Energy and Power: Power Spectral Density is units of Energy

Question

To help focus answers: The following is specific to “Energy” and “Power” as used for signal processing, not physical units of energy and power (then to ask can units of “Watts” and “Joules” be used, and if not- what then are the units that could be used that would make more sense). Further, this is specific to ergodic, stationary (if random) or otherwise deterministic discrete time signals with a finite duration, as well as discrete in frequency (via the DFT).

Specific to discrete time systems, and preferably without bringing in continuous time equivalents, I would like to confirm my understanding of the following. I'm looking both for confirmation or correction, and if I am correct I would be interested in other perhaps clearer or alternate ways to describe this or other related insights. Note that there is an interesting Time Frequency duality with Power and Energy. I summarize my questions at the end:

Confirming my understanding with some gaps as <_____>:

Given a finite duration of discrete time domain samples as $x[n]$, the conjugate product of each sample $|x[n]|^2$ would have both units of energy and units of power, and specifically it can be described as "<_____> Power" given it is the energy at that instant, over the time duration until the next sample.

Using SI units for clarity or explanatory analogy, if we provide the time duration of one sample as $T$ seconds, and $|x[n]|^2$ as energy in Joules, then we would have the <_____> power in each sample as $|x[n]|^2/T$ Joules/sec = Watts. I'll simplify this to the case where $T=1$ seconds such that, in this case, $|x[n]|^2$ is both the energy of one sample and the power over one sample duration.

If we sum all the samples for $|x[n]|^2$, we get the total energy:

$$E_T = \sum_n |x[n]|^2$$

If we average the energy of each sample over the time duration of all samples, we get the average or total power:

$$P_T = \frac{1}{N}\sum_n |x[n]|^2$$

Similarly, given a finite duration of discrete frequency domain samples as $X[k]$, the conjugate product of each sample would have both units of power and units of energy, and specifically it can be described as "<_____> Energy" given it is the power at that location in frequency over the frequency span until the next sample. The result of this is a power spectral density, as the power over some unit of BW.

Using SI units in this case, if we provide the frequency spacing of index $k$ as $B$ Hz, and $|X[k]|^2$ as power in Watts, then we would have the "<_____> Energy" in each bin as $X[k]^2/B$ Watts/Hz = Joules. I'll simplify this to the case of $B=1$ Hz, such that, in this case, $|X[k]|^2$ is both power of one sample and then energy over the frequency span of one bin.

If we sum all the samples for $|X[k]|^2$, we get the total power:

$$P_T = \sum_n |X[k]|^2$$

The total power times the total frequency span $N$ would be the total energy:

$$E_T = N\sum_n |X[k]|^2$$

Illustrating with a Simple Example:

Consider $x[n]$ as the samples for 1 cycle of a sinusoid.

$$x[n] = \cos(2\pi n/N), \space\space n = 0,1, \ldots N-1$$

If we use a scaled DFT given as:

$$X[k] = \frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{-j 2\pi n k /N}$$

The result will be $$X[k] = [0, 0.5, 0, 0, \ldots, 0, 0, 0.5]$$

The total power $P_T$ as determined from the time domain samples is the variance of a cosine which is $\sigma_x^2 = 0.5 \text{ Watts}$.

The total energy $E_T$ as determined from the time domain samples is the $P_T$ times the time duration $N$: $E_T = N/2 \text{ Joules}$.

Consistently the total power $P_T$ as determined from the frequency domain samples is given as:

$$P_T = \sum_n X[k]^2 = 0.5^2 + 0.5^2 = 0.5 \text{ Watts}$$

And the total energy $E_T$ as determined from the frequency domain samples is given as:

$$E_T = N\sum_n |X[k]|^2 = N/2 \text{ Joules}$$

My Questions

• Are my statements accurate?
  
• Are there alternate, more concise and clearer ways to explain this (ideally without continuous time analogies)?
  
• What would the best terms be for the <_____> positions above in time and frequency. In the time domain, I have seen $x[n]^2$ for each sample referred to as "instantaneous power", although it is really instantaneous energy and then the power over one sample duration (which is nearly instantaneous, and approaches the continuous time instantaneous power as $T \rightarrow 0$). I'm fine with referring to it as "instantaneous power", but then what do we refer to the frequency domain counterpart to "instantaneous", as referring to something which occurs at one frequency?

For reference, there are related posts here on Stack Exchange, some with answers from me that I may need to update if there are flaws in my thought process:

Power spectral density vs Energy spectral density

what the difference between spectral density and the power spectral density?

Inconsistency between the units of power spectral density and the definition that people often give

Is there instantaneous energy for signals? Why is $\big|x(t)\big|^2$ instantaneous power?

And on the DFT scaling used: Larger FFT vs multiple averaged FFTs for detecting small CW signals

Answer 1

This answer discusses if Joule or Watt (or any other units) are applicable to the energy and power of signals as defined in signal processing. I will argue that trying to use units for energy and power of signals in a consistent way is generally not useful or even possible. The discussion focuses on discrete signals as considered in the question, but note that most points made remain also valid for signals with a continuous argument.

Signal energy

$$E_x=\sum_{n=-\infty}^{\infty}\left|x[n]\right|^2\tag{1}$$

and signal power

$$P_x=\lim_{K\to\infty}\frac{1}{2K+1}\sum_{n=-K}^{K}\left|x[n]\right|^2\tag{2}$$

are just convenient measures for the "size" of a discrete signal. They are - if at all - just indirectly related to physical energy or power. If $x[n]$ is a voltage signal, then $E_x$ would be the energy dissipated in a load with a resistance of $1$ ohm if the voltage $x[n]$ were applied across it. Signal power $P_x$ can only be related to physical power if the index $n$ is related to time, which need not be the case at all.

Signal energy and power as defined by $(1)$ and $(2)$ are still very useful quantities when describing the "size" of signals. For example, we can judge the similarities of signals by considering the energy of their difference, or we can define signal to noise ratios by computing the ratio of the respective powers. Units are irrelevant here. The unit of $E_x$ could be anything, depending on the type of signal. Moreover, according to $(1)$ and $(2)$, the units of $E_x$ and $P_x$ would always be the same.

I don't think that trying to relate $(1)$ or $(2)$ to units is helpful in any way. We would obtain as many different units for signal energy or power as there are signal types. In my opinion, it's most useful to let units "disappear" as soon as we enter the discrete world, and let them "re-appear" (if necessary) when/if the discrete signal is converted to the analog world.

Answer 2

This is quite the question! I will write down some of my thoughts and hopefully they answer your questions and I don't make a mistake.

I would say you are right in correctly noting that "instantaneous" has to be taken with a grain of salt in the discrete case as you are instantaneous over one sample period. I would also say you are right in noting that as $T \rightarrow 1$ the "instantaneous" energy approaches the "instantaneous" power. So, the first two blanks, assuming my reading comprehension is holding up this late at night, seem to rightly be described as "instantaneous" power.

When transferring to the frequency domain, the notion of "instantaneous" doesn't make sense to me as you are accumulating values over time. You could of course take a DFT of one sample, but you would just get back out the exact same value. The DFT decomposes the signal into a set of basis functions, the complex amplitudes of which are the DFT values, so I don't see the power in a single frequency bin as a notational equivalent to the "instantaneous" power in a single time domain sample. You would likely say something like "the energy in frequency bin $k$ averaged over $N$ samples".

Another way to approach this is from a definitional standpoint of the PSD. A problem in Fourier analysis when it comes to energy vs. power lies in the fact that random signals, modeled as wide sense stationary, don't have finite energy, whereas deterministic signals do. Random signals have finite average power, and therefore can be characterized by an average power spectral density, or PSD for short. I've seen the misconception before that the energy spectral density and power spectral density are equivalent for finite duration signals, random or not, however this is not the case. The energy spectral density is defined as \begin{equation} S(\omega) = \sum_{k}\rho [k]e^{-j\omega k} \end{equation} where \begin{equation} \rho [k] = \sum_{k}x[k]x^{*}[k-n] \end{equation} The power spectral density is defined as \begin{equation} \phi(\omega) = \sum_{k}r[k]e^{-j\omega k} \end{equation} where \begin{equation} r[k] = E\lbrace x[k]x^{*}[k-n]\rbrace \end{equation} This is an important distinction because random signals have finite average power, hence the need for the expectation in order for the Fourier summation to converge.

There are two ways to compute average power. The first is expected instantaneous power. The expected instantaneous power is \begin{equation}E\lbrace x[n]x^{*}[n]\rbrace = r[0] \end{equation} Almost always an ensemble of data is not available, so the second way to obtain average power is through time averaging. So, we get the time-averaged power to be $\frac{\rho[0]}{N}$.

In order to do meaningful spectral analysis, we assume the signal being analyzed is ergodic, which means $\frac{\rho[k]}{N} = r[k]$ (for a biased estimate). So, from a theoretical standpoint, the PSD represents the expected instantaneous power as a function of frequency. However, practically speaking, most signals are not ergodic, so we get the time-averaged power as a function of frequency. This lends further evidence towards the fact that neither the power nor the energy in a given frequency bin represents an "instantaneous" value, as both the energy and power spectral densities average over time.

Hopefully that answers your third question. With respect to the second question, the definitions of power and energy in the time and frequency domains are correct, so I don't see how you could get more concise or clearer on that. I think the main issue is what do the values at each time stamp $n$ vs. each frequency bin $k$ actually mean, which I was hopefully able to clear up some in my above discussion.

As to whether or not your statements (or mine for that matter) are correct, I'll leave that up to the people to decide. Hopefully I didn't do too much rambling and actually answered some of your questions. If I didn't answer them sufficiently, let me know and I'll try to revise.

Answer 3

Summarizing what I got out of the responses and my own conclusion from this:

In signal processing we have "Energy Signals" and "Power Signals" to distinguish from signals that in their time domain cases would have either finite Energy or finite Power and have direct relationships with physical quantities where we may use units of Watts or Joules. The definitions of what is an Energy Signal and a Power Signal are well defined elsewhere explicitly (including other answers on Stack Exchange) but the main point is those definitions have no concern with what units are used. This thus leads to what total "Energy" is, in that context, for ANY discrete waveform, as given by Matt in his answer:

$$E_x=\sum_{n=-\infty}^{\infty}\left|x[n]\right|^2$$

This is the total "energy" as a word that has nothing to do with Joules, regardless if we are in time and frequency and regardless if the sample duration is finite.

For a finite duration waveform of n samples, the total Energy would be:

$$E_x=\sum_{n=0}^{N-1}\left|x[n]\right|^2$$

For a discrete frequency domain waveform, the total Energy would be:

$$E_k=\sum_{n=0}^{N-1}\left|X[k]\right|^2$$

Where this is "Signal Energy" as used in signal processing, and all the points Matt made (and Hilmar in his initial comments) about avoiding units all together for this applies (it only makes sense for a time domain waveform but it's use is not restricted to that).

---

I do have applications where I am using "signal processing" (analysis) with discrete signals and computing power spectral densities and other results where the units represented are important and do matter. Additionally using units in general can be very helpful (or necessary) in dimensional analysis. This is what led to my confusion given the naming of Energy Signals and Power Signals in looking for some dimensional consistency (as I outline in my question). I selected Matt's answer because even though I now may not necessarily like the naming choice used for "Energy Signal" and "Power Signal" - it is what it is, and if we understand the main point I summarized here and as Matt articulated - it does make sense.

Answer 4

I have a few more "philosophical" points to add. Maybe it's helpful

Signal Energy vs Physical Energy

As already stated, these are two different things and should be treated as such. Sometimes the signal energy or power can be used to estimate or approximate physical energy but that requires careful consideration. There are many cases where this would be dead wrong.

Signal power/energy is juts a bunch of numbers. These may or not be related to physical quantities and the relationship if often not trivial.

Use of units

In my experience using numbers with "assumed units" is bad practice. If you do want to model a real physical system just use the correct units and explicitly insert the appropriate impedance in the equations. Assuming you have a voltage $v[n]$ in Volts and a current $i[n]$ in Ampere you can then write the instantaneous power (maybe) as

$$p[n] = v[n] \cdot i[n] \tag{1}$$

and $p[n]$ comes out to be in Watts.

If your impedance R is real and in Ohms, than indeed you can write (maybe)

$$p[n] = \frac{v^2[n]}{R}= R \cdot i^2[n] \tag{2}$$

As long v[n] has units of a voltage (V, mV, nV, ..), i[n] has units of a current and R units of an Impedance, p[n] will have the correct units (W, mW, etc.) and the scaling will be correct. If your impedance isn't real you just have to do

$$p[n] = v[n] \cdot (v[n]*h[n]) \tag{3}$$

where h[n] is the impulse response of the admittance (inverse impedance). It's not pretty, but that's what it takes.

This approach will also force you take a good look at your units and what exactly that impedance is and whether your approximation is valid or whether you get yourself in trouble.

Being clean with units also helps you to keep track of the exact units of your "power". For example if you have a sound pressure signal x[n] in a "free enough" sound field. You can calculate

$$p[n] = \frac{x^2[n]}{\rho c} \tag{4}$$

If you bother to go through units (which you should), you'll actually find that

$$\frac{Pa^2}{kg/m^3 \cdot m/s} = \frac{N^2 \cdot m^3 \cdot s}{m^4 \cdot kg \cdot m } = \frac{kg^2 \cdot m^2 \cdot m^3 \cdot s}{s ^4 \cdot m^4 \cdot kg \cdot m } = \frac{kg}{s^3} = \frac{W}{m^2} \tag{5} $$

which is an intensity, not a power. You need to integrate over a surface to get the actual power.

Multiplying or squaring signals, Aliasing

The obvious pitfall here is that multiplications or squaring doubles the bandwidth which often results in aliasing, unless you upsample the signal first.

A trivial example would be a voltage sine wave of 15 kHz sampled at 40 kHz. If you square it in the analog domain you get the instantaneous power as

$$p(t) = \frac{1}{R}(V_0\cos(\omega_0 \cdot t)^2 = \frac{V_0^2}{2R}(1+\cos(2\omega_0 t))\tag{6}$$

The spectrum of the instantaneous power (which ironically something completely different from the power spectrum :-)) has a component at DC (average power) and component at twice the signal frequency.

If you sample the voltage at 40 kHz and just square and scale it (without up-sampling), you get the correct average power but the $2 \omega_0$ component would alias down to 10 kHz. So here the "signal power" is completely different than the "physical power".

Squaring also results in the instantaneous power always being positive. For any system that has some sort of energy storage (capacitors, inductors, masses, springs, etc.) this will generally not be the case and occasionally power and energy fill flow the other way. This just a different way of saying "the impedance is complex", but I find the physical explanation to be helpful.

Answer 5

As a physicist that works on spectrum analyzers, I struggled with this question for a long time. Eventually I came to accept that |x[n]|² is power not energy. x[n] is either a real voltage (something measured by and ADC), or a pseudo-voltage (some number in software), either "instantaneous", "average" or "integrated"(over the time of the sample). Then you can use P = V²/R to get the power. When you have real hardware ADC or DAC, the differences between thinking of it as integrated power, or average power, or just a random instantaneous sample of the power, can be calibrated out so there is not much attention paid to that. When you are in software, the differences matter even less, which may be why this does not get a lot of attention.

For an example that helps me wrap my head around things. Consider an ADC measuring a 1 V DC "signal" across a 1 Ohm resistor. If you think of the square of the samples as energy, when you double the sample rate, you get double the energy. But that can't be right, so then you have to divide by the sample period, and you get energy/time which is power.

On the other hand if you treat the samples as power, you integrate the power over time to get energy (multiply |x[n]|² by sample period).