# 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, 2017 at 5:55
• @FlorentEcochard It's a simulation software that I'm using to get the amplitude spectrums. keysight.com/main/…
– John
Aug 26, 2017 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, 2017 at 6:32

For real-valued signals, the frequency spectrum is conjugate symmetric; i.e., $$X(f) = X(-f)^*$$ This translates into an even magnitude spectrum $$|X(f)| = |X(-f)|$$ and an odd phase spectrum.
An indirect method of obtaining the upper side band is to convert the signal into an analytic signal $$x_+(t) = x(t) + j \hat{x}(t)$$, where $$\hat{x}(t)$$ is the continuous-time Hilbert transform of $$x(t)$$. The resulting spectrum is single side band: $$X_+(f) = \begin{cases} 2 X(f) ~~, &\text{ for} ~~ f > 0 \\ 0 ~~, &\text{ for} ~~ f < 0 \\ \end{cases}$$
Therefore for the given signal $$x(t) = \cos(2\pi 20000 t)$$ with a spectrum $$X(f) = 0.5 \delta(f-20000) + 0.5 \delta(f+20000)$$ the upper side band will be given by $$X_+(f) = \delta(f-20000)$$