We can use the Laplace transform to analyze an electric circuit. This is known as the Laplace transform circuit analysis, as the application of Laplace transform.

## Laplace Transform Circuit Analysis

Circuit analysis is again relatively easy to do when we are in the *s*-domain. We merely need to transform a complicated set of mathematical relationships in the time domain into the *s*-domain where we convert operators (derivatives and integrals) into simple multipliers of *s *and 1/*s*.

This now allows us to use algebra to set up and solve our circuit equations. The exciting thing about this is that *all *of the circuit theorems and relationships we developed for dc circuits are perfectly valid in the *s*-domain.

Remember,

equivalent circuits, with capacitors and inductors, only exist in the s-domain; they cannot be transformed back into the time domain.

## Laplace Transform Circuit Analysis Examples

1. Consider the circuit in Figure.(1a). Find the value of the voltage across the capacitor assuming that the value of*v _{s}(t)* = 10

*u(t)*

and assume that at

*t*= 0, –1 A flows through the inductor and +5V is across the capacitor.

**Solution:**

Figure.(1b) represents the entire circuit in the *s*-domain with the initial conditions incorporated. We now have a straightforward nodal analysis problem.

Since the value of *V*_{1} is also the value of the capacitor voltage in the time domain and is the only unknown node voltage, we only need to write one equation.

or

where *v*(0) = 5 V and *i*(0) = –1 A. Simplifying we get

Taking the inverse Laplace transform yields

2. For the circuit shown in Figure.(1), and the initial conditions used in Example 1, use superposition to find the value of the capacitor voltage.

**Solution:**

In as much as the circuit in the *s*-domain actually has three independent sources, we can look at the solution one source at a time. Figure.(2) presents the circuits in the *s*-domain considering one source at a time.

We now have three nodal analysis problems. First, let us solve for the capacitor voltage in the circuit shown in Figure.(2a).

or

Simplifying we get

or

For Figure.(2b) we get,

or

This leads to

Taking the inverse Laplace transform, we get

For Figure.(2c),

or

This leads to

Now, all we need to do is to add Equations.(2.1), (2.2), and (2.3):

or

which agrees with our answer in Example 1.

3. Assume that there is no initial energy stored in the circuit of Figure.(3) at *t *= 0 and that *i*_{s }= 10*u*(*t*) A.

(a) Find *V*_{o}(*s*) using Thevenin’s theorem.

(b) Apply the initial- and final-value theorems to find *v*_{o}(0^{+}) and *v*_{o}(∞).

(c) Determine *v*_{o}(*t*).

**Solution:**

Because there is no initial energy stored in the circuit, we assume that the initial inductor current and initial capacitor voltage are zero at *t *= 0.

(a) To find the Thevenin equivalent circuit, we remove the 5-Ω the resistor and then find *V*_{oc} (*V*_{Th}) and *I*_{sc}. To find *V*_{Th}, we use the Laplace transformed circuit in Figure.(4a).

Since *I*_{x }= 0, the dependent voltage source contributes nothing, so

To find *Z*_{Th}, we consider the circuit in Figure.(4b), where we first find *I*_{sc}. We can use nodal analysis to solve for *V*_{1} which then leads to*I*_{sc }(*I*_{sc} = *I*_{x }= *V*_{1}∕2*s*).

along with

leads to

Hence,

and

The given circuit is replaced by its Thevenin equivalent at terminals*a*–*b *as shown in Figure.(5).

From Figure.(5),

(b) Using the initial-value theorem we find

Using the final-value theorem we find

(c) By partial fraction,

Taking the inverse Laplace transform gives

Notice that the values of *v*_{o}(0) and *v*_{o}(∞) obtained in part (b) are confirmed.

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