# Clarification Needed: Power Spectral Density of Real Signal

For a real signal, the power spectrum will have Hermitian symmetry. If I consider the one-sided spectrum (as negative frequencies, or anything above the Nyquist frequency are redundant) must I then multiply my power spectrum by a factor of 2? I have seen this done but I can't understand why the "redundant", symmetrical portion of the spectrum should factor into the power density? Maybe this is a matter of convention? Please correct my assumptions if necessary.

• The power spectrum of a real-valued signal must be a real-valued nonnegative even function of $f$. Yes it has Hermitian symmetry but in a trivial sense since there are no imaginary components to worry about and the real components are even functions of $f$. – Dilip Sarwate Sep 18 '13 at 3:20

This stems from Parseval's theorem. For any signal, $\int_{-\infty}^{+\infty}|x(t)|^{2}\mathrm dt = \int_{-\infty}^{+\infty}|X(f)|^{2}\mathrm df$ and hence for a real signal, the one sided PSD must be multiplied by a factor of 2 because of Symmetry. Intuitively, the Fourier Analysis of any signal provides us with a spectrum which has Negative frequencies too because of the fact that Sines and Cosines can be constructed using Euler's theorem of using Complex Exponentials of positive and negative frequencies; and it so happens for any real signal that all you need are complex exponentials which are conjugates of each other to construct them.