7

In short: if you directly use the voltage output, your FFT amplitude should be in mV as well. If you have the sensor calibration curve, the FFT amplitude should be in Teslas per second (T/s), as you look at a derivative. If you look at a power density spectrum (squared), then the units above should be squared as well (mV$^2$ or T$^2$/s$^2$). In general, ...


7

The power spectrum is a general term that describes the distribution of power contained in a signal as a function of frequency. From this perspective, we can have a power spectrum that is defined over a discrete set of frequencies (applicable for infinite length periodic signals) or we can have a power spectrum that is defined as a continuous function of ...


3

Say you have a measurement in unit $U$ (V$^2$) per unit $u$ (Hz). I do claim (although not in public) that the logarithm is a kind of unitless transformation, see for instance What is the logarithm of a kilometer? Is it a dimensionless number? In other words, it is somehow $0$-homogeneous, while a power is $2$-homogeneous: multiplying by a constant yields a ...


2

Simply because the intention of converting to a dB scale, here, is only to scale the power density, and not the bandwidth. So, that's simply by definition. Look at it this way: if you have a PSD plot with frequency on the x-axis and power density on y, you'd usually only want to scale the y-axis logarithmically. You can, of course, also scale the ...


2

As hinted to by @user28715 in comments, the correct definition to use is, using your style of notation: $$\rm dBV_{pk} = 10\log_{10}\left({V_{pk}}^2\right) = 20\log_{10}\left(\sqrt{{V_{pk}}^2}\right) = 20\log_{10}(|V_{pk}|).$$ Clearly, $\rm dBV_{pk}$ will not contain information about the sign of $\rm V_{pk}$. You cannot safely fiddle with the sign of $\rm ...


2

You'll have to convert back into linear units before you can evaluate the SINR formula. ${\textrm{SNR}_{linear}=10^{\big(\frac{\textrm{SNR}_{dB}}{10}\big)}}$ and likewise for CIR... ${\textrm{SINR}_{dB} = 10 \log_{10}(\frac{1}{\frac{1}{\textrm{SNR}}+\frac{1}{CIR}}})$ ${BER = \frac{1}{4{\textrm{SINR}}_{linear}}}$


2

impulse responses, $h(t)$, of continuous-time systems (those with output that is the same species of animal of the input) have dimensions 1/time. consider a wire which has impulse response $$ h(t) = \delta(t) $$ the area of the $\delta(t)$ is equal to the dimensionless $1$. the width of the nascent delta is in dimension of time, so the height must be in ...


1

Short answer: your conclusion seems ok. Concept of units make sense only in the world of physics. Mathematical objects such as DFT trasforms do not have any kind of units but just their values, quasi exceptions being the planar angle meausure of radian and volumetric angle mesaure of the steradian. However, when a variable represents a physical quantity (as ...


1

In general, no. In fact, if the low-pass filter characteristics were well-chosen in the sense of getting the best post-filtering signal to noise ratio, then the amount that you might degrade any such measurements would be minimal. In part this conclusion comes from my "well-chosen" qualifier. Roughly speaking (and I can only speak roughly because your ...


1

If $x(t)$ is an action signal like voltage or current, then the square of $x(t)$ is proportional to instantaneous power and the constant of proportionality depends on what kinda animal $x(t)$ is and what load $x(t)$ is connected to. So if $\big|x(t)\big|^2$ is proportional to instantaneous power, so also is the integral of $\big|x(t)\big|^2$ over all time ...


1

Your reasoning sounds correct (with unit considerations), you can see a discussion on both continuous and discrete aspects of the Fourier transform in What are the units of my data after an FFT? What is unclear to me is why you expect the PSD to be in unit $\frac{m^2}{Hz}$? Powers are powers: if a quantity has unit $\textrm{u}$, its square has unit $\...


1

You can use the following : Consider the discrete time sequence $x[n]$ and the DTFT $X(\omega)$ related by: $$ X(\omega) = \sum_k x[n] e^{-j \omega n} \longleftrightarrow x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(\omega) e^{j\omega n} d\omega$$ Then define $X'(\mu)$ as the normalized DTFT, which is related to $X(\omega)$ as $$X'(\mu) = X(2\pi \mu) = \sum_k ...


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