# How to calculate Eb/N0 for a digital BPSK modulated multicarrier waveform

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?

• What are $A$, $f_s$, and $f_n$? – BlackMath Jun 20 at 22:14
• Amplitude, sampling rate, and $n$-th carrier, respectively. Question edited for clarity. – The Dude Jun 21 at 1:00
• So, why do have intercarrier interference? – BlackMath Jun 21 at 1:07
• @BlackMath There is no intercarrier interference. What do you mean? – The Dude Jun 21 at 10:19
• 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. – BlackMath Jun 21 at 16:33