What's the Laplace transform of an independent DC voltage or a current source?

I came across this while reading transients from a book. While solving a first order circuit in Laplace domain, it took the Laplace of a DC voltage source as V/s. I am not sure how it worked that out and there is not an explanation either.

Any help on this will be greatly appreciated as this blocking my progress on transients.

  • 1
    \$\begingroup\$ Think of a constant voltage source as a step signal jumping from 0 to V at time t=0. Then look at your favorite Laplace transforms table. \$\endgroup\$
    – Eugene Sh.
    Sep 10, 2021 at 18:56
  • \$\begingroup\$ That was my original thought too. But step starts to exist from t>0 while a DC source has always existed, I mean even for t<0 too. So I thought Laplace of a DC source would be zero considering u(-t) is for t<0 and u(t) is for t>0 and their Laplace transforms would cancel each other. \$\endgroup\$
    – User12121
    Sep 10, 2021 at 19:04
  • \$\begingroup\$ Usually t=0 is the moment the circuit is turned on, an usually Laplace transform is used on functions defined for t>=0 \$\endgroup\$
    – Eugene Sh.
    Sep 10, 2021 at 19:06

1 Answer 1


It depends on how you define the Laplace Transform. In the two-sided transform:

$$\mathcal L\{f(t)\} = \int_{-\infty}^{\infty} e^{-st}f(t)dt$$

a DC source becomes a delta function:

$$\mathcal L\{V\} = V\cdot \mathcal L\{1\} = V\cdot \delta(s)$$

because DC is the zero-frequency component of the signal, just like in the Fourier transform.

But the Laplace transform is usually used for stability analysis and control theory. And in those domains, the two-sided Laplace transform describes acausal systems -- systems that respond to a stimulus before that stimulus actually happens. This is nonphysical. So the one-sided transform is used instead:

$$\mathcal L\{f(t)\} = \int_{0}^{\infty} e^{-st}f(t) dt$$

This is equivalent to multiplying your time domain function \$f(t)\$ by a step function (usually written \$u(t)\$). And just like with the Fourier transform:

$$\mathcal L\{V\cdot u(t)\} = V\cdot\mathcal L\{u(t)\} = \frac V s$$

So what \$V/s\$ is really describing is a DC voltage source that's turned on at \$t=0\$.

  • \$\begingroup\$ This is really a satisfying explanation. Thanks a lot !!!! \$\endgroup\$
    – User12121
    Sep 22, 2021 at 16:43
  • \$\begingroup\$ Thanks! If this answered your question you can accept the answer by clicking on the check mark next to the answer. \$\endgroup\$
    – Adam Haun
    Sep 22, 2021 at 17:25
  • \$\begingroup\$ I haven't got enough reputation to vote. I joined here only recently. \$\endgroup\$
    – User12121
    Sep 23, 2021 at 2:53
  • \$\begingroup\$ It's not a vote. You're the question asker, so you should be able to accept an answer regardless of reputation by clicking the check mark below the voting buttons. \$\endgroup\$
    – Adam Haun
    Sep 23, 2021 at 5:18

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