Why aren't negative frequencies folded in reconstruction of the aliased signal?

Question

I'm working on the problem 1.9 from the book Introduction to Signal Processing by Sophocles J. Orfanidis. The pdf version and solution is freely available here.

This is the solution for part a of the exercise:

I'm a bit confused about the cancelling of two middle terms.

This is my drawing showing my understanding of what they're doing.

However, why don't you do the same for negative frequencies?

If you do this then you would get a different result.

EDIT: typo - the last term in the image should be 2sin(20πt) instead.

Answer 1

They are doing the same for the negative frequencies, implicitly.

In a problem like this, all signals are real: the input, the sampled signal, and the reconstructed signal. As a consequence, all spectrums are symmetric. If aliasing produces a frequency component at $x$ Hz, then it will also produce one at $-x$ Hz.

Let us take a closer look. We need the following concepts:

1. A sine $$\sin(2\pi f_0 t) = \frac{e^{j2\pi f_0 t}-e^{j2\pi (-f_0) t}}{2j}$$ has frequencies $f_0$ and $-f_0$.

2. A single frequency component $e^{j2\pi f_0 t}$ sampled at frequency $f_s$ produces aliases at frequencies $f_0 + Nf_s$ for all integers $N$.

3. Out of the infinite number of aliases, we're interested in those with the lowest frequency; those are the unique aliases that appear in the Nyquist range $-f_s/2$ to $f_s/2$.

Now, let's focus on $\sin(2\pi f_C t)$ with $f_C=30$. It has two frequencies, $30$ and $-30$. Let us find the aliases of each frequency separately.

• The term $e^{j2\pi 30 t}/2j$ will alias to frequencies $30+40N$, some examples of which are $-90,-50,-10,30,70,110...$. The only value within the Nyquist range corresponds to $N=-1$, so we have $30-40=-10$. We can say that $e^{j2\pi (30) t}$ aliases to $e^{j2\pi (-10) t}/2j$.
• The term $-e^{j2\pi (-30) t}/2j$ will alias to frequencies $-30+40N$. The only alias in the Nyquist range corresponds to $N=1$, where we have $-30+40=10$. We can say that $-e^{j2\pi (-30) t}/2j$ aliases to $-e^{j2\pi (10) t}/2j$.

We can conclude these two things:

1. The aliases are symmetric. This is the reason many textbooks don't calculate every single alias; as soon as you find an alias at $x$, you know there is one at $-x$. This is true only for real signals, which is what we are dealing with in this problem.

2. The sum of the two aliases we calculated is $$ \frac{e^{j2\pi (-10) t}-e^{j2\pi (10) t}}{2j}$$ which implies that $f_0 = -10$. Then, the aliased sine wave is $\sin(2\pi (-10) t) = -\sin(2\pi 10 t)$. This means that aliasing can change the sign of the aliased signal.

Coming back to the textbook problem, we see that the term $\sin(2\pi 10t)$ is cancelled by the alias $-\sin(2\pi 10t)$. So, a third conclusion is, aliases may cancel out terms in the input signal!

I guess a fourth and final conclusion might be, avoid aliasing at all costs.

Answer 2

The answer is that $\sin(-x) = -\sin(x)$, which is a standard trigonometry identity. Thus, $\sin(20\pi t) + \sin(-20\pi t) = \sin(20\pi t) - \sin(20\pi t) = 0$