Frequency Domain Modeling
Frequency Domain Modeling
- Modeling from first principles can be difficult and time-consuming
- The Bode plots of a system can be obtained through experiments.
- For a minimum phase system, an approximate transfer function can be derived from the Bode magnitude plot
How to Get the Transfer Function
- Determine an asymptotic magnitude plot by approximating the actual gain plot by a sequence of straight lines with slopes of \(20N\).
- From the low frequency plot, determine the number of integrator \(1/s^N\) and the constant term \(K\).
- The beginning slope is \(20N\) for \(s^N\).
- Calculate the constant term \(K\) accordingly
How to Get the Transfer Function
- Determine the corner frequency from the intersection of two consecutive lines.
- From the change of slopes of the consecutive lines, determine the associated basic factor.
- \(+20\) implies first order zero and \(-20\) implies first order pole;
- \(+40\) implies second order zero and \(-40\) implies second order pole.
- Determine the damping ratio for second order zeros and poles.
- \(-20\log(2\zeta)\) correction for second order poles.
- \(+20\log(2\zeta)\) correction for second order zeros.
- The phase plot can be used to verify your results.
Example 1
Given the Bode plot of a minimum phase system. Obtain an estimated transfer function.
- Approximate the magnitude plot via a sequence of straight lines with slope \(20N\).
- For the low frequency:
- The slope is 0. Hence, no integrator;
- The magnitude is 20dB. Hence, the constant term is \(K = 10\).
- Determine the first and second order terms:
- There is second order term with natural frequency \(\omega_n = 10\).
- The correction is -6dB. Hence \(-20\log(2\zeta) = -6\), \(\zeta = 1\). The system is critically damped.
- There is a first order zero at \(100\).
| Corner Frequency |
Slope Before |
Slope After |
Change of Slope |
| 10 |
0 |
-40 |
-40 |
| 100 |
-40 |
-20 |
+20 |
\[G(s) = \frac{10(0.01s+1)}{(s/10)^2+2(s/10)+1} = \frac{10(0.01s+1)}{(0.1s+1)^2}\]
- Verify the transfer function using the Bode phase plot
| Frequency |
Low |
1 |
10 |
100 |
1000 |
| Slope Change |
|
-90 |
+45 |
+90 |
-45 |
| Slope |
0 |
-90 |
-45 |
45 |
0 |
Example 2
Determine the transfer function of a minimum phase system from Bode magnitude plot:
- There is second order term on the denominator with natural frequency \(\omega_n = 0.2\).
- The correction is +6dB. Hence \(-20\log(2\zeta) = 6\), \(\zeta = 0.25\).
\[\left[(s/0.2)^2+0.5(s/0.2)+1\right]^{-1}\]
- There is second order term on the numerator with natural frequency \(\omega_n = 1\).
- The correction is +2dB. Hence \(20\log(2\zeta) = 2\), \(\zeta = 0.63\).
\[s^2+1.26s+1\]
The minimum phase transfer function corresponding to the magnitude plot is:
\[G(s) = \frac{10(s^2+1.26s+1)}{s\left[(s/0.2)^2+0.5(s/0.2)+1\right]}.\]
Example 3: Pupillary Light Reflex Dynamics
- The human eye is an organ that is easily accessible for experiments. It has a control system that adjusts the pupil opening to regulate the light intensity at the retina.
- To determine its dynamics, light intensity on the eye was varied sinusoidally and the pupil opening was measured.
- By matching the magnitude, a model can be obtained as
\[G_1(s) = \frac{0.17}{(0.08s+1)^3}.\]
- The phase of \(G_1(s)\) (plotted as the green dashed line, is significantly different from actual phase
- We can add a delay term \(\exp(-0.2s)\) to match the phase, without affecting the magnitude (why?)
\[G(s) = \frac{0.17}{(0.08s+1)^3}\exp(-0.2s).\]