Question on the definition of one-sided PSD in AWGN channel 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 $x  \sim \mathcal N(0,
&gt; \sigma_x^2), y \sim \mathcal N (0, \sigma_y^2)$. 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)?
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 $E\{|n|^2\}=\sigma^2_x+\sigma^2_y$.