Table of Contents
- 1. Minimum Phase and Non-Minimum Phase System
- 1.1. Constant K
- 1.2. Integrator \(1/s\)
- 1.3. First Order \(1/(\tau s+1)\)
- 1.4. Second Order \(1/((s/\omega_n)^2+2\zeta(s/\omega_n)+1)\)
- 1.5. Minimum Phase System
- 1.6. Non-Minimum Phase System: Unstable Zeros
- 1.7. Non-Minimum Phase System: Delay
- 1.8. Bode Plot of Minimum Phase and Non-Minimum Phase System
- 1.9. Implications of Non-Minimum Phase
- 2. Bode Plots of Complex Transfer Functions
1 Minimum Phase and Non-Minimum Phase System
1.1 Constant K
1.2 Integrator \(1/s\)
1.3 First Order \(1/(\tau s+1)\)
1.4 Second Order \(1/((s/\omega_n)^2+2\zeta(s/\omega_n)+1)\)
1.5 Minimum Phase System
- A transfer function \(G(s)\) is minimum phase if both \(G(s)\) and \(1/G(s)\) are causal and stable.
- Roughly speaking it means that the system does not have zeros or poles on the right-half plane. Moreover, it does not have delay.
- Bode discovered that the phase can be uniquely derived from the slope of the magnitude for minimum-phase system. Bode's Relation
Basic Factor | Mag Slope(Low Freq) | Phase(Low Freq) | Mag Slope(High Freq) | Phase (High Freq) |
K | 0 | 0 | 0 | 0 |
\(s^N\) | 20N | 90N | 20N | 90N |
\(1/(\tau s+1)\) | 0 | 0 | -20 | -90 |
\(1/((s/\omega_n)^2+2\zeta(s/\omega_n)+1)\) | 0 | 0 | -40 | -180 |
1.6 Non-Minimum Phase System: Unstable Zeros
Consider two transfer functions: \[G_1(s) = 1,\,G_2(s) = \frac{1-s}{1+s}\]
- \(G_1(s)\) is minimum phase since it does not have unstable zeros/poles.
- The magnitude of \(G_1(s)\) is 0dB, and the phase is \(0^\circ\).
- \(G_2(s)\) is non-minimum phase, since \(G_2(1) = 0\).
- One can check that \(1/G_2(s) = \frac{1+s}{1-s}\) is unstable.
\[|G_2(j\omega)| = \frac{\sqrt{1+\omega^2}}{\sqrt{1+\omega^2}} = 1,\,\angle G_2(j\omega) = -\tan^{-1}\omega -\tan^{-1}\omega = -2\tan^{-1}\omega.\]
1.7 Non-Minimum Phase System: Delay
- Time delay often exists in communication systems, chemical and manufacturing processes, etc.
- Time delay systems are of non-minimum phase behavior.
- The transfer function of the time delay factor is \(G_3(s) = \exp(-\tau s)\), where \(\tau\) is the delay.
\[G_3(j\omega) = \exp(-j\omega\tau) = 1\angle -\omega\tau\times\frac{180^\circ}{\pi}.\]
1.8 Bode Plot of Minimum Phase and Non-Minimum Phase System
Minimum phase system has minimum phase change amongst all transfer functions that have the same magnitude plot.
1.9 Implications of Non-Minimum Phase
- Non-minimum phase systems are much more difficult to control than minimum phase system
- Unfortunately, non-minimum phase is quite common in practice
- Flexible Structure
- Circuits
- Back a car into the parking lot
- The unstable zeros can be changed by reallocating sensors and actuators, or by introducing new sensors and actuators.
- You will learn this in EE4273 Digital Control Systems where we teach state space model.
2 Bode Plots of Complex Transfer Functions
Bode plots of a transfer function (system) can be obtained using Matlab (experiment). However, to interpret the Bode plots, we shall take a look at how basic factors of the transfer function affect the Bode plots.
2.1 Example 1
2.1.1 Step 1: Decompose the Transfer Function into Basic Factors
Factor | Corner Frequency |
---|---|
2 | |
\(1/s\) | |
\(2s+1\) | 0.5 |
\(1/(0.1s+1)\) | 10 |
\(1/(0.02s+1)\) | 50 |
By adding the Bode plots of the above factors together, we can get the Bode plot of \(G(s)\)
2.1.2 Step 2: Determine the Plot at the Low Frequency
- For low frequency, only need to look at the constant \(K\) and \(s^N\) term.
- Draw the plot of \(s^N\) first
- Magnitude: a straight line passes through \((1,0dB)\) with a slope of \(20NdB\).
- Phase: a horizontal line at \(90N^\circ\).
- Adding the constant term \(K\)
- Magnitude: add \(20\log|K|\)
- Phase: If \(K > 0\), do nothing. If \(K < 0\), subtract \(180^\circ\).
2.1.3 Step 3: Add First and Second Order Terms
Factor | Corner Frequency |
\(2s+1\) | 0.5 |
\(1/(0.1s+1)\) | 10 |
\(1/(0.02s+1)\) | 50 |
- Magnitude
For \((\tau s+1)^N\), change the slope at corner frequency \(1/\tau\) by \(20N\).
Frequency | Low | 0.5 | 10 | 50 |
Slope Change | +20 | -20 | -20 | |
Slope | -20 | 0 | -20 | -40 |
- Phase
For \((\tau s+1)^N\)
- change the slope at \(0.1/\tau\) by \(45N\).
- change the slope at \(10/\tau\) by \(-45N\).
Frequency | Low | 0.05 | 1 | 5 | 100 | 500 |
Slope Change | +45 | -45 | -45 -45 | +45 | +45 | |
Slope | 0 | 45 | 0 | -90 | -45 | 0 |
2.1.4 Step 4 (Optional): Check your result in Matlab
num = [2000,1000]; den = [1,60,500,0]; sys = tf(num, den); bode(sys);
2.2 Example 2
2.2.1 Step 1: Decompose the Transfer Function into Basic Factors
Factor | Corner Frequency |
---|---|
2.5 | |
\(1/s\) | |
\((s/2)^2+0.2s/2+1\) | 2 |
2.2.2 Step 2: Determine the Plot at the Low Frequency
The low frequency factors are \(2.5\) and \(1/s\).
- Magnitude
- Draw a straight line passing through \((1,0dB)\) with slope -20.
- Shift the magnitude plot of \(1/s\) up by \(20\log(2.5)\).
- Phase
- Draw a constant line at \(-90^\circ\).
2.2.3 Step 3: Add First and Second Order Terms
Consider a second order term:
\[\left[\left(\frac{s}{\omega_n}\right)^2+2\zeta\frac{s}{\omega_n}+1\right]^{\pm 1}.\]
- Magnitude
- change the slope at the corner frequency \(\omega_n\) by \(\pm 40dB\);
- add \(\pm 20\log(2\zeta)\) correction at the corner frequency \(\omega_n\).
- Phase
- draw the phase plot using calculators.
2.2.4 Step 4: Check your result in Matlab
num = [10]; den = [1,0.4,4,0]; sys = tf(num, den); bode(sys);
2.3 Example 3
Factor | Corner Frequency |
5 | |
\(1/s\) | |
\(1/(0.5s+1)\) | 2 |
\(0.1s+1\) | 10 |
\(1/\left[(s/50)^2+0.6s/50+1\right]\) | 50 |
- Magnitude
- Low Frequency: Draw a straight line passing through \((1,0dB)\) with slope -20. Shift this line up by \(20\log(5)=14dB\).
- Adding the first and second order term. Add \(-20\log(0.6) = 4.4dB\) correction at 50.
Frequency | Low | 2 | 10 | 50 |
Slope Change | -20 | +20 | -40 | |
Slope | -20 | -40 | -20 | -60 |
- Phase
- Phase starts at \(-90^\circ\) due to the integrator.
- Phase ends at \(-90-90+90-180 = -270^\circ\).
- Use calculator to draw the phase plot.