Network functions, poles and zeros

Why it matters

A network function packs the entire behaviour of a linear circuit into one rational function of s. From its poles and zeros you can read off stability, the transient response, the frequency response, and whether a given impedance can be built from passive parts at all. Filter design, sensor dynamic compensation and stability checks of instrumentation amplifiers all start here.

Key ideas

Network function. A network function is the ratio of a response transform to an excitation transform, with all initial conditions zero and a single source: H(s) = Response(s)/Excitation(s).

• Driving-point functions have excitation and response at the same port: the input impedance Z(s) = V1/I1 or the input admittance Y(s) = I1/V1.
• Transfer functions have the response at a different port: the voltage ratio V2/V1, the current ratio I2/I1, the transfer impedance V2/I1 and the transfer admittance I2/V1.

Rational form. For lumped linear time-invariant circuits: H(s) = K·(s − z1)(s − z2)…(s − zm)/((s − p1)(s − p2)…(s − pn)).

• The zeros z_i make H = 0. The poles p_i make H infinite. K is the scale factor.
• The poles are the natural frequencies of the circuit. The transient (natural) response is a sum of terms e^(p_i·t).
• Poles and zeros are real or occur in complex-conjugate pairs, because the coefficients are real.
• Count the poles or zeros at infinity as well (the difference n − m).

Pole location and time response.

• A pole on the negative real axis gives a decaying exponential.
• A complex pair −α ± jωd in the left half-plane gives a damped sinusoid.
• A simple pair on the jω axis gives a sustained oscillation: marginal stability.
• A right-half-plane pole, or a repeated jω-axis pole, gives a growing response: instability.
• A circuit is BIBO stable if every pole is strictly in the left half-plane.

Frequency response. Put s = jω. Draw vectors from each zero and each pole to the point jω. Then:

• |H(jω)| = K × (product of zero-vector lengths)/(product of pole-vector lengths)
• ∠H(jω) = Σ zero angles − Σ pole angles

A pole near the jω axis produces a resonant peak; a zero on the jω axis produces a notch, which is complete rejection at that frequency.

Minimum phase. A transfer function with all zeros in the left half-plane is minimum phase. Right-half-plane zeros, as in lattice and all-pass networks, add extra phase lag for the same magnitude.

Necessary conditions for driving-point functions of passive RLC networks.

• All coefficients of the numerator and denominator are real and positive. The polynomials are Hurwitz: no missing terms unless all even or all odd terms are missing.
• Poles and zeros lie in the closed left half-plane, and any on the jω axis are simple with real positive residues.
• The degrees of the numerator and denominator differ by at most 1, and so do their lowest powers of s.

Transfer functions of passive networks need only the denominator to be Hurwitz. Their zeros can lie anywhere, and numerator coefficients can be negative or missing.

Checks. At s = 0, set inductors short and capacitors open. At s → ∞, set inductors open and capacitors short. The values Z(0) and Z(∞) found this way must match the rational function.

Formulas

• H(s) = K·Π(s − z_i)/Π(s − p_j).
• Z_R = R, Z_L = sL, Z_C = 1/(sC) (Ω).
• Impulse response: h(t) is the inverse transform of H(s). Step response: the inverse transform of H(s)/s, with final value H(0) if the system is stable.
• Magnitude: |H(jω)| = K·Π|jω − z_i|/Π|jω − p_j|.
• Phase: ∠H(jω) = Σ∠(jω − z_i) − Σ∠(jω − p_j).
• Second-order denominator: s² + 2ζω0·s + ω0², giving poles −ζω0 ± jω0·√(1 − ζ²).

Worked examples

Example 1 (standard): driving-point impedance. Given: R = 2 Ω in series with L = 1 H, with that branch in parallel with C = 0.5 F. Find Z(s), its poles and zeros, and check its limits.

1. RL branch: Z1 = 2 + s. Capacitor: Z2 = 1/(0.5s) = 2/s.
2. Parallel combination: Z(s) = Z1·Z2/(Z1 + Z2) = (2 + s)(2/s)/(2 + s + 2/s) = 2(s + 2)/(s² + 2s + 2).
3. Zero: s = −2. Poles: s = −1 ± j1. A zero at s = ∞ appears because the denominator degree exceeds the numerator degree by 1.
4. Checks: Z(0) = 4/2 = 2 Ω; at DC the inductor is a short and the capacitor is open, leaving 2 Ω ✓. As s → ∞, Z → 0 because the capacitor shorts the port ✓.
5. Spot check at ω = 1 rad/s: Z(j1) = 1.6 − j1.2 Ω.
6. All coefficients are positive, and the degrees differ by 1, consistent with a passive RLC network.

Answer: Z(s) = 2(s + 2)/(s² + 2s + 2); zero at −2; poles at −1 ± j1.

Example 2 (GATE level): frequency response from poles and zeros, and the impulse response. Given: H(s) = 10(s + 1)/((s + 2)(s + 5)). Find |H| and ∠H at ω = 2 rad/s, the DC gain, and h(t).

1. Zero vector from −1 to j2: 1 + j2, with length 2.236 and angle 63.43°.
2. Pole vectors: 2 + j2 (length 2.828, angle 45°) and 5 + j2 (length 5.385, angle 21.80°).
3. |H(j2)| = 10 × 2.236/(2.828 × 5.385) = 1.468.
4. ∠H(j2) = 63.43° − 45° − 21.80° = −3.37°.
5. DC gain: H(0) = 10 × 1/(2 × 5) = 1. This is also the final value of the unit-step response.
6. Partial fractions: H(s) = −3.333/(s + 2) + 13.333/(s + 5).
7. So h(t) = −3.333·e^(−2t) + 13.333·e^(−5t) for t ≥ 0. Both poles are in the left half-plane, so the system is stable.

Answer: |H(j2)| = 1.47, ∠H = −3.4°, H(0) = 1, h(t) = −3.33e^(−2t) + 13.33e^(−5t).

Common mistakes

• Cancelling a pole against a zero without comment. A cancelled pole is a hidden mode of the circuit; say so, or avoid the cancellation.
• Calling a system stable when it has simple poles on the jω axis. It is only marginally stable.
• Measuring the angles of pole and zero vectors from the wrong reference. Measure from the positive real direction.
• Forgetting the scale factor K when evaluating |H(jω)|.
• Applying the positive-coefficient test to the numerator of a transfer function. It applies to driving-point functions, and to the denominator only for transfer functions.
• Mixing up the sum of the poles (−a_{n−1}/a_n) with the product of the poles.

For GATE IN

Expect: locating poles and zeros of an RLC network function; deciding stability from pole positions; evaluating |H(jω)| and its phase at a given frequency; identifying a filter type (low-pass, high-pass, band-pass, notch) from its pole-zero plot; testing whether a given Z(s) can be a passive driving-point impedance; finding the impulse or step response from H(s). Practise quick root-finding of quadratics, and the vector method for magnitude and phase.

Quick check

1. H(s) = (s + 3)/(s² + 4s + 13). Where are the poles?
2. Is a system with poles at −1 and ±j2 (simple) stable?
3. Can Z(s) = (s² − s + 1)/(s + 2) be the impedance of a passive RLC network?
4. What is the DC gain of H(s) = 6/((s + 2)(s + 3))?

Answers: 1. −2 ± j3. 2. Only marginally stable. 3. No; it has a negative coefficient. 4. 1.