Suppose that I have a BPSK modulated multicarrier carrier waveform $L$ samples long at a sampling rate of $f_{s}$:

$$ x_{tx}[l] = \sum\limits_{n=0}^{N-1} A \cdot \cos\left(2 \pi \frac{f_{n}}{f_{s}} l + \theta_{n}\right) \quad l=0,1,2,\ldots,L-1 ; \; \theta_{n} \in \{-1, 1\}$$

where $A$ is a constant amplitude, and $f_{n}$ is the $n$-th carrier.

I pass this into an AWGN channel where the elements $v[l] \sim \mathcal{CN}(0, \sigma_{v}^{2})$ are iid. This results in the received signal:

$$ x_{rx}[l] = x_{tx}[l] + v[l] = \sum\limits_{n=0}^{N-1} A \cdot \cos\left(2 \pi \frac{f_{n}}{f_{s}} l + \theta_{n}\right) + v[l] \quad l=0,1,2,\ldots,L-1 $$

  • Question:

    How do I calculate $\frac{E_{b}}{N_{0}}$ of the received waveform $x_{rx}[l]$?

My attempt:

$E_{b}$ stands for Energy per Bit. Therefore, I first need to calculate the total energy of the waveform. This is simple:

$$E_{t} = \sum\limits_{l=0}^{L-1} x_{tx}[l]^2$$

Then, due to the modulation type, the energy per bit is just:

$$E_{b} = \frac{E_{t}}{N}$$.

Now $N_{0}$ is the noise spectral density. Therefore if we are given a two sided bandwidth of $f_{s}$ Hz, then:

$$N_{0} = \frac{\sigma_{v}^{2}}{f_{s}}$$

Therefore, our $\frac{E_{b}}{N_{0}}$ ratio is:

$$ \frac{E_{b}}{N_{0}} = \frac{f_{s}E_{t}}{N \sigma_{v}^{2}} $$

Am I correct? Or am I missing something?

  • $\begingroup$ What are $A$, $f_s$, and $f_n$? $\endgroup$ – BlackMath Jun 20 at 22:14
  • $\begingroup$ Amplitude, sampling rate, and $n$-th carrier, respectively. Question edited for clarity. $\endgroup$ – The Dude Jun 21 at 1:00
  • $\begingroup$ So, why do have intercarrier interference? $\endgroup$ – BlackMath Jun 21 at 1:07
  • $\begingroup$ @BlackMath There is no intercarrier interference. What do you mean? $\endgroup$ – The Dude Jun 21 at 10:19
  • $\begingroup$ You have the received signal with $N$ terms, each with a different subcarrier frequency. In OFDM, the received signal can be written in the frequency domain as $$Y[k]=X[k]+N[k]$$ So, the SNR is $$\frac{\mathbb{E}[|X[k]|^2]}{\mathbb{E}[|N[k]|^2]}$$ Then by substituting for $X[k]$ and $N[k]$ is terms of $x[n]$ and $z[n]$, the signal and noise in the time domain, you can find the SNR. $\endgroup$ – BlackMath Jun 21 at 16:33

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