Laplace Transform —
Pole-Zero & System Stability
- 1. Why Laplace Transform?
- 2. LT Definition & Bilateral vs Unilateral
- 3. Region of Convergence (ROC)
- 4. Standard LT Pairs
- 5. Properties of Laplace Transform
- 6. Inverse Laplace Transform (Partial Fractions)
- 7. Transfer Function H(s) — Poles & Zeros
- 8. BIBO Stability from ROC & Poles
- 9. Initial & Final Value Theorems
- 10. GATE Questions (12)
Circuit analysis: Solving an RLC circuit with differential equations is hard. LT converts d/dt → s·, ∫dt → 1/s, turning ODEs into algebraic equations in s-domain. Solve in s → take inverse LT → get time response.
Control systems: The transfer function H(s) = Y(s)/X(s) tells you the gain, bandwidth, and stability from just the pole/zero locations — without solving the ODE. Engineers design PID controllers, compensators, and feedback loops entirely in the s-domain.
Filter design: Butterworth, Chebyshev, Bessel filters are all specified by their pole-zero pattern in the s-plane, then converted to hardware using the bilinear transform.
The s-plane is the foundation of control and signal processing. Pole locations determine system behavior: LHP poles → stable/decaying response; RHP poles → unstable/growing response; jω-axis poles → sustained oscillation (marginal stability).
1. Why Laplace Transform?
The Fourier Transform requires the signal to be absolutely integrable, which excludes many important signals like ramp, step, and growing exponentials. The Laplace Transform introduces a convergence factor e^(−σt) that forces convergence for a wider class of signals, enabling analysis of unstable systems and transient responses.
2. LT Definition & Bilateral vs Unilateral
3. Region of Convergence (ROC)
The ROC is the set of s-values for which X(s) = ∫|x(t)e^(−st)|dt converges. It is always a vertical strip in the s-plane and never contains poles.
| Signal Type | ROC | Example |
|---|---|---|
| Right-sided (causal): x(t)=0 for t<T | Re(s) > σ₀ (right half-plane) | e^(−at)u(t) → ROC: Re(s)>−a |
| Left-sided (anti-causal): x(t)=0 for t>T | Re(s) < σ₀ (left half-plane) | −e^(−at)u(−t) → ROC: Re(s)<−a |
| Two-sided (bilateral) | σ₁ < Re(s) < σ₂ (vertical strip) | e^(−a|t|) → ROC: −a<Re(s)<a |
| Finite duration | Entire s-plane (except possibly s=0,∞) | rect(t/τ) → all s |
| Right-sided with poles at s=−a,−b | Re(s) > max(−a, −b) | ROC must be to right of all poles |
1. ROC is a right half-plane for causal signals, left half-plane for anti-causal.
2. ROC never contains poles.
3. FT exists ↔ ROC includes the jω-axis (Re(s)=0).
4. For BIBO stability: ROC of H(s) must include jω-axis → all poles in LHP.
4. Standard Laplace Transform Pairs
| x(t) [causal, t≥0] | X(s) | ROC |
|---|---|---|
| δ(t) | 1 | All s |
| u(t) | 1/s | Re(s) > 0 |
| t·u(t) (ramp) | 1/s² | Re(s) > 0 |
| tⁿu(t) | n!/s^(n+1) | Re(s) > 0 |
| e^(−at)u(t) | 1/(s+a) | Re(s) > −a |
| t·e^(−at)u(t) | 1/(s+a)² | Re(s) > −a |
| tⁿe^(−at)u(t) | n!/(s+a)^(n+1) | Re(s) > −a |
| sin(ω₀t)u(t) | ω₀/(s² + ω₀²) | Re(s) > 0 |
| cos(ω₀t)u(t) | s/(s² + ω₀²) | Re(s) > 0 |
| e^(−at)sin(ω₀t)u(t) | ω₀/[(s+a)² + ω₀²] | Re(s) > −a |
| e^(−at)cos(ω₀t)u(t) | (s+a)/[(s+a)² + ω₀²] | Re(s) > −a |
| δ(t − T) | e^(−sT) | All s (T>0) |
5. Properties of Laplace Transform
| Property | Time Domain x(t) | s-Domain X(s) | ROC |
|---|---|---|---|
| Linearity | ax(t) + by(t) | aX(s) + bY(s) | ROC_x ∩ ROC_y |
| Time Shifting | x(t − t₀)u(t−t₀) | e^(−st₀)X(s) | Same ROC |
| s-Domain Shifting | e^(−at)x(t) | X(s+a) | ROC shifted by a |
| Time Scaling | x(at), a>0 | (1/a)X(s/a) | Scaled ROC |
| Time Reversal | x(−t) | X(−s) | −ROC |
| Differentiation (time) | dx/dt | sX(s) − x(0⁻) [unilateral] | ROC_x (at min) |
| Integration (time) | ∫₀^t x(τ)dτ | X(s)/s | ROC_x ∩ {Re(s)>0} |
| Differentiation (s) | −t·x(t) | dX(s)/ds | ROC_x |
| Convolution | x(t)*h(t) | X(s)·H(s) | ROC_x ∩ ROC_h |
| Initial Value Theorem | x(0⁺) | lim_{s→∞} s·X(s) | Causal x(t) |
| Final Value Theorem | x(∞) | lim_{s→0} s·X(s) | Poles of sX(s) in LHP |
Differentiation in Circuit Analysis
Series RLC circuit: KVL gives: L·di/dt + R·i + (1/C)∫i dt = v_in(t)
Taking LT (unilateral, zero initial conditions):
L·s·I(s) + R·I(s) + (1/Cs)·I(s) = V_in(s)
→ I(s) = V_in(s) / (Ls + R + 1/Cs)
Transfer function: H(s) = I(s)/V_in(s) = Cs/(LCs² + RCs + 1)
Poles of H(s) = natural frequencies of the RLC circuit. Finding poles tells you whether the circuit rings (complex poles), overdamps (real poles), or has sustained oscillation (poles on jω-axis). No differential equations needed!
6. Inverse Laplace Transform (Partial Fractions)
For rational X(s) = N(s)/D(s), the inverse LT is found by partial fraction expansion (PFE), then using the LT pair table.
Case 1 — Distinct Real Poles
Case 2 — Repeated Poles (order r)
Case 3 — Complex Conjugate Poles
7. Transfer Function H(s) — Poles & Zeros
For an LTI system with impulse response h(t), the transfer function is H(s) = LT{h(t)}. It can also be expressed as a rational function:
All 3 poles (s=-1, s=-1±j2) are in the Left Half-Plane (Re(s)=-1<0). The zero at s=-2 does not affect stability. ROC extends from σ=-1 to +∞, which includes the jω-axis — confirming BIBO stability.
8. BIBO Stability from ROC & Poles
Feedback amplifier: H(s) = K/(s+2). Pole at s=−2 (LHP) → stable for any K>0.
Feedback with loop gain: Closed-loop H_cl(s) = KG(s)/(1+KG(s)). If loop gain is too high, poles migrate to RHP → amplifier oscillates (unstable). This is why audio systems howl when the microphone is too close to the speaker.
Motor control: An electric motor's angular velocity response H(s)=1/(Js+B) has a single LHP pole at s=−B/J. Adding a proportional controller K shifts the pole to −(B+K)/J — still stable but faster response. Integral control (adds s=0 pole) → design carefully to avoid instability.
9. Initial & Final Value Theorems
FVT gives wrong answer if applied carelessly. Always check: does s·X(s) have all poles in the open LHP? If poles are on the jω-axis or RHP, the FVT cannot be used. Common GATE trap: applying FVT to 1/(s²+ω₀²) and getting 0 — but the actual signal oscillates forever.
GATE Questions — Laplace Transform (12)
Standard LT pair: t·e^(−at)u(t) ↔ 1/(s+a)² with a = 3.
Answer: (B) 1/(s+3)²
Poles of H(s) are at s = −1 and s = −3. Both poles have Re(s) < 0 (in LHP).
For a causal system, ROC: Re(s) > −1. This includes the jω-axis.
Answer: (A) Stable — all poles in open LHP, ROC includes jω-axis.
IVT: x(0⁺) = lim_{s→∞} s·X(s) = lim_{s→∞} s·(2s+5)/(s²+5s+6)
Answer: (A) x(0⁺) = 2
X(s) = 1/(s+2) has a pole at s = −2.
For a right-sided (causal) signal: ROC is to the right of the rightmost pole.
ROC: Re(s) > −2 → corresponds to x(t) = e^(−2t)u(t).
If ROC were Re(s) < −2: x(t) = −e^(−2t)u(−t) [left-sided, anti-causal].
Answer: (A) Re(s) > −2
Check FVT applicability: s·X(s) = 5/(s+5). Pole at s = −5 (LHP) ✓ → FVT valid.
x(t) = (1 − e^(−5t))u(t) → x(∞) = 1 ✓
Answer: (C) 1
PFE: (2s+3)/[(s+1)(s+2)] = A/(s+1) + B/(s+2)
X(s) = 1/(s+1) + 1/(s+2) → x(t) = (e^(−t) + e^(−2t))u(t)
Answer: (B) (e^(−t) + e^(−2t))u(t)
(A) Pole at s=−2: LHP → stable
(B) Pole at s=+2: RHP → UNSTABLE
(C) Poles at s=−1±j: LHP → stable
(D) Poles at s=−1, s=−2: LHP → stable
Answer: (B) — pole at s=+2 in RHP → unstable
e^(−at)u(t) ↔ 1/(s+a), so 1/(s+1) ↔ e^(−t)u(t)
Time shift property: multiply by e^(−sT) ↔ delay by T in time.
Signal is zero for t<2, then decays as e^(−(t−2)) for t≥2.
Answer: (B) e^(−(t−2))u(t−2)
Standard pairs: cos(ω₀t)u(t) ↔ s/(s²+ω₀²) and sin(ω₀t)u(t) ↔ ω₀/(s²+ω₀²)
Memory trick: cos starts at amplitude 1 → needs the s in numerator (like derivative). sin starts at 0 → ω₀ in numerator.
Answer: (A) s/(s²+ω₀²)
Y(s) = X(s)·H(s) = [1/(s+b)] · [1/(s+a)] = 1/[(s+a)(s+b)]
y(t) = [e^(−at) − e^(−bt)]/(b−a) · u(t)
Answer: (A)
Unilateral LT differentiation: LT{dx/dt} = sX(s) − x(0⁻)
The initial condition x(0⁻) is subtracted. This allows solving ODEs with initial conditions directly in the s-domain.
For zero initial conditions: LT{dx/dt} = sX(s) (same as bilateral).
Answer: (B) sX(s) − x(0⁻)
Check: s·X(s) = s · 3s/(s²+9) = 3s²/(s²+9)
Poles of s·X(s): s²+9=0 → s = ±j3. These poles are on the jω-axis — NOT in the open LHP.
FVT condition violated → FVT is NOT applicable.
Actual x(t) = 3·cos(3t)·u(t) — oscillates forever, no final value.
Answer: (D) FVT not applicable (poles of s·X(s) on jω-axis)