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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 ...


35

This ratchet-like Maxwell's demon has the same problem as all of the other ones: the door/coil mechanism itself will heat up, and become useless. Before thinking about this one, think about the simpler scenario where there's just a door, that opens if a fast particle hits it hard enough. Since particles have energy on the order of $kT$, the door must ...


16

The spectral density, or spectral function, describes the coupling between a small quantum system that is coupled to a larger environment. In many cases, this environment can be modelled effectively as a system of free bosonic or fermionic modes, with Hamiltonian (working in units with $\hbar = 1$) $$ H_B = \sum_k \omega_k b_k^{\dagger}b_k. $$ The mode ...


16

Here is the answer. Background: In the bottom surface of the kettle are many microscopically tiny pores, cracks and crevices. many of these retain extremely minute amounts of air after the kettle is filled with water. these air-charged pores will serve as nucleation sites, at which the boiling process will first be triggered. as the water heats up, you ...


13

Saying that $\delta(0) = 0$ is completely non-sensical since the Dirac delta function is not a function to begin with. When we physicists write $$ \int \delta(x)f(x) \mathrm{d}x = f(0) \tag{1}$$ when that's all the "definition" of the delta "function" you actually need. Formally, the $\delta$ function is a tempered distribution, something that assigns ...


12

Apart from motor and bearing noise, most of the acoustic power comes from the eddy swirls following the trailing edge of the blade after it passes by. There is also an outward pulse of air as the leading edge of each blade pushes forward cutting the air. The trailing eddies produce a broad spectrum of random noise, modulated by the fan blade frequency. ...


12

The key to this is the physical principle that the quantity you're asking about (delay between noise and noise cancelling) carries dimensional information (i.e. it's a time) and therefore it has to depend on the specific situation. The simplest case is trying to cancel out a pure note, with a sinusoidal waveform, then the delay can be as long as you want: ...


12

This is indeed a problem, which is dealt with using seismic isolation via Internal Seismic Isolation and Hydraulic External Pre-Isolators. The original LIGO isolation systems were passive - described here as shock absorbers - and included a single pendulum system, but the Advanced LIGO (aLIGO) upgrades added actuators that counteract vibrations in various ...


11

There are several ways I can interpret the question, so my main focus is going to be on the autocorrelation of an Ornstein-Uhlenbeck (O-U) process. So what is an O-U process and how is it different from regular Brownian diffusion? Brownian diffusion The stochastic differential equation (SDE) for Brownian diffusion of a particle can be written as $$\mathrm{...


10

By comparing the signal to the background. Suppose you get 10 IR photons from the camera and lens background but an extra 5 from the source then you can still detect the source. There is a whole science of signal processing to detect signals much fainter than the background. Especially in IR astronomy.


9

In astronomy, the background from the camera itself is called "dark current" and is removed by first taking an exposure with the shutter closed for, say, half an hour, and then subtract those counts from the real observations, normalized to the exposure time of a given image. Sometimes, if you're bored at the telescope due to bad weather, you can even take ...


9

In addition to the correct mathematical interpretation appearing in the other answer by ACuriousMind, perhaps a good physically minded viewpoint is to observe that objects like $\delta(t)$ have always to be interpreted in the sense of the average value, using some smearing or averaging function (in QFT we use the same interpretation regarding field ...


8

The position of the mass, as a function of time, will simply be a filtered version of the random noise 'input' signal. To see this in the frequency domain, take the (magnitude of the) Fourier transform of both sides and rearrange: $$|X(\omega)| = \frac{1}{\sqrt{\left(1 - \omega^2\right)^2 + \frac{1}{Q^2}\omega^2}}|N(\omega)|$$ For $\omega = 1$, we have $$...


7

There are a couple of main sources of intrinsic error (that is, not associated with counting photons from your source) which CCD's have. The first is as you have already mentioned called read noise. Here is a reasonable definition of read noise (taken from Romanishin's free pdf on Photometry): After an integration (exposure), the CCD must be read out to ...


7

$\newcommand{\bra}[1]{\langle #1 |}$ $\newcommand{\ket}[1]{| #1 \rangle}$ $\newcommand{\braket}[2]{\langle #1 | #2 \rangle}$ $\newcommand{\bbraket}[3]{\langle #1 | #2 | #3 \rangle}$ Although the question asks specifically about a harmonic oscillator, we can understand the meaning of the spectral density by considering a somewhat more general problem. ...


7

$\def\ii{{\rm i}} \def\dd{{\rm d}} \def\ee{{\rm e}} \def\Tr{{\rm Tr}} $As far as I know and would expect, the replacement of a quantum heat reservoir with a noisy classical field cannot be rigorously justified in general. However, for the simple problem of pure dephasing posed here, there is indeed a correspondence between the quantum and classical noise ...


7

White noise is characterised by the autocorrelation function $$\langle \eta(t) \eta(t') \rangle = D \delta(t-t'),$$ (here I'm assuming that $\eta$ is dimensionless so $[D] = [t]$). This means that (1) the noise should be uncorrelated with itself at different times, but also that (2) the variance of $\eta(t)$ must be infinite. Condition (2) is the essential ...


6

The situation you are describing is an example of Fresnel diffraction (or near-field diffraction). In general, when a wave propagates every point of the wave front can be thought of as its own source of waves traveling in all directions (called Huygens construction). It turns out that neighboring point sources along an infinite straight wave front reinforce ...


6

Interesting. In your title you say "background noise." So I was going to suggest the sea shell effect - putting a sea shell to your ear you think you hear the sea, even if it is far away. If you did think you heard music, that could have been an illusion. Your brain is wired to look for patterns - see animal shapes in the clouds, hear people talking (or ...


5

I think you've just derived the Stefan-Boltzman law for a one-dimensional system. The T^4 comes from three dimensions. The more dimensions the quanta can populate the higher power of T you get.


5

The threshold theorem says that if the error rate is below the threshold, a quantum algorithm with T locations (breadth times depth) can be made fault-tolerant with a blow-up (in both number of qubits and circuit size) by a factor which is a polynomial in the log of T. This is not enough to change BQP.


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 idea that frequency modulated signals are more resilient to noise than amplitude-modulated ones is somewhat of a myth. Both are susceptible to noise: the demodulation sequence (including the human hearing and sight senses) reacts slightly differently to the effects of noise so that. It can be shown that if there is additive Gaussian noise with ...


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 ...


5

The optimal distribution of photon frequencies for sending messages, assuming no noise but quantum shot noise, is indistinguishable from thermal (blackbody) radiation at a given temperature. So find the temperature for thermal radiation corresponding to your desired power, find its entropy, convert that to bits, and you have the theoretical maximum amount of ...


4

I didn't see the episode, but it may be referring to "Phreaking", by which the signals from a CRT monitor can be listened-in on (it uses high frequency changing currents to display the information, so these will inevitably result in some RF radiation from which this information can in principle be extracted). Wikipedia article has a bit more info.


4

It seems that the confusion is due to some unfortunate notation. As the OP states, Fano noise is due to the variance in photoelectron production per incident photon, and this should indeed be signal-dependent. However, the author also states that the total noise is given by: $$\tag{1} \sigma^2_\mathrm{TOTAL} = \sigma^2_\mathrm{READ} + \eta_i F_F + \eta_i S ...


4

A single measurement like this has a lot of noise on it - and random signal is always going to have some random correlation. You should definitely not pay too much attention to the stuff that is in the tail of the correlation distribution - it's all noise. The fact that the built in function does not produce negative values is related to you only looking at ...


4

The first term on the RHS arises from the assumption that the particle is moving through a viscous fluid (in this case, with no net flow velocity). The collisions from the front deliver a higher impulse than collisions from the rear, on average, which leads to a coherent net force directed opposite the particle velocity. The second term arises from the ...


4

I believe that it should be as you suggest since he changes the energy per degree of freedom from $k_{\rm B}T$ to $$ \frac{h\nu}{e^{h\nu/k_{\rm B}T} - 1}\,. $$ He explicitly states in the text before equation (7). We can also just use dimensional analysis: The dimensions of the first equation in the question post are \begin{align} \text{Voltage}^2 / \...


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