I'm posting here because I don't understand why when we have a signal, let's say a voltage, the units of its sampled values are Volts multiplied by Hertz and not just only Volts? It's something I've seen in my course and this is what our teacher taught us but it's doesn't make sense to me.

He also taught us then the unit of the Z-Transform of this sampled signal is in Volts (whereas to me it would have been more logical to have V.Hz for this one).

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    $\begingroup$ I never seen VHz being used as time-domain unit. $\endgroup$ Aug 27, 2022 at 8:17
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    $\begingroup$ Could be a slew rate, i.e. $V/s$ but that's admittedly a bit of a stretch $\endgroup$
    – Hilmar
    Aug 27, 2022 at 12:58
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    $\begingroup$ @Hilmar if we start trying interpreting it as that, isn't that a slippery slope (pun intended)? $\endgroup$ Aug 28, 2022 at 7:01

3 Answers 3


Referring to SI units.

Considering the mathematical representation of ideal sampling process $$ x[n] = x(nT_s) \tag{1}$$

then the unit of the samples is the unit of $x(t)$. It will be Volt if the unit of $x(t)$ is so. Furthermore, the units of DTFT $X(e^{j\omega})$ will also be Volt as indicated by the sum $$ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n] ~e^{-j\omega n} \tag{2} $$

Note that the exponential function, as well as all trigonometric and logarithm functions, returns a unitless number. So the sum in Eq.2 has units derived from unit of $x[n]$ which is Volts.

Note that the discrete-time frequency $\omega$ is a unitless value, as indicated by its normalisation relation to the physical-frequency: $$\omega = \Omega ~ T_s \tag{3}$$

Where the physical angular frequency $\Omega$ has a unit of radians per second, and the period $T_s$ has a unit of seconds; thereby yielding a unitless number $\omega$. (it's customary to call $\omega$ as radians per sample, but that's not a physical unit; and physically both radians and samples are unitless.)

And this is consistent with the inverse DTFT relation: $$ x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{j\omega}) e^{j\omega n} d\omega \tag{4} $$
where the units of $x[n]$, and $X(e^{j\omega n})$ are both Volts respectively; and $\omega$ is unitless.

Note also that if the unit of $x(t)$ is Volts, then the unit of its CTFT is Volt.Sec or V/Hz : $$X_c(\Omega) = \int_{-\infty}^{\infty} x(t) e^{ - j \Omega t} dt \tag{5}$$ And from this, you can again deduce the unit of DTFT as Volts from its relation to CTFT : $$X(e^{j\omega }) = \frac{1}{T_s} \sum_k X_c(\frac{\omega - 2\pi k}{T_s}) \tag{6} $$

One can follow a mistaken approach to deduce the unit of samples $x[n]$. For example, a reasoning based on impulse train modulation followed by integration will yield:

$$ x_s(t) = x(t) \delta_T(t) = x(t) \sum_k \delta(t - k T_S) \tag{7} $$


$$x[n] = \int_{nT_s - \epsilon}^{nT_s + \epsilon} x_s(t) dt \tag{8} $$

Now, if we assume a unit of Volts for $x_s(t)$ then we will conclude that unit of $x[n]$ based Eq.8 is Volts.Sec. The mistake is on the unit of $x_s(t)$ which should be Volt.Hz, since the time-domain impulse function has a unit of 1/s (Hz), as one might mistakenly assume it being a unitless function.

This can be easily seen by the limiting definitin of the time-domain unit impulse function as follows:

$$ \delta(t) = \lim_{\Delta \to 0} \delta_{\Delta}(t) \tag{9} $$ where the generating function $\delta_{\Delta}(t)$ is defined as

$$ \delta_{\Delta}(t) = \begin{cases} { \frac{1}{\Delta} ~~~,~~~ 0 < t < \Delta \\~0~~~~ , ~~~\text{otherwise} }\end{cases} \tag{10} $$

From Eq.10 it can be seen that since the unit of $\Delta$ is seconds, then the unit of $\delta_{\Delta}(t)$ and of $\delta(t)$ will be 1/s or Hz...

Then the unit of $x_s(t)$ of Eq.7 can be seen to be Volt.Hz and then the unit of $x[n]$ from Eq.8 will again be Volts as expected, and as also indicated by Eq.1.

NOTE if your author refers to $x_s(t)$ of Eq.7 as the sampled signal, then he is right: its unit is indeed Volt.Hz, as shown. However, unit of the samples (the sampled signal?) $x[n]$ is Volt.

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    $\begingroup$ oooh, the last paragraph is especially enlightening! Never looked at it that way. $\endgroup$ Aug 27, 2022 at 20:56
  • $\begingroup$ By Equation 10, do you mean that the unit of any rectangular signal is Hz? I don't see how the unit of $\Delta$ in $\frac{1}{\Delta}$ must be second. $\endgroup$
    – AlexTP
    Aug 27, 2022 at 21:03
  • $\begingroup$ @AlexTP for a time-domain impulse, the unit of $t$ is time (seconds). Since $\Delta$ in Eq.10 is a duration of time, its unit will be s. Then since the amplitude of the pulse function is $1/ \Delta$, its unit will be 1/s. Note that if $t$ didn't have any units then $\delta(t)$ will be unitless as indicated by a discrete-frequency impulse $\delta(\omega)$. However when $\delta(.)$ refers to a physical context, then it will have a unit. For example $\delta(x)$ will have a unit of 1/m if x refers to length. $\delta(\Omega)$ has a unit of sec for $\Omega$ has a unit of 1/s so on... $\endgroup$
    – Fat32
    Aug 27, 2022 at 21:49
  • $\begingroup$ @Fat32 Sorry I still don't get it. IMO, the definition of $\delta_{\Delta}(t)$ is exactly the same as the one of any rectangular signal whose unit can be volt, ampere, joule, etc. (let $x(t) = 0.5$ ampere within $t=0$ and $t=2$; or $y(t)=10$ volt within $0<t<0.1$second. These are possibly real-world signals, right?). Yes, you can define $\delta(t)$ to have unit Hz, but it is your definition and even if it seems compatible with several calculations, the unit cannot be justified by the definition of $\delta_{\Delta}(t)$. $\endgroup$
    – AlexTP
    Aug 27, 2022 at 22:22
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    $\begingroup$ Thanks a lot for your answer. It really helped me. I wasn't able to thank you before because I wasn't allowed to comment. $\endgroup$
    – Chizu
    Jun 19, 2023 at 8:06

I somewhat disagree with Marcus' answer but that may be just a philosophical difference.

As long as your work is not just pure but but related to an actual physical phenomenon, using units properly can be quite useful.

Real world things are typically defined as "physicals quantities". Physical quantities are the product of a number and a unit. For example, a voltage is a physical quantity and we can express this as $U_1 = 1000\,\text{mV}$ or $U_1 = 1\,\text{V}$. Numbers and units can be different, but as long as the product is the same, the physical quantities are the same.

If you properly propagate units through all your calculations, the final result will also have the correct units. That can be a somewhat tedious process, so many people just strip off all units at the beginning and re-attach them to the result. However, working with units can be quite helpful as a sanity check: if you find yourself having to take the sine or the logarithm of a quantity that has units, you've done something wrong.

As an example NASA lost a $300+ million Mars probe because they were sloppy with the units.

You can propagate units in both discrete and continuous domains, but there are differences. Sampling per se doesn't change the units. A time domain voltage has units of "Volts" regardless of whether it's continuous or discrete. However discrete Fourier transform uses sums, whereas the continuous Fourier Transforms use integrals (which do change units). A discrete Voltage spectrum has units of $\text{V}$, a continuous one has units of $\text{V}/\text{Hz}$. Similar differences exist for transfer functions and impulse responses, but the rules are well defined and everything is consistent.

when we have a signal, let's say a voltage, why the units of its sampled version are Volts multiplied by Hertz and not just only Volts?

That's just plain wrong. The sampled signal has the same units as the continuous signal. $\text{V}\cdot \text{Hz}$ is pretty unusual to start with. Can you please provide a reference on the exact context this is being used?

  • $\begingroup$ I think we might only disagree on the "sharpness" with which I call it an interpretation, I think (not sure though). Anyway, nice answer, +1. $\endgroup$ Aug 27, 2022 at 20:54

So, first of all, samples are unitless; they are numbers. It's the ability and need to attach a physical meaning to them that makes us "lug around" physical units with them. The mathematical functions applied to samples can't deal with units – what's the sine of a volt?

it's really just an interpretation. I never seen your teacher's interpretation, and it strikes me as a bit unusual, as that would be a unit I'd expect from something in the frequency domain, not the time domain. But maybe there's an application context in which it makes sense!

The statement about the z-transform seems consistent with his interpretation though. If we take a summation in time direction as the discrete-time equivalent to an integral, we get a unit factor of time in there, cancelling the Hertz. This highlights how much this is interpretation, not strict math per se: If I add distances, say $\sum d_i= 1\text{ m} + 1\text{ m}$, I get 2m, not 2m·Hz, unless I give magical properties to the $\sum$ symbol, which it really doesn't have.

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    $\begingroup$ I agree with this answer. Samples are unitless, and so are discrete-time signals. The units matter only when the aforementioned mathematical elements are used to reconstruct continuous-time signals that represent physical phenomena. $\endgroup$
    – AlexTP
    Aug 27, 2022 at 21:06

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