The continuous-time Fourier Transform (CTFT) of a signal $x(t)$ (with unit $unit$) is:

$$X(\omega)=\int_{-\infty}^{\infty} x(t)e^{-i\omega t}dt$$

which should be in $unit\cdot sec$ or $\frac{unit}{Hz}$.

The Energy Spectral Density (ESD) of $x(t)$ is defined as:


which should have unit $\frac{unit^2}{Hz^2}$. But that doesn't seem right and certainly doesn't jive with the textbooks.

So what gives?

This question has also been asked here: Power Spectral Density computation and units, but I found the answers unsatisfactory.


2 Answers 2


If you want correct dimensions, the signal $x(t)$ must have dimension $\sqrt{W}$ because then its energy


has the correct dimension $W\cdot s=J$. E.g., if you have a voltage signal, you need to normalize it by some load impedance to get the correct dimension $V/\sqrt{\Omega}=\sqrt{W}$.

With the dimension of $x(t)$ being $\sqrt{W}$, the dimension of its Fourier transform $X(\omega)$ is $\sqrt{W}\cdot s$, and, consequently, the energy density $|X(\omega)|^2$ has dimension $W\cdot s^2=W\cdot s/Hz=J/Hz$, as it should be.

In signal processing it is often the case that the correct dimensions are ignored, e.g., we would not normalize voltage by the square root of impedance. In that case energy and power are defined by the square of the signal, which is correct in the sense that this quantity is proportional to the actual energy or power, and a simple normalization gives the correct value, including the correct dimension. So for a signal with dimension $[unit]$, we would obtain $[unit]^2\cdot s/Hz$ for the energy density, and $[unit]^2/Hz$ for the power density. So in practice you would often see, e.g., $V^2\cdot s/Hz$ for energy density, and $V^2/Hz$ for power density.

See also the explanations given here and here.

  • $\begingroup$ That's a solid argument. However, it still doesn't jive with the generalized ESD that we are more used to (e.g. en.wikipedia.org/wiki/Spectral_density#Energy_spectral_density) with the unit $unit^2/Hz$. I'm very used to seeing $V^2/Hz$ or $lbf^2/Hz$ in ESD. Is this just hand-waving, or am I missing something? Surely Wikipedia can't be wrong? $\endgroup$
    – JZYL
    Jun 13, 2019 at 2:08
  • $\begingroup$ @Jimmy: If you do without the appropriate normalization (e.g., divide voltage by square root of impedance), then you'll get $V^2$ for "power". So for energy density you get $V^2\cdot s / Hz$. It is for power spectral density that you get $V^2/Hz$ in that case. This is explained here. $\endgroup$
    – Matt L.
    Jun 13, 2019 at 8:04
  • $\begingroup$ I've also added a paragraph to my answer explaining this issue. $\endgroup$
    – Matt L.
    Jun 13, 2019 at 8:12
  • $\begingroup$ Sorry Matt, i don't like how you deal with dimensionality of things. Really, I don't downvote you much (as if that makes a measurable difference to your rep). This is like only my second time. $\endgroup$ Jun 13, 2019 at 9:16
  • 1
    $\begingroup$ @MattL. You are right. PSD has the extra time-window division (1/T). I was interchanging ESD and PSD too liberally. Very convincing explanation. $\endgroup$
    – JZYL
    Jun 13, 2019 at 14:50

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 proportional to energy (and having the same constant of proportionality). So you can call this "energy":

$$E_x = \int\limits_{-\infty}^{\infty}\big|x(t)\big|^2 \, dt$$

as long as you remember that it's really proportional to energy and the constant of proportionality depends on what dimension is $x(t)$ and what load that signal is connected to.

For example, if $v(t)$ is a voltage, then $|v(t)|^2$ is proportional to instantaneous power and if $v(t)$ is connected to a resistor having resistance $R$, then instantaneous power is

$$ p(t) = \frac{1}{R}\big|v(t)\big|^2 $$

and the total energy is

$$E_v=\int\limits_{-\infty}^{\infty}p(t) \, dt$$


$$E_v=\frac{1}{R}\int\limits_{-\infty}^{\infty}\big|v(t)\big|^2 \, dt$$

Clearly the constant of proportionality is $\frac{1}{R}$.

If $x(t)$ is a dimensionless value in a DSP or computer, then you have to scale it with the reference voltage $V_\mathrm{ref}$ of the D/A converter to make this number a voltage that delivers power to a load. Then the constant of proportionality becomes $\frac{V^2_\mathrm{ref}}{R}$ and the energy really is (with correct scaling):

$$E_x = \frac{V^2_\mathrm{ref}}{R} \int\limits_{-\infty}^{\infty}\big|x(t)\big|^2 \, dt$$


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