# Circuit Application of The Laplace Transform

Having mastered how to obtain the Laplace transform and its inverse, we are now prepared to employ the Laplace transform to analyze circuits. This usually involves three steps.

#### Steps in applying the Laplace transform:

• Transform the circuit from the time domain to the $s$ domain.
• Solve the circuit using nodal analysis, mesh analysis, source transformation, superposition, or any circuit analysis technique with which we are familiar.
• Take the inverse transform of the solution and thus obtain the solution in the time domain.
Only the first step is new and will be discussed here. As we did in phasor analysis, we transform a circuit in the time domain to the frequency or $s$ domain by Laplace transforming each term in the circuit. For a resistor, the voltage-current relationship in the time domain is $$v(t)=R i(t) \tag{1}$$ Taking the Laplace transform, we get $$V(s)=R I(s) \tag{2}$$ For an inductor, $$v(t)=L \frac{d i(t)}{d t} \tag{3}$$ Taking the Laplace transform of both sides gives $$V(s)=L\left[s I(s)-i\left(0^{-}\right)\right]=s L I(s)-L i\left(0^{-}\right) \tag{4}$$ or $$I(s)=\frac{1}{s L} V(s)+\frac{i\left(0^{-}\right)}{s} \tag{5}$$ The $s$-domain equivalents are shown in Fig. 1, where the initial condition is modeled as a voltage or current source.
Fig. 1: Representation of an inductor: (a) time-domain, (b,c) s-domain equivalents.
For a capacitor, $$i(t)=C \frac{d v(t)}{d t} \tag{6}$$ which transforms into the $s$ domain as $$I(s)=C\left[s V(s)-v\left(0^{-}\right)\right]=s C V(s)-C v\left(0^{-}\right) \tag{7}$$ or $$V(s)=\frac{1}{s C} I(s)+\frac{v\left(0^{-}\right)}{s} \tag{8}$$ The $s$-domain equivalents are shown in Fig. 2.
Fig. 2: Representation of a capacitor: (a) time-domain, (b,c) s-domain equivalents.
With the $s$-domain equivalents, the Laplace transform can be used readily to solve first and second-order circuits. We should observe from Eqs. (3) to (8) that the initial conditions are part of the transformation. This is one advantage of using the Laplace transform in circuit analysis. Another advantage is that a complete response - transient and steady state - of a network is obtained. We will illustrate this with Examples $1$ and 2.
If we assume zero initial conditions for the inductor and the capacitor, the above equations reduce to:
Resistor: $\quad V(s)=R I(s)$
Inductor: $\quad V(s)=s L I(s)$
Capacitor: $\quad V(s)=\frac{1}{s C} I(s)$
The $s$-domain equivalents are shown in Fig. 3.
Fig. 3: Time-domain and s-domain representations of passive elements under zero initial conditions.
We define the impedance in the $s$-domain as the ratio of the voltage transform to the current transform under zero initial conditions, that is, $$Z(s)=\frac{V(s)}{I(s)}$$ Thus the impedances of the three circuit elements are
Resistor: $\quad Z(s)=R$
Inductor: $\quad Z(s)=s L$
Capacitor: $Z(s)=\frac{1}{s C}$
The admittance in the $s$ domain is the reciprocal of the impedance, or $$Y(s)=\frac{1}{Z(s)}=\frac{I(s)}{V(s)}$$ The use of the Laplace transform in circuit analysis facilitates the use of various signal sources such as impulse, step, ramp, exponential, and sinusoidal.
Example 1: Find $v_{o}(t)$ in the circuit in Fig. 4, assuming zero initial conditions.
Fig. 4: Example 1.
Solution: We first transform the circuit from the time domain to the $s$ domain.
$$\begin{array}{l}u(t) \quad \Longrightarrow \quad \frac{1}{s} \\1 \mathrm{H} \quad \Longrightarrow \quad s L=s \\\frac{1}{3} \mathrm{~F} \quad \Longrightarrow \quad \frac{1}{s C}=\frac{3}{s} \\\end{array}$$
The resulting $s$-domain circuit is in Fig. 5.
Fig. 5: Mesh analysis of the frequency domain equivalent of the same circuit.
We now apply mesh analysis. For mesh 1,
$$\frac{1}{s}=\left(1+\frac{3}{s}\right) I_{1}-\frac{3}{s} I_{2}$$
For mesh 2 ,
$$0=-\frac{3}{s} I_{1}+\left(s+5+\frac{3}{s}\right) I_{2}$$
or
$$I_{1}=\frac{1}{3}\left(s^{2}+5 s+3\right) I_{2}$$
Substituting this into Eq. (15.12.1),
$$\frac{1}{s}=\left(1+\frac{3}{s}\right) \frac{1}{3}\left(s^{2}+5 s+3\right) I_{2}-\frac{3}{s} I_{2}$$
Multiplying through by $3 s$ gives
$$\begin{array}{c}3=\left(s^{3}+8 s^{2}+18 s\right) I_{2} \quad \Longrightarrow \quad I_{2}=\frac{3}{s^{3}+8 s^{2}+18 s} \\V_{o}(s)=s I_{2}=\frac{3}{s^{2}+8 s+18}=\frac{3}{\sqrt{2}} \frac{\sqrt{2}}{(s+4)^{2}+(\sqrt{2})^{2}}\end{array}$$
Taking the inverse transform yields
$$v_{o}(t)=\frac{3}{\sqrt{2}} e^{-4 t} \sin \sqrt{2} t \mathrm{~V}, \quad t \geq 0$$
Example 2: Find $v_{o}(t)$ in the circuit of Fig. 6. Assume $v_{o}(0)=5 \mathrm{~V}$.
Fig. 6: Example 2.
Solution: We transform the circuit to the $s$ domain as shown in Fig. 7. The initial condition is included in the form of the current source $C v_o(0)=$ $0.1(5)=0.5$ A. [See Fig. 2(c).]
Fig. 7: Nodal analysis of the equivalent of the circuit in Fig. 6.
We apply nodal analysis. At the top node,
$$\frac{10 /(s+1)-V_o}{10}+2+0.5=\frac{V_o}{10}+\frac{V_o}{10 / s}$$
or
$$\frac{1}{s+1}+2.5=\frac{2 V_o}{10}+\frac{s V_o}{10}=\frac{1}{10} V_o(s+2)$$
Multiplying through by 10 ,
$$\frac{10}{s+1}+25=V_o(s+2)$$
or
$$V_o=\frac{25 s+35}{(s+1)(s+2)}=\frac{A}{s+1}+\frac{B}{s+2}$$
where
\begin{aligned} &A=\left.(s+1) V_o(s)\right|_{s=-1}=\left.\frac{25 s+35}{(s+2)}\right|_{s=-1}=\frac{10}{1}=10 \\ &B=\left.(s+2) V_o(s)\right|_{s=-2}=\left.\frac{25 s+35}{(s+1)}\right|_{s=-2}=\frac{-15}{-1}=15 \end{aligned}
Thus,
$$V_o(s)=\frac{10}{s+1}+\frac{15}{s+2}$$
Taking the inverse Laplace transform, we obtain
$$v_o(t)=\left(10 e^{-t}+15 e^{-2 t}\right) u(t)$$