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 ...


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 ...


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 ...


5

OP clarified that the question in the comments as follows: If we ignore any modulation for now and assume that we are receiving pure tones plus the band limited noise and we try to improve the SNR in post processing how much improvement can we expect by oversampling and is there a limit to it? My original question was about this aspect. First consider the ...


5

So for frequency hopping spread spectrum (FHSS),when the receiver is unaware of the transmitter's hopping pattern, does that mean the ADC sample rate needs to be higher than twice the entire hopping bandwidth? Yes. If I use a 2Msps (4x data BW) ADC it looks like I will still recover my data; sure it under-samples some of MHz modulated frequency but it does ...


5

That is above the Nyquist rate, so why is the signal degraded? It's not degraded in any way form or shape. The perceived degradation is purely cosmetic but not functional. See for example: How is sampling affecting this sine wave? But the time domain plot still shows something unexpected No it doesn't. It looks exactly as it should. If that's unexpected, ...


4

In order to resample without aliasing, the resampling process needs to apply an anti aliasing filter at the new Nyquist Frequency (or thereabouts). So chances are, your resampling process applied a low pass filter at 256 Hz or so and that removed A LOT of the energy.


4

The claim is wrong. Sampling of a pure sinusodial whose frequency is below but arbitrarily close to the Nyquist frequency (half the sampling frequency) is a perfectly valid operation, as long as you can create ideal (zero width transition band) brickwall lowpass filters to be used at the reconstruction interpolation of the continuous waveform from its ...


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 ...


4

My input signal is a dc signal (sensor output) and im getting kind of a headache to understand why oversampling can increase the resolution of dc signals. The ADC puts out integers. So, let $x$ be the integer that would come out of the ADC (100.3, say, or -333.3). Now let $y = \lfloor r \rfloor$ be the quantization operation, where you get straight ...


4

There really is no practical difference between the two. The important thing to understand about aliasing is not the exact definition of the word, but rather the concept: when two frequency bands alias and there is a signal in one of these bands, after sampling you can't tell which band the signal came from. In a sense, by the time you're asking what is or ...


4

Apart from practical errors related to the truncation of the infinitely-long energy signal when discretized to digital, can we find a UNIQUE mapping between the two signal forms? Yes. Say you're starting with $x(t)$, which is perfectly bandlimited to less than $f_B = \frac{1}{2 T_s}$, and you take samples at every $t = T_s k$: $x_k = x(T_s k)$. All you need ...


4

You are unlikely to be able to evaluate correctly an ideal sinc interpolation filter in real-time (or even with a small time lag), since the future samples of your data are unknown. Plus, when you received your last sample $x[n]$, you generally don't know whether you are on a peak. So, it somehow boils down to how real-time you want to be; how fast you ...


4

Unlike the Gardner Loop, the M&M synchronizer should be performed after the RRC filter in the receiver for best performance. With cases of high RRC alpha, the M&M won't work as expected without the complete Raised-Cosine filtering (RRC in transmitter followed by RRC in receiver) as the slope of the error term will reverse, with high self-noise, as I ...


3

The following graphic should clear up why a useful answer is 5 seconds, consistent with conversions between the discrete-time and continuous-time domains. A good example of this are discrete time approximations to continuous time integration or differentiation where the inclusion of $\Delta T$ (sample period) is made. With all such approximations, the time ...


3

I believe that $|\Delta_l|$ in that equation is a typo and should actually be $|\Lambda_l|$, because from the first equation in $(2)$ we have $$|\Lambda_l|=\frac{\psi_H-\psi_l}{k}$$


3

Given $ \left\{ x \left[ n \right] \right\}_{n \in M} $ where $ M $ is the set of indices given for the samples of $ x \left[ n \right] $. The trivial solution (Which it would be great to have a faster more efficient solution is what I'm looking for) would be: $$ \arg \min_{y} \frac{1}{2} \left\| \hat{F}^{T} y - x \right\|_{2}^{2} $$ Where $ \hat{F} $ is ...


3

It is easier to prove it by using atoms. The 1st atom is the Absolute Value function $ \left| \cdot \right| $ which is convex. Then you have linear operation by the subtraction which is convex (Also concave). Then you linear combination which is also Convex. Hence the function is Convex.


3

I agree with your result for $Z(j\omega)$. Apart from scaling, what you have is the original spectrum with positive and negative frequencies swapped and multiplied with factors $1-j$ and $1+j$, respectively. In order to restore the original signal, we need to add right-shifted and left-shifted versions of the spectrum, while getting rid of the complex ...


3

Is this an entirely stupid idea? No, but you've just came to the conclusion that instead of sampling complex, with Nyquist rate being the bandwidth, you should do twice as many samples. That simply means you're not doing IQ sampling, but low-IF or direct-RF sampling. Mix the signal with a higher intermediate frequency and filter it in such a way that ...


3

Does summing two audio samples affect sampling rate? No. Sorry this is one tortured piece of code and it's difficult to read. A few things to check Are you using "ping pong" buffers for your DMA? You should have two set of buffers that alternate on each frame: one for receiving/sending data and one for working on it. Make sure you I2S ...


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. $\omega(t_N)= \pi$ $$t_N = \frac{0.5-a}{2b} = 300$$


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

We have a coherent and consistent theory of sampling that proves that band-limited signals can be sampled without loss of information. We have a coherent and consistent theory of quantization that proves that quantization always creates an error The question of "why" is meaningless. Things are what they are. In any real world application you need ...


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).


2

Can we sample the Dirac function? Strictly speaking: "sampling" would be taking the instantaneous value. Since the Dirac Delta doesn't have a value at $t=0$ (it is not really a function!), NO. Realistically speaking: an ADC can't measure instantaneous values. That's impossible, because it would require infinite bandwidth of the conversion system (...


2

There are many advantages, but the most obvious to me Advantage 1 : Oversampling followed by decimation allows you use to simpler and smaller anti-aliasing filters. These filters cost less, take up less space on a PCboard, draw less power, etc. Advantage 2 : In multi-channel applications, the tolerance and variation of the analog components of your anti-...


2

And additional consideration not mentioned that comes up in radio design is in the decision to use quadrature sampling of a baseband signal (as in "Zero-IF receivers") over a "Digital-IF" receiver that is achievable when the signal can be sampled at a much higher rate as a real signal. The Digital IF signal avoids the quadrature imbalance ...


2

With downsampling you have complete control over the process and it comes down to what compromise of processing complexity, delay, aliasing and loss of passband you can accept. With a lower rate A/D you are pretty much at the mercy of someone elses spectral trade-offs and in addition you get the quantization/noise of one analog pass. -k


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