49

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 ...


45

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+...


33

Preamble The lesson here is that graphical intuition isn't always the best choice. For example, you can say that the intuition for derivatives is that they're slopes, and for integrals is that they're areas. But why would "slopes" be that useful in physics? I mean, besides inclined planes, you don't see that many literal slopes in a physics class. And why "...


28

An capacitor has one intuitive property: Its voltage can't change instantly since its voltage is dependent on the charge it has stored, and charge doesn't move at infinite speeds (there is always resistance somewhere), therefore you can't instantly charge up a capacitor without infinite current. More capacitance means less voltage for the same amount of ...


13

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 ...


13

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 ...


13

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 $...


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)-\...


11

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 $-p_0$, ...


11

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 ...


10

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 ...


10

Whenever I think of a convolution I imagine a moving average: Suppose we have a function $f(x)$ and we like to calculate the moving average using the weight function $w(x)$. What we calculate is $$ \bar{f}(x_0) = \int w(x) f(x_0-x) dx $$ which is a convolution. A physical example can be found e.g. in optics: Suppose we have a structured surface, which is ...


10

What is the physical behaviour which allows a capacitor to act as a high or low pass filter? A capacitor alone cannot act as either. To create a filter you need a combination of resistance and capacitance or inductance and capacitance (or RL). You need two immittances, at least one of which is reactive. Let's take a practical example, an RC circuit. This ...


9

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 ...


9

Imagine electricity as water in a pipe. The current can flow in either direction (direct current, DC) or one way then the other way (alternating current, AC). Now put a rubber membrane in the pipe. This is the capacitor. Now it will slow and then stop DC, but AC can still keep wobbling back and forth. In this way, capacitors block DC but enable AC. ...


8

If you want to use this as a link between standard computers or microprocessors you will have to design and build a encoder to turn voltage levels into signals on the rope at the transmitting end and a decoder to turn rope signals into voltage levels at the receiving end. For two-way communication you will need two encoders and two decoders. The encoders and ...


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 ...


7

Sometimes for physical intuition, it's nice to think about the extreme cases. For instance, a zero frequency signal is just a DC voltage. If we send it through the RC high pass filter, the capacitor is just like a break in the circuit, and prevents any current from flowing. Slightly more quantitatively, the capacitor equation $Q = CV$ implies that if we ...


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

Periodicity Some excellent answers on the $\sin x$ and $\cos x$ functions and how they're solutions to the relevant differential equations were already given, but an important point can still be mentioned: Sine and cosine are used because they are periodic and signals/waves are usually considered to be or are approximated by periodic functions. In many ...


6

The Fourier transform (and hence $S$) are defined on negative frequencies, while spectrometers only output values at positive frequencies. For any real-valued signal, $f(\omega) = f(-\omega)^*$, which implies that the negative frequencies don't carry any information not already in the positive frequencies. Therefore, whether you say the domain is all ...


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

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 ...


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 ...


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