Table of Contents
1 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
1.1 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
- 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.
1.1.1 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 |
1.1.2 Example 2
The following magnitude plot has been obtained from experiments of a minimum phase system. Determine its transfer function. A straight line mag. approximation is provided.
- For the low frequency:
- The slope is -20, which corresponds to an integrator \(1/s\).
- The magnitude of the low frequency asymptote is 20dB at frequency 1. Hence, \(K = 10\).
Determine the first and second order terms:
- 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 \(20\log(2\zeta) = 2\), \(\zeta = 0.63\).
\[s^2+1.26s+1\]
| Corner Frequency | Slope Before | Slope After | Change of Slope |
| 0.2 | -20 | -60 | -40 |
| 1 | -60 | -20 | +40 |
\[G(s) = \frac{10(s^2+1.26s+1)}{s\left[(s/0.2)^2+0.5(s/0.2)+1\right]}.\]
1.1.3 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?)
\[G(s) = \frac{0.17}{(0.08s+1)^3}\exp(-0.2s).\]