- Asked 2/27
- 2071 Chaitra · 5 marks
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
- Draw straight-line asymptotes on the measured magnitude curve, using slopes that are multiples of ±20 dB/decade.
- 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.
- Find the gain K:
- Type 0: the low-frequency level is .
- Type 1: the −20 dB/dec line (extended if needed) meets 0 dB at ω = K. Equivalently, at any point ω on it, where M is the magnitude in dB.
- Type 2: the −40 dB/dec line meets 0 dB at .
- Corner frequencies: at every break in slope:
- −20 dB/dec change: simple pole ;
- +20 dB/dec change: simple zero ;
- −40 dB/dec change: double pole or quadratic , with = corner.
- Damping of a quadratic factor: measure the resonant peak above the asymptote at . Then , or approximately dB at .
- Check with the phase plot: for a minimum-phase system the phase must agree with the factors found. For example, the phase approaches at high frequency. Extra phase lag with no extra magnitude change indicates a transport delay (the slope of phase vs ω gives T) or an RHP zero.
- 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:
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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