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I was trying to find the transfer function of the following circuit, given the voltage on the inductor as output, and V as voltage input:

schematic

simulate this circuit – Schematic created using CircuitLab

and, if I'm not mistaken, it should be:

$$ G(s)=\frac{V_L(s)}{V(s)}=\frac{1}{2}\frac{s}{s+\frac{R}{2L}} $$

Then I tried to work out the output time-expression using Laplace and assuming an impulse input; in s it should be:

$$ V_L(s)=\frac{1}{2}\frac{s}{s+\frac{R}{2L}}{V(s)}=\frac{1}{2}\frac{s}{s+\frac{R}{2L}} $$

since \$V(s)=\mathcal{L}[δ(t)]=1\$. Using Laplace the output should then be:

$$ v_L(t)=\frac12\left(δ(t)-\frac{R}{2L}e^{-\frac{R}{2L}t}\right) $$

and here I'm confused: if there is an error, I can't find it, but since that should be a voltage, the part \$\frac{R}{2L}\$ confuses me, since it seems to me it messes with the unit of measurement, \$\frac{R}{2L}\$ not being a dimensionless quantity, so the exponential part doesn't look like a voltage.

What I mean in saying that the exponential part doesn't look like a voltage is this: I'm calculating a voltage, so both sides of the equivalence should be voltages. Since \$\frac{R}{2L}\$ is measured as 1/s[econds], for the exponential part to be a voltage, there should be somewhere a V ⋅ s, so that the product between it and \$\frac{R}{2L}\$ is a voltage.

Moreover, I was under the impression that \$δ(t)\$ isn't measured as 1/s[econds], but to add it to \$\frac{R}{2L}\$ it should be.

I'm not sure what I'm missing, if there is some mistake in my calculations, or if I'm missing something in analyzing the result; and since the calculation were confirmed as correct, I'm not sure how \$\frac12\left(δ(t)-\frac{R}{2L}e^{-\frac{R}{2L}t}\right)\$ is a voltage, with \$\frac{R}{2L}\$ being \$second^{-1}\$.

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    \$\begingroup\$ Why not strip the problem down to one resistor and one inductor then, rinse and repeat. Then, you’ll find dozens of examples on the internet that you can compare. You can also solve it for a step function by multiplying the Laplace formula by 1/s and the, you’ll probably find scores of answers and waveforms that give you a very sound comparison. \$\endgroup\$
    – Andy aka
    Dec 17, 2021 at 23:46
  • \$\begingroup\$ I already solved it with the step, but it didn't help with this case. And the analysis to the impulse I found are all with current as output; I know I can get the tension by deriving the current, but I was trying to understand if there is any error in my reasoning without passing via the current, and if not why that expression is a voltage. \$\endgroup\$
    – Mauro
    Dec 18, 2021 at 8:44
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    \$\begingroup\$ @Mauro Why do you think R/(2L) should be dimentionless? It's a time constant, like RC. The answer is correct, BTW. \$\endgroup\$ Dec 18, 2021 at 8:45
  • \$\begingroup\$ Because I'm calculating a voltage, so since \$\frac{R}{2L}\$ is 1/s, then somewhere there should be a V ⋅ s, otherwise the exponential wouldn't be a voltage. Moreover, I was under the impression that \$δ(t)\$ isn't measured in 1/s, so either I'm misunderstanding it, or I don't understand how can it be added to something measured as 1/s. \$\endgroup\$
    – Mauro
    Dec 18, 2021 at 10:09

4 Answers 4

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There are two important facts to realize to understand this:

  1. The units of \$\delta (t)\$ are \$seconds^{-1}\$, because of the defining property that \$\int \delta (t) dt = 1\$
  2. The units of the Laplace Transform \$V(s)\$ of the voltage \$V(t)\$ are Volt-seconds because of the definition of the Laplace transform \$L\{V\}(s)\equiv \int_0^{\infty} V(t)e^{-st}dt\$

In your expression for \$v_L(t)\$, your two terms in fact do have the same units of \$seconds^{-1}\$. So then the question becomes why don't they appear to have units of voltage?

The answer to that is that when you write your input function as \$V(s) = L[\delta(t)]=1\$ you dropped some units. The Laplace transform of a voltage impulse is not unitless, but rather has the usual units of Volt-seconds. Perhaps it would clarify if you wrote your unit impulse function as \$V(t)=W\delta(t)\$, where \$W = 1 \$ Volt-second (remember the units of \$\delta(t)\$ are \$seconds^{-1}\$). The units on \$W\$ are necessary so that \$V(t)\$ comes out in Volts. We aren't changing anything here, just being explicit about the units. When you perform the Laplace Transform of this function, you get \$W\$ Volt-seconds, as you should. We choose \$W=1\$ Volt-second to make this the ‘unit’ impulse function.

If you carry that factor of \$W\$ through your calculation, but knowing that you intend to set it equal to \$W = 1 \$ Volt-second, order is restored and your final units end up as Volts.

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  • \$\begingroup\$ I was cosidering \$\mathcal{L}[δ(t)]\$ as just volt, thanks. Just to be sure I'm understanding the general idea, what about using a step or a ramp? With \$v(t)=Eu(t)\$, \$\mathcal{L}[v(t)]=\frac{E}{s}\$, and the output would be \$v_L(t)=\mathcal{L}\left[V_L(s)=\frac12\frac{E}{s+\frac{R}{2L}}\right]=E⋅e^{-\frac{R}{2L}t}\$; am I right in thinking that E is in volt? And since applying \$\mathcal{L}[r(t)]=\frac{1}{s^2}\$ gives a coefficient \$\frac{2L}{R}\$ in the inverse transform, then the ramp is in seconds? (Which would be coherent with the impulse being the second derivative of the ramp.) \$\endgroup\$
    – Mauro
    Dec 19, 2021 at 0:37
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    \$\begingroup\$ The step is correct with regards to dimensions. The ramp as you describe is missing dimensions. The Laplace Transform of \$V(t)\$ should always have units of Volts-seconds. Doing the inverse transform should always end up taking it back to Volts. So the transform of a voltage ramp has to have a constant multiplying the \$1/s^2\$ that has dimensions Volts/seconds, since \$s\$ has units of \$seconds^{-1}\$ \$\endgroup\$
    – rpm2718
    Dec 19, 2021 at 3:21
  • \$\begingroup\$ I think I was saying the same thing, but I wasn't clear: since the ramp has to have a constant in volt/second, so \$R(s)=E/s^2\$ with E in volt/seconds, this means that - since \$v(t)=Er(t)\$ must be in volt - \$r(t)\$ is in seconds. This is also coherent with the fact that using the ramp as input in the circuit gives \$v_L(t)=E\frac{L}{R}\left(1-e^{-\frac{R}{2L}t}\right)\$, and since \$v_L(t)\$ must be in volt and \$\frac{L}{R}\$ is in seconds, E must be in volt/seconds, so again since \$v(t)=Er(t)\$ must also be in volt then \$r(t)\$ must be in seconds. Am I right? \$\endgroup\$
    – Mauro
    Dec 19, 2021 at 8:52
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    \$\begingroup\$ Yep, you got it! \$\endgroup\$
    – rpm2718
    Dec 19, 2021 at 10:27
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I'm not sure what I'm missing, if there is some mistake in my calculations, or if I'm missing something in analysing the result.

You can prove this by using an online inverse Laplace solver: -

enter image description here

And yes, your calculations are correct.

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  • \$\begingroup\$ Then how's that a voltage? As far as I understand, \$\frac{R}{2L}\$ is 1/s, so unless there is somewhere a V ⋅ s the exponential can't be a voltage. And since the exponential part is measured in 1/s, is also \$δ(t)\$ measured as 1/s? I'd say so, otherwise the addition would be off since it'd be an addition of two different dimensions, but then I don't understand why is 1/s, and how those 1/s simplifies to give a voltage. \$\endgroup\$
    – Mauro
    Dec 18, 2021 at 10:13
  • \$\begingroup\$ R/2L is is not 1/s. What voltage do you refer to? \$\endgroup\$
    – Andy aka
    Dec 18, 2021 at 10:36
  • \$\begingroup\$ Since \$\frac{L}{R}\$ is a time constant measured in seconds, the reverse \$\frac{R}{L}\$ should be 1/s, shouldn't it? If I'm not mistaken, \$\left[\frac{R}{L}\right]=\frac{Ω}{H}=\frac{V⋅A}{A⋅Wb}=\frac{V}{Wb}=\frac{V}{V⋅s}=\frac{1}{s}\$. As voltage, I'm referring to the \$v_L\$ I'm calculating: since the result is \$v_L(t)=\$, then the right member should be a voltage, and I don't understand how it can be, since - as far as I understand, \$\frac{R}{2L}\$ is 1/s. \$\endgroup\$
    – Mauro
    Dec 18, 2021 at 10:48
  • \$\begingroup\$ Sorry, maybe I understood where I wasn't clear, I'll edit the question to be sure this isn't a misunderstanding and to clarify my question: while saying \$\frac{R}{2L}\$ is 1/s I didn't mean Laplace s, I meant s as seconds. \$\endgroup\$
    – Mauro
    Dec 18, 2021 at 10:57
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    \$\begingroup\$ The term s is a variable (unless you mean s = seconds); L/R is a time constant i.e. it is fixed and invariable. You can look at the dimensions but, L/R is a fixed value whereas s (not seconds) is a Laplace variable. \$\endgroup\$
    – Andy aka
    Dec 18, 2021 at 11:20
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Well, the transfer function of the circuit is given by:

$$\mathscr{H}\left(\text{s}\right):=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}=\frac{\text{R}\space\text{||}\space\text{sL}}{\text{R}+\left(\text{R}\space\text{||}\space\text{sL}\right)}=\frac{\text{sL}}{\text{R}+2\text{sL}}\tag1$$

Where \$\alpha\space\text{||}\space\beta=\frac{\alpha\beta}{\alpha+\beta}\$ and I used the well-known Laplace transform.

The impulse response of a system is given by the output response when a the input function is a dirac delta function, so:

$$\text{V}_\text{i}\left(\text{s}\right)=\mathscr{L}_t\left[\delta\left(t\right)\right]_{\left(\text{s}\right)}=1\tag2$$

The Laplace transform can be found using the definition of the Laplace transform or by examining a table of selected Laplace transforms.

Now, the output time response of this given by:

\begin{equation} \begin{split} \text{v}_\text{o}\left(t\right)&=\mathscr{L}_\text{s}^{-1}\left[\underbrace{\text{V}_\text{i}\left(\text{s}\right)}_{=\space1}\cdot\frac{\text{sL}}{\text{R}+2\text{sL}}\right]_{\left(t\right)}\\ \\ &=\mathscr{L}_\text{s}^{-1}\left[\frac{\text{sL}}{\text{R}+2\text{sL}}\right]_{\left(t\right)}\\ \\ &=-\frac{\text{R}}{4\text{L}}\cdot\exp\left(-\frac{\text{R}t}{2\text{L}}\right) \end{split}\tag3 \end{equation}


We can find the general output by using the convolution property of the Laplace transform:

\begin{equation} \begin{split} \text{v}_\text{o}\left(t\right)&=\mathscr{L}_\text{s}^{-1}\left[\text{V}_\text{i}\left(\text{s}\right)\cdot\frac{\text{sL}}{\text{R}+2\text{sL}}\right]_{\left(t\right)}\\ \\ &=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\text{V}_\text{i}\left(\text{s}\right)\right]_{\left(\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\text{sL}}{\text{R}+2\text{sL}}\right]_{\left(t-\tau\right)}\space\text{d}\tau\\ \\ &=\int_0^t\text{v}_\text{i}\left(\tau\right)\cdot\left(-\frac{\text{R}}{4\text{L}}\cdot\exp\left(-\frac{\text{R}\left(t-\tau\right)}{2\text{L}}\right)\right)\space\text{d}\tau\\ \\ &=\frac{\text{R}}{4\text{L}}\int_t^0\text{v}_\text{i}\left(\tau\right)\cdot\exp\left(\frac{\text{R}\left(\tau-t\right)}{2\text{L}}\right)\space\text{d}\tau \end{split}\tag4 \end{equation}

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  • \$\begingroup\$ The impulse response will have a \$\delta(t)\$ due to the \$s\$ in the numerator (which will need a partial fraction expansion, first). (wxMaxima) \$\endgroup\$ Dec 19, 2021 at 14:20
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I think the mistake is in your first expression. Here is the correct one.

Transfer function

The inductor is in parallel with one of the resistors. Using Maple to simplify.

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    \$\begingroup\$ It should just be written in another form: by factoring out \$2L\$ in the denominator, you get my expression: \$G(s)=\frac{sL}{2sL+R}=\frac{1}{2L}\frac{sL}{s+\frac{R}{2L}}=\frac{1}{2}\frac{s}{s+\frac{R}{2L}}\$. \$\endgroup\$
    – Mauro
    Dec 18, 2021 at 12:43

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