doing a matlab simulation of a VLC (visible light communication) system that uses OOK for the modulator.

I have to explain the basic knowledge behing the simulation and I´m having lots of doubts when I´m trying to understand what really the Eb/N0 is and why not SNR is used. The task is to compare on a BER graph the system using no codification, using Manchester codification and using two types of Miller codifications. The conditions are AWGN with everything ideal or basic. On the modulator and demodulator, LEDs are used for the transmiter and receiver. My tutor explained me the relation between bits transmited and power transmitted, wich can be resumed in the following image: VLC diodes image

R(λ) is a linear and positive function, so Prx is proportional to Ptx enter image description here

And to calculate the SNR that the simulation uses, you use the relation enter image description here in wich Pm is enter image description here and the noise power is σ^2, as it is an AWGN channel. In the simulation I didn´t had much problems with this because what I did was the inverse process, meaning for each SNR value on the graph range (-1 to 15 dB) and having Pm equal to 0.5 W, I calculate the range of σ and then for each σ I create a random noise vector and add it to the sequence of bits that the transmiter sended. I do the same process for each codification using the same data bits and with all the results I obtain the BER graphs.

All my doubts start now with the Eb/N0 relation that is used everywhere for the BER graphs instead of SNR. I have looked in many books but they all use it with more complex modulations and codification, not a simple BASK. My questions are:

  1. The Eb is of the informatio/data bits or the bits transmited? Because if I use a coder like Manchester or Miller there are two bits for every bit of data, so I dont know if I have to use Tb or 2xTb. My knowledge of how to calculate Eb in this case is, being g(t) a rectangular shape in OOK, I suppose amplitude of bit "1" to be 1 and amplitude of bit "0" to be 0, so the energy for "1" would be 1/Tb and 0/Tb for "0".

  2. The OOK transmited signal is a passband signal, with the visible light being the carrier frecuency. Here on the other hand it says that the EB/N0 is analyzed after it has changed back to baseband. That I have not seen explained in the books but it is good to know. It also says that it is measured after the matched filter, ¿what would be the matched filter in this case for an OOK in a VLC system? Here it says that the energy is the convolution between the transmited signal and the matched filter. In my matlab simulation I dont need a matched filter because I just take the value of the vector of the sum of the transmited and noise vectors. Is it correct to say that the matched filter is a square pulse with amplitude one and duration Tb?

  3. What is the bandwith B in this case? Is it 2/Tb? (a sinc in the frecuency domain with the main lobe between -1/tb and +1/Tb)

  4. In Communications System Engineering by Proakis and Salehi, it says

In general, the probability of error is a function of the code characteristics, the types of waveforms used to transmit the information over the channel, the transmitter power, the characteristics of the channel; i.e., the amount of noise, the nature of the interference, etc., and the method of demodulation and decoding.

But here it says that

BER depends only on the bit energy and noise PSD.

Isn´t BER a measure of the error probability function too?

  1. To relate the noise added by the channel to the signal transmited (σ) and the N0/2 dsp for the BER/N0 ratio, is it correct to say σ^2 = N0/2*B?

  2. If a FEC codification is also used adding redundant bits for error correction, what bits are counted? The transmited ones only or all?


1 Answer 1

  1. $E_b$ refers to the energy spent per information bit. That's the whole point!
  2. You're technically right that you've got a visible light carrier, but you're using this as a baseband system, to be honest – your receiver doesn't try to mix down your signal, it just detects intensity. You must always look at statements in context!
  3. Define "bandwidth": That's not unambigous. Technically, your drawing shows rectangular waves, which have infinite support in frequency domain. In reality, your system, intentionally or unintentionally, has limited physical ability to change fast. What you state as bandwidth is, however, up to how you mathematically define that - is the bandwidth what contains 50% of the transmitted power? Or is it the point in frequency after which your (low-pass) system never reaches -6dB of the DC gain? Or...?
  4. (since I wrote that answer): The answer there and Matt's comment assume your receiver is fixed, your question is about designing the receiver system, so these answers do not concern the same thing :) Context really matters! You're "shopping" statements and definitions from different contexts, and that's really just going to confuse you, if you don't understand their background.
  5. Your statement is a bit confusing, because $\sigma$ is neither the signal transmitted nor the noise added. It's usually the symbol for a standard deviation – need to be a bit careful of what. An index can help clarify that. Generally, for white noise (as in, AW.N), yes, the noise spectral density level $N_0$ is a constant over all frequencies, so the noise power is the integral of that over the noise-equivalent bandwidth; which might be a hint on how you'd want to define your bandwidth (3.). But if you use a noise-equivalent bandwidth, then that's a practical property of the receiver, and then the formula you stated is almost certainly wrong (unless you specifically designed your receive filtering so that it becomes right).
  6. same question as 1., same answer as 1.
  • $\begingroup$ Thank you very much Marcus for your help, it clarifies many doubts I have. I´ll process your answer now, and I´m sure more doubts will come about the matter, but for now you answered all my questions. $\endgroup$
    – Elias4l
    Commented May 16, 2022 at 21:51

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