# Difference between single sided and double sided amplitude spectrum

I've recently gotten stuck with a small issue. I was asked to plot the single sided amplitude spectrum of the signal $x(t)=\cos(2\pi\cdot20000t)$.

I know that $X(f) = \frac{1}{2}\cdot\left[\delta(f-20000) + \delta(f+20000) \right]$. So there should be 2 impulses with amplitudes of $\frac{1}{2}$ at $\pm 20\,\text{kHz}$.

But I tried using Keysight's Advanced Design System to plot the single sided amplitude spectrum, and only got one impulse at $20\,\text{kHz}$ with an amplitude of $1$.

Why is this so?

• What's "the advanced design system"? Aug 26 '17 at 5:55
• @FlorentEcochard It's a simulation software that I'm using to get the amplitude spectrums. keysight.com/main/…
– John
Aug 26 '17 at 5:58
• IDK about this software but it probably normalized the amplitude. Check the documentation maybe. Also, if you were asked to plot the single-sided spectrum I would suspect that the negative frequency peak should not appear on your plot... Aug 26 '17 at 6:32

For baseband real signals the frequency spectrum is conjugate symmetric; i.e., $$X(f) = X(-f)^*$$ This translates to an even magnitude spectrum; i.e., $$|X(f)| = |X(-f)|$$ and an odd phase spectrum.

Because of the symmetry it's also called as the double side band spectrum in commnnications terminology. For the real basband signal, a single side band is also sufficient to represent it. THere are different approaches to obtain the single side band as the uppersideband and the lower side band.

One method of obtaining the upper side band is to convert the signal into the analytic signal which is $x_+(t) = x(t) + j \hat{x}(t)$ where $\hat{x}(t)$ is the Hilbert transform of $x(t)$. The resulting spectrum is: $$X_+(f) = \begin{cases} 2 X(f) ~~, &\text{ for} ~~ f > 0 \\ 0 ~~, &\text{ for} ~~ f < 0 \\ \end{cases}$$

Therefore for the given signal $x(t) = \cos(2\pi 20000 t)$ whose spectrum is $$X(f) = 0.5 \delta(f-20000) + 0.5 \delta(f+20000)$$ the upper side band will be given by $$X_+(f) = \delta(f-20000)$$

Note that you can adjust the amplitude of single side band if you wish to do so.

I think the term "baseband real signals" could be misleading here, because there is no carrier frequency given in the question and therefore no down conversion to baseband.

If you calculate the FT of any real signal, the negative frequencies are always complex conjugate to the positive frequencies (as written above), so there is no additional information in the negative frequencies. Therefore, when showing the spectrum of real bandpass signal, one usually only shows a single side spectrum.