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

Question

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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Answer 1

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).