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Determining transfer function from Bode plot

Short answer

Fit straight-line asymptotes to magnitude plot; low-frequency slope gives system type, vertical level gives gain K. Each slope change indicates a pole or zero corner frequency; quadratic factors identified by resonant peak. Phase plot confirms minimum-phase assumption. Combine factors to write transfer function.

Discuss how Bode plot can be used to determine transfer function of the system.

A Bode plot can be read "backwards" to find the transfer function of a minimum-phase system. The magnitude asymptotes give the gain, type and corner frequencies; the phase plot confirms them.

Procedure

  1. Draw straight-line asymptotes on the measured magnitude curve, using slopes that are multiples of ±20 dB/decade.
  2. Initial (low-frequency) slope gives the type:
    • 0 dB/dec: type 0 (no pole at the origin);
    • −20 dB/dec: one pole at the origin;
    • −40 dB/dec: two poles at the origin.
  3. Find the gain K:
    • Type 0: the low-frequency level is 20log⁡K20\log K.
    • Type 1: the −20 dB/dec line (extended if needed) meets 0 dB at ω = K. Equivalently, K=ω⋅10M/20K = \omega \cdot 10^{M/20} at any point ω on it, where M is the magnitude in dB.
    • Type 2: the −40 dB/dec line meets 0 dB at ω=K\omega = \sqrt K.
  4. Corner frequencies: at every break in slope:
    • −20 dB/dec change: simple pole 1/(1+s/ωc)1/(1 + s/\omega_c);
    • +20 dB/dec change: simple zero (1+s/ωc)(1 + s/\omega_c);
    • −40 dB/dec change: double pole or quadratic 1/(1+2ζs/ωn+s2/ωn2)1/(1 + 2\zeta s/\omega_n + s^2/\omega_n^2), with ωn\omega_n = corner.
  5. Damping of a quadratic factor: measure the resonant peak above the asymptote at ωn\omega_n. Then Mr=1/(2ζ1−ζ2)M_r = 1/(2\zeta\sqrt{1 - \zeta^2}), or approximately 20log⁡12ζ20\log\frac{1}{2\zeta} dB at ωn\omega_n.
  6. Check with the phase plot: for a minimum-phase system the phase must agree with the factors found. For example, the phase approaches −90∘×(number of poles−zeros)-90^\circ \times (\text{number of poles} - \text{zeros}) at high frequency. Extra phase lag with no extra magnitude change indicates a transport delay e−sTe^{-sT} (the slope of phase vs ω gives T) or an RHP zero.
  7. Write the result in time-constant form, then convert to pole-zero form if needed.

Short example

The initial slope is −20 dB/dec, and its extension crosses 0 dB at ω = 10; the slope breaks to −40 dB/dec at ω = 5:

G(s)=10s(1+s/5)=50s(s+5)G(s) = \frac{10}{s(1 + s/5)} = \frac{50}{s(s+5)}

Limitations

The method assumes minimum-phase behaviour, and asymptotes give only approximate corner frequencies. Accurate values come from fitting the exact curve (e.g. −3 dB at a simple-pole corner).

✓ Answer checked by Claude Opus 5.5 (high effort)

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