# Tag Info

95

The sampling rate of a real signal needs to be greater than twice the signal bandwidth. Audio practically starts at 0 Hz, so the highest frequency present in audio recorded at 44.1 kHz is 22.05 kHz (22.05 kHz bandwidth). Perfect brickwall filters are mathematically impossible, so we can't just perfectly cut off frequencies above 20 kHz. The extra 2 kHz is ...

73

44,100 was chosen by Sony because it is the product of the squares of the first four prime numbers. This makes it divisible by many other whole numbers, which is a useful property in digital sampling. 44100 = 2^2 * 3^2 * 5^2 * 7^2 As you've noticed, 44100 is also just above the limit of human hearing doubled. The just above part gives the filters some ...

28

Does the Nyquist frequency of the Cochlear nerve impose the fundamental limit on human hearing? No. A quick run-through the human auditory system: The outer ear (pinnae, ear canal), spatially "encodes" the sound direction of incidence and funnel the sound pressure towards the ear drum, which converts sound into physical motions, i.e. mechanical ...

21

It is actually not distorted, it is sampled at high enough rate. What fools you is the straight lines drawn between sample points, it gives you a false impression of the waveform. It shows you a linear interpolation of the signal. It does not represent how the signal would actually look like. A sampled signal exists only at the sample points, and to convert ...

15

Harry Nyquist invented/discovered/proved a lot of things; it can be hard to keep track of them all. The three most important for signal processing and communications are probably these: If you sample a (real) signal $s(t)$ at $f_s>2B$ samples per second, then $s(t)$ can be reconstructed from its samples, where $B$ is the bandwidth of $s(t)$. The lower ...

15

Once you start changing the amplitude you are increasing the bandwidth of the signal. That's called "amplitude modulation" and the highest frequency is now the sum of the original frequency and the highest frequency in the modulation signal. The sampling theorem still holds. You still only need twice the bandwidth but the bandwidth has increased ...

13

The Nyquist rate is above twice the bandlimit of a baseband signal that you want to capture without ambiguity (e.g. aliasing). Sample at a lower rate than twice 20kHz, and you won't be able to tell the difference between very high and very low frequencies just from looking at the samples, due to aliasing. Added: Note that any finite length signal has ...

10

Basically, twice the bandwidth is a common requirement for signal sampling, thus $2\times 20 = 40$ kHz is a minimum. Then, a little more is useful to cope with imperfect filtering and quantization. Details follow. What you need in theory is not what is required in practice. This goes along the quote (attributed to many): In theory there is no difference ...

9

The actual requirement is to sample at GREATER then twice the bandwidth, not at a rate equal to it... So only your 80Hz same set actually meets the requirement, because the 60Hz case is ambiguous in general, consider if you were sampling sin (2PiFt) instead then you would get a flat line at zero amplitude.... And changing the angle between sin and cos would ...

8

Harry Nyquist made so many contributions that it's easy to get confused. Related to sampling, Nyquist proved that a signal $s(t)$ bandlimited to $B$ Hz can be reconstructed from samples taken at a rate larger than $2B$. This is unrelated to Faster than Nyquist (FTN) signaling, though. Related to communications, Nyquist also showed that the maximum ISI-...

8

The OP's opening statement is incorrect: $f_s > f_{max}/2$ prevents frequency aliasing for a bandlimited signal, but not amplitude aliasing $f_s > 2 f_{max}$ prevents aliasing. It's as simple as that. There is no such distinction as "amplitude aliasing". Since the OP has stated the signal is band-limited; as long as we can assume that means ...

7

The Nyquist criteria refers not to the frequency, but to the bandwidth, which is related to information density in a signal. A very high frequency signal, of approximately known frequency, with a sufficiently small bandwidth, will still be aliased or folded down with baseband frequencies by undersampling. But if the bandwidth (or other known ...

7

I don't have a real answer but I have the feeling that this result will help you out: Bernstein's inequality says that, if the signal $x(t)$ is bandlimited to $|f|\leq B$, then $$\left| \frac{\textrm{d}x(t)}{\textrm{d}t}\right|\leq 4\pi B \,\textrm{sup}_{\tau\in\mathbb{R}}|x(\tau)| ,\,\,t\in\mathbb{R}$$ where $\textrm{sup}$ stands for "least upper bound". I ...

6

The sample rate needs to be GREATER than (NOT just equal to) twice the highest non-zero frequency content of the signal being sampled. Just a little bit greater might work, but the closer the sample rate is to twice the signal frequency, the longer in time you may need to sample to raise the signal above the noise and complex conjugate image in a DFT/FFT ...

6

The sampling theorem states that $f_\mathrm{S} \geq 2f_\mathrm{max}$, where $f_\mathrm{S}$ and $f_\mathrm{max}$ are the sampling and maximum signal freuqency, respectively. But there's an additional condition: The equal sign only holds if the signal spectrum does not contain a dirac impulse at $f_\mathrm{S}/2$ which is clearly the case in your example. ...

6

you don't have to set $X\left(\frac{N}{2} \right)=0$ if you don't want to. it will correspond to this component: $$X(k)\frac{1}{N}e^{j 2 \pi \frac{nk}{N}}\bigg|_{k=\frac{N}{2}} = X\left(\frac{N}{2} \right)\frac{1}{N}(-1)^n$$ but when you sample some $x[n]$, FFT it and find that $X\left(\frac{N}{2} \right) \ne 0$, you do not know the phase of that ...

6

You're correct, it has to do with the Cut Off frequency of the Gaussian Blur Filter in its Frequency Domain. In order to see it, just apply a DFT (Using MATLAB it can be achieved by fft / fft2) and look on the absolute value. Look for the -3dB point and you'll see. There is also an intuitive explanation on the original article which say that blurring ...

6

These terms are indeed named in a confusing manner, as frequency and rate are pretty much synonyms. Either way: Nyquist frequency is the maximum frequency in a signal that can be well recorded given a certain sampling rate. Nyquist rate is the sampling rate needed to record signal well given a certain maximum frequency in a signal. given sampling rate = ...

6

In the most general case, if you want to sample a continuous-time signal without loss of information, the minimum sampling rate is independent of any choice basis functions. The faster the signal changes with respect to the independent variable (which doesn't need to be time), the faster you have to sample. And if the signal is not band-limited or if it can'...

6

First to clear up the OP's misunderstanding: the Nyquist Stability Criteria involves clockwise encirclements of -1, not the origin, and this would be the polar plot for the open-loop gain specifically. I've included some details below for those that are more interested. First a review of the basic equation relating Open Loop gain and Closed Loop Gain for a ...

6

Approaching The Sampling Theorem as Inner Product Space Preface There are many ways to derive the Nyquist Shannon Sampling Theorem with the constraint on the sampling frequency being 2 times the Nyquist Frequency. The classic derivation uses the summation of sampled series with Poisson Summation Formula. Let's introduce different approach which is more ...

6

There is no aliasing as 𝑓 = 30 Hz is less than or equal to the folding frequency, 30 Hz and 40 Hz, respectively. Yes and no. There isn't significant aliasing when you're sampling at 80Hz, because the resulting signal has frequency components at 30Hz and 50Hz. The result is thus unambiguous as long as you take that 50Hz signal into account. There is ...

5

I think you are mixing two things that are actually not related. "Beating" happens if you add two sine waves that are close in frequency. What you describe is sampling sine wave close to the Nyquist Frequeny. If you plot the samples, it looks like there is beating going on, but that's not actually the case. All information is properly preserved and if you ...

5

The concept of reconstruction has nothing to do with the application, rather it has to do with the question: did I get the same signal that is really there. If you cannot recreate the signal back, that means the conversion process is loosing/modifying underlying information, which in most cases you do not want to happen. So the confidence on the Fourier ...

5

You may not need to explicitly reconstruct. But if you did reconstruct a waveform using the samples that you have, and end up with something different from the actual input, your controller is controlling as if that new different reconstructed waveform was really the input. Depending on what your controller is doing, you may have wanted it to do something ...

5

In addition to @hotpaw2 explanation, a graphic. There are two analog square waves (red and green), with different lengths. They are depicted with a fine sampling, denoted by crosses. Their actual sampling is denoted by circles. The red one is shorter than the green one, as can be seen in the interval $]0.7\;0.8[$. Yet, the sample points are the same. Thus, ...

5

Observations I have used +1 and -1 in the sequence instead of your 1 and 0. With $\alpha=1$, the band-limited continuous function $f_m(T)$ in your first two figures (with the above mentioned modification) is: $$f_m(T) = \sum_{k=1-m}^m \operatorname{sign}\left(\operatorname{sinc}(\pi k - \pi/2)\right)\operatorname{sinc}(\pi T-\pi k),\tag{1}$$ where: \...

5

My answer is related to this question Why root raised cosine filter can eliminate intersymbol interference (ISI) ? A classical system with Nyquist pulse $p(t)$ is: The equivalent baseband signal has bandwidth limited in $[-1/2T,1/2T]$ and a sampling rate $F_s > 1/T$ is enough to avoid aliasing. We normally sample at rate $t=kT$. I wonder what would ...

5

Yes your understanding is basically correct. The 1st paragraph (2 lines) expresses the fundamental relation between the DFS and the DFT of a finite-length sequence $x[n]$ while the 2nd paragraph tries to put down the relation between the DFT $X[k]$ of a sequence and the DTFT $X(e^{j\omega})$ of it (assuming it exists). However this 2nd paragraph shall better ...

5

Is the rate of 2B exclusive? Yes. The sampling theorem states that the signal must be band limited to half the sample rate. That implies that the energy at the Nyquist frequency must be zero. In practice you need a healthy margin between the highest usable frequency and the Nyquist frequency. There is always some "transition band" that you need to get the ...

Only top voted, non community-wiki answers of a minimum length are eligible