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89

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 ...


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 ...


11

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 ...


11

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 ...


8

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 ...


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-...


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

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

Note that the Nyquist criterion applies to bandlimited signals; but the so-limited band does not need to be baseband, or even contiguous, just not overlapping (aliased) with any other sub-bands after folding. Periodic sampling fan-folds an infinite spectrum in to "fans" with a width of half the sampling rate. Any frequency bands, after this fan-folding, ...


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

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

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

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. ...


5

Applying a non-linear function will always introduce harmonics, and mixing non-linear functions with sampled versions of continuous signals does add the wrinkle you note above (where high-frequency harmonics are aliased to low frequencies.) I can think of a few ways to proceed: You can use an oversampling factor high enough to capture the extra harmonics (...


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

Your understanding is correct. I don't have the full text of the paper, but it sounds like they aren't being very precise with their description. As you pointed out, downsampling doesn't improve frequency resolution in and of itself, it merely reduces the amount of bandwidth that can be represented unambiguously. However, I can take a guess at what they ...


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

This question, and a number of similar others, from DTSP book can be a little tricky to recognise the fact that it's not actually asking alias free operation (which would require 8 khz cutoff signal as you expected), rather it is actually asking how much aliasing (due to initial sampling block) is tolerable: you can allow aliasing in those regions of the DT ...


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 ...


4

A few approaches to alias-free nonlinear distortion (in increasing order of difficulty): Subband distortion: Use a low pass filter to extract the lower end of the signal. If you choose a cutoff frequency of $\frac{f_s}{2N}$ you can apply any non-linear transfer function $f$ with derivatives starting at $f^{N+1}$ vanishing to avoid aliasing. Add just the ...


4

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 ...


4

Your approach seems conceptually sound. I suspect that the problem is one of two things: 1) you have implemented it incorrectly, 2) Your filter bandwidth is too wide. You are implementing the "smart" approach in that you are doing it the computationally efficient way. I would try doing it the dumb way (insert I-1 zeroes between every sample, filtering ...


4

The need for modulation: Your voice signals generally lies in the frequency range 1kHz to 4kHz and music lies in the range 20Hz to 20Khz. Say your locality has 4-5 AM broadcast stations. Now all the AM broadcast stations cannot transmit their content as is because if all the stations transmit at the same time there will be signal interference. Also, for ...


4

The channel frequency controls the local oscillator on your SDR which is the frequency about which it covers. So you are receiving signals from 16KHz below 107.5MHz to 16KHz above. The concept you want to look up is called heterodyning. When you modulate (multiply) a signal by a sine wave you end up with two copies shifted in frequency space by the ...


4

Interpolation using a Sinc function kernel. This assumes that the sampled signal was perfectly bandlimited to below half the sampling rate. Note that perfectly bandlimited signal are infinite in extent. For finite-length "real world" signals, using a windowed Sinc interpolation kernel (thus a finite computation with a noise floor) is a common method to ...


4

Let $Y(\omega)=X(\omega)+X(\omega)e^{-j\omega}$ with $X(\omega)$satisfying $X(\omega)=0$ for $|\omega|>\omega_m$. Then, by its definition, $Y(\omega)$ also satisfies $Y(\omega)=0$ for $|\omega|>\omega_m$, i.e., the maximum frequency remains unchanged, and, consequently, also the Nyquist frequency remains unchanged. Note that if $x(t)$ is the signal ...


4

you actually have to sample at more than twice the highest frequency. if $B$ is the bandwidth or highest frequency and $f_\text{s}$ is the sampling frequency, then to satisfy the Sampling Theorem (for normal baseband sampling): $$ 2B < f_\text{s} $$ it is not $$ 2B \le f_\text{s} $$ sampling at exactly twice the frequency of a sinusoid is "critically ...


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