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The time signal which i'm trying to find the aliases for is:

$$x:{\mathbb R}\rightarrow {\mathbb R}\\\ x(t)=\cos(50t) +2\cos(70t).$$

If the sample period is $T_s = \frac{\pi}{60}$ then according to Nyquist -Shannon sampling theorem (which btw my professor failed to prove) there is/are a signal(s) which after sampling will be equal to the sampled version of the above signal, if we sample those with the same sample frequency in the frequency band $[-55, 55]$.

I don't understand the meaning of the last sentence , what does a frequency band means here?

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Let's use a simple example: A sine wave of 400 Hz. The Fourier spectrum has two components: one at 400 Hz and one at -400 Hz (since it's a real valued time signal).

Now let's sample this at 1 kHz. Sampling in one domain is periodic repetition in the other domain. So the sampled signal consists of your two original frequencies, -400 Hz and +400 Hz repeated again and again in 1 kHz intervals. So you get components at +-400 Hz, +-600 Hz, +-1400 Hz, +-1600 Hz, etc.

Any of these mirror frequencies (600, 1400, 1600, 2400, 2600, ...) will give you the exact same samples as the original 400 Hz.

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  • $\begingroup$ Hi, I'm still struggling here, does the frequency band make sense in frequency domain or i can give it some meaning in time domain as well? So i need to look at its spectrum to find out what's happening? And i don't understand when you say So the sampled signal consists of your two original frequencies, -400 Hz and +400 Hz repeated again and again in 1 kHz intervals, to me the sampled signal is just a bunch of discrete points over time. And i'm sorry if my questions appears to be very idiotic, but i'm very confused. $\endgroup$
    – Sam B
    Jan 1 '19 at 20:43
  • $\begingroup$ I remember a part of the conversation between me and my professor, he told me if you sample the signal, you'll see impulses at -70, 70, 50, -50, which is completely vague to me. $\endgroup$
    – Sam B
    Jan 1 '19 at 20:47
  • $\begingroup$ @Hilmar: You answered above a while ago but somehow I came upon this thread recently. Would you mind explaining where you get the +-600, +-1400 and +- 1600 etc. Thanks. $\endgroup$
    – mark leeds
    Feb 22 at 3:52
  • $\begingroup$ @markleeds: it's simply $+-400 + n \cdot 1000, n \in \mathbb{N}$ : 400 Hz and -400Hz plus any integer multiple of the sample rate. It's easy to see when you draw it. $\endgroup$
    – Hilmar
    Feb 22 at 17:18
  • $\begingroup$ okay. got it. thanks. $\endgroup$
    – mark leeds
    Feb 23 at 1:29
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In Fourier analysis we are trying to see how sinusoids over all frequencies are scaled and added to define a time signal $x(t)$. Usually we have a time signal, do Fourier analysis, and we get a new function $X(f)$ that shows how each sinusoid at a frequency $f$ contributes to creating the time signal. This is called the spectrum of $x(t)$.

When first learning this theory, there is something that is often not stressed enough about Fourier analysis, and the Fourier transform specifically: it assumes that the sinusoids used for the analysis are complex. That is of the form

$$e^{j2{\pi}ft}$$

Any other details about signals being real or complex and how they show up on the spectrum stems from this assumption.

You can see this from the definition of the Fourier transform where the input signal is multiplied by the complex sinusoids over all frequencies

$$X(f) = \int_{-\infty}^{\infty}x(t)e^{-j2{\pi}ft}dt$$

Let's take the example of $x(t) = e^{-j2{\pi}f_0t}$. Taking the Fourier transform we get

$$X(f) = \int_{-\infty}^{\infty}e^{-j2{\pi}f_0t}e^{-j2{\pi}ft}dt = \delta(f-f_0)$$

This should be intuitively satisfying. Since our time signal is a complex sinusoid at some frequency $f_0$, our spectrum results in a single "spike" that only exists when $f = f_0$. In other words, the result tells us that only the complex sinusoid of the frequency $f_0$ contributes to the signal.

Now let's take the example where our time signal is now the real sinusoid $x(t) = cos({2{\pi}f_0t})$. A transform table will tell you that the Fourier transform is

$$X(f) = \frac{1}{2}[\delta(f - f_0) + \delta(f + f_0)]$$

Why do the two "spikes" show up? Why is it that this result looks like the one above, but now includes a mirrored version on the other side of the spectrum?

Recall the result

$$cos(x) = \frac{e^{-jx} + e^{jx}}{2}$$

A real sinusoid can be seen as the sum of two complex sinusoids.

Using our signal $x(t)$ we get

$$cos(2{\pi}f_0t) = \frac{e^{-j2{\pi}f_0t} + e^{j2{\pi}f_0t}}{2}$$

Treating each term separately and taking the Fourier transform, you will yield the sum of the two Dirac deltas.

Applying this to your example, which assumes the angular frequency $\omega$:

$$x(t) = cos(50t) + 2cos(70t)$$ $$ = \frac{e^{-j50t} + e^{j50t}}{2} + ({e^{-j70t} + e^{j70t}})$$

Following the same logic as above and applying the angular-frequency Fourier transform you should see why you get Dirac deltas at $\omega = -70, -50, 50, 70$.

Sampling

Using the sampling theorem, it states that when sampling at a rate of $f_s$, the resulting discrete signal now has a spectrum that contains copies of the original spectrum centered around all multiples of $f_s$. In your example, you are actually under-sampling and therefore aliasing.

Your sampling period results in a angular-frequency of $\omega_s = 1/T_s \approx 19 \space rads/s$. When the frequency band is referenced, the problem is asking you to only consider the chunk of the spectrum between $\omega = -55$ and $\omega = 55$.

Since you will be aliased, copies from the $cos(70t)$ term will actually appear at $\omega = -50$ and $\omega = 50$. This is the same result as if you sampled the original signal correctly, that is, you met the Nyquist criteria.

You should really try and simulate this. It will accelerate mastering and becoming comfortable with these concepts.

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