44

This problem can be solved with noise-shaping. Since the shape of the spectrum is known, it can be used as a base for the power spectral density: $$ P(f,T)=\frac{ 2 h f^3}{c^2} \frac{1}{e^\frac{h f}{k_\mathrm{B}T} - 1} $$ where $k_\mathrm{B}$ is the Boltzmann constant, $h$ is the Planck constant, and $c$ is the speed of light. This outputs the relative ...


42

While Sine and Cosine functions were originally defined based on right angle triangles, looking at that point of view in the current scenario isn't really the best thing. You might have been taught to recognize the Sine function as "opposite by hypotenuse", but now it's time to have a slightly different point of view. Consider the unit circle $x^2+y^2=1$ on ...


14

Fourier transform is a linear operation. This means that the infinite sinusoidal signal can be written as the sum of the sinus in the window plus the sinus outside the window. If $f(t)$ is your window function this means $$ \underbrace{\sin(t)}_{g_0(t)} = \underbrace{\sin(t) f(t)}_{g_+(t)} + \underbrace{(1-f(t)) \sin(t)}_{g_-(t)} $$ or in the fourier domain $...


13

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, ...


12

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 ...


12

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 ...


12

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. ...


12

OK, so you start off with a monochromatic sinusoidal function at frequency $\omega_0$ and period $T=2\pi/\omega_0$, $$f(t)=A\sin(\omega_0t)$$ whose Fourier transform is a pair of delta functions: $$ \tilde f(\omega) =\mathcal F[f](\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t)e^{i\omega t}\mathrm dt = \frac{A}{2i}\left(\delta(\omega+\omega_0)-\...


10

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 ...


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

The fundamental reason for this is that a truly monochromatic waveform $$ f(t) = f_0e^{-i\omega t} $$ is active for all real times $t$ ─ it doesn't start, and it doesn't stop ─, and this means that you need an infinite time to produce one, and you need an infinite time to detect it. Since the energy density of the wave is constant, the infinite duration also ...


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 $$\frac{\sum_{i=1}^n(F_{i+1}-F_i)}{...


9

The trigonometric functions form a basis for the space of "reasonable signals". (For the purposes of this answer, "reasonable signals" are continuous functions having finite energy and bounded power.) The word "basis" here is meant exactly the way it is used in linear algebra. (This is explicitly discussed on the linked page.) Why would anyone use this ...


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

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 ...


7

Note: Emilio Pisanty wrote an answer that is probably a better fit for the question and site, but I'm leaving this answer around because I feel it can contribute to an understanding of how this works in practice. For one thing, you'd need to be able to differentiate between the signals inside the frequency band. As an example, I'm going to use a Morse code ...


7

This follows from classical Fourier analysis. The frequency spread and time duration of a pulse are related by $$ \Delta \omega \Delta t \approx 2 \pi $$ so to make a truly monochromatic pulse where $\Delta \omega$ is basically $0$ implies this pulse is infinite in duration. Thus, any pulse with a finite duration cannot be truly monochromatic.


7

Mathematics has progressed from geometry to calculus and differential equations. It is established that differential equations whose solutions describe waves have sinusoidal functions in those solutions. This should not be surprising, as waves are periodic in time or space, and sines and cosines are periodic functions. As a consequence differential equations ...


6

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 ...


6

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 ...


6

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 ...


6

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 ...


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

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 ...


5

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 ...


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

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, it is ...


5

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 ...


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 ...


5

If you are cooling your object that you wish to hear, then the exact sound will depend on the exact temperature (as given by yuki96's answer at 17nK). However, any temperature above the nanoKelvin temperature scale will sound the same, but the volume will increase with temperature (according to the Stefan-Boltzmann law). The sound of a warm blackbody (such ...


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