Transfer function from Bode plot with example

A Bode plot obtained from a frequency-response test can be used to identify the transfer function. The asymptotic magnitude curve gives the gain, type and corner frequencies, and the phase curve confirms them (for a minimum-phase system).

Steps

1. Fit asymptotes with slopes of 0, ±20, ±40 … dB/decade to the magnitude curve.
2. Low-frequency slope gives the system type: 0, −20 or −40 dB/dec means 0, 1 or 2 integrators.
3. Gain K from the low-frequency segment:
   • Type 0: $M_{low} = 20\log K$.
   • Type 1: the extended −20 dB/dec line crosses 0 dB at ω = K.
   • Type 2: the −40 dB/dec line crosses 0 dB at $\omega = \sqrt K$.
4. Each change of slope is a corner frequency $\omega_c$:
   • −20 dB/dec change: pole $1/(1 + s/\omega_c)$;
   • +20 dB/dec change: zero $(1 + s/\omega_c)$;
   • −40 dB/dec change: a quadratic (complex) pole pair or a double real pole. For a quadratic, find ζ from the height of the resonant peak: peak above asymptote ≈ $-20\log 2\zeta$.
5. Verify with phase: at very low ω the phase should be $-90^\circ \times$ (number of integrators), and at very high ω it should be $-90^\circ \times$ (poles − zeros). Unexplained extra lag means a time delay or a non-minimum-phase zero.

Worked example

Suppose the asymptotic magnitude plot is:

• a horizontal line at 20 dB up to ω = 2 rad/s;
• then −20 dB/dec up to ω = 10 rad/s;
• then −40 dB/dec beyond 10 rad/s.

The phase goes from 0° to −180°.

   30 |
      |******************
      |                  ******
    0 |-----------------------****-------------------
      |                          ***
      |                            ****
  -30 |                               ****
      |                                  ****
      |                                     ****
  -60 |                                        ***
      |                                           ***
      |
  -90 |
      +----------------------------------------------
       0.1        1          10          100        1000
Magnitude asymptotes (dB) vs w (rad/s)

Reading it:

1. Initial slope 0 dB/dec, so the system is type 0.
2. $20\log K = 20$ dB, so K = 10.
3. At ω = 2 the slope changes by −20 dB/dec: a pole $1/(1 + s/2)$.
4. At ω = 10 the slope changes by a further −20 dB/dec: a pole $1/(1 + s/10)$.
5. Phase check: two poles give 0° → −180°, which is consistent.

$$G(s) = \frac{10}{\left(1 + \dfrac{s}{2}\right)\left(1 + \dfrac{s}{10}\right)} = \frac{200}{(s+2)(s+10)}$$

Check of one point: at ω = 10 the asymptote gives $20 - 20\log(10/2) = 6.0$ dB. From the factors of the TF: $20\log10 - 20\log5 = 6.0$ dB. ✓ (The true curve is about 3 dB lower at each corner.)