0
$\begingroup$

This my elaboration of the aliasing issue: a continuous signal can be represented by factors of :

$e^{(i2{\pi}ft)}$ if we sample this signal then I will get:

$e^{(i2{\pi}fk/N)}$ where $k=0,1,2.., N-1$

this point can be represented by the below factor as well:

$e^{(i2{\pi}f_pt)}$

if we put all together:

$e^{(i2{\pi}ft)} = e^{(i2{\pi}f_pt)}$

$e^{(i2{\pi}(f_p-f)t)} = 1$

this is satisfied only if :

$(f_p-f) 2{\pi} /N = 2{\pi}m,$ this lead to

$f_p-f = N*m$

now here it comes the analysis

$N$ is actually the sampling frequency, let say $N$ = 100 so if a say my signal is a single frequency $f$ = 100 then my aliasing frequency (or the other signal frequency) would be:

$f_p = f + m*N = 100 + m*100 = 100(1+m),$

$f_p(m=-1) = 0,$ issue,

$f_p(m=0) = 100,$ no issue,

$f_p(m=1)= 200,$..issue,

now if my $f = 50$ $f_p = f + m*N = 50 + m*100 = 100(0.5+m),$

$f_p(m=-1) = -50,$ issue

$f_p(m=0) = 50,$ no issue

$f_p(m=1)= 150,$..issue

now if my $f = 25$

$f_p = f + m*N = 25 + m*100 = 100(0.25+m),$

$f_p(m=-1) = -75,$ no issue

$f_p(m=0) = 25,$ no issue

$f_p(m=1)= 125,$.no .issue

now if my $f = 75$

$ff = f + m*N = 75 + m*100 = 100(0.75+m),$

$ff(m=-1) = -25,$ issue

$ff(m=0) = 75,$ no issue

$ff(m=1)= 175,$.no .issue

so then after tabulating all possible frequencies I can say that the best frequencies where $f_p$ doesn't exist is when $f < N / 2$ so What do you think about this reasoning. is there other way to explain the 1/2 term that multiplies $N$ to give us the right signal frequency ( that can be correctly sampled with $N$ samples)?

Appreciate your opinion

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Can I please ask you to clarify the notation a little bit? Two points: a) please consider using the latex way of writing expressions, b) please try to stick to one "domain" (Either continuous or discrete). When it comes to notation it might be the difference in one letter but it is important conceptually too. $\endgroup$ – A_A Jun 30 '20 at 9:30
  • $\begingroup$ A couple of points: 1) A signal can be represented by a sum of sinusoidals, not a product, so the terms aren't "factors". 2) For a complex valued tone as you are modelling, a negative frequency is perfectly valid, you should have no issue with that. It is the same as a left handed screw vs a right handed one. When two opposing same sized corkscrews are added togehter they form a real valued signal. If you reveres both and add them together again, you get the same signal, so no way to discriminate. Congratulations on working out the meaning of aliasing for yourself. $\endgroup$ – Cedron Dawg Jun 30 '20 at 11:37
  • $\begingroup$ I hope to add some intuition to accompany the link from @dilip-sarwate. In light of the Nyquist (Shannon) Sampling Theorem, once a user has samples, the user "pretends" that the original continuous-time signal was bandlimited, with $\pm\frac{1}{2}f_{\textrm{samp}}$ as the boundaries of the frequency band. This is a better assumption if some low-pass filtering was performed before sampling. For a properly bandlimited signal, all of its "frequency content" comes from this frequency band. $\endgroup$ – Joe Mack Jul 1 '20 at 18:54
1
$\begingroup$

From the definition of sampling, and the definition of periodicity follows that any signal is identical to signal at a frequency higher by an integer amount of sampling frequencies.

Since for real signals, negative frequencies aren't distinguishable from positive ones, it directly follows that for every $0.5f_\text{sample}\le f \le f_\text{sample}$, there is a negative frequency $-0.5f_\text{sample}\ge \tilde f=1- f \ge f_\text{sample}$ that looks the same as a positive frequency.

Hence, only frequencies lower than $|f| < 0.5f_\text{sample}$ can be uniquely represented.

Improve this answer
$\endgroup$
0
$\begingroup$

I find this topic easiest to understand in the frequency domain. Please see my explanation there for more detail with illustrations:

Sampled and aliasing signal

Basically, sampling is a modulation. In fact, we call the usual form of digital sampling “PCM”—Pulse Code Modulation. On the analog side, up to the point of being en-“Code”-d into a digital value, it’s fundamentally Pulse Amplitude Modulation—amplitude modulation (AM) by a pulse train at the sampling frequency.

AM is very simple to understand—it produces sum and different frequencies of the input signal and modulating signal. That is, a sum and difference frequency is produced in the output, for input and modulator frequency component.

The input frequencies are not known in advance for arbitrary input, but the modulator’s frequency spectrum is well known—a pulse train is the sum of all cosines that are of integer multiples of the sampling frequency. So, a pulse train of 100 Hz is the sum of a cosine of 0 Hz (a constant), plus a cosine of 100 Hz, 200 Hz, 300 Hz… all at the same amplitude. You can see this from the Fourier Transform, or just play with it intuitively by summing up cosines:

https://www.desmos.com/calculator/4uzpvenl13

Fortunately, we can mostly just look at the first two frequencies—0 Hz and the sampling frequency. For example, if your signal is a sinusoid of frequency 300 Hz, and the sample rate is 48 kHz, the 0 Hz component in the pulse train produces 300 Hz in the output (0 + 300, plus 0 - 300—negative frequencies are another topic, but we can ignore them). And also the frequencies 48000 + 300 and 48000 - 300. We’re mainly concerned about what happens near the Nyquist frequency—half the sample rate—so disregard the higher frequency for now. So, with 300 Hz sampled at 48 kHz, the lowest two frequencies we get are 300 Hz and 47700 Hz. When we play it back, everything above 24 kHz is removed by the lowpass filter in the DAC, and we hear 300 Hz. Perfect.

But as we move that input signal up in frequency, it should be apparent that the next higher frequency is moving down. For examples, at 20 kHz, the two frequencies produced are 20000 Hz and 28000 Hz. If we keep moving the input frequency up, the two meet at half the sample rate. And we we go above that, the frequency above dips below half the sample rate. That’s why if you sweep a signal upward, it seems to turn around and go down once it passes through the Nyquist frequency. And this is aliasing—at 30 kHz sinusoid samples at 48 kHz becomes 18 kHz plus 30 kHz—plus 66 kHz (96 - 30) plus 78 kHz (48 + 30), and many others. When we play it back the lowpass filter eliminates all by 18 kHz, but that's an alias of our 30 kHz input.

That's why everything from half the sample rate up is a no-go—we won't get back exactly what we put in.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.