Unit 3 - Notes

ELE205

Unit 3: Transient Analysis and Laplace Transforms

1. Basics of Response: Steady State and Transient

When a circuit containing storage elements (Inductors $L$ , Capacitors $C$ ) is subjected to a change in excitation (switching actions), the response does not change instantaneously. The total response is divided into two parts:

1.1 Transient Response (Natural Response)

Definition: The behavior of a circuit from the time of switching ( $t=0$ ) until it reaches a steady value.
Cause: The inability of inductors to change current instantaneously and capacitors to change voltage instantaneously due to stored energy.
Characteristics: It decays to zero as $t \to \infty$ (in stable circuits). Mathematically, it is the solution to the homogenous differential equation.
Notation: $i_n(t)$ or $v_n(t)$ .

1.2 Steady State Response (Forced Response)

Definition: The behavior of the circuit as $t \to \infty$ after the transient dies out.
Cause: Driven by the independent sources connected to the circuit.
Characteristics: It mirrors the form of the input source (e.g., if the source is DC, steady state is constant; if AC, steady state is sinusoidal).
Notation: $i_f(t)$ or $v_f(t)$ .

Total Response:

x(t) = x_{transient}(t) + x_{steady\_state}(t)

2. DC Response of RL, RC, and RLC Circuits

Analysis assumes a switch closes or opens at $t=0$ involving a DC voltage source $V$ .

2.1 DC Response of an RL Circuit

Circuit: Series Resistor ( $R$ ) and Inductor ( $L$ ).
Equation: By KVL: $V = Ri(t) + L\frac{di(t)}{dt}$
Initial Condition: $i(0^-) = i(0^+) = I_0$ (Current through inductor cannot change instantly).
Solution (Current):
- If initial current $I_0 = 0$ : $i(t) = \frac{V}{R}(1 - e^{-t/\tau})$
Time Constant ( $\tau$ ): $\tau = \frac{L}{R}$ (seconds).
Voltage across Inductor: $v_L(t) = L\frac{di}{dt} = V e^{-t/\tau}$ .

2.2 DC Response of an RC Circuit

Circuit: Series Resistor ( $R$ ) and Capacitor ( $C$ ).
Equation: By KVL: $V = Ri(t) + v_c(t)$ . Since $i = C\frac{dv_c}{dt}$ , then $V = RC\frac{dv_c}{dt} + v_c$ .
Initial Condition: $v_c(0^-) = v_c(0^+) = V_0$ .
Solution (Capacitor Voltage):
- If uncharged initially ( $V_0=0$ ): $v_c(t) = V(1 - e^{-t/\tau})$
Time Constant ( $\tau$ ): $\tau = RC$ (seconds).

2.3 DC Response of an RLC Circuit

Circuit: Series R, L, and C.
Equation: $L\frac{d^2i}{dt^2} + R\frac{di}{dt} + \frac{1}{C}i = 0$ (Source-free for simplicity of characteristic roots).
Characteristic Equation: $s^2 + \frac{R}{L}s + \frac{1}{LC} = 0$ .
Roots:
- Neper Frequency (Damping Factor): $\alpha = \frac{R}{2L}$
- Resonant Frequency: $\omega_0 = \frac{1}{\sqrt{LC}}$

Response Types based on Damping:

Overdamped ( $\alpha > \omega_0$ ): Two real, distinct roots.
- $i(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t}$ (Slow decay, no oscillation).
Critically Damped ( $\alpha = \omega_0$ ): Two real, equal roots.
- $i(t) = (A_1 + A_2 t)e^{-\alpha t}$ (Fastest decay without oscillation).
Underdamped ( $\alpha < \omega_0$ ): Complex conjugate roots.
- $i(t) = e^{-\alpha t} (B_1 \cos \omega_d t + B_2 \sin \omega_d t)$
- Damped Frequency: $\omega_d = \sqrt{\omega_0^2 - \alpha^2}$ . (Oscillatory decay).

3. Sinusoidal Response of RL, RC, and RLC Circuits

Excitation source: $v(t) = V_m \sin(\omega t)$ .

3.1 RL Circuit (Sinusoidal)

Differential Equation: $L\frac{di}{dt} + Ri = V_m \sin(\omega t)$
Total Response: Contains an exponential transient term ( $Ke^{-Rt/L}$ ) and a sinusoidal steady state term.
Steady State Impedance: $Z = R + j\omega L = |Z|\angle \phi$ .
Current: $i_{ss}(t) = \frac{V_m}{|Z|} \sin(\omega t - \phi)$ , where $\phi = \tan^{-1}(\frac{\omega L}{R})$ .
Note: Current lags voltage.

3.2 RC Circuit (Sinusoidal)

Differential Equation: $RC\frac{dv_c}{dt} + v_c = V_m \sin(\omega t)$
Steady State Impedance: $Z = R - j\frac{1}{\omega C}$ .
Current: $i_{ss}(t) = \frac{V_m}{|Z|} \sin(\omega t + \phi)$ , where $\phi = \tan^{-1}(\frac{1}{\omega RC})$ .
Note: Current leads voltage.

3.3 RLC Circuit (Sinusoidal)

Impedance: $Z = R + j(\omega L - \frac{1}{\omega C})$ .
Reactance: $X = X_L - X_C$ .
Phase Angle: $\phi = \tan^{-1}\left(\frac{\omega L - 1/\omega C}{R}\right)$ .
Resonance: Occurs when $X_L = X_C$ , i.e., $\omega = \frac{1}{\sqrt{LC}}$ . At resonance, $Z$ is minimum (purely resistive), and current is maximum.

4. The Laplace Transform

4.1 Definition

The Laplace transform converts a time-domain function $f(t)$ ( $t \ge 0$ ) into a complex frequency-domain function $F(s)$ .

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

Where

s = \sigma + j\omega

is the complex frequency variable.

4.2 Laplace Transform of Useful Functions

Function $f(t)$	Laplace Transform $F(s)$	Region of Convergence
Unit Impulse $\delta(t)$	$1$	All $s$
Unit Step $u(t)$	$\frac{1}{s}$	$Re(s) > 0$
Unit Ramp $t$	$\frac{1}{s^2}$	$Re(s) > 0$
Exponential $e^{-at}$	$\frac{1}{s+a}$	$Re(s) > -a$
Sine $\sin(\omega t)$	$\frac{\omega}{s^2 + \omega^2}$	$Re(s) > 0$
Cosine $\cos(\omega t)$	$\frac{s}{s^2 + \omega^2}$	$Re(s) > 0$
Damped Sine $e^{-at}\sin(\omega t)$	$\frac{\omega}{(s+a)^2 + \omega^2}$	$Re(s) > -a$
Damped Cosine $e^{-at}\cos(\omega t)$	$\frac{s+a}{(s+a)^2 + \omega^2}$	$Re(s) > -a$
Power $t^n$	$\frac{n!}{s^{n+1}}$	$Re(s) > 0$

4.3 Properties of Laplace Transform

Linearity: $\mathcal{L}\{af(t) + bg(t)\} = aF(s) + bG(s)$
Time Shifting: $\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s)$
Frequency Shifting: $\mathcal{L}\{e^{-at}f(t)\} = F(s+a)$
Time Scaling: $\mathcal{L}\{f(at)\} = \frac{1}{a}F(\frac{s}{a})$
Differentiation in Time Domain:
- $\mathcal{L}\{f'(t)\} = sF(s) - f(0^-)$
- $\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0^-) - f'(0^-)$
Integration in Time Domain: $\mathcal{L}\{\int_0^t f(\tau) d\tau\} = \frac{F(s)}{s}$
Differentiation in s-Domain: $\mathcal{L}\{-t f(t)\} = \frac{dF(s)}{ds}$

4.4 Laplace Transform Theorems

Initial Value Theorem

Used to find the value of $f(t)$ at $t=0^+$ directly from $F(s)$ .

\lim_{t \to 0^+} f(t) = \lim_{s \to \infty} sF(s)

Final Value Theorem

Used to find the steady state value ( $t \to \infty$ ) directly from $F(s)$ . Valid only if poles of $sF(s)$ are in the left half-plane.

\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)

4.5 Laplace Transform of Periodic Functions

If $f(t)$ is periodic with period $T$ (i.e., $f(t) = f(t+T)$ ), then:

F(s) = \frac{1}{1 - e^{-sT}} \int_0^T f(t) e^{-st} dt

5. Inverse Laplace Transform

The process of converting $F(s)$ back to $f(t)$ .

\mathcal{L}^{-1}\{F(s)\} = f(t)

5.1 Partial Fraction Method

Used when $F(s)$ is a rational function $\frac{N(s)}{D(s)}$ .

Real Distinct Roots:

$F(s) = \frac{K_1}{s+p_1} + \frac{K_2}{s+p_2} + \dots$
$f(t) = K_1 e^{-p_1 t} + K_2 e^{-p_2 t} + \dots$
Calculate residues $K_i$ by: $K_i = [(s+p_i)F(s)]|_{s=-p_i}$ .
Repeated Real Roots:
For a pole $(s+p)^n$ , expand as:

$\frac{K_1}{s+p} + \frac{K_2}{(s+p)^2} + \dots + \frac{K_n}{(s+p)^n}$
Inverse involves terms like $t e^{-pt}, t^2 e^{-pt}$ , etc.
Complex Roots:
Usually result in damped sinusoidal terms. Complete the square in the denominator to match forms $(s+a)^2 + \omega^2$ .

5.2 Convolution Integral

Convolution in the time domain corresponds to multiplication in the s-domain.

\mathcal{L}^{-1}\{F(s)G(s)\} = f(t) * g(t) = \int_0^t f(\tau)g(t-\tau) d\tau

This is powerful for finding the response of a system

H(s)

to an input

X(s)

where

Y(s) = H(s)X(s)

6. Applications of Laplace Transforms to Electrical Circuits

This method converts the entire circuit into the s-domain (Algebraic equations instead of differential equations), solves for the unknown, and then applies the Inverse Laplace Transform.

6.1 Transformed Circuit Models

Resistor ( $R$ ):
- Time domain: $v(t) = R i(t)$
- s-domain: $V(s) = R I(s)$
- Representation: Resistance $R$ remains $R$ .
Inductor ( $L$ ):
- Time domain: $v(t) = L \frac{di}{dt}$
- Transform: $V(s) = sL I(s) - L i(0^-)$
- s-domain Models:
  - Series: Impedance $sL$ in series with a voltage source $L i(0^-)$ (opposing current direction).
  - Parallel: Impedance $sL$ in parallel with a current source $\frac{i(0^-)}{s}$ .
Capacitor ( $C$ ):
- Time domain: $i(t) = C \frac{dv}{dt}$
- Transform: $I(s) = sC V(s) - C v(0^-) \Rightarrow V(s) = \frac{1}{sC}I(s) + \frac{v(0^-)}{s}$
- s-domain Models:
  - Series: Impedance $\frac{1}{sC}$ in series with a voltage source $\frac{v(0^-)}{s}$ (aiding current flow out of positive terminal).
  - Parallel: Impedance $\frac{1}{sC}$ in parallel with a current source $C v(0^-)$ .

6.2 General Procedure for Solution

Calculate Initial Conditions: Determine $i_L(0^-)$ and $v_C(0^-)$ using the circuit state at $t < 0$ .
Draw the s-Domain Circuit: Replace $R, L, C$ with s-domain impedances and initial condition sources. Transform independent sources (e.g., $V_{DC} \to V/s$ ).
Analyze Circuit: Apply KVL (Mesh), KCL (Nodal), or Network Theorems in the s-domain.
Solve Algebraically: Find the desired variable $V(s)$ or $I(s)$ .
Inverse Transform: Perform Partial Fraction Expansion and apply $\mathcal{L}^{-1}$ to obtain $v(t)$ or $i(t)$ .

6.3 Concepts of Transfer Function

Definition: Ratio of the Laplace transform of the output (response) to the Laplace transform of the input (excitation), assuming zero initial conditions.
$H(s) = \frac{Y(s)}{X(s)}$
Poles: Roots of the denominator of $H(s)$ . Determine the stability and the nature of the transient response (damped, oscillatory, etc.).
Zeros: Roots of the numerator of $H(s)$ . Determine the magnitude of the response components.