Chapter 3 • 7 Hours • Expected Marks: 7M
Always Tested in Q4 of All Exams (6M)
Transient Analysis Using Laplace Transform
Algebraic transformation of dynamic circuits, s-domain equivalent impedance networks with initial-condition sources, and partial fraction inversion techniques.
3.1 & 3.2 Laplace Transform Principles & s-Domain Models
Concept & Operational Advantages
While the classical method requires solving differential equations and separately calculating complementary functions, particular integrals, and constants from boundary values, the unilateral Laplace transform converts linear differential equations into simple algebraic equations. Furthermore, initial conditions (\(i_L(0^-)\) and \(v_C(0^-)\)) are automatically incorporated from the very first step! \[\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty f(t) e^{-st} dt\]
Core Operational Properties
| Operation | Time Domain \(f(t)\) | s-Domain \(F(s)\) |
|---|---|---|
| First Derivative | \(\frac{df(t)}{dt}\) | \(s F(s) - f(0^-)\) |
| Second Derivative | \(\frac{d^2f(t)}{dt^2}\) | \(s^2 F(s) - s f(0^-) - f'(0^-)\) |
| Definite Integral | \(\int_0^t f(\tau) d\tau\) | \(\frac{F(s)}{s}\) |
| Exponential Shift | \(e^{-at} f(t)\) | \(F(s + a)\) |
s-Domain Transformed Circuit Elements
| Element | Time Domain Equation | Laplace Transform | s-Domain Thevenin Model (Series) | s-Domain Norton Model (Parallel) |
|---|---|---|---|---|
| Resistor \(R\) | \(v(t) = R i(t)\) | \(V(s) = R I(s)\) | Impedance \(R\) | Admittance \(1/R\) |
| Inductor \(L\) | \(v(t) = L \frac{di}{dt}\) | \(V(s) = sL I(s) - L i(0^-)\) | Impedance \(sL\) in series with voltage source \(L i(0^-)\) opposing \(I(s)\) | Admittance \(\frac{1}{sL}\) in parallel with current source \(\frac{i(0^-)}{s}\) |
| Capacitor \(C\) | \(i(t) = C \frac{dv}{dt}\) | \(I(s) = sC V(s) - C v(0^-)\) | Impedance \(\frac{1}{sC}\) in series with voltage source \(\frac{v(0^-)}{s}\) aiding \(V(s)\) | Admittance \(sC\) in parallel with current source \(C v(0^-)\) aiding \(I(s)\) |
3.3 & 3.4 Response of RL, RC & RLC Circuits via Laplace Transform
Master Step-by-Step Procedure
- Analyze circuit at \(t = 0^-\) under pre-switch steady-state conditions to find initial inductor current \(i_L(0^-)\) and capacitor voltage \(v_C(0^-)\).
- Redraw the entire circuit directly in the s-domain for \(t > 0\):
• DC sources \(V_0 \to \frac{V_0}{s}\), exponential sources \(V_0 e^{-at} \to \frac{V_0}{s+a}\).
• Replace \(R \to R\), \(L \to sL\) with series initial generator \(L i_L(0^-)\), \(C \to \frac{1}{sC}\) with series generator \(\frac{v_C(0^-)}{s}\). - Write standard mesh KVL or nodal KCL directly in the s-domain using Ohm's law.
- Solve algebraically for the desired variable \(I(s)\) or \(V(s)\).
- Apply Partial Fraction Expansion (PFE):
• Distinct real poles: \(\frac{A}{s - p_1} + \frac{B}{s - p_2}\).
• Complex conjugate poles: Complete the square in the quadratic denominator: \((s + \alpha)^2 + \omega^2\). - Take the inverse Laplace transform to obtain the complete time-domain response \(i(t)\) or \(v(t)\) for \(t \ge 0\).
Master Worked Example 1 (Exponential Excitation to Series RLC)
Series RLC with Exponential Voltage Source (Laplace Method)
PYQ 2082 Baishakh Q4
Problem Statement:
An exponential voltage \(v(t) = 20e^{-4t}\) is suddenly applied at \(t = 0\) to a series RLC circuit comprising \(R = 2\,\Omega, L = 0.5\text{H}\), and \(C = 1\text{F}\). Initial conditions are zero. Obtain the expression for current \(i(t)\) using the Laplace transform method.
Step 1: s-Domain Transformation
\[V(s) = \mathcal{L}\{20e^{-4t}\} = \frac{20}{s + 4}\]
Total series impedance:
\[Z(s) = R + sL + \frac{1}{sC} = 2 + 0.5s + \frac{1}{s} = \frac{0.5s^2 + 2s + 1}{s} = \frac{s^2 + 4s + 2}{2s}\]
Since initial conditions are zero (\(i_L(0^-) = 0, v_C(0^-) = 0\)):
\[I(s) = \frac{V(s)}{Z(s)} = \frac{\frac{20}{s+4}}{\frac{s^2 + 4s + 2}{2s}} = \frac{40s}{(s + 4)(s^2 + 4s + 2)}\]
Step 2: Roots of Quadratic Denominator
\[s^2 + 4s + 2 = 0 \implies s = \frac{-4 \pm \sqrt{16 - 8}}{2} = -2 \pm \sqrt{2}\]
Let \(s_1 = -2 + \sqrt{2} \approx -0.586\), and \(s_2 = -2 - \sqrt{2} \approx -3.414\).
Denominator has three distinct real poles: \(s = -4\), \(s = s_1\), \(s = s_2\).
\[I(s) = \frac{40s}{(s+4)(s - s_1)(s - s_2)} = \frac{K_0}{s+4} + \frac{K_1}{s - s_1} + \frac{K_2}{s - s_2}\]
Step 3: Evaluating Residues via Heaviside Method
\[K_0 = \left. \frac{40s}{s^2 + 4s + 2} \right|_{s = -4} = \frac{40(-4)}{(-4)^2 + 4(-4) + 2} = \frac{-160}{16 - 16 + 2} = \frac{-160}{2} = -80\]
\[K_1 = \left. \frac{40s}{(s+4)(s - s_2)} \right|_{s = s_1}\]
Note that \(s_1 - s_2 = 2\sqrt{2}\), and \(s_1 + 4 = 2 + \sqrt{2}\):
\[K_1 = \frac{40(-2+\sqrt{2})}{(2+\sqrt{2})(2\sqrt{2})} = \frac{40(-2+\sqrt{2})}{4\sqrt{2} + 4} = 10 \frac{\sqrt{2}-2}{\sqrt{2}+1} = 10 (\sqrt{2}-2)(\sqrt{2}-1) = 10(2 - 3\sqrt{2} + 2) = 10(4 - 3\sqrt{2}) \approx -2.426\]
\[K_2 = \left. \frac{40s}{(s+4)(s - s_1)} \right|_{s = s_2}\]
Note that \(s_2 - s_1 = -2\sqrt{2}\), and \(s_2 + 4 = 2 - \sqrt{2}\):
\[K_2 = \frac{40(-2-\sqrt{2})}{(2-\sqrt{2})(-2\sqrt{2})} = \frac{40(2+\sqrt{2})}{(2-\sqrt{2})(2\sqrt{2})} = 10 \frac{2+\sqrt{2}}{2\sqrt{2}-2} = 10(4 + 3\sqrt{2}) \approx 82.426\]
Check sum of residues: \(K_0 + K_1 + K_2 = -80 + 10(4 - 3\sqrt{2}) + 10(4 + 3\sqrt{2}) = -80 + 40 - 30\sqrt{2} + 40 + 30\sqrt{2} = 0\) (Valid!).
Step 4: Inverse Laplace Transform
\[i(t) = K_0 e^{-4t} + K_1 e^{s_1 t} + K_2 e^{s_2 t}\]
\[i(t) = -80 e^{-4t} + 10(4 - 3\sqrt{2}) e^{(-2+\sqrt{2})t} + 10(4 + 3\sqrt{2}) e^{(-2-\sqrt{2})t}\text{ A} \quad (t \ge 0)\]
Final Answer:
\[i(t) = -80 e^{-4t} - 2.43 e^{-0.586 t} + 82.43 e^{-3.414 t}\text{ A}\]
3.6 Comparison Table: Classical Method vs. Laplace Transform Method
| Attribute | Classical Differential Equation Method | Laplace Transform Method |
|---|---|---|
| Domain of Analysis | Time domain \(t\) throughout. | Complex frequency domain \(s = \sigma + j\omega\). |
| Mathematical Tools | Calculus (differentiation, integration, ODE solving). | Algebra (linear algebraic equations, rational fractions). |
| Initial Conditions | Evaluated at the very end to determine arbitrary constants (\(A_1, A_2\)). | Incorporated at the very beginning into s-domain generator sources. |
| Non-Homogeneous Forcing | Requires separate trial solutions for Particular Integral (undetermined coefficients). | Directly transformed into \(F(s)\) via standard transform tables. |
| High-Order Complexity | Becomes tedious and error-prone for orders \(n \ge 2\) and multi-mesh networks. | Handled systematically using standard matrix methods \([Z(s)][I(s)] = [V(s)]\). |