I have a continuous signal x(t) such that

$x(t)=12cos(6\pi t)+6cos(24\pi t)+3cos(30 \pi t)$

and is asked to sketch a 1-sided Amplitude Spectrum of the signal x(t) if sampled above the minimum sampling rate.

because $w=2\pi$, i worked out that the three signals are 3Hz, 12Hz and 15 Hz.

I'm just wondering, when I plot the Amplitude Spectrum should the Amplitude just be the corresponding coefficients? ie. 12 for 3Hz, 6 for 12Hz and 3 for 15Hz?

EDIT: Additionally, what's the difference between 1-sided Amplitude Spectrum and 2-sided Amplitude Spectrum? Does one offer any more benefit over the other?

  • $\begingroup$ For real signals, $X(f) = X(-f)^\ast$, so no need to show both sides. Also, for power spectral density, $G_x(f)$ is non-negative, real, and symmetric. $\endgroup$
    – Juancho
    Commented Jun 20, 2012 at 13:32

2 Answers 2


To complement @wrapperapps answer, you can express sinusoids as sum of exponentials: $\cos(2 \pi f_0 t) = (e^{j 2 \pi f_0 t} + e^{-j 2 \pi f_0 t})/2$.

Thus, the fourier transform yields $\delta(f+f_0)/2 + \delta(f-f_0)/2$: each frequency delta has amplitude $1/2$.

  • $\begingroup$ So If i'm plotting the 1-sided amplitude spectrum, the amplitude would be half of the coefficient. Does that mean if i plot the 2-sided amplitude spectrum, they would have amplitude 1/4 of the coefficient? $\endgroup$
    – davidx1
    Commented Jun 20, 2012 at 21:39

For 2-sided spectrum, the amplitude will be halved since there is a negative and positive component for each frequency. Since the components are symmetrical about 0 Hz when sampled above Nyquist, sometimes the 1-sided spectrum is used (which is still half amplitude I think?). In this case there isn't really much benefit other than clarity of seeing the mirrored frequency deltas.


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