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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?

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  • $\begingroup$ What's "the advanced design system"? $\endgroup$
    – Florent
    Aug 26, 2017 at 5:55
  • $\begingroup$ @FlorentEcochard It's a simulation software that I'm using to get the amplitude spectrums. keysight.com/main/… $\endgroup$
    – John
    Aug 26, 2017 at 5:58
  • $\begingroup$ 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... $\endgroup$
    – Florent
    Aug 26, 2017 at 6:32

2 Answers 2

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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.

Because of this symmetry, it's also called as a double side band spectrum. However, for a real-valued signal, a single side band is also sufficient to represent it. Different methods exist in obtaining those upper or lower side bands individually, such as half band filtering as a direct method.

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 $$X_+(f) = \delta(f-20000)$$

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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.

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