# Tag Info

11

When your phone says that no network is available, it actually means that your network is not available. Fortunately though, mobile phones can make emergency calls on any network, and so if it finds another network then it can make emergency calls on that. If no network is available whatsoever (e.g. out in the desert) then you cannot make emergency calls. ...

10

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 $-p_0$, ...

10

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 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 ...

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

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, ...

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 ...

7

From a physics perspective, the fundamental reason for this is something called the bandwidth theorem (and also the Fourier limit, bandwidth limit, and even the Heisenberg uncertainty principle). In essence, it says that the bandwidth $\Delta\omega$ of a pulse of signal and its duration $\Delta t$ are related: $$\Delta\omega\,\Delta t\gtrsim 2\pi.$$ A ...

6

The DFT is used when all you have available are samples of the function, rather than the function itself. If you are doing an FT on experimental data, it's always (as far as I know) recorded in discrete numbers: an array of floating point numbers, for example. There are a few times when the DFT has some applicability to real systems, for example simple ...

5

An intuitive dimensional reason why it couldn't work: a state vector in $\mathbb C^{N+1}$ is described by 2N real coordinates (one complex dimension is irrelevant), and so is its Fourier transform. If we only consider the normalized squared moduli of the components, we have 2N real numbers as well, so if these would actually be independent we should be able ...

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 $100\: \mathrm{MHz}$ ...

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

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

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

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

Consider a single value of $m$. The Fourier series for just that $m$ gives $$a_m \cos(2\pi m t / T) + b_m \sin(2\pi m t / T) \, .$$ This can be rewritten as $$M_m \cos (2\pi m t / T + \phi_m)$$ where $$M_m = \sqrt{a_m^2 + b_m^2} \qquad \text{and} \qquad \phi_m = \tan^{-1}(-b_m/a_m) \, .$$ So, you can see that $a_m$ and $b_m$ are just the cartesian coordinate ...

4

You do know that your phone will transmit only enough power to reach the nearest tower right? Most of the time that is much less than the max it is capable of. But when there are buildings between you and the tower or you are on the open road you'll be happy not to be dropping so many calls... So there are two things to notice here. First - every 3 dB ...

4

As far as I understand, aliasing comes from the fact, that you use a bad sampling rate Aliasing can also come from a 'bad' anti-aliasing filter. So why is it you just don't use a fast sampling rate all the time For the same reason that we don't use a sledge-hammer to crack a nut. The problem isn't so much that the signal of interest is aliased, ...

4

You are not required to. Functions can be decomposed into a wide array of orthogonal basis functions, including the Bessel functions (in the Hankel transform) and the Legendre functions. The sine function just happens to be the overall simplest to deal with in the general case.

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

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

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

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

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

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

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 ...

4

Matlab's silent output is correct. Physically, sound is a fluctuation of the molecules in some medium. If your waveform is perfectly constant, it corresponds to constant pressure: no fluctuations, meaning no sound. If it's very nearly constant, you will probably still be unable to hear the corresponding pressure wave without the aid of significant ...

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

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. ...

3

To be sure, it's the continuous (time) Fourier transform versus the discrete time Fourier transform (DTFT). The former is a continuous transformation of a continuous signal while the later is a continuous transformation of a discrete signal (a list of numbers). The discrete Fourier transform (DFT), on the other hand, is a discrete transformation of a ...

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