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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 ...


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 ...


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 ...


4

In contrast to common misunderstanding, aliasing does not affect the process of sampling, rather it affects the process of reconstruction. i.e., the samples themselves do not contain error, but their interpretation at the reconstruction (or anything related with it) will be in error. So, if you sample a sine wave $$x(t) = A \sin( 2 \pi f_0 t + \theta) $$ at ...


3

Perfect recovery is one thing, niceness is another. Sampling above x2 Nyquist is sufficient for perfect recovery, after which we can FFT-upsample to make it look nice - which is more efficient than directly sampling at a higher rate (though it has its advantages).


3

Your approach confuses frequency with phase; the correct formulation is $$ \sin(2\pi \phi(t)) $$ where $\phi(t) = \int \omega(t)dt$. Related post. I derived the most general form for a linear chirp here; Python code: def lchirp(N, fmin=0, fmax=None, tmin=0, tmax=1): fmax = fmax if fmax is not None else N/2 t = np.linspace(0, 1, N) a = (fmin - ...


3

The phase of your signal is $$\phi(t) = 2\pi c\dot (a + b\cdot t) \cdot t = 2\pi \cdot (a\cdot t + b\cdot t^2) $$ The frequency is the derivative of the phase with respect to time NOT phase divided by time. So we get $$\omega(t) = \frac{d \phi}{dt} = 2\pi \cdot (a + 2b\cdot t) $$ Solving for Nyquist, i.e. $\omega(t_N)= \pi$ $$t_N = \frac{0.5-a}{2b} = 300$$


2

The auditory system encodes sound in frequency domain, i.e. the activation levels of auditory nerve fibers represent the amplitude or energy in a frequency band assigned to that particular fiber. The ear itself does the transformation from time to frequency domain. If you somehow modified the ear itself to be sensitive to higher frequencies, the output axons ...


2

Spectrogram given by MATLAB is right. The frequency axis shows frequency from 0 to 110mHz which is equivalent to 0 to 0.110Hz. mHz stands for milli-Hertz. There is no problem with that plot. Now the question is why we are seeing high energy components at all frequencies in the range 0 to 110mHz. That I think is because of the Aliasing effect. The sampling ...


2

A few additional things to note: the original version of the Shannon theorem was later corrected by replacing "greater than B hertz" with "greater than or equal to B hertz", and, the theorem only applies to signals of infinite duration (which a laser's signal isn't), albeit various algorithms have been developed to control levels of ...


2

So I took a look at the slides and it seems to be giving a cursory overview of how to measure Doppler with multiple pulses, so it's very generic. Hopefully this is a pretty straight forward thing to discuss given that your questions are suspicious of their statements, that's good! First Question: why is the Nyquist theorem applied to the doppler frequency $...


2

B): $f_s \geq 2B = 2\cdot(2015-1612) = 2\cdot 403 = 806$. "Second-order" bandpass sampling is described in this paper (or older, here). It's sampling $x(t)$ at a lower sampling rate, $M$ times, each with a different offset - then keeping only (carefully selected) 2 of the $M$ sequences: $$ \begin{align} x_A(n) &= x(n / f_0) \\ x_B(n) &= x(n ...


2

You don't get a nice signal because you don't complete the D/A recovery process. After going through the DAC, you need to pass the signal from a lowpass filter which eliminates the extra frequency components. Ideally the lowpass filter should be flat in a bandwidth equal to sampling frequency and zero outside. Since there is no such filter we have to ...


2

It's the spectrum of a discrete signal: sampling in time $\Leftrightarrow$ periodizing in frequency - explained in detail here. Overlap means there's aliasing, and we require a higher sampling rate. FFT returns one period of this spectrum since it contains all the information. (More precisely, your image shows DTFT, and DFT (which FFT implements) is a ...


1

What's the limit? I think it's primarily "how long are you willing to wait". As far as I know there is no theoretical limit on how narrow you can make a bandpass, but you'll run into the limit of the time/frequency "uncertainty" principle: the more something is defined in frequency, the less it is defined in time. Roughly speaking: a ...


1

0.11Hz is 110mHz Vertical lines are due to the zoom; they're the bottom row ... zoomed: This is a purely visual effect, no data is added. To make the data zoom between 0 and 2mHz, increase nfft, then zoom the spectrogram. Why Matlab Spectrogram of slow and rarely sampled signal shows high frequencies The highest frequency it's showing is 110mHz, which is ...


1

Sinc interpolation can exactly reconstruct an above-Nyquist-sampled strictly bandlimited signal from noiseless samples. See the Whittaker-Kotelnikov-Shannon reconstruction or resampling theorem: https://en.wikipedia.org/wiki/Whittaker–Shannon_interpolation_formula and https://ccrma.stanford.edu/~jos/resample/Theory_Ideal_Bandlimited_Interpolation.html For ...


1

To implement a digital-to-analog converter, all you need is an ideal low-pass filter to filter out the periodic frequency response that $\varOmega \geq \varOmega_s/2$. $$ H(j\varOmega) =\left\{ \begin{aligned} T,\ \ \ \ |\varOmega|<\varOmega_s/2 \\ 0,\ \ \ \ |\varOmega|\geq\varOmega_s/2 \end{aligned} \right. $$ The impulse response of an ideal low-pass ...


1

Actually, that formula is specifically for an unmodulated sine wave at full scale just prior to clipping. This SNR result provides a reference point as to what the total quantization noise power will relative to that full scale sine wave at the input to the ADC, and that noise power will be at that similar total power level relative to that full scale ...


1

Is it possible to perform a dft looking after the Nyquist frequency. Of course. Every DFT will do this. Here is why: Applying a DFT requires the signal to be time discrete, which means that it is periodic in the frequency domain with the sample rate being the period. Let's say you are sample rate is 40 kHz. Then the value of the DFT at 1kHz will be the same ...


1

At Nyquist the signal goes $[1, -1, 1, -1,...]$ - it's the fastest possible discrete variation for any input length. Zero padding won't help: it'll only lower the lowest possible (non-zero) frequency. Going beyond Nyquist thus necessarily implies increasing the physical sampling rate, or "rate of information", such that the same discrete variation $...


1

Is there any general thumb rule as to what should be the optimal sampling rate (4x,5x...10x)? This depends a lot on your application, your specific requirements and the typical spectrum of the signals. 9kHz sounds like audio, where something like 2.5x-3x would be a good starting point. A non trivial aspect of sample rate selection is "what can the HW ...


1

According to the lowpass sampling theorem, the required minimum sampling-rate (a.k.a Nyquist rate) is twice the highest frequency present in the frequeqncy spectrum of the bandlimited signal. Note however that, the lowpass sampling theorem assumes there is no impulse (no pure sinusoidal component) at the highest frequency (a.k.a Nyquist frequency) of the ...


1

Following your edit, the first challenge in creating a controller is the task to create a stabilizing controller. After that, performance can be tuned. In order to find if the created controller stabilizes the closed-loop, the nyquist plot can be used. The nyquist stability criterion states that the amount of encirclements of the point -1 (hereby are counter-...


1

I'm a bit puzzled with your Bode plot, here's what I get with your transfer function and Matlab 2019b. My phase starts at almost -180 degrees while yours start at 0 degree. Is it possible that you've made a mistake somewhere in your analysis? Your transfer function is a bit complex, you could simplify it by removing the fast poles and zeroes. For example, ...


1

Suggest you take a look at a similar question I posted Sample rates, Samples per Symbol, and Digital Pulse Shaping In general, the Positive BW for an RRC filter is $$BW_{pos} = (1+a)\frac{R_b}{2\log_2(M)} = (1+a)\frac{R_s}{2}$$ where $R_b$ is the bit rate, $a$ is the excess bw, $R_s$ is the symbol rate, and $M$ is the constellation size. Since you're dealing ...


1

Nyquist says that you can send up to twice as many pulses (symbols) per second as the channel bandwidth $B$ with zero ISI, so you need $R_s \leq 2B$. That is all there is to it. The sinc pulse has zero excess bandwidth so the bandwidth of the signal is equal to the symbol rate, $B_s = R_s$. About sampling, Nyquist says that you need to sample at least twice ...


1

The wavetable is ignorant or agnostic about the playback rate and the sample rate in Hz. What the wavetable knows about is how many points or samples define the waveform, $N$. All it knows is the amplitude and phase of every harmonic (which are two numbers for each harmonic), and with a little bit of nuance regarding the DC component and, if the number of ...


1

"You simply can't..." - it isn't true. Let me explain by pictures... Base band signal with bandwith B: S(f) sampled with Fp > 2*B: S(f) sampled with Fp < 2*B: Analytic representation of S(f) (complex signal, has no band mirror on negative frequencies): Analytic signal sampled with B < Fp < 2*B: As you can see information isn't lost....


1

You simply can't. The analytic signal contains the exact same amount of information as the original real-valued signal, and if either is undersampled, you get the loss of information that undersampling gets you. That's it. Unless you've got any secondary information about the signal that you forgot to mention, there's nothing you can do.


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