# Tag Info

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

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

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 ...

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 ...

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

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) ...

4

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, ...

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 ...

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

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 ...

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 ...

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 ...

2

It's possible to split this combined signal into the original components again. You can do that because the sine and cosine functions form a base of a Hilbert space, a space called $L^2(\mathbb{R})$. Now what does this mean? The word "space" is perhaps confusing, a mathematical space is essentially just a set of mathematical objects, in this case the space ...

2

The Fourier transform of a periodic function is a delta function at every integer position with coefficient equal to the corresponding Fourier series value. You can show this by multiplying the function by a very wide Gaussian and taking the limit. The mathematical theory is made rigorous in the subject of tempered distributions.

2

If x(t) is a random process it is quite unlikely that the derivative xdot(t) exists. So your description looks somewhat problematic. It seems that you have a Wiener process (= random walk, Brownian motion). See http://en.wikipedia.org/wiki/Wiener_process Here the changes in x are Gaussian and uncorrelated with x itself. Then x itself also follows a ...

2

My impression would be that the lower frequencies are more apparent in the male spectrum than the female spectrum. If you want to build a nice test, my approach would be to determine some average male and average female spectrum. Then you can see which of your average or most common spectrum correlates best the test person. However, you should take are ...

2

We have the following relationship $$\Gamma(\tau)=\int_0^\infty \bar{S}(\nu)e^{-2\pi i\tau\nu}d\nu$$ where $\Gamma(\tau)$ is the temporal coherence function, which can be measured with a Michelson interferometer, and $\bar{S}(\nu)$ is the real normalized power spectral density function (PSD). As can be readily proven, the above is a Fourier transform ...

2

Your question is specifically about how the concept of frequency can be applied to aperiodic signals. Simplest example is a finite width rectangular pulse - the signal is zero outside the pulse. This is certainly aperiodic. Now for periodic signals f(t), you can, for each frequency which is a multiple of the period, compute the n'th Fourier coefficient as ...

2

Yes, Fourier transforming the data is certainly the best way of finding what the frequency of the periodic signal is. I'm not sure what the units of the y axis are on your graph above - it looks like the might be in dB. Change the scale to linear (not logarithmic) and you should see much, much clearer indications of what the frequency is (if it exists). ...

1

From your description of the experiment (please correct me if my assumptions are wrong), it sounds like your apparatus consists of the application of a controlled stress to the sample (and the sensor), and the resulting strain in the sensor is measured. Whenever the stress applied by your apparatus changes, it will take some time for the system to settle to ...

1

For ideal device the observed signal is identical to the true signal. So if the true signal is a single point $$T_\text{point}(x) = T_0 \; \delta(x-x_0)$$ then the observed signal is also single point $$O_\text{ideal}(y)\bigl[T_\text{point}(x)\bigr] = \alpha T_0 \; \delta(y-x_0),$$ where $\alpha$ is the sensitivity. Non-ideal device blurs the point ...

1

For the Fourier transform of a function to exist, its absolute value must be integrable, $\int\limits_{-\infty}^\infty |f(x)|\mathrm{d}x<\infty$. The absolute value of a periodic function is not integrable on an infinite domain, so no Fourier transform. [To enjoy the full power of Fourier analysis, the function should be square integrable, ...

1

I needed to ensure that the pattern itself was long enough to get the correct $\Delta f$ in the frequency domain, since $\Delta f = \frac{1}{N}$ where $N$ is the number of points in the pattern signal. If this $\Delta f$ does not match that of the transformed data, then there is no real way to combine the two.

1

Without being able to see the data, I can only suggest two things that worked for me in a project about turbulence in water. The first is, if turbulence is a significant factor, it will show up in the auto-correlation function as violent oscillations getting worse and worse in the graph, basically exceeding the accuracy of the software to calculate it after ...

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