Consider a bandpass signal $s(t)$ with bandwidth $W$.

After bandpass filtering, let the output signal be $r(t)=s(t)+n(t)$

I have read a paper that denotes $n(t)$ as Gaussian noise with one-sided power spectral density $N_0$. Therefore, the noise power is $\sigma^2=\mathbb{E}\{n^2(t)\}=N_0W$.

What would be the purpose of denoting the noise as single-sided?

It seems that if we consider the noise as double-sided with power spectral density $\frac{N_0}{2}$, noise power is still $\sigma^2=\mathbb{E}\{n^2(t)\}=N_0W$ since we have to integrate over the negative frequencies and the positive frequencies.

What is the purpose of describing AWGN noise as single-sided versus double-sided? Considering the case of real signals, do both end up giving the same results?


1 Answer 1


Both descriptions give the same result. We often use the one-sided noise power spectral density (PSD) because for real-valued processes the negative frequencies are redundant, so defining the PSD for positive frequencies is sufficient. You just have to scale the noise power spectrum such that integrating the one-sided power spectral density (PSD) over the positive frequencies gives the same result as integrating the two-sided PSD over positive and negative frequencies. I.e., for white noise with a constant PSD, defining the one-sided PSD as $N_0$ means that we would need to define the two-sided PSD as $N_0/2$ such that the noise power remains the same: $N_0W=N_0/2\cdot 2W$


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