10.2 Circuit analysis using Laplace transforms

Transformed Circuit Elements

Impedance and Admittance Transformations

The core idea behind Laplace-domain circuit analysis is replacing derivatives and integrals with algebraic expressions in $s$. Each passive element gets a new "impedance" that you can treat just like resistance in a DC circuit, except now it's a function of $s$.

Impedances in the s-domain:

Admittance is simply the reciprocal of impedance: $Y(s) = \frac{1}{Z(s)}$. You'll use impedances in series combinations and mesh analysis, and admittances in parallel combinations and nodal analysis, just like you would with phasors in AC analysis.

Circuit Element Transformations and Initial Conditions

Sources also transform into the s-domain. A time-domain voltage source $v(t)$ becomes its Laplace transform $V(s)$, and a current source $i(t)$ becomes $I(s)$. For a DC source of value $V_0$ switched on at $t = 0$, that's $V_0/s$.

The real payoff of Laplace analysis is how it handles initial conditions. Instead of solving a differential equation and then fitting constants to initial values, you build the initial conditions directly into the circuit model.

Inductor with initial current $i(0)$:

• Series model: An impedance $sL$ in series with a voltage source $Li(0)$ (polarity opposes the current direction).
• Parallel model: An impedance $sL$ in parallel with a current source $i(0)/s$.

Capacitor with initial voltage $v(0)$:

• Series model: An impedance $\frac{1}{sC}$ in series with a voltage source $v(0)/s$ (polarity matches the initial voltage).
• Parallel model: An impedance $\frac{1}{sC}$ in parallel with a current source $Cv(0)$.

Which model you pick (series or parallel) depends on whether you're doing mesh or nodal analysis. The series model fits naturally into mesh analysis; the parallel model fits nodal analysis.

S-Domain Circuit Analysis

Once every element is transformed, the circuit is purely algebraic. You apply the same techniques you already know from resistive circuit analysis, just with $s$-dependent expressions.

Mesh Analysis in S-Domain

1. Redraw the circuit in the s-domain. Replace every element with its s-domain impedance and include initial-condition sources using the series model.

2. Identify independent meshes. The number of independent meshes equals the number of branches minus the number of nodes plus one ($b - n + 1$).

3. Assign mesh currents and write a KVL equation around each mesh. Each equation will be in terms of $s$ and the unknown mesh currents $I_1(s), I_2(s), \ldots$

4. Solve the system of equations for the mesh currents. This is straightforward linear algebra, though the expressions can get lengthy.

5. Apply the inverse Laplace transform to get the time-domain currents.

Mesh analysis works best for circuits with several series-connected loops and current sources (which you handle with supermeshes, same as before).

Nodal Analysis in S-Domain

1. Redraw the circuit in the s-domain. Replace every element with its s-domain admittance and include initial-condition sources using the parallel model.
2. Choose a reference node (ground) and identify the remaining independent nodes.
3. Write a KCL equation at each independent node. Express branch currents using admittances: $I = Y(s) \cdot V(s)$.
4. Solve the system of equations for the node voltages $V_1(s), V_2(s), \ldots$
5. Apply the inverse Laplace transform to recover the time-domain voltages.

Nodal analysis tends to be more efficient for circuits with many parallel branches or multiple voltage sources (handled with supernodes).

Inverse Laplace Transform and Solution Interpretation

After solving in the s-domain, you need to get back to $v(t)$ or $i(t)$. The standard approach:

1. Simplify the s-domain expression into a ratio of polynomials $\frac{N(s)}{D(s)}$.
2. Factor the denominator to find the poles of the expression.
3. Perform partial fraction expansion to break the expression into simpler terms like $\frac{A}{s + a}$, $\frac{As + B}{s^2 + bs + c}$, etc.
4. Use a Laplace transform table to convert each term back to the time domain.

For example, $\frac{5}{s(s+2)}$ expands to $\frac{5/2}{s} - \frac{5/2}{s+2}$, which inverse-transforms to $\frac{5}{2}(1 - e^{-2t})u(t)$.

Interpreting the result:

• Transient response comes from poles with negative real parts. These are the decaying exponential and damped sinusoidal terms that die out over time.
• Steady-state response comes from poles at the origin ($s = 0$) or on the imaginary axis ($s = \pm j\omega$). These persist indefinitely.
• The total response is the sum of both. The Laplace method gives you the complete response in one shot, which is a major advantage over classical differential equation methods where you solve for the natural and forced responses separately.