# Fourier Transforms, Convolution, Cross-correlation: what is their physical unit exactly?

Let us assume we are talking about real, deterministic, electrical signals $$x(t)$$ and $$y(t)$$ (magnitude in Volts).

There are different kind of Fourier Transforms. I made a table to summarize: NB: By the "<-" symbol, I mean variable substitution.

I tried to figure the physical unit of the output. The $$\mathrm{V}\cdot\mathrm{s} = \mathrm{V/Hz}$$ for the FT is fine, but I am not satisfied with what I get with the alternative transforms... is that right?

Also, analyzing the units of convolution product of $$x(t)$$ and $$y(t)$$: $$(x \ast y)(t) = \int\limits_{-\infty}^{+\infty}x(t-u)y(u) \ \mathrm{d}u$$ ... or the cross-correlation of $$x(t)$$ and $$y(t)$$: $$(x \star y)(d) = \int\limits_{-\infty}^{+\infty}\overline{x(t-d)}y(t) \ \mathrm{d}t$$ ...it would yield $$\mathrm{V}^2 \cdot \mathrm{s}$$ unit, which is... I don't know... somewhat closer to an energy?

But, as I understand it, the convolution product is actually a signal, typically the output of a filter bank from the original signal.

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In electronics, for the convolution product, the $$y$$ function should rather be seen as a pattern. Typically $$y$$ is an impulse response (often noted $$h$$), its unit is $$s^{-1}$$ and then the unit of the convolution product is $$V$$, which is legit for a signal.

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On the other hand, the cross-correlation is rather meant to be some kind of inner-products series, so it makes more sense to see it as some kind of energy (the auto-correlation of a signal at $$d=0$$ multiplied by some coefficient is actually its energy).

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In the cross-correlation product, both $$x(t)$$ and $$y(t)$$ are indeed signals. The above definition is for finite-energy signals, but it changes for finite-power signals: $$(x \star y)(d) = \lim_{T\to\infty}\frac{1}{T}\int\limits_{-\frac{T}{2}}^{+\frac{T}{2}}\overline{x(t-d)}y(t) \ \mathrm{d}t$$

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So physically speaking, maybe there are some unitary normalization coefficients missing in those formulas?

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For finite-energy deterministic signal $$x(t)$$, we get: $$R_0.E_x = \int\limits_{-\infty}^{+\infty}|x(t)|^2 \ \mathrm{d}t <{+\infty}$$ For finite-power, we get: $$R_0.P_x = \lim_{T\to\infty}\frac{1}{T}\int\limits_{-\frac{T}{2}}^{+\frac{T}{2}}|x(t)|^2 \ \mathrm{d}t <{+\infty}$$

We may divide those by $$R_0 = 1 \Omega$$ to get respectively energy or power quantities (hypothetical, relative to 1 Ohm).

Same goes for cross-correlation, depending on whether the signals are finite-energy and/or finite-power.

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• There was an interesting article: Can One Take the Logarithm or the Sine of a Dimensioned Quantity or a Unit? Dimensional Analysis Involving Transcendental Functions, pubs.acs.org/doi/10.1021/ed1000476 Sep 7, 2020 at 1:01
• Thank you for answering. I cannot easily access that article, however it reminds me of something. Let's assume $P$ is a power variable and $P_0$ is a power constant (both are positive), then the decibel level of P, relative to $P_0$, is: $L_p = 10.log_{10}(\frac{P}{P_0}))$ (in dB). However for signal x (in V), I've seen many electrical engineers write $X_{dB} = 20.log_{10}(|x|)$ when it should definitely be $X_{dB} = 20.log_{10}(\frac{|x|}{V_0})$ where $V_0 = 1 V$. So there might be something here. Plus, normalizing discrete signals seems harder for some reason. Sep 7, 2020 at 15:47
• The main gist of the article was all these transcendental functions cannot have units. In your example, I think Lp should be dimensionless. Think about Taylor expansion of a function. Imagine what will happen to the units after expansion as derivatives and powers. You cannot even add them. Sep 7, 2020 at 15:56
• The way you are thinking correct. Each quantity is divided by its dimension. This is exactly what the author said. Sep 7, 2020 at 16:46
• Transcendental function inputs and outputs should indeed be dimensionless, but as @Dan N. said, of course you can always multiply them by a quantity which does have a unit. For instance: $x(t) = V_0.sin(wt)$ is legit, its unit is V, even though $wt$ and $sin(wt)$ are dimensionless. Sep 8, 2020 at 12:50

## 1 Answer

I agree that the units for Fourier transform are V/Hz. Most of my experience with the convolution integral dealt with a convolution of the impulse response of a linear network with the voltage input to the network. The impulse response for, say, a simple RC network will have the product RC in the denominator. Thus, the unit of impulse response is per second. So, the units of a convolution would be volts-seconds * per second = volts. For correlation, either auto or cross-, in the case of power signals (as opposed to energy signals), you should divide the integral by the period, T. If it is not periodic, still divide by T, but take the limit as T goes to infinity. Now the period, T, has units of seconds and you are integrating over time, which has units of seconds. The time in the numerator and denominator cancel, and you are left with volts squared, which gives power when divided by resistance (assuming that the volts are rms).

• I don't think FT can have any units because exponential functions have to be dimensionless. Sep 8, 2020 at 1:36
• It is not true that functions with exponents are dimensionless, but it is true that the exponents themselves should be dimensionless. The voltage across a capacitor discharging through a resistor has an "e" raised to -t/(RC), which is an exponential, but the multiplier of the exponential is the voltage across the capacitor at time t = 0. Thus, the units are that of the multiplier, which is volts. Sep 8, 2020 at 12:37
• Thank you very much for your input! Helps a lot. So for signal $x$ in $V$ and impulse response $h$ in $s^{-1}$ we get: $(x \ast h)(t) = \int_{-\infty}^{+\infty}x(t-u)h(u)du$ and it works out, we get output in $V$. Now, for correlation, if I get it, what you say it depends on whether signals are finite energy or finite power. And also I agree the resistance is always omitted (or it's assumed it is $R_0=1$ ohm, but it does not make much sense physically). I guess correlation is more of a mathematical tool for analysis, so units and constant coefficients may be overlooked. Sep 8, 2020 at 13:11
• By the way, if you're new to Stack Exchange you might want to check out the LateX formatting for mathematical formulas: dsp.stackexchange.com/editing-help#comment-formatting Sep 8, 2020 at 13:14
• //"It is not true that functions with exponents are dimensionless, but it is true that the exponents themselves should be dimensionless."// @DanN. can you be more clear with this. you are saying that $x$ in $e^x$ must be dimensionless? but that "it is not true that" $e^x$ is dimensionless?? Sep 8, 2020 at 18:03