Question on the definition of one-sided PSD in AWGN channel

Question

I am currently simulating the following in a telecommunications exercise in Python:

Consider a binary sequence $m[n]$ that is being modulated with QPSK (Gray-mapped) that is being transmitted via an AWGN (additive white gaussian noise) channel that adds complex noise of the form $$n = x + jy$$ such that $x,y$ are independent with $x \sim \mathcal N(0, > \sigma_x^2), y \sim \mathcal N (0, \sigma_y^2)$. The variances should be selected in a way that the one-sided PSD of the white noise is $N_0 / 2$.

I am a little bit confused by the definition of one-sided PSD:

Since the one-sided PSD is $N_0 / 2$ the two sided PSD must be

$$S_{n} (f) = \frac {N_0} 4$$

The inverse-DTFT transform gives us the discrete autocorellation function $$R_n [m] = \frac {N_0} 4 \delta [m]$$ which means that $\mathbb E[n^2] = N_0 / 4$

$$\mathbb E[n^2] = \mathbb E[x^2 + 2jxy + y^2] = \sigma_x^2 + \sigma_y^2 = \frac {N_0} 4$$

So we can select $$\sigma_x^2 = \frac {N_0} 4, \sigma_y^2 = \frac {N_0} 4$$ for simulating $n = x + jy$

My question is: Have I understood the meaning of the one-sided PSD correctly (and therefore are my assumptions correct)?

Answer 1

The relation between one-sided and two-sided noise PSD is correct. But note that the power of a complex random variable $n$ is given by $E\{|n|^2\}$, and not by $E\{n^2\}$. So in your case you obtain $E\{|n|^2\}=\sigma^2_x+\sigma^2_y$.