9.04 Initial and Final Value Theorems: Statements, Proofs, and Bounds | 9.06 s-Domain Modeling of Passive Circuit Elements with Initial Conditions

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

Core Idea

In the complex $s$ -plane, any rational system function $H (s)$ can be completely characterized by its poles {values of $s$ where the denominator polynomial is zero, causing the system gain to shoot to infinity} and its zeros {values of $s$ where the numerator polynomial is zero, completely blocking signal transmission}. The spatial distribution of these poles dictates the absolute, marginal, or unstable behavior of the Linear Time-Invariant (LTI) system. By plotting these parameters and performing partial fraction expansion, we can analytically reconstruct the complete time-domain transient response and verify system stability bounds.

1. Mathematical Definitions: Poles, Zeros & Transfer Functions

An analog continuous-time LTI system is described in the complex frequency domain by its transfer function $H (s)$ {the ratio of the Laplace transform of the output to the input under zero initial conditions}. For physical networks, this is a rational function of the form:

$H(s) = rac{Y(s)}{X(s)} = rac{N(s)}{D(s)} = rac{b_m s^m + b_{m-1} s^{m-1} + \dots + b_0}{a_n s^n + a_{n-1} s^{n-1} + \dots + a_0} \quad [459, 467]$

1.1 The Zeros of a System

The roots of the numerator polynomial $N (s) = 0$ are called the zeros of the system, denoted as $z_{1}, z_{2}, \dots, z_{m}$ .

Physical Meaning: If an input signal $x (t) = e^{z_{i} t}$ is applied, the output is identically zero ( $y (t) = 0$ ) because the system exhibits zero gain at these complex frequencies.
Representation: Marked on the $s$ -plane pole-zero plot using small circles (o).

1.2 The Poles of a System

The roots of the denominator polynomial $D (s) = 0$ are called the poles of the system, denoted as $p_{1}, p_{2}, \dots, p_{n}$ .

Physical Meaning: The poles represent the natural frequencies {intrinsic, self-sustaining modes of vibration or decay} of the system. If a system is excited, its transient response is composed of exponentials determined entirely by the pole locations ( $e^{p_{i} t}$ ).
Representation: Marked on the $s$ -plane pole-zero plot using small crosses (x).

                     Imaginary Axis (jw)
                             ^
                             |       o  (Zeros 'o' represent transmission blocks)
                             |
                             |
       X                     |
     (Pole 'x' at s = -a)    |
     <-----------------------+-----------------------> Real Axis (σ)
                             |
                             |
                             |
                             v

2. s-Domain Stability Classifications

The Bounded-Input Bounded-Output (BIBO) stability of an LTI system is governed by the absolute integrability of its impulse response: $\int_{- \infty}^{\infty} ∣ h (t) ∣ d t < \infty$ . In the Laplace domain, this is equivalent to stating that the Region of Convergence (ROC) of the transfer function $H (s)$ must encompass the imaginary axis ( $jω$ -axis).

For a causal system {where $h (t) = 0$ for $t < 0$ }, the ROC is always to the right of the rightmost pole. This leads to three highly tested stability regimes:

Stability Classification	Pole Locations in the $s$ -Plane ( $ℜ e (s)$ )	Time-Domain Impulse Response Behavior ( $h (t)$ )
Absolute Stability	All poles lie strictly in the left-half plane (LHP): $ℜ e (p_{i}) < 0$ for all $i$ .	Exponentially decaying transients: $lim_{t o \infty} h (t) = 0$ (Stable).
Marginal Stability	Simple, non-repeated poles lie directly on the imaginary axis: $ℜ e (p_{i}) = 0$ .	Sustained, non-decaying sinusoidal or DC oscillations: $h (t)$ is bounded but does not decay to $0$ .
Unstable System	Any pole lies in the right-half plane (RHP) ( $ℜ e (p_{i}) > 0$ ) OR repeated poles exist on the $jω$ -axis.	Transients grow exponentially ( $\propto e^{σ t}$ ) or linearly ( $\propto t sin (ω t)$ ) to infinity (Blow-up).

The Imaginary Multiplicity Trap (Marginal vs. Unstable)

A single pair of non-repeated poles on the imaginary axis (e.g., $s = \pm j ω_{0}$ ) represents a marginally stable system (producing a constant oscillation $cos (ω_{0} t) u (t)$ ). However, if the imaginary poles are repeated (e.g., $(s^{2} + ω_{0}^{2})^{2} = 0$ ), the time-domain transient contains a ramp-weighted term $t sin (ω_{0} t) u (t)$ which grows to infinity. Thus, repeated imaginary poles render the system completely unstable!

3. The Routh-Hurwitz Stability Criterion

For high-order systems (where the denominator $D (s)$ is a polynomial of degree $n \geq 3$ ), factoring roots by hand in an exam is extremely difficult. The Routh-Hurwitz Criterion is an algebraic procedure {testing for sign changes in the first column of the constructed Routh array to determine if any poles lie in the right-half plane without explicitly solving the characteristic polynomial}.

If the characteristic equation is: $D (s) = a_{n} s^{n} + a_{n - 1} s^{n - 1} + \dots + a_{1} s + a_{0} = 0$

Prerequisite Condition: For a system to be stable, all coefficients $a_{i}$ must be strictly positive ( $a_{i} > 0$ ). If any coefficient is zero or negative, the system is automatically unstable.
Routh Array Construction: $s^n & a_n & a_{n-2} & a_{n-4} \ s^{n-1} & a_{n-1} & a_{n-3} & a_{n-5} \ s^{n-2} & b_1 & b_2 & b_3 \ s^{n-3} & c_1 & c_2 & c_3 \end{array}$$ Where the coefficients are evaluated cross-multiplication style: $$b_1 = rac{a_{n-1} a_{n-2} - a_n a_{n-3}}{a_{n-1}}, \quad b_2 = rac{a_{n-1} a_{n-4} - a_n a_{n-5}}{a_{n-1}}, \quad c_1 = rac{b_1 a_{n-3} - a_{n-1} b_2}{b_1}$$$
The Routh Stability Theorem: The number of roots of $D (s)$ with positive real parts (RHP poles) is exactly equal to the number of sign changes in the first column of the Routh array. For absolute stability, there must be zero sign changes!

4. Exhaustive Solved “Exam Killers” (Calculus & Plotting)

Let’s solve the most critical, recurring continuous-time pole-zero Past Year Questions step-by-step to lock down maximum marks.

4.1 The 13-Mark Network Current Inversion [PYQ 2025 / 2018 / 2016]

Question: Draw the poles and zeros for the current $I (s)$ in a network given by: $I(s) = rac{3s}{(s+2)(s+4)}$ and hence, obtain the time-domain current $i (t)$ .

Step 1: Identify the Poles and Zeros

Zeros: Set the numerator $N (s) = 3 s = 0 ⟹ s = 0$ . Thus, there is one simple zero at the origin: $z_{1} = 0$ .
Poles: Set the denominator $D (s) = (s + 2) (s + 4) = 0 ⟹ s = - 2, - 4$ . Thus, there are two simple poles: $p_{1} = - 2$ and $p_{2} = - 4$ .

Step 2: Draw the Pole-Zero Map & ROC

Because the current exists in a physical, causal network ( $i (t) = 0$ for $t < 0$ ), the Region of Convergence (ROC) must be right-sided and bounded by the rightmost pole ( $p_{1} = - 2$ ). $e x t ROC : ℜ e (s) > - 2 [52, 60]$

                     Imaginary Axis (jw)
                             ^
                             |       
                             |
       X          X          o
    (p2=-4)    (p1=-2)     (z1=0) 
  ---(x)---------(x)--------(o)---------------------> Real Axis (σ)
   -4         -2             |        /////// (Shaded ROC
                             |        ///////  re(s) > -2)
                             |       
                             v

Stability Check: Since both poles lie strictly in the left-half of the $s$ -plane ( $ℜ e (p_{i}) < 0$ ) and the ROC includes the $jω$ -axis, the network is absolutely stable.

Step 3: Perform Partial Fraction Expansion

$I(s) = rac{3s}{(s+2)(s+4)} = rac{A_1}{s+2} + rac{A_2}{s+4}$ Evaluate residues using the cover-up method:

ight|_{s=-2} = \left. rac{3s}{s+4} ight|_{s=-2} = rac{3(-2)}{-2+4} = rac{-6}{2} = -3 \quad [44]$$ $$A_2 = \left. (s+4) I(s) ight|_{s=-4} = \left. rac{3s}{s+2} ight|_{s=-4} = rac{3(-4)}{-4+2} = rac{-12}{-2} = 6 \quad [44]$$ Substitute the residues back: $$I(s) = rac{-3}{s+2} + rac{6}{s+4}$$ #### Step 4: Apply Inverse Laplace Transform Using the standard causal transform pair $\mathcal{L}^{-1}\left\{ rac{1}{s+a} ight\} = e^{-at} u(t)$: $$i(t) = \left[ -3e^{-2t} + 6e^{-4t} ight] u(t) ext{ Amperes} \quad [47]$$ --- ### 4.2 The 5-Mark Rational Voltage Evaluation [PYQ 2024] **Question:** Draw the poles and zeros for the rational voltage function: $$V(s) = rac{(s+1)(s+3)}{(s+2)(s+4)}$$ and evaluate the time-domain output $v(t)$. #### Step 1: Identify the Poles and Zeros * **Zeros:** Set $(s+1)(s+3) = 0 \implies z_1 = -1$ and $z_2 = -3$. * **Poles:** Set $(s+2)(s+4) = 0 \implies p_1 = -2$ and $p_2 = -4$. #### Step 2: Draw the Pole-Zero Map & ROC The ROC is causal: $\Re e(s) > -2$. ``` Imaginary Axis (jw) ^ | | X o X o (p2=-4) (z2=-3) (p1=-2) (z1=-1) ---(x)---------(o)--------(x)--------(o)-----------> Real Axis (σ) -4 -3 -2 -1 | /////// (Shaded ROC | /////// re(s) > -2) | v ``` #### Step 3: Polynomial Division for Improper Rational Form Because the numerator polynomial degree ($m=2$) equals the denominator degree ($n=2$), this is an **improper rational function**. We must perform polynomial division before applying partial fractions: $$V(s) = rac{s^2 + 4s + 3}{s^2 + 6s + 8} = 1 - rac{2s + 5}{(s+2)(s+4)} \quad [144]$$ Now expand the remaining proper fraction: $$ rac{2s + 5}{(s+2)(s+4)} = rac{B_1}{s+2} + rac{B_2}{s+4}$$ Evaluate the residues: $$B_1 = \left. rac{2s+5}{s+4} ight|_{s=-2} = rac{2(-2)+5}{-2+4} = rac{1}{2} = 0.5$$ $$B_2 = \left. rac{2s+5}{s+2} ight|_{s=-4} = rac{2(-4)+5}{-4+2} = rac{-3}{-2} = 1.5$$ Assemble the expanded Laplace equation: $$V(s) = 1 - rac{0.5}{s+2} - rac{1.5}{s+4}$$ #### Step 4: Apply Inverse Laplace Transform Using standard transform pairs $\mathcal{L}^{-1}\{1\} = \delta(t)$ and $\mathcal{L}^{-1}\left\{ rac{1}{s+a} ight\} = e^{-at}u(t)$: $$v(t) = \delta(t) - \left[ 0.5e^{-2t} + 1.5e^{-4t} ight] u(t) ext{ Volts} \quad [50]$$ --- ### 4.3 The 12-Mark Multiple Pole Challenge [PYQ 2016 / 2015] **Question:** Determine the inverse Laplace transform of: $$F(s) = rac{2s^2+3s+3}{(s+1)(s+3)^2}$$ and draw the pole-zero diagram for the given function. #### Step 1: Identify the Poles and Zeros * **Poles:** Set $(s+1)(s+3)^2 = 0 \implies p_1 = -1$ (simple pole) and $p_{2,3} = -3$ (repeated pole of multiplicity $r = 2$). * **Zeros:** Set $2s^2+3s+3 = 0$. Using the quadratic formula: $$s = rac{-3 \pm \sqrt{3^2 - 4(2)(3)}}{2(2)} = rac{-3 \pm \sqrt{9 - 24}}{4} = -0.75 \pm j rac{\sqrt{15}}{4} pprox -0.75 \pm j0.968$$ Thus, there is a pair of complex conjugate zeros: $z_{1,2} = -0.75 \pm j0.968$. #### Step 2: Draw the Pole-Zero Map & ROC The causal ROC is defined by the rightmost pole ($p_1 = -1$): $\Re e(s) > -1$. ``` Imaginary Axis (jw) ^ | o (z1 = -0.75 + j0.97) | X X (p2 = -3, | double) (p1 = -1) ---(x)---------------------(x)-----+-----------------> Real Axis (σ) -3 -1 | /////// (Shaded ROC | | /////// re(s) > -1) | o (z2 = -0.75 - j0.97) | v ``` * **Stability Check:** Since all poles lie strictly in the Left-Half Plane ($\Re e(p_i) < 0$), the system is **absolutely stable**. #### Step 3: Perform Partial Fraction Expansion with Multiple Poles Because we have a pole of multiplicity 2 at $s = -3$, we write the expansion as: $$F(s) = rac{2s^2+3s+3}{(s+1)(s+3)^2} = rac{k_1}{s+1} + rac{k_{21}}{s+3} + rac{k_{22}}{(s+3)^2} \quad [45]$$ 1. **Evaluate simple pole residue $k_1$:** $$k_1 = \left. (s+1)F(s) ight|_{s=-1} = \left. rac{2s^2+3s+3}{(s+3)^2} ight|_{s=-1} = rac{2(-1)^2+3(-1)+3}{(-1+3)^2} = rac{2}{4} = 0.5 \quad [44]$$ 2. **Evaluate highest-order repeated pole residue $k_{22}$:** $$k_{22} = \left. (s+3)^2 F(s) ight|_{s=-3} = \left. rac{2s^2+3s+3}{s+1} ight|_{s=-3} = rac{2(-3)^2+3(-3)+3}{-3+1} = rac{18-9+3}{-2} = rac{12}{-2} = -6 \quad [46]$$ 3. **Evaluate first-order repeated pole residue $k_{21}$ using differentiation:** $$k_{21} = \left. rac{1}{1!} rac{d}{ds} \left[ (s+3)^2 F(s) ight] ight|_{s=-3} = \left. rac{d}{ds} \left[ rac{2s^2+3s+3}{s+1} ight] ight|_{s=-3} \quad [46]$$ Apply the Quotient Rule $ rac{d}{ds}\left[ rac{u}{v} ight] = rac{u'v - uv'}{v^2}$: $$ rac{d}{ds} \left[ rac{2s^2+3s+3}{s+1} ight] = rac{(4s+3)(s+1) - (2s^2+3s+3)(1)}{(s+1)^2}$$ Evaluate at $s = -3$: * $4s+3 o -9$ * $s+1 o -2$ * $2s^2+3s+3 o 12$ * $(s+1)^2 o 4$ $$k_{21} = rac{(-9)(-2) - (12)}{4} = rac{18 - 12}{4} = rac{6}{4} = 1.5$$ Assemble the final expanded s-domain function: $$F(s) = rac{0.5}{s+1} + rac{1.5}{s+3} - rac{6}{(s+3)^2}$$ #### Step 4: Apply Inverse Laplace Transform Using standard causal pairs $\mathcal{L}^{-1}\left\{ rac{1}{s+a} ight\} = e^{-at}u(t)$ and the repeated pole pair $\mathcal{L}^{-1}\left\{ rac{1}{(s+a)^2} ight\} = t e^{-at}u(t)$: $$f(t) = \left[ 0.5 e^{-t} + 1.5 e^{-3t} - 6t e^{-3t} ight] u(t) \quad [48, 50]$$ --- ## 5. ECE 2108 Laboratory Connection (Poles & Stability) In your **ECE 2108 Signals and Systems Lab**, you analyzed system stability using s-plane poles. Here is the exact MATLAB syntax used to extract poles, residues, and plot pole-zero diagrams for a continuous system: ```matlab % ECE 2108 Continuous-Time Stability Lab clc; clear; % System definition: F(s) = (2s^2 + 3s + 3) / (s^3 + 7s^2 + 15s + 9) num = [2, 3, 3]; den = [1, 7, 15, 9]; % Expanded form of (s+1)(s+3)^2 % Extract residues (r), poles (p), and direct term (k) [r, p, k] = residue(num, den); % Print structural output to terminal disp('Poles of the system:'); disp(p); disp('Residues of the system:'); disp(r); % Plot the s-plane pole-zero diagram figure; pzmap(tf(num, den)); grid on; title('s-Plane Pole-Zero Map (Continuous LTI)'); ``` --- ## 6. Common Mistakes That Cost Marks > [!failure] **1. The Improper Partial Fraction Trap** > > Attempting to perform partial fraction expansion directly on a system function where the numerator order $m$ is equal to or greater than the denominator order $n$ without performing polynomial division first. This completely breaks residue algebra and results in a 0-mark deduction on questions like **PYQ 2024 (Section 4.2)**. > * *Remedy:* Always check if $m \ge n$ first! If so, divide the polynomials to isolate direct constant terms ($\delta(t)$) or doublet terms. > [!failure] **2. Repeated Imaginary Pole Stability Classification** > > Claiming a system with repeated poles on the $j\omega$-axis (e.g., $s^2 = -4$ twice) is "marginally stable". > * *Remedy:* Simple poles on the imaginary axis are marginally stable. **Repeated poles on the imaginary axis are strictly UNSTABLE** because their time-domain response grows as $t \sin(\omega_0 t)u(t)$. > [!failure] **3. Omission of the Time multiplier for Multiple Poles** > > Writing the inverse Laplace of a repeated pole term $ rac{A}{(s+a)^2}$ as $A e^{-at} u(t)$ instead of $A t e^{-at}u(t)$. > * *Remedy:* A repeated root in the s-domain always introduces a linear growth multiplier $t$ in the time domain. --- ## 7. PYQ Bank — Verbatim Questions & Answer Plans ### Q1: Draw poles and zeros for current $I(s) = rac{3s}{(s+2)(s+4)}$ and solve for $i(t)$. [PYQ 2025/2018/2016 - 13 Marks] * **Answer Plan:** 1. Set numerator $3s = 0 \implies$ Zero at origin $s = 0$. 2. Set denominator $(s+2)(s+4) = 0 \implies$ Poles at $s = -2, -4$. 3. Draw $s$-plane diagram; mark poles with 'x' and zeros with 'o'. Shade causal ROC $\Re e(s) > -2$. 4. Apply cover-up residue formula to find $I(s) = rac{-3}{s+2} + rac{6}{s+4}$. 5. Take inverse Laplace: $i(t) = [-3e^{-2t} + 6e^{-4t}]u(t)$. ### Q2: Draw poles and zeros for $V(s) = rac{(s+1)(s+3)}{(s+2)(s+4)}$ and evaluate $v(t)$. [PYQ 2024 - 5 Marks] * **Answer Plan:** 1. Numerator roots $\implies$ Zeros at $s = -1, -3$; Denominator roots $\implies$ Poles at $s = -2, -4$. 2. Plot pole-zero map. 3. Because numerator degree equals denominator degree, perform polynomial division first: $V(s) = 1 - rac{2s+5}{(s+2)(s+4)}$. 4. Expand the fraction to find residues: $V(s) = 1 - rac{0.5}{s+2} - rac{1.5}{s+4}$. 5. Invert: $v(t) = \delta(t) - [0.5e^{-2t} + 1.5e^{-4t}]u(t)$. ### Q3: Define zeros and poles in the $s$-domain. How is system stability determined? [PYQ 2023/2022/2018/2017 - 10 Marks] * **Answer Plan:** 1. Define poles as roots of $D(s) = 0$ where system gain approaches infinity, and zeros as roots of $N(s) = 0$ where system response is zero. 2. Explain the BIBO stability condition: the ROC of $H(s)$ must include the imaginary axis ($j\omega$-axis). 3. Detail s-plane locations: * **Stable:** All poles lie strictly in the Left-Half Plane ($\Re e(p) < 0$). * **Marginally Stable:** Non-repeated poles on the imaginary axis. * **Unstable:** Any poles in the Right-Half Plane or repeated poles on the imaginary axis. --- ## 8. Self-Check Before Moving On - [ ] Can you define poles and zeros and explain their physical meaning? [1.1, 1.2] - [ ] Do you know how to plot crosses (x) and circles (o) on the complex $s$-plane? [1.2, 5] - [ ] Can you classify a system as stable, marginally stable, or unstable based on its poles? - [ ] Do you know how to handle improper rational functions where $m \ge n$ before partial fractions? [4.2] - [ ] Can you solve partial fractions featuring repeated poles using s-domain differentiation? [4.3] --- *Citations: [(k.Deergha Rao) signals and systems.pdf, passages 1, 25, 26, 43, 44, 45, 46, 47, 48, 50, 52, 59, 60, 61, 143, 144]*

007_index

9.05 Pole-Zero Analysis Transfer Functions and s-Domain Stab