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9

Say if I transmit: $\sin(2\pi x)$ And separately: $\sin(2\pi x\times 2)$ Does it end up as a single wave of: $\sin(2\pi x)+\sin(2\pi x\times 2)$? Yes, that's exactly how it works. This is called superposition. There are electromagnetic waves at hundreds of different frequencies all filling the air simultaneously. The way something like a ...


9

This is a good question with a lot of deep math and physics behind it (information theory). I will try to give you a casual answer. Signal to noise ratio: First, you should ask yourself what a "signal" is. For example, when you listen to the radio, especially AM radio, you hear the sounds / music / voices just fine even though there is static / noise in ...


8

This has been extensively studied in linguistics and acoustics. Humans and other primates predict speaker gender through a combination of fundamental frequency $F_0$ ("pitch") and Vocal-Tract-Length estimates ($VTL$) which are a proxy for body size. Sometimes "formant dispersion" is used for $VTL$. It is usually defined as ...


8

The sound that reaches your ear is just air pressure fluctuating over time. You can use a transducer of some sort to convert the value of air pressure to some other form - for example: to the depth of a groove being cut into a helical track on a layer of wax on a rotating drum to the depth of a groove being cut into a spiral track on a circular disc of ...


8

No. Consider any state with a momentum wavefunction symmetric about zero. It's position-space and momentum-space norm-squared probability distributions are not changed by time-reversal, even though the wavefunction clearly is. Here is an explicit example. Take the four Gaussian wavepacket of mean positions $x_0$ or $-x_0$, mean momenta $p_0$ or $-p_0$, ...


7

We know exactly where the spacecraft is, and it knows pretty well where we are. Distance does not aggravate the accuracy of aim problem, indeed the further apart the less relative motion, so aim gets easier. The problem is signal attenuation by dispersal. i.e. at twice the distance, the signal will be a quarter of the strength. The solution, for Voyager, ...


5

Treating the signals as time series: If the first signal $S_1$ has a noise component $N_1$ added to it, then the noisy signal is $S_1+N_1$, similarly the second signal is $S_2+N_2$, so the difference signal would be $(S_1+N_1)-(S_2+N_2)$ and its signal to noise ratio would be $\langle(S_1-S_2)^2\rangle\over\langle(N_1-N_2)^2\rangle$ If the signals are ...


5

As always, a communication via electromagnetic radiation depends on both ends. Uplink from earth can be done with a lot of power and big dishes, of course. Downlink is limited to the power of the nuclear battery on board but has a rather impressive 2.7 meters dish!. On top of that they use a rather slow bitrate, I think with a lot of redundancy. All this ...


5

The problem you describe is (mathematically) similar to blind deconvolution. Given a signal which is the result of blurring an image (a linear operation) and adding noise, blind deconvolution tries to estimate the blur and the image. As described here, the blind deconvolution process consists roughly of: Guess the blurring function (transfer function) ...


5

Unless someone is signing a sustained note, human voice sounds aren't going to be regularly repeating. That means you can't really declare something as the fundamental frequency with everything else being a series of harmonics. Instead, it makes more sense to think of voice in the context of the continuous spectrum. If you do that you will see most of the ...


5

In signal processing, the Nyquist–Shannon sampling theorem says you need at least 2 samples of a frequency to be able to perfectly reconstruct it. So in your question, a sampling rate of $200\: \mathrm{MHz}$ means you can perfectly reconstruct frequencies in the range of $0 - 100\: \mathrm{MHz}$. So what happens when frequencies above $100\: \mathrm{MHz}$ ...


4

Did you see the experimental setup? I believe that a leak in a gas pipeline generally makes a whistling sound whereas leaking liquid will be very quiet. Without sound, the autocorrelation approach doesn't make sense. To find the leak in a pipeline for liquids, one could use a marker to find the leak (e.g. add some color). Or one could seal segments of the ...


4

Yes, it is simple to prove using moment generating functions. And yes, the mathematics is very closely related to that of quantum field theory. You compute $G(j) = <exp(\sum j_i x_i)>$ where each $j_i$ is a "source" for the corresponding $x_i$. This is easily shown to be something like $G(j) = exp(\sum j_i \mu_{ij}^{-1} j_j)$ To get expectation ...


4

It doesn't look that much like a normal distribution to me - particularly on the x axis, the right-hand tail looks heavier than the left, whereas the left one is much longer. But, generally speaking, normal distributions tend to arise when lots of small, independently distributed random numbers (of any distribution) are added together. (The theorem that ...


4

An harmonic oscillator. When evolving with time, its joint distribution in (p,x) is given by the Boltzman distribution: $e^{-H(p,x)}$, but the energy along a trajectory is constant. Nevertheless if write explicitly the hamiltonian you will find that $e^{-H} = e^{-p^2/2 - x^2/2}$ and although the energy is constant the individual distributions of $x$ and ...


4

Human voices tend to average around middle C - male voices average an octave below this and female voices an octave above. Middle C is 261.6Hz. If you have an amplitude-time graph the way to measure the frequencies contained in it is to Fourier transform it. This gives you a plot of amplitude against frequency. If you take some reasonable clear signal, like ...


4

Comments to the question (v1): I) Reconstruction of phases from modulus$^1$ $|f(x)|$ of a signal $f(x)$ and modulus $|\tilde{f}(k)|$ of its Fourier transformed (FT) signal $$\tag{1} \tilde{f}(k) ~:=~ \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \!dx~ e^{-ikx} f(x)$$ is an interesting and likely a well-studied engineering problem, either for continuous or ...


4

Internet propagates with radio waves. Radio waves take advantage of a wave guide generated by the charged ionosphere and the ground for long distance propagation. Storm fronts with lightning and charged clouds do interfere with the propagation of a signal. Sudden changes in the atmosphere's vertical moisture content and temperature profiles can on ...


3

There are several ways to understand this result. One is to think how $I(x,y)$ is defined. It's the signal your instrument gives you at point $y$ when your source is a delta function (that is, a point source) at point x. In mathspeak, $O_\delta(y) = \int \delta(x) I(x,y) \mathrm{d}x$. If your source is composed of many different points, you need to sum over ...


3

You can make an arbitrary sound (or any waveform) by adding together a bunch of pure tones at different frequencies. So a sound, unless it happens to be a pure tone, does not contain a single frequency component, rather a range of frequencies. The mathematics behind this is called Fourier analysis and you can see many examples on Wikipedia or by searching ...


3

It a widely known and experimentally useful fact in nuclear and particle physics that the position and momentum distributions of bound systems are related to one another by a Fourier transform. Is the system you are inspecting bound? The tails in the data that Nathaniel notes suggest that it is not fully bound, which means the Fourier relationship between ...


3

It's used a lot in cosmology. Often, to a decent approximation, the quantities we try to measure in cosmology (e.g., CMB temperature and polarization maps, galaxy distributions) are realizations of Gaussian random processes to a decent approximation, but have (or are predicted to have) interesting non-Gaussian features at some low level. People estimate the ...


3

It can be done analytically, but numerical results depend on what conventions you use to define the Fourier transform. You have $\eta(t)$ a random variate. I assume that you mean short range correlated noise, so that $\langle \eta(t) \eta(t') \rangle = \sigma^2 \delta(t-t')$ ($\langle . \rangle$ indicates an ensemble average). The spectral amplitude is a ...


3

I think a simpler explanation than some kind of sound filtering as a result of the speed of rotation is that when you hit the coin with more force (resulting in a faster speed of rotation), it increases the amplitude of the coin's vibrations, enabling higher modes of vibration, and thus changing the spectrum. I'm skeptical about your claim that the spectrum ...


3

There are three ways I can imagine time resolution being limited: integration time, dispersion, and intrinsic width. Integration Time One of these ways you already eliminated: Dim sources will need longer integration times to overcome Poisson statistics/read noise/etc. in order to trigger any detection at all. Dispersion Theory The second way is the ...


3

I'm assuming that your filter is linear and time invariant, then we know that in the frequency domain $Y(\omega)=G(\omega)X(\omega)$ and thus conceptually: $G(\omega)=Y(\omega)/X(\omega)$. The (first) hard part is that we know that $X$ is band-limited so you'll only be able to evaluate $G(\omega)$ for those frequencies that lie within the frequency ...


3

Yes and no. Typically your input signal bandwidth needs to be larger than your expected bandwidth that you are trying to measure otherwise you won't find the edges. A complex input waveform can make it difficult to characterize the system properly. Deconvolution can be very error prone and time consuming to try to invent the filter response that gives you ...


3

You are correct that it's impossible to change the frequency of every component of a waveform while (a) preserving all the phase relationships between the frequencies and (b) preserving the length of the waveform. However, while it's mathematically impossible to do this precisely, it is possible to do things that sound quite a lot like it to the human ear. ...


2

Just to check that we're on the same page: Multiplication of a constant-amplitude complex signal $s=\exp{\left(i f(t)\right)}$ by another complex function with constant amplitude won't give a time-varying amplitude: $\exp{\left(i f(t)\right)} \cdot \exp{\left(i g(t)\right)} = \exp{\left(i h(t)\right)}$ Since $f$ and $g$ are real, $h$ is real, and ...


2

Losses in coaxial cable are resistive. For low frequencies, one uses the full thickness of the coaxial cable and resistance is low. As frequencies increase, the signal is unable to penetrate as deeply into the conductor. This is called the skin effect. So as frequencies increase, the amount of metal that is used to carry the signal decreases. The result is ...



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