I was trying to convert the units of the following current in Time Domain and Complex Frequency domain (s-domain): $$ i(t)=20\cdot e^{-t}H(t)\,\,\,\,\,\text{[mA,ms]} $$ Where \$\text{[mA, ms]}\$ indicates that in this formula time must be entered in \$\text{ms}\$ and the calculated current will be in \$\text{mA}\$.

Time Domain: $$ \bbox[8px,border:1px solid black] { i_1(t)=20\cdot e^{-t}H(t)\,\,\,\text{[mA,ms]}\,\, \require{AMScd} \begin{CD} @>>> \end{CD} \,\, i_2(t)=0.02 \cdot e^{-1000t}H(1000t)\,\,\,\text{[A,s]} } $$

If, for example, \$t=1\,\text{ms}=0.001\,\text{s}\$, we have: $$ \begin{alignat}{1} i_1(1)&=20\cdot e^{-1}H(1)=\frac{20}{e}\,\text{mA} &= \frac{0.02}{e}\,\text{A} \\[6pt] i_2(0.001)&=0.02\cdot e^{-1000\,\cdot\, 0.001}H(1)\,\,\,\,\,\,\,\,\,\, &=\frac{0.02}{e}\,\text{A}\,\,\,\,\,\,✔️ \end{alignat} $$ Here no big deal, trivial conversion. The problem is in s-domain.

S-Domain: $$ \bbox[4px,border:1px solid black] { \begin{alignat}{1} I_1(s)=\mathscr{L}[\,i_1(t)\,]=\frac{20}{s+1} \,\,\,\left[\text{mA,}\frac{1}{\text{ms}}\right] \\[6pt] I_2(s)=\mathscr{L}[\,i_2(t)\,]=\frac{0.02}{s+1000}\,\,\,\left[\text{A,}\frac{1}{\text{s}}\right] \end{alignat} } $$

If, for example, \$s=(e-1)\,\frac{1}{\text{ms}}=1000 \cdot (e-1) \,\frac{1}{\text{s}}\$, we have: $$ \begin{alignat}{1} I_1(e-1)&=\frac{20}{e-1+1}=\frac{20}{e}\,\text{mA} &= \frac{0.02}{e}\,\text{A} \\[4pt] I_2\left(1000 \cdot (e-1)\right) &=\frac{0.02}{1000e-1000+1000}\,\,\,\,\,\,\,\,\,\, &=\frac{0.02}{1000e}\,\text{A}\,\,\,\,❌ \end{alignat} $$

Here the conversion didn't work. Results differ by a factor of \$10^{-3}\$ and I can't figure out why. The transforms and operations are right, I did them in Mathematica to make sure they weren't wrong:

Wolfram Mathematica Operations

A note about the units of the complex frequency \$ s \$ - the product \$s \cdot t\$ must be dimensionless, so:

  • Since the time in \$i_1(t)\$ is measured in \$\text{milliseconds}\$, the complex frequency \$s\$ in \$I_1(s)\$ must be in \$\frac{1}{\text{milliseconds}}\$;
  • Similarly, \$s\$ in \$I_2(s)\$ is in \$\frac{1}{\text{seconds}}\$ because \$t\$ in \$i_2(t)\$ is measured in \$\text{seconds}\$.
  • \$\begingroup\$ What is [mA, ms]? \$e^{-t}\$ and H(t) are unitless. \$i(t)\$ is a current so the unit is A or mA, etc. \$\endgroup\$
    – Chu
    Commented Sep 3, 2019 at 8:50
  • \$\begingroup\$ @Chu \$\text{[mA, ms]}\$ indicates that in this formula time must be entered in \$\text{ms}\$ and the calculated current will be in \$\text{mA}\$ \$\endgroup\$ Commented Sep 3, 2019 at 10:13
  • 1
    \$\begingroup\$ OK, I've not seen that notation previously. Now, what is \$s=(e-1)\frac{1}{ms}\$ and why are you doing it? \$\endgroup\$
    – Chu
    Commented Sep 3, 2019 at 10:52
  • 1
    \$\begingroup\$ s =(e-1)1/ms doesn't make sense cz it belongs to another domain. Not time domain to denote it in ms or s. \$\endgroup\$
    – Mitu Raj
    Commented Sep 3, 2019 at 11:25
  • 1
    \$\begingroup\$ I think I get your question now... Will try to put an answer \$\endgroup\$
    – Mitu Raj
    Commented Sep 3, 2019 at 16:56

3 Answers 3


Nothing to do with Laplace. Just some mathematics. I found that you have dimensional inconsistency while you converted units. For instance in this equation :- $$I_1(s)=\frac{20}{s+1}$$ The dimension of current is milliAmperes, hence the whole RHS should evaluate to 'milliAmperes' units as well.

\$s\$ has units 1/ms as you explained. Okay, so in the term \$(s+1)\$, '1' should be also be representing a quantity of units 1/ms, otherwise we cannot add them dimensionally. ie., \$(s+1)\$ has the units 1/ms.

Which means, whatever the numerator '20' is representing, is not of units milliAmperes as you assumed, but of units 'milliAmperes per millisecond' or mA/ms. Otherwise the whole RHS becomes dimensionally incorrect.

So when you evaluate the expression for \$I_2(s)\$ , where \$s\$ is in terms of 1/s now, you not only have to multiply each term in the denominator by 1000 (which you have done correctly), but also have to multiply the numerator by 1000 because of the 'per millisecond' thing which you missed out there. And also to convert milliAmperes to Amperes, divide the numerator by 1000 as well. ie.,

$$I_2(s) = \frac{(\frac{20}{1000}.1000)}{1000e} A = \frac{0.02}{e}A$$

I hope this clears why your answer differs by a factor of \$10^-3\$

  • \$\begingroup\$ I think you have inverted things. The frequency unit of \$I_1(s)\$ is \$\frac{1}{\text{ms}}\$ and of \$I_2(s)\$ is \$\frac{1}{\text{s}}\$. It seems that this make the dimensional inconsistency disappear. \$\endgroup\$ Commented Sep 3, 2019 at 17:31
  • 1
    \$\begingroup\$ Either way, you missed out 'per second' or 'per millisecond'. \$\endgroup\$
    – Mitu Raj
    Commented Sep 3, 2019 at 17:35
  • 1
    \$\begingroup\$ Corrected the 'inverted' thing to avoid confusion. \$\endgroup\$
    – Mitu Raj
    Commented Sep 3, 2019 at 17:55
  • \$\begingroup\$ What you said makes sense to me. But if I anti-transform this new \$I_2(s)\$, now the problem will manifest itself into the time domain: $$ i_2(t)=20\cdot e^{-1000t}H(1000t) \rightarrow i_2(0.001)=\frac{20}{e}\,\text{A} $$ \$\endgroup\$ Commented Sep 3, 2019 at 18:21
  • 2
    \$\begingroup\$ No. your time domain is right as in the question. the exponent of e which is of the form (t or t/1) should be dimensionless hence your calculations are correct there. \$\endgroup\$
    – Mitu Raj
    Commented Sep 3, 2019 at 18:24

In signal theory, the Laplace Transformation is linear with respect to units like voltage and current. If you have a function/signal like f(kI, t) and I is the input current level, the LT is of the form kF(I,s), k being a constant, I is the current, t is time, s is complex frequency.

But the Laplace Transformation of a function of the form f(t) is not linear with respect to the time unit. It means that in general L{f(I,pt)} does not equal qF(I,s), p and q being constants. There are special formulas for LTs, e.g. L{f(I,ct)} = 1/c (F(I,(s/c))), c >0. With this last formula the result will be the same.

  • \$\begingroup\$ "With this last formula the result will be the same." I tried to go that way before, but it didn't work either, look: $$i_1(t)=20\cdot e^{-t}H(t)\,\,\,\text{[mA,ms]}\,\, \\i_1(1000t)=20\cdot e^{-1000t}H(1000t)\,\,\,\ \text{[} \color{red}{mA}{,s]} \\i_2 = i_1(1000t) \cdot 10^{-3}\,\,\,\ \text{[} \color{red}{A}{,s]} \rightarrow I_2(s)=\frac{1}{1000}\frac{20}{\frac{s}{1000}+1}\cdot 10^{-3}=\frac{0.02}{s+1000}\,\,\,\left[\text{A,}\frac{1}{\text{s}}\right] $$ The current values given by \$I_1(s)\$ and \$I_2(s)\$ will still be different, just as happened in the question example. \$\endgroup\$ Commented Sep 3, 2019 at 16:55
  • \$\begingroup\$ You have a product of 2 functions in the time domain, f(t) is the product 20 exp(-t) H(t). If H(t) is not changed into H(ct), the calculation is easy since H(t) can be ignored in many cases. But in this case here you have a product of 2 functions that needs to be transformed. How do you transform a product of 2 functions in the time domain? Maybe convolution in the frequency domain rings the bell. \$\endgroup\$
    – xeeka
    Commented Sep 3, 2019 at 19:41

I decided to investigate how changes in time domain affect the s-domain to see if I could find something that would help to solve my question. And I found something! Let's start with \$i_1(t)\$ and investigate the values in both domains that produce the same current:

$$ \bbox[8px,border:1px solid black] { i_1(t)=20\cdot e^{-t}H(t)\,\,\,\text{[mA,ms]}\,\, \require{AMScd} \begin{CD} @>>> \end{CD} \,\, I_1(s)=\frac{20}{s+1} \,\,\,\left[\text{mA,}\frac{1}{\text{ms}}\right] } $$

$$ \newcommand{\tl}[3]{#1 & #2 & #3 \\\hline} \newcommand{\bs}[0]{\,\,\,\,\,} \begin{array}{|c|c|c|} \hline \rule[0]{0pt}{3ex} \rule[0]{0pt}{-1.5ex} \tl {\color{green}{\mathrm{\bs t\,[ms] \bs}}} {\color{green}{\mathrm{\bs s\,[1/ms] \bs}}} {\color{green}{\mathrm{\bs i_1(t)=I_1(s)\,\,[mA] \bs}}} \tl {1} {e-1} {20/e} \tl {2} {e^2-1} {20/e^2} \tl {3} {e^3-1} {20/e^3} \tl {4} {e^4-1} {20/e^4} \tl {5} {e^5-1} {20/e^5} \end{array} $$

At this point we can conclude that: \$I_1(e^t-1)=i_1(t)\$. Doing the same for \$i_2(t)\$:

$$ \bbox[8px,border:1px solid black] { i_2(t)=0.02 \cdot e^{-1000t}H(1000t)\,\,\,\text{[A,s]}\,\, \require{AMScd} \begin{CD} @>>> \end{CD} \,\,I_2(s)=\frac{0.02}{s+1000}\,\,\,\left[\text{A,}\frac{1}{\text{s}}\right] } $$

$$ \newcommand{\tl}[3]{#1 & #2 & #3 \\\hline} \newcommand{\bs}[0]{\,\,\,\,\,} \begin{array}{|c|c|c|} \hline \rule[0]{0pt}{3ex} \rule[0]{0pt}{-1.5ex} \tl {\color{green}{\mathrm{\bs \,\, t\,\,[s] \,\, \bs}}} {\color{green}{\mathrm{\bs \,\, s\,[1/s] \,\, \bs}}} {\color{green}{\mathrm{\bs \,\,\, i_1(t)=I_1(s)\,\,[A] \,\,\, \bs}}} \tl {0.001} {e-1000} {0.02/e} \tl {0.002} {e^2-1000} {0.02/e^2} \tl {0.003} {e^3-1000} {0.02/e^3} \tl {0.004} {e^4-1000} {0.02/e^4} \tl {0.005} {e^5-1000} {0.02/e^5} \end{array} $$ From the table, we can conclude that: \$I_2(e^{1000t}-1000)=i_2(t)\$. Now we can realize something important that @xeeka said in his answer:

But the Laplace Transformation of a function of the form f(t) is not linear with respect to the time unit.

Summing it all up in one diagram:

$$ \require{AMScd} \begin{CD} i_1(t) @>\mathrm{\bs s-domain \bs}>> I_1(e^t-1) \\ @V \text{linear} V V @VV \text{nonlinear}V\\ i_2(t) @>\mathrm{\bs s-domain \bs}>> I_2(e^{1000t}-1000) \end{CD} $$

For \$t=1 \,\mathrm{ms}\$, we have:

$$ \require{AMScd} \begin{CD} i_1(1) @>\mathrm{\bs s-domain \bs}>> I_1(e-1) \\ @V \text{linear} V V @VV \text{nonlinear}V\\ i_2(0.001) @>\mathrm{\bs s-domain \bs}>> I_2(e-1000) \end{CD} $$

So, now everything is clear. I erroneously assumed the linearity of the complex frequency. When I calculated \$I_2\left(1000 \cdot (e-1)\right)\$ I wasn't calculating what I was expecting.

Instead, the value obtained there corresponds to \$t \approx0.00790776 \,\mathrm{s}\$ in time domain (i.e. \$0.02/1000e = i_2(0.00790776)\$) and not \$t=0.001\,\mathrm{s}\$. Finally, calculating with the correct input:

$$ I_2\left(e-1000\right) =\frac{0.02}{e-1000+1000}=\frac{0.02}{e}\,\text{A}\,\,\,\,✔️ $$


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