Laplace Transform of DC Circuits

Last edited: 2021-05-06 19:51:46


When studying transient processes, like when a switch is changed, in electrical circuits we have to make us of the Laplace transform. Here is a quick rundown on how to Laplace transform the resistor, capacitor and inductor.

First we need to establish what happens when we Laplace transform the current and voltage. We can see the voltage and current as "unknown" functions dependent on time, therefore their Laplace transforms become:

i(t)I(s)u(t)U(s).\begin{aligned} & i(t) \Rightarrow I(s) \\ & u(t) \Rightarrow U(s) \end{aligned} .

Laplace Transform of the Resistor

So lets start with the easy component: the resistor. We know from Ohm's law that:

u(t)=Ri(t).u(t) = R i(t).

If we Laplace transform this we get:

U(s)=RI(s),U(s) = R I(s),

since the resistor is time independent there is not much happening when moving from the t-domain to the s-domain. The resistor remains the same in both domains.

Laplace Transform of a Derivative

The inductor and capacitor is time dependent, so they get a little bit more trickier. First we need to see what happens when we apply the Laplace transform to a derivative of a function f(t)f(t). The Laplace transform of an arbitrary function f(t)f(t) is given by:

F(s)=0estf(t)dt.F(s) = \int_{0-}^\infty e^{-st} f(t) dt.

The 00- can be looked at more or less as what happened exactly before significant change happened to the function f(t)f(t) (we'll get back to this). So lets preform the Laplace transform on f(t)f'(t):

0estf(t)dt.\int_{0-}^\infty e^{-st} f'(t) dt.

With a little partial integration we get:

[estf(t)]00sestf(t)dt=f(0)+sF(s).\bigg[e^{-st} f(t) \bigg]_{0-}^\infty - \int_{0-}^\infty -se^{-st} f(t) dt = - f(0-) + sF(s) .

Laplace transform of the Inductor

With that out of the way lets preform the Laplace transform on the inductor first. We know that the potential over an inductor is given by:

u(t)=Ldi(t)dt.u(t) = L \frac{d i(t)}{dt}.

So with our knowledge of Laplace transforming a derivative we get the following:

U(s)=L(sI(s)i(0)).U(s) = L (s I(s) - i(0-)).

The inductor is said to be energized if there is an initial current i(0)i(0-). If there is no initial current we get a simpler relation of:

U(s)=sLI(s).U(s) = sL I(s).

We can also rewrite the relation to get the current:

I(s)=U(s)sL+i(0)s.I(s) = \frac{U(s)}{sL} + \frac{i(0-)}{s}.

We can make circuit diagrams with these relations:

Image of Laplace transform of the inductor

Note the Thevenin/Norton equivalents under "Initial current".

Laplace transform of the Capacitor

Lets move on to the capacitor. The procedure of Laplace transforming the capacitor is the same as the inductor. The current through a capacitor is given by:

i(t)=Cdu(t)dt.i(t) = C \frac{du(t)}{dt}.

So the Laplace transform of the current is:

I(s)=C(sU(s)u(0)).I(s) = C(sU(s) - u(0-)).

We can see here that if there is no initial voltage over the capacitor we get the simpler relation:

I(s)=sCU(s).I(s) = sC U(s).

Lastly the the Laplace transformed voltage is:

U(s)=I(s)sC+u(0)s.U(s) = \frac{I(s)}{sC} + \frac{u(0-)}{s}.

The circuit diagrams then become:

Image of Laplace transform of the capacitor

And that is it! You now know how to Laplace transform the resistor, inductor and capacitor!

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