I'm not quite sure how keep track of physical units when differentiaion is involved. I find it important to keep track of units because I want to be able to check the units afterwards.

I will discuss a specific case, to illustrate my confusion with regards to the units.

Consider an RC filter low pass filter. Its transfer function is given by $$ H(\omega) = \frac{1}{1+j\frac{\omega}{\omega_{RC}}}\ , $$ with $\omega_{RC} = 1/RC$.

Say, we want to know its step response function $s = s(t)$; i.e. the voltage $V_{out}(t)$ that we measure over the capacitor if we hook up a Heaviside unit input voltage $V_{in}(t) = \theta(t) = \begin{cases} 0 & \quad \text{if } \quad t \leq 0\\ 1 & \quad \text{if }\quad t >0 \ . \end{cases}$

It can be shown that $s$ is given by

$$ s(t) = RC(1-e^{-t/RC} \theta(t))\ . $$

Now, since $s(t)$ is in units volts, I would suspect that the terms $1$ and $e^{-t/RC} \theta(t)$ are in units volts per time because $RC$ is in units time.

In another case, say, we want to know the impulse response function $h = h(t)$; i.e. the voltage $V_{out}$ that we measure over the capacitor if we hook up an impulse input voltage $V_{in}(t) = \delta(t)$, where $\delta$ is the Dirac delta function.

It can be shown that $h$ is given by the derivative of the step response $s$:

\begin{align} h(t) &= \frac{d}{dt} s(t) \\ &= e^{-t/RC}\big(1-RC\delta(t)\big) \ . \end{align}

$RC$ has units time.

Both $s$ and $h$ are in units volts.

How on earth would I be able to use dimensional analysis in this case to check my results?

In the last expression $\delta(t)$ should now be in units volts per time for the equation to hold.

  • 2
    $\begingroup$ Why are you assuming that $s(t)$ has units of volts? It doesn't, but if you explain why you think that then it will be easier to provide a useful answer. $\endgroup$ Commented Jun 24, 2018 at 18:45
  • $\begingroup$ @EmilioPisanty Because the output signal of the system is measured in voltage. If I'm not mistaken, $s$ represents the voltage that we're measuring given a Heaviside step function as the input voltage. $\endgroup$ Commented Jun 24, 2018 at 19:31

2 Answers 2


Your understanding of the step response function is pretty flawed, though it's hard to pin down exactly how as you've not given a formal definition for what you're using.

To start with, the formula you've given is completely wrong, as it produces a discontinuous output, i.e. if you plot $(1-e^{-t}\theta(t))$ what you get is this:

Mathematica graphics

This makes it pretty clear that the correct form for the response function should be $$ s(t) = \mathrm{const}\times (1-e^{-t/RC})\theta(t), $$ with the theta function as a global multiplier to make everything vanish before the voltage ramps up.

Now, as to the value of the constant, that's really up to how you define it, but there's really no reasonable understanding under which that constant should equal $RC$. You say that "it can be shown" that that is the correct form, but you don't give any details that would enable us to help you pinpoint where the mistake is inside your derivation, but it's simply not a workable relation.

Instead, you have two different options for the normalization of the step response function

  • One is to have a dimensional forcing, $V(t) = V_0 \theta(t)$, with $V_0$ a constant with dimensions of voltage, in which case the response function is $$ s_\mathrm{dim} (t) = V_0 (1-e^{-t/RC})\theta(t) $$ and has the dimensions of voltage.

  • The other one is to have a dimensionless forcing, $V(t) = \theta(t)$, with a unit (and therefore dimensionless) driving term at $t>0$, which will produce a response function of the form $$ s_\mathrm{d.less} (t) = (1-e^{-t/RC})\theta(t) $$ which, like the driving, is dimensionless.

Generally speaking, since the response function is best thought of as a property of the system, to be used in a convolution (or other similar mathematical device) with the actual driving voltage applied to the system, it is normally better to use the second understanding, and to include the units of volts on the actual physical driver and the actual physical response, while leaving the system's response unit-neutral.

Your second question is a bit trickier, because there the natural "unit-neutral" driver isn't actually dimensionless: that is, $$ V(t) = \frac{\mathrm d}{\mathrm dt} \theta(t) = \delta(t) $$ has dimensions of $[V(t)] = [1/T]$, i.e. inverse time. As you correctly point out, the impulse response function is given by the derivative of the step response function, so therefore it reads \begin{align} s_\mathrm{impulse} (t) & = \frac{\mathrm d}{\mathrm dt} s_\mathrm{step} (t) \\ & = \frac{\mathrm d}{\mathrm dt} (1-e^{-t/RC})\theta(t) \\ & = (1-e^{-t/RC})\delta(t) + \frac{1}{RC}e^{-t/RC}\theta(t) \\ & = \frac{1}{RC}e^{-t/RC}\theta(t), \end{align} where the delta-function term vanishes because it only cares about the behaviour at $t=0$, i.e. it pulls in the behaviour $$ f(t) \delta(t) = f(0)\delta(t), $$ which is obvious from the definition of the delta function, and in this case $f(t) = 1-e^{-t/RC}$ has $f(0)=0$.

Dimensionally, here both $e^{-t/RC}$ and $\theta(t)$ are dimensionless, giving $s(t)$ the dimensions of $[s(t)] = [1/RC] = [1/RC]$, i.e. an inverse time, which, as it should, matches the driver that produces it.

And finally, just to emphasize this: the number $1$ is always dimensionless, as is $\theta(t)$. If your formalism indicates that that's not the case, then you're doing things wrong: there's no such thing as "one, but with units of volts", there's only "you had a constant $V_0 = 1\:\rm V$ and you were sloppy with your constants", which you should never do.

  • $\begingroup$ I'm aware that the delta-function term vanishes, but the units of the two terms $\frac{1}{RC} e^{-t/RC} \theta(t)$ and $(1-e^{-t/RC})\delta(t)$ should still agree. The second term explicitly has units $[1/T]$. Hence, I assume the unit $[1/T]$ is implicit in the first term. $\endgroup$ Commented Jun 25, 2018 at 5:47
  • $\begingroup$ No, it isn't (and, again, there is no such thing as an "implicit" unit, what you're describing is better known as a mistake in the algebra). In this case, the unit is carried by the delta function, which, as noted above, has dimensions of inverse time. $\endgroup$ Commented Jun 25, 2018 at 10:15
  • $\begingroup$ If the unit is carried by the delta function then it would make sense to keep explicit track of the units in brackets; i.e. $V(t) = \frac{\operatorname d}{\operatorname{d} t} \theta(t) = \delta(t) (\text{per unit time})$, just for the sake of clarity and keeping track of the units. $\endgroup$ Commented Jun 25, 2018 at 11:17
  • $\begingroup$ Also, to emphasize why $V(t)$ has units $[1/T]$ instead of volts, we define the impulse input $\delta(t)$ --- of which $V(t)$ is the response --- as a unit-less quantity. If that's not the case, then I really don't know what we're assuming from the start. $\endgroup$ Commented Jun 25, 2018 at 11:31
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    $\begingroup$ The Dirac delta always has units inverse to those of its argument, as a natural consequence of its definition, since $$\int \delta(t) \mathrm dt = 1$$ requires $[\delta(t)\mathrm dt] = [\delta(t) t] = 1$ and therefore $[\delta(t)]=[1/t]$. The units of the Dirac delta are already kept track of using unambiguous notation, and there is no need of additional markers; you can use them if you want for your personal comfort but demanding that the community add redundant notation just so you can use it as a crutch is not going to make much headway. $\endgroup$ Commented Jun 25, 2018 at 18:11

If $f(x)$ has units of $X$, and $x$ units of $Y$, then $\frac{\delta f}{ \delta x}$ will have units of $X/Y$.

There seems to be bit of a confusion concerning $s(t)$ in your example. The characteristic equation of the system is $s(t) = -(RC)^{-1}$, and therefore, $s(t)$ is in units of time.

There is a nice tutorial explaining how to obtain the voltage step-response of the system here.

  • $\begingroup$ It is worth using a definition of differentiation to show that this must be the case. $\endgroup$ Commented Jun 24, 2018 at 19:18
  • $\begingroup$ Yes, this is what I assumed to be true. The differentiation operator $\frac{d}{dx}$ gives rise to a unit $x$. I get confused when we, for instance, have $f = f(x/c)$, with $c$ in the same "unit" as $x$. In that case we explicitly see the unit coming out of the differentiation operation: $\frac{df}{dx} = \frac{1}{c} \frac{df}{d(x/c)}$. Hence, my confusion as stated in the post. When invoking the chain rule on $\frac{d}{dt}\bigg( e^{-t/RC} \theta(t) \bigg)$ the first factor gives rise to $\frac{1}{RC}$, the second factor does not give rise to such unit constant; it is implicit. $\endgroup$ Commented Jun 24, 2018 at 19:45

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