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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.

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This is the solution for part a of the exercise:

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

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

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

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

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

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2 Answers 2

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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.

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  • $\begingroup$ Can you show explicitly how to get the same result they got taking into account of folding negative frequencies as well? I got a different result. $\endgroup$
    – hana
    May 23 at 13:38
  • $\begingroup$ There are eight frequencies in the input: $f_A=5$, $f_B=10$, $f_C=30$, and $f_D=45$, and their negatives. $f_A$ and $f_B$ are not aliased. $\pm f_C$ produces aliases at 10 and -10 as explained in my answer. $f_D$ produces one alias at $45-40=5$ and $-f_D$ produces $-45+40=-5$, corresponding to a sine wave of frequency 5. $\endgroup$
    – MBaz
    May 23 at 13:58
  • $\begingroup$ I think what may be confusing you is the sign of the sine wave? $\endgroup$
    – MBaz
    May 23 at 13:59
  • $\begingroup$ I don't have any problem calculating aliased frequencies like that. I want to know how would you reconstruct the aliased signal ya(t) from these aliased frequencies. In the image I sent I did calculate all folding frequencies and then reconstruct the signal but got a different result than their answer. $\endgroup$
    – hana
    May 23 at 14:00
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    $\begingroup$ Just want to mention that I probably wrongly blamed the book on drawing fourier plot wrong. Examples in the book is for cosine instead of sine and the drawing is good for cosine. $\endgroup$
    – hana
    May 23 at 19:10
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The answer is that $\sin(-x) = -\sin(x)$, which is a standard trigonometry identity. Thus, $\sin(20\pi t) + \sin(-20\pi t) = \sin(20\pi t) - \sin(20\pi t) = 0$

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  • $\begingroup$ well, I think you misundestood my quesiton! I, of course, know baisc math like that. My question is more about the concept rather than math. $\endgroup$
    – hana
    May 23 at 13:35
  • $\begingroup$ Sorry, when you said you were confused by the canceling of the middle terms, I took it literally. MBaz's answer is correct. $\endgroup$
    – Gillespie
    May 23 at 13:46
  • $\begingroup$ No worries, MBaz's answer doesn't show how to get the result taking negative frequencies folding into account. $\endgroup$
    – hana
    May 23 at 13:50

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