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Poles and zeros of a network function

Short answer

Poles and zeros are the roots of denominator and numerator of the network function H(s). Poles determine the natural response terms (e.g., e^(p t)), stability, and damping; zeros set amplitudes and block specific frequencies. A pole-zero diagram on the s-plane directly reveals time-domain behavior.

What do you understand by poles and zeros of a network function? State their significance in analyzing the time domain response of a network.

A network function H(s) (driving point or transfer function) of a lumped linear network is a ratio of polynomials in the complex frequency s = σ + jω:

H(s) = N(s)/D(s) = K (s − z1)(s − z2)...(s − zm) / [(s − p1)(s − p2)...(s − pn)]
  • Zeros z1, z2, ... are the values of s for which H(s) = 0 (roots of N(s)).
  • Poles p1, p2, ... are the values of s for which H(s) → ∞ (roots of D(s)). They are also the natural frequencies of the network.

They are shown on the s-plane (poles as ×, zeros as ○) in a pole-zero diagram. Poles and zeros are either real or occur in complex-conjugate pairs.

Significance in the time domain

  1. Natural response. Each pole pi contributes a term Ki·e^(pi·t) to the response (partial fractions followed by the inverse Laplace transform). Example: H(s) = (s + 2)/[(s + 1)(s + 3)] = 0.5/(s + 1) + 0.5/(s + 3), so the impulse response is 0.5e^(−t) + 0.5e^(−3t).
  2. Stability. Poles in the left half plane (σ < 0) give decaying terms, so the network is stable. Poles on the jω axis give sustained oscillation (marginal stability). Poles in the right half plane give growing terms (unstable).
  3. Speed and damping. A real pole at −a gives e^(−at): the farther from the origin, the faster the decay. A complex pair −α ± jβ gives e^(−αt) cos(βt + θ): α sets the decay rate and β the oscillation frequency.
  4. Zeros do not create new time functions, but they set the amplitudes (residues Ki) and the phase of each term. A zero at s = z means that an input e^(zt) produces no output; a zero on the jω axis blocks that sinusoidal frequency completely.
  5. The forced response depends on the poles of the input as well; the poles of H(s) decide the transient (natural) part.

✓ Answer checked by Claude Opus 5.5 (high effort)

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