Perhaps someone can help me resolve something - this is my understanding:

In deterministic signal analysis, for a continuous signal $x(t)$ the signal energy is defined by $$E_{\textrm{s}} = \int^{+\infty}_{-\infty} |x(t)|^2\mathrm dt \hspace{1cm} \textrm{Units:}\hspace{0.3cm}[\textrm{signal}^2\cdot \textrm{time}]$$ where the subscript $s$ is to indicate explicitly that we are talking about "signal energy", and not real physical energy (which would be in units of Joules if you were to divide signal energy by some load impedance).

Similarly, the average power of a signal is defined by $$P_{\textrm{s}}^\textrm{ av} = \lim_{T\to\infty}\frac{1}{T} \int^{+T/2}_{-T/2} |x(t)|^2\mathrm dt \hspace{1cm} \textrm{Units:}\hspace{0.3cm}[\textrm{signal}^2]$$

This makes sense because it is the same unit as the rate of signal energy transferred, which is signal power.

Therefore, the units of power spectral density should be [signal power per frequency band], or $[\textrm{signal}^2 / \textrm{Hz}]$.

My problem is that I have seen many times now people who seem to know what they are talking about saying that the power spectral density is given by

$$ S_{xx}(f) = |X(f)|^2 $$

where $X(f)$ is the Fourier transform of $x(t)$. BUT, the units of this quantity are not correct. Since the units of the Fourier transform $X(f)$ are $[\textrm{signal}\cdot \textrm{time}]$, then the units of $S_{xx}(f)$ written above are $[\textrm{signal}^2\cdot \textrm{time}^2] = [\textrm{signal}^2\cdot \textrm{time} /\textrm{Hz}]$, which are the units of energy spectral density, not power spectral density. Am I missing something fundamental here? Why do people often write this simple definition of $S_{xx}(f)$?

See these answers for some examples:

  • $\begingroup$ Thanks Dan, but no it doesn't. I have read many of these questions on DSP (you should see how many tabs I've got going!). I am still confused about the incosistency with the units. It seems to me that for PSD you always have to divide by some time interval to get the correct units (and in most definitions this also has the T to infinity limit). $\endgroup$
    – teeeeee
    Mar 31, 2020 at 12:53
  • $\begingroup$ It seems someone else had the same question a while ago dsp.stackexchange.com/q/22829/38419 but it didn't receive any meaningful answer with regards to the units. $\endgroup$
    – teeeeee
    Mar 31, 2020 at 12:57
  • $\begingroup$ For a strict physical interpretation your initial assumption is wrong. Energy units are NOT [signal^2 seconds]. Energy over time is power (Watts) so the signal units would have to be $\sqrt{W}$ which doesn't exist . There is always some sort of impedance involved or two complementary signals (voltage/current, force/velocity, etc) $\endgroup$
    – Hilmar
    Mar 31, 2020 at 18:18
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    $\begingroup$ Thats why I said "signal energy", not physical energy. The load which is being driven (no matter what it is) will only appear at the end as a scaling factor. The units of signal energy are [signal^2 seconds]. $\endgroup$
    – teeeeee
    Mar 31, 2020 at 18:19
  • $\begingroup$ @Hilmar You will notice that I also denoted signal energy by $E_s$, with the subscript to be explicitly clear that we are dealing with signal energy. $\endgroup$
    – teeeeee
    Mar 31, 2020 at 18:30

5 Answers 5

  • The OP is correct in their dimensional analysis
  • $|X(f)|^2$ is NOT the power spectral density, despite what other authors might claim. Other authors probably call this the power spectral density because it is close to right and it captures most of the important features without having to delve into technicalities.

Power has dimensions of $[\text{signal}^2]$. Energy has dimensions of $[\text{power}\cdot\text{time}] = [\text{signal}^2\cdot\text{time}]$. The spectral density of anything has dimensions of $[\text{thing}\cdot \text{frequency}^{-1}]$. Thus, power spectral density has dimensions of $[\text{signal}^2 \cdot \text{frequency}^{-1}] = [\text{signal}^2\cdot \text{time}]$. Note that it is coincidental that power spectral density has the same dimensions as energy and it should be understood that power spectral density is power per frequency. Also note that the Fourier transform of anything always has dimensions of $[\text{thing}\cdot\text{frequency}^{-1}]$.

The power spectral density is more nicely defined as follows. We define the windowed signal

$$ x_{\Delta t}(t) = \begin{cases} x(t) \text{ for } |t|< \frac{\Delta t}{2}\\ 0 \text{ for } |t| \ge \frac{\Delta t}{2} \end{cases} $$

The windowed Fourier transform is then

$$ X_{\Delta t}(f) = \int_{t=-\infty}^{+\infty} x_{\Delta t}(t) e^{-i2\pi f t} dt = \int_{t=-\frac{\Delta t}{2}}^{\frac{\Delta t}{2}} x(t) e^{-i2\pi f t} dt $$

The power spectral density is then defined by

$$ S_{xx}(f) = \lim_{\Delta t\rightarrow \infty} \frac{1}{\Delta t} |X_{\Delta t}(f)|^2 $$

More properly when dealing with random signals one might take an expectation value of the squared windowed transform.

This can be expressed another way. We can define a window function

$$ w_{\Delta t}(t) = \frac{1}{\sqrt{\Delta t}} \theta\left(t-\frac{\Delta t}{2}\right)\theta\left(\frac{\Delta t}{2} - t\right) $$

Here $\theta$ is the Heaviside function. And a windowed version of $x(t)$ given by

$$ x_{w_{\Delta t}}(t) = x(t)w_{\Delta t}(t) $$

Note that this is the exact same as the windowed function defined above but with a factor of $\frac{1}{\sqrt{\Delta t}}$ built in. The Power spectral density can then be defined equivalently as

$$ S_{xx}(f) = \lim_{\Delta t \rightarrow \infty} |X_{w_{\Delta t}}(f)|^2 $$

The reason we must work with $x_{w_{\Delta t}}(t)$ rather than $x(t)$ is that $x(t)$ is that, if $x(t)$ has constant power or at least finite power for infinite time, then $x(t)$ has infinite energy. However, even if $x(t)$ has infinite energy, $x_{w_{\Delta t}}(t)$ has finite energy. Note that the window function is not dimensionless but acts so that the finite total energy in $x_{w_{\Delta t}}(t)$ given by $\int |x_{w_{\Delta t}}(t)|^2 dt$ is in fact the average finite energy in $x(t)$.

We also have the fact that infinite length signals do not have well behaved Fourier transforms, for example, the Fourier transform of a pure tone $e^{+i2\pi f_0 t}$ is a dirac delta function, i.e. not well behaved. The windowed version of this will have a well-behaved Fourier transform.

@Dan Boschen expresses some confusion about reconciling the dimensions of $S_{xx}(f)$ with the Fourier transform of the autocorrelation function. There is no need for confusion. The units agree.

$$ S_{XX}(f) = \tilde{R}_{xx}(f) = \int R_{xx}(t) e^{-i2\pi ft} dt = \int \langle x(t)x(0)\rangle e^{-i2\pi ft}dt $$

The expression on the right has dimensions of $[\text{signal}^2\cdot \text{time}]$ which is the same as the units of power spectral density expressed above. This should hint that the Fourier transform of the auto-correlation function is NOT given by $|X(f)|^2$...

$R_{xx}(t)$ (for stationary $x(t)$) is defined as

ensemble average: \begin{align} R_{xx}(t) = \langle x(t)x(0) \rangle = \int yz f_{x(t),x(0)}(y,z) dy dz \end{align}

$f_{x(t),x(0)}(y,z)$ is the joint probability density function for the random variables $x(t)$ and $x(0)$ so it has dimensions of $[\text{signal}^{-2}]$.

time average: \begin{align} R_{xx}(t) = \langle x(t)x(0) \rangle = \lim_{\Delta t \rightarrow \infty} \frac{1}{\Delta t} \int_{t'=-\frac{\Delta t}{2}}^{\frac{\Delta t}{2}} x(t'+t)x(t') dt' \end{align}

  • $\begingroup$ Thanks for the answer, would you mind clarifying your symbols for signal and its Fourier transform? (For example, I used $x$ to denote singal, and $X$ for it transform). Just so it is easier to distinguish. $\endgroup$
    – teeeeee
    Apr 1, 2020 at 9:28
  • $\begingroup$ Yes, I can do that, I thought I was copying your notation and thought it was a bit unclear to use $X(t)$ and $X(f)$, my bad! $\endgroup$
    – Jagerber48
    Apr 1, 2020 at 9:28
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    $\begingroup$ @teeeeee it's updated. $\endgroup$
    – Jagerber48
    Apr 1, 2020 at 9:34
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    $\begingroup$ @teeeeee ah, you're right. Ok, the text in the answer is correct and has always been correct. That comment is incorrect. I think I meant to say $\tilde{R}_{xx}(f)$ has dimensions of $[V^2\cdot s]$. What you are suspecting is correct. $R_{xx}(t)$ should definitely have dimensions of $[V^2]$ so its Fourier transform should have dimensions of $[V^2 \cdot s]$. This is what has always been reflected in my answer. $\endgroup$
    – Jagerber48
    Apr 1, 2020 at 10:02
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    $\begingroup$ Great, I was just remarking that this I think is where Dan's confusion was arising from - he was not considering the time averaged $R_{xx}$ with the $1/T$ factor, which then messed up dimensions. $\endgroup$
    – teeeeee
    Apr 1, 2020 at 10:03

THIS IS NOT YET A COMPLETE ANSWER BUT THE CONTINUATION OF THE OP's QUESTION IN MY ATTEMPT TO ANSWER and like the OP, I would welcome a short concise bottom line answer that addresses this.

Update: After the OP pushed me through the mud on this I ended up siding with his same level of questioning and concluding that in a strict sense $|X(f)|^2$ is en energy spectral density as given by its units. When $x(t)$ is a "power signal" meaning it has infinite energy (such as a sine wave for all time), the second reference linked below states that in this case the PSD is actually the limit of the ESD as in

$$S_x(f) = lim_{T \rightarrow \infty }\frac{1}{2T}|X(f)|^2 $$

What I still can't resolve is the formal definition of PSD as the Fourier Transform of the autocorrelation function given that I clearly see from this how the PSD is $|X(f)|^2$, but then has the conflict with units as the OP has stated, further detailed below.

Leaving my notes and references below but the clear answer is still elusive.

See https://en.wikipedia.org/wiki/Spectral_density where the distinction between Energy Spectral Density and Power Spectral Density is detailed for $S_{xx}(f)$ generally where when $S_{xx}(f)$ is determined using $|X(f)|^2$ for a signal having finite energy it is indeed a "Energy Spectral Density". However for signals of infinite energy $S_{xx}(f)$ is defined using the Fourier Transform of the Autocorrelation function (which can also be shown to be equal to $|X(f)|^2$), resulting in a "Power Spectral Density" with consistent units in both cases. So it is a matter of defining carefully what $S_{xx}(f)$ is for the signal of interest.

This was also a useful reference:


Consider first that for continuous signals over all time (infinite energy), the "Power Spectral Density" is given as the Fourier Transform of the Autocorrelation function:

$$S_{xx}(f) = \mathscr{F}\{R_{xx}(\tau)\} = \frac{1}{2\pi}\int_{-\infty}^{\infty}R_{xx}(\tau)e^{j\omega \tau}d\tau$$

Also consider that the Fourier property relating multiplication of two functions in one domain to cross-correlation in the other, such that the inverse Fourier Transform of the product of two frequency domain functions is the cross-correlation of those functions in time. Therefore

$$R_{xx}(\tau) = \mathscr{F}^{-1}\{ X(f)X^*(f) \} = \mathscr{F}^{-1}\{|X(f)|^2 \}$$

From which we see that:

$$\mathscr{F}\{R_{xx}(\tau)\} = |X(f)|^2 = S_{xx}(f)$$

As for units- Generally, units under integration follow the property such that for

$$y = \int f(x)dx $$

The units of $y$ are the units of $f$ multiplied by the units of $x$.

Consider one layer down from PSD in how the Fourier Transform of a voltage actually has units of inverse frequency (as in volts/Hz):

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

If the units of $x(t)$ are in volts then mathematically the units of $H(\omega)$ are volts-seconds. So the Fourier Transform itself is in units of $\frac{1}{Hz}$ such that when we multiply the resulting function by a frequency would return us to our original units (as done in the inverse Fourier Transform.)

Now consider the units of the autocorrelation function:

$$R_{xx}(\tau) = \int x(\tau)x(t-\tau)d\tau$$

If $x$ was in units of volts, we see that the autocorrelation function has units of $v^2 \cdot \sec$ ... which is units of energy!

  • $\begingroup$ Okay, if I consider a $1\Omega$ resistor driven by cosine with $A$ and $f_0$, the PSD is given by $\frac{A^2}{4R}[\delta(f-f_0) + \delta(f+f_0)]$, with units $[V^2\cdot\Omega^{-1}\cdot\textrm{Hz}^{-1}]$. Integrating over frequency gives total average power to be $\frac{A^2}{2R}$ with units $[V^2\cdot\Omega^{-1}]=[W]$ - just the rms power delivered to the resistor. $\endgroup$
    – teeeeee
    Mar 31, 2020 at 13:42
  • $\begingroup$ However, if I calculate the Fourier transform $X(f)=\frac{A}{2}[\delta(f-f_0) + \delta(f+f_0)]$ this has units of $[V\cdot s]$, and so $|X(f)|^2$ has units of $[V^2\cdot s\cdot\Omega^{-1}\cdot\textrm{Hz}^{-1}]$ when normalising by the load resistance. This is not the same as the PSD units. $\endgroup$
    – teeeeee
    Mar 31, 2020 at 13:42
  • $\begingroup$ The units are wrong with the formula you gave, and don't work out. If the units of $|X(f)|^2$ is $[V^2/\textrm{Hz}^2]$, then the units of $R_{xx}(\tau)$ become $[V^2/\textrm{Hz}]$ (because we perfomed the inverse transform, which multiples the units by another $\textrm{Hz}$). Finally, taking the FT of $R_{xx}$ to get spectral density again introduces another factor of $[s]$ to the units, to give $[V^2\cdot s /\textrm{Hz}]$ - again the units of energy spectral density, not power spectral density. $\endgroup$
    – teeeeee
    Mar 31, 2020 at 16:36
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    $\begingroup$ I do agree that the PSD is the transform of the autocorrelation. But, maybe the difference is that, for deterministic power signals I would define the autocorrelation as $$R_{xx}(\tau) = \lim_{T\rightarrow\infty} \frac{1}{T} \int_{-T/2}^{T/2} x^\ast(t)x(t+\tau) dt$$ Then the units of this are $[V^2]$ and if you then take the FT of $R_{xx}$ you would have the power spectral density with correct units of $[V^2/\textrm{Hz}]$. $\endgroup$
    – teeeeee
    Mar 31, 2020 at 17:26
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    $\begingroup$ The system is flagging this for too many comments. Please take any further discussion to chat or Post-Processing (if able). $\endgroup$
    – Peter K.
    Apr 1, 2020 at 20:55

I am curious about the same issue the OP raised: sometimes the units of power spectral density seem not quite right. I assume it is just me, so I always go back to my touchstone reference 1. Blinchikoff and Zverev use the definitions of Fourier transform and inverse transforms I have always used and preferred [1, p. 294]:

Fourier transform and inverse

and they give the units of the autocorrelation function and its Fourier transform [1, p. 304]:

Autocorrelation units

Since Dan Boschen has not yet got this one nailed to the wall, I started looking at books on my shelves. McGillem and Cooper 2 say this:

Parseval's theorem

with units as per the last sentence: $V^2 s/Hz$, i.e., $J/Hz$.

Bracewell, in his classic book 3, discusses this issue on pages 46-47:

"We shall refer to the square modulus of a transform as the energy spectrum; that is, $|F(s)|^2$ is the energy spectrum of $f(x)$. The term is taken directly from the physical fields where it is used." And quite a bit more.

I could check more books, but there seems to be no point.

Bottom line: The OP is correct about the units issue and my guess is that the issue arises if care is not taken to distinguish among the three types of functions given in the table by Blinchikoff and Zverev.

1 H.J. Blinchikoff, A.I. Zverev, "Filtering in the Time and Frequency Domains", Wiley-Interscience, John Wiley & Sons, NY, ©1976.

2 C.D. McGillem, G.R. Cooper, "Continuous and Discrete Signal and System Analysis", 2nd Ed., Holt, Rinehart and Winston, NY, ©1984, p. 126.

3 R.N. Bracewell, "The Fourier Transform and Its Applications", 2nd Ed., revised, McGraw-Hill Book Co., NY, ©1965, 1978, and 1986.

  • $\begingroup$ Hi Ed, yes all those definitions you presented make perfect sense to me. My problem occurs when people say that power spectral density is $|X(f)|^2$. However, it seems to me that this quantity is actually energy spectral density. $\endgroup$
    – teeeeee
    Mar 31, 2020 at 15:42
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    $\begingroup$ Based on the energy integral you gave in your question, it makes sense for an aperiodic signal, such as a pulse. So the table in my answer indicates that you are correct: the FT of the autocorrelation function is an energy spectral density. I see no problem with this being the case and wonder if experts are simply doing something mathematicians wryly call “abuse of language”. In any event, I think you have raised a valid issue and this also appears to pop up whenever someone inquires about white noise: the “delta function” comes and goes, for no solid reason, as far as I can tell. $\endgroup$
    – Ed V
    Mar 31, 2020 at 16:20
  • $\begingroup$ Nice contributions. I liked where @Hilmar was going in making the point that it is meaningless to assign units to $f(t)$ in a general expression to conclude J or W since $f(t)$ could and often is other magnitude quantities (such as counts). These would hold as J or W if we specifically say f(t) is in units of volts or amps and through or across a 1 ohm resistor (otherwise it is simply an "energy qty" or a "power qty". Most texts seem to totally ignore that but I think it leads to the very confusion the OP brought up, and more discipline with units is of practical interest. $\endgroup$ Apr 1, 2020 at 0:05
  • $\begingroup$ Thanks for your generous remark and I agree with what you say in your comment! This is really an informative and thought provoking site and I like being able to occasionally contribute little bits while watching the actual experts, such as you, pulverize problems. ;-) $\endgroup$
    – Ed V
    Apr 1, 2020 at 0:18

It has more to do with the simple nature of analysis of the data present. For example: given a sequence, it is easier to calculate the DTFT to get the Fourier transform then to calculate the autocorrelation and calculate the Fourier transform of the autocorrelation to get the PSD.

One could easily interpolate the DTFT, assuming a sampling time to get the PSD in continuous time.

Essentially, the characterization of the spectrum power is easily seen by the DTFT although in the unitless frequency domain.

  • $\begingroup$ I don't mention DTFT - I am talking about only continuous deterministic signals. Still, the question remains - the units of this often-quoted expression seem wrong to me. $\endgroup$
    – teeeeee
    Mar 31, 2020 at 12:25
  • $\begingroup$ Energy spectral density is essentialy the PSD integrated for an interval of time. Since in practice there is no infinte interval we are essentially measuring the energy spectral density and assuming the signal is periodic $\endgroup$ Mar 31, 2020 at 12:56
  • $\begingroup$ Do you mean that the expression for $S_x(f)$ is assuming periodic with period $T=1$? $\endgroup$
    – teeeeee
    Mar 31, 2020 at 13:01
  • $\begingroup$ It assumes periodicity with the length of the signal x(t) I would say $\endgroup$ Mar 31, 2020 at 18:39

This is not really an answer, but a different angle:

Physically speaking, the power of a single signal is not well defined. Physical power (or intensity) is always the product of two root-power quantities (which used to be called field quantities). Voltage times current, force times velocity, etc.

Let's say you have an impedance with a circuit with a voltage and a current. The power is given by the product of voltage and current, not by the square of the voltage or the square of the current. If the impedance happens to be a resistor, then voltage and current are proportional and the power is indeed proportional to the square of each root power quantity. It's not the same, but at least it's proportional. However, if the circuit is an ideal capacitor, the power is simply zero despite the fact that current and voltage are non-zero.

Defining energy as

$$E = \int x^2(t) dt ,[W/Hz=J]$$

is just wrong. That would imply that x(t) has units of $\sqrt{W}$ which doesn't exist since it would require roots of SI base units.

You can certainly define a formula this way, but it's not physical energy. Sometimes there is just a proportionality factor missing, but often it's way more sophisticated than that.

  • 1
    $\begingroup$ Ofcourse it is NOT physical energy, but that is indeed the standard definition for signal energy in every textbook I've seen. You can definite the power in the same way for non-resistive elements though - using instead the complex impedance? $\endgroup$
    – teeeeee
    Mar 31, 2020 at 18:39
  • $\begingroup$ If you have a complex impedance, you can't just square the voltage. In the time domain you need to calculate the current by convolving the current with an equivalent impulse response and than multiply the two. Convolution and multiplication are not commutative. In this case it's easier in the frequency domain $\endgroup$
    – Hilmar
    Mar 31, 2020 at 19:55
  • $\begingroup$ So continuing your point in case you are leading somewhere. The last formula is correct if we normalize the resistance to 1 ohm and define the units of x to be current or voltage (as is often the case since we are generally interested in a proportional power amount anyway, such as when dealing with SNR). But generally we could use any units and still allow $x^2$ to be a "power quantity" for most of our purposes in that it is a scalar quantity that is proportional to power, no? Your capacitor example is a good one. $\endgroup$ Mar 31, 2020 at 23:58
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    $\begingroup$ @teeeeee If it's not physical energy you shouldn't give it units like Watt or Joule which the text book cited in Ed V's answer clearly does. $\endgroup$
    – Hilmar
    Apr 1, 2020 at 11:26
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    $\begingroup$ "Physically speaking, the power of a single signal is not well defined. Physical power ... is always the product of two ... quantities". This really bugs me but I cannot find any meaningful counterexamples. It appears if it is describable by one signal, it's either already power, or assumes something else is constant. Maybe it can be shown as, since $P(t) = d/dt E(t)$, and (I posit) energy is always aggregated, i.e. $\int f(t) dt$, then it's always chain rule under the hood and $P(t)$ becomes some quantity in $E(t)$ times its rate of change, irreducible to one signal. $\endgroup$ Mar 4, 2023 at 17:15

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