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

Question

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?

Answer 1

Peter said it all in his comment, but I'll try to explain it here in a bit more detail.

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 $0.7$.