# 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 – M. Farooq Sep 7 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. – lostdatum Sep 7 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. – M. Farooq Sep 7 at 15:56
• The way you are thinking correct. Each quantity is divided by its dimension. This is exactly what the author said. – M. Farooq Sep 7 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. – lostdatum Sep 8 at 12:50

• 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. – lostdatum Sep 8 at 13:11
• //"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?? – robert bristow-johnson Sep 8 at 18:03