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?
