Transfer function analysis of LTI systems

Why it matters

Every instrument has dynamics: a thermocouple in a sheath lags, a pressure transducer rings, an anti-aliasing filter attenuates. The transfer function packs those dynamics into one rational function H(s), from which you can read the steady-state gain, the speed of response, overshoot, stability and the response to any sinusoid — the core of dynamic calibration and of instrument selection.

Key ideas

Definition. For an LTI system initially at rest, the transfer function is the ratio of the Laplace transform of the output to that of the input: H(s) = Y(s)/X(s). It equals the Laplace transform of the impulse response h(t) and does not depend on the input applied.

From a differential equation. Take the Laplace transform with zero initial conditions; each dᵏ/dtᵏ becomes sᵏ. For a2 y'' + a1 y' + a0 y = b1 x' + b0 x, H(s) = (b1 s + b0)/(a2 s² + a1 s + a0). The denominator set to zero is the characteristic equation.

Poles, zeros and gain

• Poles: roots of the denominator. They set the natural modes e^(pt) and therefore stability and speed.
• Zeros: roots of the numerator. They do not change stability but shape the amplitudes of the modes and the frequency response (a zero on the jω-axis blocks that frequency completely).
• DC (steady-state) gain: H(0), the output/input ratio for a constant input when the system is stable.
• H(s) is proper if the degree of the numerator ≤ that of the denominator. Physical systems are proper; an improper H(s) (for example an ideal differentiator s) has unbounded gain at high frequency.

Stability. A causal LTI system is BIBO stable if and only if every pole has a negative real part. Simple poles on the jω-axis (and none in the right half-plane) give a marginally stable system: the natural response neither grows nor decays, but some bounded inputs (at the pole frequency) produce unbounded outputs, so it is not BIBO stable. The Routh–Hurwitz test checks pole locations without solving the polynomial.

First-order instrument H(s) = K/(τs + 1): step response K(1 − e^(−t/τ)), reaching 63.2 % at t = τ and about 98 % at 4τ; corner frequency ω = 1/τ.

Second-order instrument H(s) = K ωn²/(s² + 2ζωn s + ωn²): natural frequency ωn, damping ratio ζ. Underdamped (0 < ζ < 1) poles are −ζωn ± jωd; critically damped at ζ = 1; overdamped above. Instruments such as accelerometers and galvanometers are typically designed for ζ ≈ 0.6–0.7 to balance speed and overshoot.

Frequency response. For a stable system, put s = jω. A sinusoidal input A sin(ωt) gives, in steady state, A·|H(jω)| sin(ωt + ∠H(jω)). The phase lag can be read as a time delay −∠H/ω.

Interconnections. Cascade: H1·H2. Parallel: H1 + H2. Negative feedback: G/(1 + GH).

Formulas

• H(s) = Y(s)/X(s) = L{h(t)} — zero initial conditions.
• H(s) = K (s − z1)(s − z2)…/((s − p1)(s − p2)…) — pole-zero form.
• y_ss(t) = A |H(jω)| sin(ωt + ∠H(jω)) — sinusoidal steady state.
• H(s) = K/(τs + 1): |H(jω)| = K/√(1 + (ωτ)²), ∠H = −tan⁻¹(ωτ).
• H(s) = K ωn²/(s² + 2ζωn s + ωn²): ωd = ωn √(1 − ζ²).
• Mp = e^(−ζπ/√(1 − ζ²)) — peak overshoot (fraction), underdamped step.
• tp = π/ωd — time to first peak.
• ts ≈ 4/(ζωn) — 2 % settling time.
• Closed loop = G(s)/(1 + G(s)H(s)) — negative feedback.

Symbols: s complex frequency (s⁻¹); K static sensitivity/DC gain (output unit per input unit); τ time constant (s); ωn undamped natural frequency (rad/s); ζ damping ratio (dimensionless); ωd damped frequency (rad/s); Mp overshoot (fraction or %); tp, ts (s); A input amplitude.

Worked examples

Example 1 (standard). A temperature sensor behaves as H(s) = 1/(0.1s + 1). The measured temperature varies as 2 sin(10t) °C about its mean. Find the indicated amplitude, the phase lag and the equivalent time lag.

1. τ = 0.1 s, ω = 10 rad/s, so ωτ = 1.
2. |H(j10)| = 1/√(1 + 1²) = 0.707. Indicated amplitude = 2 × 0.707 = 1.414 °C.
3. ∠H(j10) = −tan⁻¹(1) = −45° = −0.785 rad.
4. Time lag = 0.785/10 = 0.0785 s.
5. The sensor shows 1.41 °C amplitude (29 % amplitude error), lagging by 45°, i.e. about 78.5 ms.

Example 2 (GATE level). A pressure transducer has H(s) = 100/(s² + 12s + 100). Find ωn, ζ, the DC gain, the poles, and for a unit step the peak overshoot, peak time and 2 % settling time.

1. Compare with ωn²/(s² + 2ζωn s + ωn²): ωn² = 100 ⇒ ωn = 10 rad/s; 2ζωn = 12 ⇒ ζ = 0.6.
2. DC gain H(0) = 100/100 = 1.
3. ωd = 10 √(1 − 0.36) = 8 rad/s; poles at −ζωn ± jωd = −6 ± j8 (left half-plane ⇒ stable).
4. Mp = e^(−0.6π/0.8) = e^(−2.356) = 0.0948.
5. tp = π/8 = 0.393 s; ts ≈ 4/(0.6 × 10) = 0.667 s.
6. ωn = 10 rad/s, ζ = 0.6, poles −6 ± j8, Mp ≈ 9.5 %, tp ≈ 0.393 s, ts ≈ 0.667 s.

Common mistakes

• Including initial conditions when forming H(s) — the transfer function assumes the system is at rest.
• Concluding stability from zeros. Only poles decide stability.
• Calling jω-axis poles "stable". They are marginally stable, which is not BIBO stable.
• Using ωn instead of ωd for the ringing frequency or peak time.
• Reading ts = 4/(ζωn) as exact; it is an envelope approximation.
• Evaluating |H(jω)| with s² → ω² instead of −ω².
• Applying the final value theorem or H(0) as "steady-state gain" to an unstable system.

For GATE IN

Expect: finding H(s) from a differential equation or circuit, pole-zero locations and stability, the steady-state output for a sinusoidal input (magnitude and phase at one frequency), and first- or second-order instrument parameters (τ, ωn, ζ, overshoot, settling time) from H(s) or a step response. Practise evaluating H(jω) cleanly and the standard second-order formulas.

Quick check

1. What is H(s) for y' + 5y = 10x?
2. H(s) = (s − 2)/((s + 1)(s + 3)) — stable?
3. DC gain of H(s) = 20/(s² + 4s + 5)?
4. A first-order sensor with τ = 2 s — percentage of a step reached at t = 2 s?
5. Damping ratio of s² + 4s + 16? Answers: 1. 10/(s + 5). 2. Yes — poles at −1, −3 (the right-half-plane zero does not affect stability). 3. 20/5 = 4. 4. 63.2 %. 5. ωn = 4, 2ζ·4 = 4 ⇒ ζ = 0.5.