9.00 Laplace Transform and s-Domain Circuit Applications Com

Chapter 9: Laplace Transform & s-Domain Circuit Applications - Compact Review

9.01 Foundation of Laplace Transform & s-Plane Representation | 9.02 Laplace Transforms of Singularity & Elementary Functions | 9.03 Mathematical Properties of the Laplace Transform | 9.04 Initial and Final Value Theorems | 9.05 Pole-Zero Analysis & Stability | 9.06 s-Domain Modeling of Passive Circuit Elements | 9.07 Transient Response of RL & RC Circuits | 9.08 Transient Response of RLC Networks

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9.01 Foundation of Laplace Transform & s-Plane Representation

9.01.1 Bilateral & Unilateral Laplace Transform Definitions

*(Target: Theory Descriptive / Formulation)*

• Bilateral Laplace Transform: Integrates over the entire real timeline; maps non-causal systems: $F(s)={\int }_{-\infty }^{\infty }f(t)e^{-st}dt$
• Unilateral Laplace Transform: Integrates from the instantaneous left boundary ($t=0^{-}$) to account for switching transients and initial energy states; maps causal signals: $F(s)={\int }_{0^{-}}^{\infty }f(t)e^{-st}dt$
• Inverse Laplace Transform: Integrates along a vertical Bromwich contour in the s-plane: f(t) = rac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} F(s) e^{st} \, ds

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9.01.2 Comparison: Fourier Transform vs. Laplace Transform

*(Target: Theory Descriptive / Conceptual Comparison)*

Feature / Dimension	Continuous-Time Fourier Transform (CTFT) [8.01]	Laplace Transform (Bilateral / Unilateral) [9.01]
Complex Frequency Variable	Purely imaginary phasor: $s=j\omega$ (frequency $\omega$ in rad/s)	Complex frequency: $s=\sigma +j\omega$ ($\sigma$: attenuation, $\omega$: oscillation)
Existence / Convergence	Requires signal to be absolutely integrable: $\int_{-\infty}^{\infty} \lvert f(t)
vert dt < \infty$	Converges for exponentially growing signals using decay factor $e^{-\sigma t}$
Applicability Limit	Limited to stable systems and steady-state signals	Generalizes to transient, unstable, and switched systems
Physical Viewpoint	Maps a signal directly to its harmonic sinusoidal components	Maps a signal to a plane of damped sinusoidal oscillators

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9.01.3 Comparison: Bilateral vs. Unilateral Laplace Transform

*(Target: Theory Descriptive)*

Feature / Dimension	Bilateral Laplace Transform [9.01]	Unilateral Laplace Transform [9.01]
Lower Integration Limit	$-\infty$	$0^{-}$ (specifically captures impulses and derivatives at origin)
Switching Transients	Incapable of modeling initial states before switches activate	Explicitly designed for initial conditions and transient circuit analysis
Signal Domain	Applicable to two-sided, left-sided, and right-sided signals	Strictly restricted to causal, right-sided signals ($f(t)=0$ for $t<0$)
ROC Boundary	Can be a vertical strip, half-plane, or empty	Always a right-half plane to the right of the rightmost pole

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9.01.4 Relationship Between Laplace and Fourier Transforms

*(Target: Mathematical Proof / 5-Mark Derivation)*

• Derivation Summary: Substituting the complex frequency $s=\sigma +j\omega$ into the bilateral Laplace transform integral:

ight] e^{-j\omega t} , dt \quad ext{(derivation)}$\ast \ast \ast Last-LineBoxedEquivalent:\ast \ast$F(s)\Big|_{s = \sigma + j\omega} = \mathcal{F}\left{ f(t) e^{-\sigma t} ight}$$

• The CTFT Projection: If the Region of Convergence (ROC) includes the imaginary axis ($\sigma =0$), the CTFT is obtained by direct projection: $F(j\omega )=F(s){|}_{s=j\omega }$.

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9.01.5 Region of Convergence (ROC) Properties

*(Target: Theory Descriptive)*

1. Strip Symmetries: The ROC consists of vertical strips parallel to the $j\omega$-axis because convergence depends only on the real part $\sigma =\Re e(s)$.
2. Pole Exclusion: The ROC cannot contain any poles, as $|F(s)|o\infty$ at a pole.
3. Finite-Duration Signals: If $f(t)$ is finite-duration ($t_1\le t\le t_2$) and convergent for at least one $s$, the ROC is the entire s-plane (excluding possibly $s=\pm \infty$).
4. Right-Sided Signals: If $f(t)$ is right-sided ($f(t)=0$ for $t<t_1$), the ROC is a right-half plane: $\Re e(s)>{\sigma }_{\max }$.
5. Left-Sided Signals: If $f(t)$ is left-sided ($f(t)=0$ for $t>t_2$), the ROC is a left-half plane: $\Re e(s)<{\sigma }_{\min }$.
6. Two-Sided Signals: If $f(t)$ is two-sided, the ROC is a vertical strip bounded by poles on both sides: ${\sigma }_r<\Re e(s)<{\sigma }_l$.

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9.02 Laplace Transforms of Singularity & Elementary Functions

9.02.1 Sufficient Condition for Unilateral Existence

*(Target: Theory Descriptive)*

• A unilateral Laplace transform exists if the signal $f(t)$ is piecewise continuous on every finite interval in $t\ge 0$ and of exponential order ($|f(t)|\le M{e}^{kt}$ for positive constants $M$ and $k$): ${\lim }_{so\infty }|F(s)|=0extfor\Re e(s)>kext(derivation)$

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9.02.2 Master Laplace Transform Pairs Table

*(Target: Numerical Solving / Reference)*

Signal $f(t)$ ($t\ge 0^{-}$)	Laplace Transform $F(s)$	Region of Convergence (ROC)
Unit Impulse $\delta (t)$	$1$	Entire s-plane
Unit Step $u(t)$	rac{1}{s}	$\Re e(s)>0$
Ramp Function $tu(t)$	rac{1}{s^2}	$\Re e(s)>0$
Power Function rac{t^{n-1}}{(n-1)!} u(t)	rac{1}{s^n}	$\Re e(s)>0$
Decaying Exponential e^{-lpha t} u(t)	rac{1}{s+lpha}	\Re e(s) > -lpha
Growing Exponential e^{lpha t} u(t)	rac{1}{s-lpha}	\Re e(s) > lpha
Modulated Power rac{t^{n-1}}{(n-1)!} e^{-lpha t} u(t)	rac{1}{(s+lpha)^n}	\Re e(s) > -lpha
Sine Wave $\sin ({\omega }_{0}t)u(t)$	rac{\omega_0}{s^2 + \omega_0^2}	$\Re e(s)>0$
Cosine Wave $\cos ({\omega }_{0}t)u(t)$	rac{s}{s^2 + \omega_0^2}	$\Re e(s)>0$
Damped Sine e^{-lpha t} \sin(\omega_0 t) u(t)	rac{\omega_0}{(s+lpha)^2 + \omega_0^2}	\Re e(s) > -lpha
Damped Cosine e^{-lpha t} \cos(\omega_0 t) u(t)	rac{s+lpha}{(s+lpha)^2 + \omega_0^2}	\Re e(s) > -lpha

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9.02.3 The Power Function Derivation

*(Target: Mathematical Proof / Heavily Tested)*

• Theorem: Prove that \mathcal{L}\left\{ t^n ight\} = rac{n!}{s^{n+1}} for integer $n\ge 0$.
• Derivation Steps:
  1. Base case $n=0$: \mathcal{L}\{1\} = \int_0^{\infty} e^{-st} dt = rac{1}{s}.
  2. Set up integration by parts recursively: $I_n(s)={\int }_0^{\infty }{t}^n{e}^{-st}dt$.
  3. Let $u=t^n⟹du=n{t}^{n-1}dt$, and dv = e^{-st} dt \implies v = -rac{e^{-st}}{s}.
  4. Apply limits and evaluate boundary terms:

ight]0^{\infty} + rac{n}{s} \int_0^{\infty} t^{n-1} e^{-st} , dt = rac{n}{s} I{n-1}(s) \quad ext{(derivation)}$\ast \ast \ast Last-LineBoxedEquivalent:\ast \ast$F(s) = rac{n!}{s^{n+1}}$$

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9.03 Mathematical Properties of the Laplace Transform

9.03.1 Master Laplace Properties Table

*(Target: Numerical Solving)*

Property	Time-Domain $f(t)$	s-Domain $F(s)$	ROC constraint
Linearity	$a{f}_1(t)+b{f}_2(t)$	$a{F}_1(s)+b{F}_2(s)$	At least $R_1\cap R_2$
Time Shifting	$f(t-t_0)u(t-t_0)$	$e^{-s{t}_0}F(s)$	Unchanged
s-Domain Shifting	$e^{s_{0}t}f(t)$	$F(s-s_0)$	Shifted by $\Re e(s_0)$
Time Scaling	$f(at)$	$rac{1}{	a
ight)$	Scaled by $a$
First Derivative	rac{df(t)}{dt}	$sF(s)-f(0^{-})$	At least $R$
n-th Derivative	rac{d^n f(t)}{dt^n}	$s^{n}F(s)-{\sum }_{k=1}^n{s}^{n-k}{f}^{(k-1)}(0^{-})$	At least $R$
Time Integration	${\int }_{0^{-}}^{t}f(au)dau$	rac{F(s)}{s}	At least $R\cap \{\Re e(s)>0\}$
s-Differentiation	$-tf(t)$	rac{dF(s)}{ds}	Unchanged
Time Division	rac{f(t)}{t}	${\int }_s^{\infty }F(u)du$	Converges if limit at $to0$ exists
Time Convolution	$f_1(t)\ast f_2(t)$	$F_1(s)F_2(s)$	At least $R_1\cap R_2$

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9.03.2 s-Domain Differentiation Property Proof

*(Target: Mathematical Proof / 5-Mark Derivation)*

• Theorem: Prove that \mathcal{L}\{-t \cdot f(t)\} = rac{dF(s)}{ds}.
• Derivation Steps:
  1. Express the definition integral of the Laplace transform: $F(s)={\int }_{0^{-}}^{\infty }f(t)e^{-st}dt$.
  2. Differentiate both sides with respect to the complex parameter $s$ under the integral sign:

ight] = \int_{0^-}^{\infty} f(t) \left[ rac{\partial}{\partial s} e^{-st} ight] , dt \quad ext{(derivation)}$\ast \ast \ast Last-LineBoxedEquivalent:\ast \ast$rac{dF(s)}{ds} = \int_{0^-}^{\infty} \left[ -t f(t) ight] e^{-st} , dt = \mathcal{L}{-t f(t)}$$

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9.03.3 Even and Odd Waveform Laplace Symmetries

*(Target: Mathematical Proof / 5-Mark Derivation)*

• Even Symmetries: If $f(t)=f(-t)$, then $F(s)=F(-s)$ (derivation).
• Odd Symmetries: If $f(t)=-f(-t)$, then $F(s)=-F(-s)$ (derivation).
• ROC Requirement: For an even or odd signal to have a valid Laplace representation, the s-plane poles must occur with symmetric mirror-image alignments about the imaginary axis, forcing a bilateral strip ROC.

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9.04 Initial and Final Value Theorems: Statements, Proofs, and Bounds

9.04.1 Initial Value Theorem (IVT)

*(Target: Mathematical Proof / Heavily Tested)*

• Statement: For a causal signal $f(t)$ with unilateral Laplace transform $F(s)$: ${\lim }_{to{0}^{+}}f(t)={\lim }_{so\infty }sF(s)$
• Proof Derivation:
  1. Write the Laplace transform of the first derivative: \mathcal{L}\left\{ rac{df(t)}{dt} ight\} = sF(s) - f(0^-).
  2. Take the limit as $so\infty$ on both sides: \lim_{s o \infty} \int_{0^-}^{\infty} rac{df(t)}{dt} e^{-st} \, dt = \lim_{s o \infty} [sF(s) - f(0^-)] \quad ext{(derivation)}
  3. Because the decaying term $e^{-st}o0$ as $so\infty$, the integral on the LHS vanishes.
  • Last-Line Boxed Equivalent: $0={\lim }_{so\infty }sF(s)-f(0^{-})⟹f(0^{+})={\lim }_{so\infty }sF(s)$

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9.04.2 Final Value Theorem (FVT)

*(Target: Mathematical Proof / Heavily Tested)*

• Statement: For a causal signal $f(t)$ with unilateral Laplace transform $F(s)$: ${\lim }_{to\infty }f(t)={\lim }_{so0}sF(s)$
• Proof Derivation:
  1. Write the Laplace transform of the first derivative: \mathcal{L}\left\{ rac{df(t)}{dt} ight\} = sF(s) - f(0^-).
  2. Take the limit as $so0$ on both sides: \lim_{s o 0} \int_{0^-}^{\infty} rac{df(t)}{dt} e^{-st} \, dt = \lim_{s o 0} [sF(s) - f(0^-)] \quad ext{(derivation)}
  3. Evaluating the LHS at $s=0$ simplifies $e^0=1$: \int_{0^-}^{\infty} rac{df(t)}{dt} dt = f(\infty) - f(0^-).
  • Last-Line Boxed Equivalent: $f(\infty )-f(0^{-})={\lim }_{so0}sF(s)-f(0^{-})⟹f(\infty )={\lim }_{so0}sF(s)$

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9.04.3 Practical Value Theorem Application Guidelines

*(Target: Theory Descriptive / Stability Check)*

• IVT Restriction: The numerator degree of $F(s)$ must be strictly less than the denominator degree. If the degree is equal, an impulse $\delta (t)$ exists at $t=0$, causing the IVT limit to represent only the continuous transient boundary.
• FVT Stability Boundaries: The Final Value Theorem fails completely if $sF(s)$ contains poles on or to the right of the imaginary axis (e.g., imaginary poles at $s=\pm j{\omega }_0$ cause sustained sinusoidal oscillations that do not settle to a single steady-state value as $to\infty$). Always verify that all poles of $sF(s)$ lie strictly in the stable Left-Half Plane (LHP).

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9.05 Pole-Zero Analysis, Transfer Functions & s-Domain Stability

9.05.1 Definitions: Zeros and Poles

*(Target: Theory Descriptive)*

• Zeros ($z_k$): The roots of the numerator polynomial $N(s)=0$ where the transfer function drops to zero magnitude: $H(z_k)=0$.
• Poles ($p_k$): The roots of the denominator polynomial $D(s)=0$ where the system gain approaches infinity: $H(p_k)o\infty$.

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9.05.2 Continuous LTI System Transfer Function $H(s)$

*(Target: Numerical Solving / System Mapping)*

• The ratio of the output Laplace transform to the input Laplace transform under zero initial conditions: H(s) = rac{Y(s)}{X(s)} = rac{b_m s^m + b_{m-1} s^{m-1} + \dots + b_0}{a_n s^n + a_{n-1} s^{n-1} + \dots + a_0}

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9.05.3 Stability Classifications on the s-Plane

*(Target: Theory Descriptive / Pole Mapping)*

                     jω-axis (Imaginary)
                            ▲
                            │     Unstable Region
       Stable Region        │     (RHP)
       (LHP)                │
                            │         X (sp)
            X (sp)          │
                            │
    ────────────────────────┼────────────────────────► σ (Real)
                            │
            X (sp)          │
                            │         X (sp)
                            │
                            ▼

1. Absolute Stability: All system poles lie strictly in the open Left-Half Plane (LHP) ($\Re e(p_k)<0$). The impulse response decay is bounded and absolutely integrable: ${\int }_0^{\infty }|h(t)|dt<\infty$.
2. Marginal Stability: Simple, non-repeated poles lie directly on the imaginary $j\omega$-axis (e.g., $s=\pm j{\omega }_c$), with all other poles in the LHP. This yields sustained, constant-amplitude oscillations.
3. Instability: Any pole lies in the Right-Half Plane (RHP) ($\Re e(p_k)>0$) OR repeated/multiple poles lie on the imaginary axis (yielding quadratically growing, unbounded oscillations).

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9.05.4 The Routh-Hurwitz Stability Criterion

*(Target: Theory Descriptive / Stability Test)*

• Purpose: Evaluates system stability without explicitly factoring higher-order characteristic equations.
• Procedure: Compiles the coefficients of $a_n{s}^n+a_{n-1}{s}^{n-1}+\cdots +a_0=0$ into a Routh array. System stability requires that all elements in the first column of the Routh array have the same sign. The number of sign changes in the first column equals the exact count of unstable poles lying in the RHP.

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9.06 s-Domain Modeling of Passive Circuit Elements with Initial Conditions

9.06.1 Component s-Domain Modeling Equivalents

*(Target: Numerical Solving / Circuit Modeling)*

Circuit Component	Time-Domain Relationship	s-Domain Impedance ($Z(s)$)	s-Domain Series Equivalent (KVL Form)	s-Domain Parallel Equivalent (KCL Form)
Resistor ($R$)	$v_R(t)=R\cdot i_R(t)$	$R$	$V_R(s)=R\cdot I_R(s)$	I_R(s) = rac{1}{R} V_R(s)
Inductor ($L$)	v_L(t) = L rac{di_L(t)}{dt}	$Ls$	$V_L(s)=Ls{I}_L(s)-L{i}_L(0^{-})$	I_L(s) = rac{V_L(s)}{Ls} + rac{i_L(0^-)}{s}
Capacitor ($C$)	i_C(t) = C rac{dv_C(t)}{dt}	rac{1}{Cs}	V_C(s) = rac{I_C(s)}{Cs} + rac{v_C(0^-)}{s}	$I_C(s)=Cs{V}_C(s)-C{v}_C(0^{-})$

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9.06.2 Equivalent Circuit Topologies (with Initial Conditions)

*(Target: Circuit Formulation / Analysis)*

1. Inductor s-Domain Model:

• Series Model (KVL): An impedance $Ls$ in series with an independent voltage source of value $Li(0^{-})$ pointing in a direction that opposes KVL:

          i(t)                      I(s)           + -
        o--►--[ L ]--o    ======►  o--►--[ Ls ]----( L*i(0^-) )----o
                                                    Voltage Source

• Parallel Model (KCL): An impedance $Ls$ in parallel with a current source of value rac{i(0^-)}{s} pointing in the direction of the initial current.

2. Capacitor s-Domain Model:

• Series Model (KVL): An impedance rac{1}{Cs} in series with a step voltage source of value rac{v(0^-)}{s} pointing in a direction that supports KVL:

          i(t)   + -                I(s)          +   -
        o--►--[   C   ]--o  ====►  o--►--[ 1/Cs ]----[ v(0^-)/s ]----o
                                                    Step Voltage

• Parallel Model (KCL): An impedance rac{1}{Cs} in parallel with an impulsive current source of value $Cv(0^{-})$ pointing opposite to the capacitor voltage polarity.

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9.07 Transient Response of RL & RC Circuits using Laplace Transform

*(Target: Numerical Solving / Operational Steps)*

1. Redraw the Circuit in the s-Domain: Replace all passive components ($R,L,C$) with their s-domain impedances (R, Ls, rac{1}{Cs}). Account for initial conditions using series or parallel step sources.
2. Write Mesh/Loop or Nodal Equations: Set up s-domain algebraic loop or node voltage equations using KVL and KCL.
3. Group and Isolate the Target Variable: Solve for the target variable (e.g., $I(s)$ or $V_o(s)$) by collecting and simplifying rational polynomial fractions.
4. Perform Partial Fraction Expansion: Split the target expression into simpler poles: Y(s) = rac{k_1}{s - p_1} + rac{k_2}{s - p_2} + \dots
5. Compute the Inverse Unilateral Laplace Transform: Convert each term back to the time-domain using the master pair lookup tables.

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9.08 Transient Response of Series & Parallel RLC Networks

9.08.1 Comparison: Series RLC vs. Parallel RLC Parameters

*(Target: Numerical Solving / Parameter Analysis)*

Feature / Parameter	Series RLC Circuit Transient	Parallel RLC Circuit Transient
Governing Time-Domain ODE	rac{d^2v_C}{dt^2} + rac{R}{L}rac{dv_C}{dt} + rac{1}{LC}v_C = rac{v_{in}}{LC}	rac{d^2i_L}{dt^2} + rac{1}{RC}rac{di_L}{dt} + rac{1}{LC}i_L = rac{1}{RC}rac{di_s}{dt}
s-Domain Characteristic Equation	s^2 + rac{R}{L}s + rac{1}{LC} = 0	s^2 + rac{1}{RC}s + rac{1}{LC} = 0
Attenuation Constant (lpha)	lpha = rac{R}{2L}	lpha = rac{1}{2RC}
Resonant Frequency (${\omega }_0$)	\omega_0 = rac{1}{\sqrt{LC}}	\omega_0 = rac{1}{\sqrt{LC}}
Characteristic Roots ($s_1,s_2$)	s_{1,2} = -lpha \pm \sqrt{lpha^2 - \omega_0^2}	s_{1,2} = -lpha \pm \sqrt{lpha^2 - \omega_0^2}

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9.08.2 Damping State Classifications

*(Target: Theory Descriptive / Parameter Check)*

Damping State	Mathematical Condition	Root Characteristics on s-Plane	Waveform Expression $x(t)$
Over-damped	lpha > \omega_0	Two distinct, real, negative roots on LHP	$x(t)=A_1{e}^{s_{1}t}+A_2{e}^{s_{2}t}$
Critically-damped	lpha = \omega_0	Two repeated, real, negative roots: s_{1,2} = -lpha	x(t) = (A_1 + A_2 t) e^{-lpha t}
Under-damped	lpha < \omega_0	Complex conjugate LHP roots: s_{1,2} = -lpha \pm j\omega_d	$x(t) = e^{-lpha t} \left[ B_1 \cos(\omega_d t) + B_2 \sin(\omega_d t)
ight]$

Where the damped frequency of oscillation is defined as: \omega_d = \sqrt{\omega_0^2 - lpha^2}

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9.09 Common Mistakes That Cost Marks

Critical Exam Pitfalls

• The Swapped Capacitor Series Polarities Trap: When modeling capacitor initial conditions in the s-domain with a series voltage source, the source polarity must match the polarity of the initial voltage $v_C(0^{-})$ (pointing in the direction of the electric field). For inductors, the series initial current source opposes KVL, acting as a generator.
• Applying the Final Value Theorem to Oscillatory Systems: Do not apply the FVT to marginally stable systems or oscillatory systems (e.g., F(s) = rac{\omega_0}{s^2 + \omega_0^2}). The limit ${\lim }_{so0}sF(s)=0$ is a mathematically valid calculation, but the steady-state final value of a sinusoidal wave does not exist as $to\infty$.
• The Time Delay Step Function Omission: When applying the Time Shifting Property ($f(t-t_0)\leftrightarrow e^{-s{t}_0}F(s)$), ensure that the time-domain signal is explicitly multiplied by the shifted unit step $u(t-t_0)$. If the step function is not shifted (i.e., $f(t-t_0)u(t)$), the property fails.
• Forgetting to Multiply by s Before Applying Theorems: Students often calculate ${\lim }_{so\infty }F(s)$ or ${\lim }_{so0}F(s)$ directly instead of evaluating the correct theorem limits: $\lim sF(s)$. This leads to an immediate loss of all theorem-solving marks.

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9.10 PYQ Bank — Verbatim Questions & Answer Plans

9.10.1 PYQ 2025: Series Second-Order RLC Transient (Section B, Q6c)

• Question: In the circuit of Fig. 6(c), find the currents $i_1(t)$ and $i_2(t)$ and output voltage across $5\Omega$ resistor when the switch is closed, and also determine the initial and final value of current. (09 Marks)
• Answer Plan:
  1. Convert the circuit components to s-domain: $R=10\Omega o10$, $L=0.02extHo0.02s$, $R_L=5\Omega o5$.
  2. Model initial conditions: Circuit is initially relaxed, so initial current $i(0^{-})=0$. The DC source $100extVo100/s$.
  3. Write KVL equations for Loop 1 and Loop 2 to solve for the currents $I_1(s)$ and $I_2(s)$ in the s-domain.
  4. Apply partial fraction expansion to solve for $i_1(t)$ and $i_2(t)$ in the time domain.
  5. Compute output voltage: $v_o(t)=R_L\cdot i_2(t)=5\cdot i_2(t)$.
  6. Compute initial and final current values using ${\lim }_{so\infty }sI(s)$ and ${\lim }_{so0}sI(s)$ to verify time-domain boundaries.

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9.10.2 PYQ 2023: s-Domain Differentiation Property Proof (Section B, Q6b)

• Question: If x(t) is a signal with Laplace transform X(s) then prove rac{dX(s)}{ds} = \mathcal{L}\{-tx(t)\}. (05 Marks)
• Answer Plan: Use the complete calculus proof detailed in Section 9.03.2.

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9.10.3 PYQ 2024: Value Theorems & s-Domain Definitions (Section B, Q6a)

• Question: What is the main difference between Fourier transform and Laplace transform? Find the Laplace transform of $t^n$ function. (12 Marks)
• Answer Plan:
  1. Tabulate the core differences as detailed in Section 9.01.2.
  2. Write out the complete mathematical induction and integration-by-parts proof for $\mathcal{L}\{t^n\}$ detailed in Section 9.02.3.

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9.10.4 PYQ 2016: Inverse Laplace of Repeated Multiplicity Poles (Section B, Q7a)

• Question: Determine the inverse Laplace transform of F(s) = rac{2s^2+3s+3}{(s+1)(s+3)^2}. Draw the pole-zero diagram for the given function. (12 Marks)
• Answer Plan:
  1. Formulate the partial fraction expansion with multiple poles: F(s) = rac{k_1}{s+1} + rac{k_{21}}{s+3} + rac{k_{22}}{(s+3)^2}
  2. Calculate residues using derivative evaluations: $k_1=0.5$, $k_{22}=-6$, $k_{21}=1.5$.
  3. Compute the inverse Laplace transform term-by-term: f(t) = \left[ 0.5e^{-t} + 1.5e^{-3t} - 6te^{-3t} ight]u(t).
  4. Draw the pole-zero plot on the s-plane: a simple pole at $s=-1$ and a repeated pole of multiplicity 2 at $s=-3$.

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9.11 Self-Check Before Moving On

• Can you define the unilateral Laplace transform and list its existence conditions?
• Do you know how to prove the s-domain differentiation property starting from the transform definition?
• Can you solve both the initial and final value theorems, stating the poles stability constraints?
• Can you draw the s-domain equivalent models for capacitors and inductors, showing the initial condition source polarities?
• Can you calculate series and parallel second-order attenuation constants (lpha) and damping ratios?

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Source: ECE 2107 Syllabus, (K. Deergha Rao) Signals and Systems (Ch 4), Senior Lecture Notes (lec 8, lec 11, lec 12).