In this chapter we show how the magnitude of the frequency response of the open-loop system can be used to determine the performance of the closed-loop feedback system. A general feedback loop is shown in Fig. 16.1. In this feedback system, \(r(t)\) is the reference input to be tracked by \(y(t)\), \(d_{in}(t)\) is an unknown input disturbance, \(d_{out}\) is an unknown output disturbance, and \(n(t)\) is an unknown noise signal. Note that because of the presence of \(n\), the error signal \(e=r-y\) is no longer the input to \(C(s)\). The input to \(C(s)\) is now \(r-y-n\). Note that this definition differs from the one used in Chapter 12 and will result in a different transfer function between the error and the noise. To find the transfer function from the inputs \(r\), \(d_{in}\), \(d_{out}\), and \(n\) to the error \(e(t)\) we start by finding the transfer functions to \(y\). Starting at \(Y(s)\) and following the loop backward until returning to \(Y\) we get \begin{align} & Y(s) = -D_{out}(s) + P \left( - D_{in}(s) + C\left(R(s) - N(s) - Y(s)\right) \right) \notag \\ \implies & (1+PC) Y(s) = PC R(s) - PC N(s) - D_{out}(s) - P D_{in}(s) \notag \\ & Y(s) = \frac{PC}{1+PC}R(s) - \frac{PC}{1+PC} N(s) % - \frac{1}{1+PC} D_{out}(s) - \frac{P}{1+PC} D_{in}(s). \tag{16.1} \end{align} Defining the error to be \(E(s) \stackrel{\triangle}{=}R(s)-Y(s)\) we get \begin{align} E(s) &= R(s) - Y(s) \\ % &= \frac{1}{1+PC}R(s) + \frac{PC}{1+PC} N(s) % + \frac{1}{1+PC} D_{out}(s) + \frac{P}{1+PC} D_{in}(s), \tag{16.3} \end{align} Therefore,
The transfer function from the reference \(r\) to the error \(e\) is \(\frac{1}{1+P(s)C(s)}\),
The transfer function from the noise \(n\) to the error \(e\) is \(\frac{P(s)C(s)}{1+P(s)C(s)}\),
The transfer function from the output disturbance \(d_{out}\) to the error \(e\) is \(\frac{1}{1+P(s)C(s)}\), and
The transfer function from the input disturbance \(d_{in}\) to the error \(e\) is \(\frac{P(s)}{1+P(s)C(s)}\).
Chapter 16 activity
Select one input signal and follow how it appears in the error equation. The highlighted path reinforces why different specifications use different closed-loop transfer functions.
Transfer to error:
Specification connection:
For typical applications, we have that
The reference signal \(r(t)\) is a low frequency signal with frequency content below \(\omega_r\),
The noise signal \(n(t)\) is a high frequency signal with frequency content above \(\omega_{\mathit{no}}\),
The output disturbance signal \(d_{out}(t)\) is a low frequency signal with frequency content below \(\omega_{d_{out}}\),
The input disturbance signal \(d_{in}(t)\) is a low frequency signal with frequency content below \(\omega_{d_{in}}\).
The objective for the control design is to find \(C(s)\) so that the following specifications are satisfied:
Tracking: Track reference signals \(r(t)\) with frequency content below \(\omega_r\) so that the error due to the reference signal satisfies \[\left|e(t)\right|\leq \gamma_r\left|r(t)\right|,\]
Noise Attenuation: Attenuate noise signals \(n(t)\) with frequency content above \(\omega_{no}\) so that the error due to the noise signal satisfies \[\left|e(t)\right|\leq \gamma_n\left|n(t)\right|,\]
Reject Output Disturbances: Reject output disturbance signals \(d_{out}(t)\) with frequency content below \(\omega_{d_{out}}\) so that the error due to the output disturbance signal satisfies \[\left|e(t)\right|\leq \gamma_{d_{out}}\left|d_{out}(t)\right|,\]
Reject Input Disturbances: Reject input disturbance signals \(d_{in}(t)\) with frequency content below \(\omega_{d_{in}}\) so that the error due to the input disturbance signal satisfies \[\left|e(t)\right|\leq \gamma_{d_{in}}\left|d_{in}(t)\right|.\]
The objective in the remainder of this chapter is to show how each of these specifications can be translated into a requirement on the magnitude of the (open) loop gain \(20\log_{10}\left|P(j\omega)C(j\omega)\right|\).
The transfer function from the reference signal \(r(t)\) to the error \(e(t)\) is \[E(s) = \frac{1}{1+P(s)C(s)}R(s).\] Therefore, \[\left|E(j\omega)\right| \leq \left|\frac{1}{1+P(j\omega)C(j\omega)}\right|\left|R(j\omega)\right|.\] The objective is to track reference signals \(r(t)\) with frequency content below \(\omega_r\) so that the error due to the reference signal satisfies \[\left|e(t)\right|\leq \gamma_r \left|r(t)\right|.\] Therefore, the controller \(C(s)\) needs to be designed so that \[\left|\frac{1}{1+P(j\omega)C(j\omega)}\right| \leq \gamma_r,\] or equivalently so that \[\left|1+P(j\omega)C(j\omega)\right| \geq \frac{1}{\gamma_r}.\] If \(\left|P(j\omega)C(j\omega)\right|\gg 1\), then a sufficient condition is that \[\left|P(j\omega)C(j\omega)\right| \geq \frac{1}{\gamma_r},\] or equivalently \[\tag{16.4} 20\log_{10}\left|P(j\omega)C(j\omega)\right| \geq 20\log_{10}1/\gamma_r,\] which needs to be satisfied for all \(\omega\leq\omega_r\). On a Bode plot, the constraint represented by Equation 16.4 is shown graphically in Fig. 16.2.
Alternatively, from the Bode plot of \(PC\) we can compute the tracking characteristics of the closed-loop system. For example, suppose that on the Bode plot \[20\log_{10}\left|P(j\omega)C(j\omega)\right|> B_r\text{~dB}\] for all \(\omega<\omega_r\), then \[\left|P(j\omega)C(j\omega)\right|> 10^{B_r/20},\] for all \(\omega<\omega_r\). If \(\left|P(j\omega)C(j\omega)\right|\gg 1\) for all \(\omega<\omega_r\), then \[\left|\frac{1}{1+P(j\omega)C(j\omega)}\right|\lessapprox 10^{-B_r/20},\] which implies that \[\left|E(j\omega)\right| \leq 10^{-B_r/20} \left|R(j\omega)\right|\] for all \(\omega<\omega_r\), or in other words that \[\tag{16.5} \left|e(t)\right| \leq 10^{-B_r/20} \left|r(t)\right|,\] for all \(r(t)\) with frequency content below \(\omega_r\), which implies that \[\gamma_r = 10^{-B_r/20}.\]
The transfer function from the noise signal \(n(t)\) to the error \(e(t)\) is \[E(s) = \frac{P(s)C(s)}{1+P(s)C(s)}N(s).\] Therefore, \[\left|E(j\omega)\right| \leq \left|\frac{P(j\omega)C(j\omega)}{1+P(j\omega)C(j\omega)}\right|\left|N(j\omega)\right|.\] The objective is to attenuate noise signals \(n(t)\) with frequency content above \(\omega_{no}\) so that the error due to the noise signal satisfies \[\left|e(t)\right|\leq \gamma_n \left|n(t)\right|.\] Therefore, the controller \(C(s)\) needs to be designed so that \[\left|\frac{P(j\omega)C(j\omega)}{1+P(j\omega)C(j\omega)}\right| \leq \gamma_n.\] If \(\left|P(j\omega)C(j\omega)\right|\ll 1\), then a sufficient condition is that \[\left|P(j\omega)C(j\omega)\right| \leq \gamma_n,\] or equivalently \[\tag{16.6} 20\log_{10}\left|P(j\omega)C(j\omega)\right| \leq 20\log_{10}\gamma_n,\] which needs to be satisfied for all \(\omega\geq\omega_{no}\). On a Bode plot, the constraint represented by Equation 16.6 is shown graphically in Fig. 16.3.
Alternatively, from the Bode plot of \(PC\) we can compute the noise attenuation characteristics of the closed-loop system. For example, suppose that on the Bode plot \[20\log_{10}\left|P(j\omega)C(j\omega)\right|< -B_n\text{~dB}\] for all \(\omega>\omega_{no}\), then \[\left|P(j\omega)C(j\omega)\right|< 10^{-B_n/20},\] for all \(\omega>\omega_{no}\). If \(P(j\omega)C(j\omega)\ll 1\) for all \(\omega>\omega_{no}\), then \[\left|\frac{P(j\omega)C(j\omega)}{1+P(j\omega)C(j\omega)}\right|\lessapprox 10^{-B_n/20},\] which implies that \[\left|E(j\omega)\right| \leq 10^{-B_n/20} \left|N(j\omega)\right|\] for all \(\omega>\omega_{no}\), or in other words that \[\left|e(t)\right| \leq 10^{-B_n/20} \left|n(t)\right|,\] for all \(n(t)\) with frequency content above \(\omega_{no}\), which implies that \[\gamma_n = 10^{-B_n/20}.\]
The transfer function from the output disturbance signals \(d_{out}(t)\) to the error \(e(t)\) is \[E(s) = \frac{1}{1+P(s)C(s)}D_{out}(s).\] Therefore, \[\left|E(j\omega)\right| \leq \left|\frac{1}{1+P(j\omega)C(j\omega)}\right|\left|D_{out}(j\omega)\right|.\] The objective is to attenuate output disturbance signals \(d_{out}(t)\) with frequency content below \(\omega_{d_{out}}\) so that the error due to the output disturbance signal satisfies \[\left|e(t)\right|\leq \gamma_{d_{out}} \left|d_{out}(t)\right|.\] Therefore, the controller \(C(s)\) needs to be designed so that \[\left|\frac{1}{1+P(j\omega)C(j\omega)}\right| \leq \gamma_{d_{out}},\] or equivalently so that \[\left|1+P(j\omega)C(j\omega)\right| \geq \frac{1}{\gamma_{d_{out}}}.\] If \(\left|P(j\omega)C(j\omega)\right|\gg 1\), then a sufficient condition is that \[\left|P(j\omega)C(j\omega)\right| \geq \frac{1}{\gamma_{d_{out}}},\] or equivalently \[\tag{16.7} 20\log_{10}\left|P(j\omega)C(j\omega)\right| \geq 20\log_{10}1/\gamma_{d_{out}},\] which needs to be satisfied for all \(\omega\leq\omega_{d_{out}}\). On a Bode plot, the constraint represented by Equation 16.7 is shown graphically in Fig. 16.4.
Alternatively, from the Bode plot of \(PC\) we can compute the output disturbance rejection characteristics of the closed-loop system. For example, suppose that on the Bode plot \[20\log_{10}\left|P(j\omega)C(j\omega)\right|> B_{d_{out}}\text{~dB}\] for all \(\omega<\omega_{d_{out}}\), then \[\left|P(j\omega)C(j\omega)\right|> 10^{B_{d_{out}}/20},\] for all \(\omega<\omega_{d_{out}}\). If \(\left|P(j\omega)C(j\omega)\right|\gg 1\) for all \(\omega<\omega_{d_{out}}\), then \[\left|\frac{1}{1+P(j\omega)C(j\omega)}\right|\lessapprox 10^{-B_{d_{out}}/20},\] which implies that \[\left|E(j\omega)\right| \leq 10^{-B_{d_{out}}/20} \left|D_{out}(j\omega)\right|\] for all \(\omega<\omega_{d_{out}}\), or in other words that \[\left|e(t)\right| \leq 10^{-B_{d_{out}}/20} \left|d_{out}(t)\right|,\] for all \(d_{out}(t)\) with frequency content below \(\omega_{d_{out}}\), which implies that \[\gamma_{d_{out}} = 10^{-B_{d_{out}}/20}.\]
The transfer function from the input disturbance signals \(d_{in}(t)\) to the error \(e(t)\) is \[E(s) = \frac{P(s)}{1+P(s)C(s)}D_{in}(s).\] Therefore, \[\left|E(j\omega)\right| \leq \left|\frac{P(j\omega)}{1+P(j\omega)C(j\omega)}\right|\left|D_{in}(j\omega)\right|.\] The objective is to attenuate input disturbance signals \(d_{in}(t)\) with frequency content below \(\omega_{d_{in}}\) so that the error due to the input disturbance signal satisfies \[\left|e(t)\right|\leq \gamma_{d_{in}} \left|d_{in}(t)\right|.\] Therefore, the controller \(C(s)\) needs to be designed so that \[\left|\frac{P(j\omega)}{1+P(j\omega)C(j\omega)}\right| \leq \gamma_{d_{in}},\] or equivalently so that \[\frac{\left|1+P(j\omega)C(j\omega)\right|}{\left|P(j\omega)\right|} \geq \frac{1}{\gamma_{d_{in}}}.\] If \(\left|P(j\omega)C(j\omega)\right|\gg 1\), then a sufficient condition is that \[\frac{\left|P(j\omega)C(j\omega)\right|}{\left|P(j\omega)\right|} \geq \frac{1}{\gamma_{d_{in}}},\] or equivalently \[\tag{16.8} 20\log_{10}\left|P(j\omega)C(j\omega)\right| - 20\log_{10}\left|P(j\omega)\right| \geq 20\log_{10}1/\gamma_{d_{in}},\] which needs to be satisfied for all \(\omega\leq\omega_{d_{in}}\). On a Bode plot, the constraint represented by Equation 16.8 is shown graphically in Fig. 16.5, where it can be seen that the loop gain \(\left|PC\right|\) must be above the magnitude of the plant \(\left|P\right|\) by the specified amount.
Alternatively, from the Bode plot of \(PC\) we can compute the input disturbance rejection characteristics of the closed-loop system. For example, suppose that on the Bode plot \[20\log_{10}\left|P(j\omega)C(j\omega)\right|-20\log_{10}\left|P(j\omega)\right|> B_{d_{in}}\text{~dB}\] for all \(\omega<\omega_{d_{in}}\), then \[\frac{\left|P(j\omega)C(j\omega)\right|}{\left|P(j\omega)\right|}> 10^{B_{d_{in}}/20},\] for all \(\omega<\omega_{d_{in}}\). If \(\left|P(j\omega)C(j\omega)\right|\gg 1\) for all \(\omega<\omega_{d_{in}}\), then \[\left|\frac{P(j\omega)}{1+P(j\omega)C(j\omega)}\right|\lessapprox 10^{-B_{d_{in}}/20},\] which implies that \[\left|E(j\omega)\right| \leq 10^{-B_{d_{in}}/20} \left|D_{in}(j\omega)\right|\] for all \(\omega<\omega_{d_{in}}\), or in other words that \[\tag{16.9} \left|e(t)\right| \leq 10^{-B_{d_{in}}/20} \left|d_{in}(t)\right|,\] for all \(d_{in}(t)\) with frequency content below \(\omega_{d_{in}}\), which implies that \[\gamma_{d_{in}} = 10^{-B_{d_{in}}/20}.\]
Recall that the loop gain \(P(s)C(s)\) is said to be
Type 0 if there are no free integrators,
Type 1 if there is one free integrator,
Type 2 if there are two free integrators, etc.
Note that for a type 0 system \[\begin{aligned} & \lim_{s\to 0} P(s)C(s) = \text{~constant} \\ \implies & \lim_{\omega\to 0}\left|P(j\omega)C(j\omega)\right| = \text{~constant} \\ \implies & \lim_{\omega\to 0} 20\log_{10}\left|P(j\omega)C(j\omega)\right| = \text{~constant}. \end{aligned}\] In other words, at low frequency, the Bode magnitude plot is as shown in Fig. 16.6.
Recall that for a step input to a type 0 system where \(R(s) = A/s\), the steady-state error is \[e_{ss} = \lim_{t\to \infty} e(t) = \lim_{s\to 0} sE(s) = \lim_{s\to 0} \frac{A}{1+P(s)C(s)} = \frac{A}{1+M_p},\] where \[M_p = \lim_{s\to 0}P(s)C(s) = \lim_{\omega\to 0} \left|P(j\omega)C(j\omega)\right|.\] As shown in Fig. 16.6, \(20\log_{10}M_p\) is the constant limit of the loop gain as \(\omega\to 0\).
Therefore, from the Bode plot of \(PC\) for a type 0 system, we can compute the tracking error for a step input of size \(A\). Suppose that on the Bode plot \[\lim_{\omega\to 0} 20\log_{10}\left|P(j\omega)C(j\omega)\right| = B_0\text{~dB}.\] Then \[M_p = 10^{B_0/20},\] and \[\tag{16.10} \lim_{t\to\infty}\left|e(t)\right| \leq \frac{1}{1+M_p} A.\]
For a type 1 system since \(P(s)C(s)\) has one free integrator \[\begin{aligned} & \lim_{s\to 0} sP(s)C(s) = \text{~constant} \\ \implies & \lim_{\omega\to 0}\left|(j\omega)P(j\omega)C(j\omega)\right| = \text{~constant} \\ \implies & \lim_{\omega\to 0}\frac{\left|P(j\omega)C(j\omega)\right|}{\left|\frac{1}{j\omega}\right|} = \text{~constant} \\ \implies & \lim_{\omega\to 0} \left[ 20\log_{10}\left|P(j\omega)C(j\omega)\right| - 20\log_{10}\left|\frac{1}{j\omega}\right|\right] = \text{~constant}. \end{aligned}\] In other words, at low frequency, the Bode magnitude plot of \(20\log_{10}\left|P(j\omega)C(j\omega)\right|\) has a slope of \(-20\) dB/decade as \(\omega\to 0\), as shown in Fig. 16.7.
Recall that for a ramp input to a type 1 system where \(R(s) = A/s^2\), the steady-state error is \[e_{ss} = \lim_{t\to \infty} e(t) = \lim_{s\to 0} sE(s) = \lim_{s\to 0} \frac{A}{s+sP(s)C(s)} = \frac{A}{M_v},\] where \[M_v = \lim_{s\to 0}sP(s)C(s) = \lim_{\omega\to 0} \left|(j\omega)P(j\omega)C(j\omega)\right|.\] As shown in Fig. 16.7, \(20\log_{10}M_v\) is the amount that the loop gain exceeds the Bode plot of \(20\log_{10}\left|1/j\omega\right|\) as \(\omega\to 0\).
Therefore, from the Bode plot of \(PC\) for a type 1 system, we can compute the tracking error for a ramp input with slope \(A\). Suppose that on the Bode plot \[\lim_{\omega\to 0} \left[ 20\log_{10}\left|P(j\omega)C(j\omega)\right| - 20\log_{10}\left|\frac{1}{j\omega}\right| \right] = B_1\text{~dB},\] then \[M_v = 10^{B_1/20},\] and \[\tag{16.11} \lim_{t\to\infty}\left|e(t)\right| \leq \frac{1}{M_v} A.\]
For a type 2 system where \(P(s)C(s)\) has two free integrators we have \[\begin{aligned} & \lim_{s\to 0} s^2P(s)C(s) = \text{~constant} \\ \implies & \lim_{\omega\to 0}\left|(j\omega)^2P(j\omega)C(j\omega)\right| = \text{~constant} \\ \implies & \lim_{\omega\to 0}\frac{\left|P(j\omega)C(j\omega)\right|}{\left|\frac{1}{j\omega}\right|^2} = \text{~constant} \\ \implies & \lim_{\omega\to 0} \left[ 20\log_{10}\left|P(j\omega)C(j\omega)\right| - 20\log_{10}\left|\frac{1}{j\omega}\right|^2\right] = \text{~constant}. \end{aligned}\] In other words, at low frequency, the Bode magnitude plot of \(20\log_{10}\left|P(j\omega)C(j\omega)\right|\) has a slope of \(-40\) dB/decade as \(\omega\to 0\), as shown in Fig. 16.8.
Recall that for a parabolic input to a type 2 system where \(R(s) = A/s^3\), the steady-state error is \[e_{ss} = \lim_{t\to \infty} e(t) = \lim_{s\to 0} sE(s) = \lim_{s\to 0} \frac{A}{s^2+s^2P(s)C(s)} = \frac{A}{M_a},\] where \[M_a = \lim_{s\to 0}s^2P(s)C(s) = \lim_{\omega\to 0} \left|(j\omega)^2P(j\omega)C(j\omega)\right|.\] As shown in Fig. 16.8, \(20\log_{10}M_a\) is the amount that the loop gain exceeds the Bode plot of \(20\log_{10}\left|1/j\omega\right|^2\) as \(\omega\to 0\).
Therefore, from the Bode plot of \(PC\) for a type 2 system, we can compute the tracking error for a parabola input with curvature \(A\). Suppose that on the Bode plot \[\lim_{\omega\to 0} \left[ 20\log_{10}\left|P(j\omega)C(j\omega)\right| - 20\log_{10}\left|\frac{1}{j\omega}\right|^2 \right] = B_2\text{~dB}.\] Then \[M_a = 10^{B_2/20},\] and \[\tag{16.12} \lim_{t\to\infty}\left|e(t)\right| \leq \frac{1}{M_a} A.\]
Chapter 16 activity
Use two markers to measure the low-frequency slope of |L(jω)|. The slope tells you how many free integrators are in the loop: 0, 1, or 2. After classifying the type, choose an input class to estimate the corresponding steady-state error constant.
Step 1: measure the low-frequency slope
Place both markers on the low-frequency straight-line part of the Bode magnitude curve.
slope = (gain2 - gain1) / (log10ω2 - log10ω1) dB/decade
Step 2: estimate the steady-state error
The input size A scales the final error. It does not move the Bode curve. After the type is classified, compare |L(jω)| with the selected input-class guide to estimate Mp, Mv, or Ma.
Measured slope
Type classification
Measured constant
Steady-state result
Interpretation
Example Problem
3.1.19For the single-link robot arm, use the bode command
(from MATLAB or Python) to create a graph that simultaneously displays
the Bode plots for (1) the plant, (2) the plant under PID control, using
the control gains calculated in
HW 3.1.13.
(a) To what percent error can the closed-loop system under PID control track the desired input if all of the frequency content of \(\theta_r(t)\) is below \(\omega_r = 0.4\) radians per second?
(b) If the reference input is \(\theta_r(t) = 5t^2\) for \(t\geq 0\), what will the steady-state tracking error to this input be when using PID control?
(c) If all of the frequency content of the input disturbance \(d_{in}(t)\) is below \(\omega_{d_{in}} = 0.01\) radians per second, what percentage of the input disturbance shows up in the output \(\theta\) under PID control?
(d) If all of the frequency content of the noise \(n(t)\) is greater than \(\omega_{no} = 100\) radians per second, what percentage of the noise shows up in the output signal \(\theta\)?
Design Study A guided exercise
Read tracking, disturbance, and noise bounds as frequency-specific magnitude requirements.
Use the marked Bode plot to read the frequency and magnitude values used in the specification checks.
Answer: The acceptable tracking region is specified by a frequency \(\omega_r\) and a magnitude bound \(\gamma_r\).
Answer: A magnitude error bound \(\gamma_r\) corresponds to approximately \(100\gamma_r%\).
Answer: Disturbance rejection is a low-frequency requirement, while sensor-noise attenuation is a high-frequency requirement.
(a) The Bode plots of the plant \(P(s)\) and the loop gain with PID control \(P(s)C_{PID}(s)\) are shown in Fig. 16.9.
From Fig. 16.9 we see that below \(\omega_r = 0.4\) rad/sec, the loop gain is above \(B_r=44.5\) dB. Therefore, from Equation 16.5 we have that \[\left|e(t)\right| \leq \gamma_r \left|r(t)\right|,\] where \(\gamma_r = 10^{-44.5/20} = 0.006\), which implies that the tracking error will be 0.60% of the magnitude of the input.
(b) Suppose now that the desired reference input is \(\theta_r(t) = 5t^2\). Under PID control, the slope of the loop gain as \(\omega\to 0\) is \(-40\) dB/dec, which implies that the system is type 2 and will track a step and a ramp with zero steady-state error. For a parabola, there will be a finite error. Since the loop gain under PID control is \(B_2=28\) dB above the \(1/s^2\) line, from Equation 16.12 the steady-state error satisfies \[\lim_{t\to\infty}\left|e(t)\right| \leq \frac{A}{M_a} = 5\cdot 10^{-28/20} = 0.2.\]
(c) If the input disturbance is below \(\omega_{d_{in}} = 0.01\) rad/s, then the difference between the loop gain and the plant is \(B_{d_{in}}=28\) dB at \(\omega_{d_{in}} = 0.01\) rad/s, therefore, from Equation 16.9 we see that \[\gamma_{d_{in}} = 10^{-28/20} = 0.04,\] implying that 4% of the input disturbance will show up in the output.
(d) For noise greater than \(\omega_{no}=100\) rad/sec, we see from Fig. 16.9 that \(B_n = 32.6\) dB. Therefore, \(\gamma_n = 10^{-32.6/20} = 0.023\) which implies that \(2.3\)% of the noise will show up in the output signal.
Python code to generate the relevant plots can be downloaded at http://controlbook.byu.edu.
Example Problem
3.2.19For the inner loop of the inverted pendulum, use the
bode command (in Python or MATLAB) to create a graph that
simultaneously displays the Bode plots for (1) the plant, and (2) the
plant under PD control, using the control gains calculated in
Homework 3.2.13.
(a) To what percent error can the closed-loop system track the desired input if all of the frequency content of \(\theta_r(t)\) is below \(\omega_r = 1.0\) radians per second?
(b) If all of the frequency content of the sensor noise on the inner \(n(t)\) is greater than \(\omega_{no} = 200\) radians per second, what percentage of the noise shows up in the output signal \(\theta\)?
For the outer loop of the inverted pendulum, use the MATLAB
bode command to create a graph that simultaneously displays
the Bode plots for the plant, and the plant under PID control, using the
control gains calculated in
Homework 3.2.13.
(c) If the reference signal \(y_r(t)\) has frequency content below \(\omega_r=0.001\) radians/second, what is the tracking error under PID control if \(\left|r(t)\right|\leq 50\)?
Design Study B guided exercise
Use Bode-plot bounds to estimate tracking and noise attenuation.
Use the markers to connect dB bounds to tracking and noise percentages.
Answer: The inner-loop tracking specification is evaluated below omega_r = 1 rad/s.
Answer: The inner-loop noise specification is evaluated above omega_no = 200 rad/s.
Answer: The outer-loop tracking bound is evaluated below omega_r = 0.001 rad/s.
(a) The Bode plots of the plant \(P(s)\) and the loop gain with PD control \(P(s)C_{PD}(s)\) are shown in Fig. 16.10.
From Fig. 16.10 we see that below \(\omega_r = 1.0\) rad/sec, the loop gain is above \(B_r=6.5\) dB. Therefore, from Equation 16.5 we have that \[\left|e(t)\right| \leq \gamma_r \left|r(t)\right|,\] where \(\gamma_r = 10^{-6.5/20} = 0.47\), which implies that the tracking error will be 47% of the magnitude of the input.
(b) For noise greater than \(\omega_{no}=200\) rad/sec, we see from Fig. 16.10 that \(B_n = 32.2\) dB. Therefore, \(\gamma_n = 10^{-32.2/20} = 0.0245\) which implies that \(2.45\)% of the noise will show up in the output signal.
(c) The Bode plots of the outer-loop plant \(P(s)\) and the loop gain with PID control \(P(s)C_{PID}(s)\) are shown in Fig. 16.11.
From Fig. 16.11 it can be seen that the loop gain under PID control for the outer loop is above \(B_r=154\) dB for all \(\omega<\omega_r=0.001\) radians/second. Therefore, from Equation 16.5 the tracking error satisfies \[\left|e(t)\right|\leq \gamma_r \left|r(t)\right|,\] where \(\gamma_r = 10^{-154/20}=2.0e-08\). Therefore, if \(\left|r(t)\right|\leq 50\), then \(\left|e(t)\right|\leq 100e-08\).
Python code to generate the relevant plots can be downloaded at http://controlbook.byu.edu.
Example Problem
3.3.19For the inner loop of the satellite attitude control, use the
bode command to create a graph that simultaneously displays
the Bode plots for (1) the plant, and (2) the plant under PD control,
using the control gains calculated in
Homework 3.3.13.
(a) To what percent error can the closed-loop system under PD control track a step in \(\theta_r(t)=20 t^2\) for \(t\geq0\)?
(b) If the input disturbance has frequency content below \(\omega_{d_{in}}=0.1\) radians per second, what percentage of the input disturbance appears in the output?
For the outer loop of the satellite attitude control, use the
bode command to create a graph that simultaneously displays
the Bode plots for (1) the plant, (2) the plant under PID control using
the control gains calculated in
Homework 3.3.13.
(c) If all of the frequency content of the noise \(n(t)\) is greater than \(\omega_{no} = 10\) radians per second, what percentage of the noise shows up in the output signal \(\phi\), using PI control?
Design Study C guided exercise
Use Bode-plot offsets to estimate tracking error, disturbance rejection, and noise transmission.
Answer: The inner loop is type 2 for tracking, so parabolic inputs have finite steady-state error.
Answer: Evaluate input-disturbance rejection below \(\omega_{d_{in}}=0.1\) rad/s using the dB separation between loop gain and plant.
Answer: Noise above \(10\) rad/s is evaluated from the high-frequency loop behavior; reducing it trades against bandwidth and response speed.
(a)
The Bode plots of the plant \(P(s)\) and the loop gain with PD control \(P(s)C_{PD}(s)\) are shown in Fig. 16.12.
Suppose that the desired reference input is \(\theta^r(t) = 20t^2\). Under PD control, the slope of the loop gain as \(\omega\to 0\) is \(-40\) dB/dec, which implies that the system is type 2 and will track a step and a ramp with zero steady-state error. For a parabola, there will be a finite error. Since the loop gain under PID control is \(B_2=20\) dB above the \(1/s^2\) line, from Equation 16.12 the steady-state error satisfies \[\lim_{t\to\infty}\left|e(t)\right| \leq \frac{A}{M_a} = 20\cdot 10^{-22.3/20} = 1.53.\]
(b) If the input disturbance is below \(\omega_{d_{in}} = 0.1\) rad/s, then the difference between the loop gain and the plant is \(B_{d_{in}}=38\) dB, therefore, from Equation 16.9 we see that \[\gamma_{d_{in}} = 10^{-38/20} = 0.0126,\] implying that 1.26% of the input disturbance will show up in the output.
(c) The Bode plots of the outer-loop plant \(P_{out}(s)\) and the loop gain with PI control \(P_{out}(s)C_{PI}(s)\) are shown in Fig. 16.13.
For noise greater than \(\omega_{no}=10\) rad/sec, we see from Fig. 16.13 that \(B_n = 8.4\) dB. Therefore, \(\gamma_n = 10^{-8.4/20} = 0.380\) which implies that \(38\)% of the noise will show up in the output signal.
Python code to generate the relevant plots can be downloaded at http://controlbook.byu.edu.
Chapter 16 activity
Select a specification, choose a base plant, then set the specification frequency and dB level. The shaded region marks the acceptable part of the loop-gain plot for that frequency-domain requirement.
The amplitude A scales the steady-state error calculation for the system-type cases. It does not move the Bode curves; it changes the computed error for step, ramp, and parabolic reference inputs.
At the selected frequency:
Computed bound:
Interpretation:
Tracking error is small when low-frequency loop gain is:
Noise attenuation generally requires loop gain to be:
If a specification requires |S| <= 0.1, enter that bound in dB.
Answer: -20 dB because 20 log10(0.1) = -20.
Approximately how many dB of loop gain are needed for |S| near 0.1 when L is large?
Answer: About 20 dB. When L is large, S is approximately 1/L.
Which closed-loop function is most directly associated with output disturbance rejection?
Answer: Sensitivity, S = 1/(1+PC), is the key function for output disturbance and tracking error.
If high-frequency sensor noise should contribute less than 1 percent to the output, what magnitude bound should the relevant noise transfer path satisfy?
Which closed-loop function is most directly used to bound low-frequency reference tracking error in unity feedback?
Why do input-disturbance and output-disturbance specifications use different transfer paths?
What low-frequency loop-gain slope indicates one free integrator?
Why not simply make the loop gain very large at every frequency?