# Calculate aliasing of $x_a(t) = \cos{(2\pi300t)} + \cos(2\pi600t)$ when sampled with $F_s = 1000$

I'm asked to sample the signal $$x_a(t) = \cos{(2\pi300t)} + \cos(2\pi600t)$$ with sampling frequency $$F_s = 1000$$ and plot the magnitude spectrum for the resulting sampled signal.

My thinking is that the frequency of $$300$$ does not change, so it just results in the normalized frequency $$\frac{3}{10} = 0.3$$.

Since $$600$$ is above the nyquist frequency, aliasing arises. So we get the frequencies $$\frac{6}{10} = 0.6$$ and $$\frac{6}{10} - 1 = - 0.4.$$

So in total I would plot peaks at the frequencies $$\pm0.3, \pm0.4, \pm0.6$$, but the answer is supposedly:

Why? Where is my thinking wrong? And what is the method to always get the right aliasing peaks?

• Welcome to SE.SP! Ask yourself: why should I include $\pm 0.6$ in the list of frequencies? Isn't 0.6 above Nyquist? So the only interesting frequencies should be between -0.5 and +0.5. Outside that range, the response repeats. – Peter K. May 30 '19 at 17:59

Your thinking is correct, but incomplete. Of course, the component at $$f=300$$ (unit depends on normalization of $$t$$) will not cause aliasing, so there will be no additional component below Nyquist, but what you forget is that sampling makes the spectrum periodic with period $$f_s$$ (sampling frequency). So the $$f=\pm 300$$ component results in components at $$nf_s\pm 300$$, $$n\in\mathbb{Z}$$. For $$n=1$$ you get the components at $$f=1000\pm 300$$, one of which (the one at $$f=700$$) is the one you see in the figure at normalized frequency $$0.7$$.