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?
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:
$$ X_+(f) = 
\begin{cases}
2 X(f) ~~, &amp;\text{ for} ~~ f &gt; 0 \\
0 ~~, &amp;\text{ for} ~~ f &lt; 0 \\
\end{cases}
$$

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.