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I know that a Fourier transform (FT) outputs a distribution of how much of each frequency is present in a signal, but what would be the units of that distribution? It's not a proportion or a percent because it looks like you'd have to sum all the FT outputs to get the total and divide each FT output by the total to get the percent. From the mathematical formula, the units would appear to be (signal units)x(time).

Also, when I did a fast Fourier transform (FFT), the magnitudes of all the complex FFT outputs sum to the number of bins before normalizing. Why is this?

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    $\begingroup$ It it's volts going in and they're a function of time, then what comes out are volt-second or volts per Hertz. $\endgroup$
    – robert bristow-johnson
    Commented Apr 16 at 19:40
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    $\begingroup$ Now the FFT is a fast method of doing the DFT and the DFT does not have the $\times$(time). It just adds up numbers. To relate the DFT to the FT, you gotta consider the Fourier integral as a Riemann Sum. The $\mathrm{d}t$ becomes $\Delta t$ which is the reciprocal of the sample rate. $\endgroup$
    – robert bristow-johnson
    Commented Apr 16 at 19:43
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    $\begingroup$ there's other good answers, like dsp.stackexchange.com/questions/69186/… $\endgroup$
    – Marcus Müller
    Commented Apr 16 at 19:43
  • $\begingroup$ It makes sense that the units would be (signal units)x(time), eg: (volts)x(time) or (signal units)/(frequency units). It follows from the definitional form of the FT either in continuous integral form or discrete Riemann sum form. $\endgroup$
    – Data2Dollars
    Commented Apr 16 at 23:48

2 Answers 2

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There's no special rule of unit assignment within DSP or Fourier Transform operations. You simply use the scientific conventions.

A Continuous-Time Fourier Transform is this integral:

$$ X(j\Omega) = \int_{-\infty}^{\infty} x(t) e^{-j \Omega t } \ \mathrm{d}t \tag{1} $$

and if $x(t)$ is a time-function with a unit of Volt, then the transform $X(j\Omega)$ has a unit of Volt $\times$ sec; abbrv as: V s, or stated equivalently V/Hz.

A Discrete-Time Fourier Transform is this sum:

$$ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n] e^{-j \omega n } \tag{2} $$

and if $x[n]$ is a time-sequence with a unit of Volt, then the transform $X(e^{j\omega})$ has a unit of Volt; i.e., same unit of $x[n]$.

Note that the logarithmic, exponential, and all trigonometric functions (including their inverses) are unitless...

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  • $\begingroup$ Ain't V/Hz if you're using angular frequency. It's V/(rad/sec) . $\endgroup$
    – robert bristow-johnson
    Commented Apr 16 at 23:12
  • $\begingroup$ Actually, because of that $\frac{1}{2 \pi}$ factor in $$ x(t) = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} X(j\Omega) e^{j\Omega t} \mathrm{d} \Omega $$ it's volts per Hertz. Even for $X(j\Omega)$. Ain't that weird? $\endgroup$
    – robert bristow-johnson
    Commented Apr 16 at 23:16
  • $\begingroup$ @robertbristow-johnson when I said "V/Hz" I have just translated "V.s = V/(1/s) = V/Hz" for convenience. I didn't even need to make a reference to radians there. In the inverse transform, $d \Omega$ refers to "rad/sec" and that's cancelled by $2\pi$ term implying a radian unit, as you state and it's not weird but quite normal, why? Btw, radian is not a physical unit (though included in SI base units!), so, no need to mess with it anyway :-) $\endgroup$
    – Fat32
    Commented Apr 17 at 2:16
  • $\begingroup$ It's just that while Hz and rad/sec are dimensionally identical, they are different units $\endgroup$
    – robert bristow-johnson
    Commented Apr 17 at 12:18
  • $\begingroup$ @robertbristow-johnson yes they are different units. That's because there are more units than necessary, there is an abundance of redundant nomenclature... :-) $\endgroup$
    – Fat32
    Commented Apr 17 at 13:28
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I'm not aware of literature defining this in general, although for some special cases (e.g. PSD, as commented in the question), there may be some convention/definition generally accepted. (If someone is, please do enlighten us.)

But I'd share a concrete example here, which may help to shed light on this: Suppose we have a signal $x(t)=\cos(2\pi t)+2\cos(4\pi t)$, representing the output voltage of an electric circuit at different time instant. So the unit of $x(t)$ is just Volt.

Since $x(t)$ is periodic with fundamental period $T=1$, it has a Fourier series representation, i.e.

$$x(t) = \sum_{k=-\infty}^{\infty} a_k e^{j2\pi kt}$$

For this simple $x(t)$, it's easily verified that $a_{-1} = a_{1}=\frac{1}2$, $a_2=a_{-2}=1$, and $a_k=0$ for all other values of $k$, and

$$x(t)=a_{-2} e^{-j4\pi t} + a_{-1} e^{-j2\pi t} + a_1 e^{j2\pi t} + a_2 e^{j4\pi t}.$$

So, what should the unit of $a_k$ be?

It may help to look at it this way: The Fourier series simply decomposes the time function $x(t)$ into four constituent time functions, e.g. $e^{-j4\pi t}, e^{-j2\pi t},$ etc, with $a_k$'s being their respective scaling constants. In this regard,

A natural & simple interpretation may be to have each of the constituent time functions $e^{j2\pi kt}$ assume unit of Volt, and to let $a_k$'s be unit-less. This way we're consistent physically.

Remarks:

  1. Note that $a_k$'s are not percentage, as the OP also observed, as they can be negative or even non-real. Nonetheless, they do represent weightings of the constituent functions, and hence are strongly correlated with the concept of percentage or proportion.

  2. Fourier Transform may, of course, be viewed as the limiting form of Fourier series, and hence may also be viewed as unit-less.

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