To achieve the objectives discussed in Chapter 19, the closed-loop system must be stable. Recall that tracking performance and disturbance rejection characteristics are determined by the loop shape at low frequencies when \(\left|P(j\omega)C(j\omega)\right|\gg 1\). Similarly, noise attenuation properties are determined by the loop shape at high frequencies when \(\left|P(j\omega)C(j\omega)\right|\ll 1\). In both cases we ignored the phase of \(P(j\omega)C(j\omega)\). It turns out that stability of the closed-loop system is determined by the phase of \(P(j\omega)C(j\omega)\) when \(P(j\omega)C(j\omega)=1\), as depicted in Fig. 17.1.
Define the crossover frequency \(\omega_{co}\) to be the frequency where \[\left|P(j\omega_{co})C(j\omega_{co})\right|=1,\] or equivalently where \[20\log_{10}\left|P(j\omega_{co})C(j\omega_{co})\right| = 0\text{~dB}.\] The phase \(\angle P(j\omega_{co})C(j\omega_{co})\) at the crossover frequency plays a critical role in determining the stability of the system. To better understand why this is the case, consider the transfer functions from \(R\), \(D_{in}\), \(D_{out}\), and \(N\), to \(E\) in Equation 16.3. The denominator in each transfer function is \(1+P(s)C(s)\). Therefore, the closed-loop poles are given by the closed-loop characteristic equation \[1+P(s)C(s) = 0,\] or equivalently where \[P(s)C(s) = -1.\]
Recall that for any \(s\), \(P(s)C(s)\) is a complex number with magnitude and phase. Therefore, given any \(s_0\) we can draw \(1+P(s_0)C(s_0)\) in the complex plane as shown in Fig. 17.2.
Since the physical response of the system is determined by \(P(s)C(s)\) evaluated on the \(j\omega\)-axis, there are stability problems if there exists an \(\omega_0\) such that \(P(j\omega_0)C(j\omega_0)=-1\), or in other words when \[\left|P(j\omega_0)C(j\omega_0)\right|=1 \qquad\text{and}\qquad \angle P(j\omega_0)C(j\omega_0) = -180^\circ,\] as shown in Fig. 17.3.
Since the magnitude of the loop gain is equal to one at the crossover frequency \(\omega_{co}\), it turns out that the quality of stability is determined by the phase of the loop gain at the crossover frequency. Define the phase margin \(PM\) as \[PM = \angle P(j\omega_{co})C(j\omega_{co}) - (-180)^\circ.\] Fig. 17.4 depicts the phase margin on the Bode plot.
Chapter 17 interactive figure
Move the loop gain and added phase lag. The same loop shape is used in the gain and phase plots, so the crossover markers and margin measurements update together.
Increasing gain usually moves the gain crossover. Adding phase lag directly reduces the phase margin and can reduce the gain margin.
Gain crossover
Phase crossover
Phase margin
Gain margin
Robustness note
Another way to understand the phase margin is shown in Fig. 17.5. The phase margin indicates the angular distance that \(P(j\omega_{co})C(j\omega_{co})\) is from \(-1\).
A related quantity is the Gain Margin \(GM\), which measures how far the magnitude \(\left|P(j\omega)C(j\omega)\right|\) is from unity when the phase \(\angle P(j\omega)C(j\omega) = -180\) degrees. The gain margin is depicted on the Bode plot in Fig. 17.4. An alternative view is shown in Fig. 17.6.
A good control design will have large phase and gain margins. A phase margin of \(PM=60\) degrees is considered to be very good and is optimal for many applications. A gain margin of \(GM=10\) dB would be considered very good for most applications.
From Equation 16.1, the transfer function from the reference \(R(s)\) to the output \(Y(s)\) is \[Y(s) = \frac{P(s)C(s)}{1+P(s)C(s)}R(s)\] Note that
when \(\left|P(j\omega)C(j\omega)\right|\gg 1\), then \(\left|\frac{P(j\omega)C(j\omega)}{1+P(j\omega)C(j\omega)}\right|\approx 1\),
when \(\left|P(j\omega)C(j\omega)\right|\ll 1\), then \(\left|\frac{P(j\omega)C(j\omega)}{1+P(j\omega)C(j\omega)}\right|\approx \left|P(j\omega)C(j\omega)\right|\).
When \(\left|P(j\omega)C(j\omega)\right|=1\), i.e., when \(\omega=\omega_{co}\), then \(\left|\frac{P(j\omega_{co})C(j\omega_{co})}{1+P(j\omega_{co})C(j\omega_{co})}\right|\) is a complex number divided by a complex number, and the magnitude will depend on the phase of the \(P(j\omega_{co})C(j\omega_{co})\), or in other words the phase margin \(PM\). As shown in Figure 17.5, \(PM\) will determine the magnitude of \(1+P(j\omega_{co})C(j\omega_{co})\). When \(PM\) is small, \(\left|1+P(j\omega_{co})C(j\omega_{co})\right|\) will be small and \(\left|\frac{P(j\omega_{co})C(j\omega_{co})}{1+P(j\omega_{co})C(j\omega_{co})}\right|\) will be larger than one. When \(PM\) is larger than \(90\) degrees, \(\left|1+P(j\omega_{co})C(j\omega_{co})\right|\) will be larger than \(\left|P(j\omega_{co})C(j\omega_{co})\right|\) and \(\left|\frac{P(j\omega_{co})C(j\omega_{co})}{1+P(j\omega_{co})C(j\omega_{co})}\right|\) will be less than one.
Fig. 17.7 shows the closed-loop Bode plot superimposed on the open-loop Bode plot for different values of the phase margin. Note that when the phase margin is \(PM=60\) degrees, the closed-loop frequency response looks very similar to the closed-loop frequency response for a second-order system when \(\zeta=0.707\), as shown in Fig. 15.9. Phase margins smaller than \(60\) degrees have a peaking response similar to a second-order system with \(\zeta<0.707\). Note also that the bandwidth of the closed-loop system is approximately the crossover frequency of the open-loop system. In general, the crossover frequency \(\omega_{co}\) plays a similar role for general systems that the natural frequency \(\omega_n\) plays for second-order systems.
Chapter 17 interactive figure
Predict what will happen before moving the slider: if the phase margin decreases, should the closed-loop frequency-response peak and the step-response overshoot increase or decrease? The plots use a second-order approximation to show the trend described in the text.
Damping proxy
Closed-loop peak
Step overshoot
Design interpretation
Approximation used
Fig. 17.7 also shows the corresponding step response for the closed-loop system. The step response looks similar to the step response for second-order systems shown in Fig. 8.3, where again, the step response for \(PM=60\) degrees in Fig. 17.7 is roughly equivalent to the step response for \(\zeta=0.707\) in Fig. 8.4.
For a given reference signal, the size of the control signal \(u(t)\) is determined by the crossover frequency. A large crossover frequency will result in a larger control signal \(u(t)\). Fig. 17.8 shows the open and closed-loop transfer functions for a plant equal to \(P(s) = \frac{1}{s(s+1)}\) and for three different controllers that are tuned for equivalent phase margins but with crossover frequency at \(2\), \(20\), and \(200\) respectively. The corresponding control signal when the reference is a unit step is also shown in Fig. 17.8. Note that as \(\omega_{co}\) increases, the magnitude of the control signal increases and the speed of the response also increases. In fact, the integral of the control signals will be the same for each \(\omega_{co}\). Therefore, saturation constraints on \(u(t)\) will necessarily limit the size of the crossover frequency \(\omega_{co}\).
Chapter 17 interactive figure
Move the crossover frequency while holding a similar phase-margin trend. The response gets faster as ωco increases, but the required control signal grows.
Settling-time proxy
Peak control effort
Design warning
In Bode’s original work, he proved the following theorem that relates the slope of the magnitude plot to the phase.
. For any stable, minimum phase (i.e., zeros in the LHP), linear time-invariant system \(G(s)\), the phase of \(G(j\omega)\) is uniquely related to \(20\log_{10}\left|G(j\omega)\right|\) by \[\tag{17.1} \angle G(j\omega_0) = \frac{1}{\pi}\int_{-\infty}^{\infty} \left(\frac{dM}{du}\right)W(u)\,du,\] where \[\begin{aligned} M &= 20\log_{10}\left|G(j\omega)\right|, \\ u &= \log_{10}\left|\frac{\omega}{\omega_0}\right|, \\ W(u) &= \log_{10}\left(\coth \frac{\left|u\right|}{2}\right). \end{aligned}\]
The term \(\frac{dM}{du}\) in Equation 17.1 is the slope of the magnitude of \(G(j\omega)\) on the Bode plot. A plot of \(W(u)\) is shown in Fig. 17.9, where it can be seen that \(W(u)\) heavily weights the slope of the Bode magnitude around \(\omega_0\), but also includes bleed-over from the slope for roughly a decade before and after \(\omega_0\). This implies that the phase of \(P(j\omega)C(j\omega)\) at \(\omega=\omega_0\) is determined by the slope of the Bode magnitude plot in a region about \(\omega_0\).
Recall that the phase for an integrator \(1/s\) is \(-90\) degrees and that the phase for a differentiator is \(90\) degrees. Also, recall that the phase for a constant is \(0\) degrees. Therefore, we can approximate Equation 17.1 as follows:
\(\text{Slope of~} 20\log_{10}\left|G(j\omega_0)\right| \approx +20 \text{~dB/dec~} \\ \implies \angle G(j\omega_0) \approx +90 \text{~deg}\),
\(\text{Slope of~} 20\log_{10}\left|G(j\omega_0)\right| \approx 0 \text{~dB/dec~} \\ \implies \angle G(j\omega_0) \approx 0 \text{~deg}\),
\(\text{Slope of~} 20\log_{10}\left|G(j\omega_0)\right| \approx -20 \text{~dB/dec~} \\ \implies \angle G(j\omega_0) \approx -90 \text{~deg}\),
\(\text{Slope of~} 20\log_{10}\left|G(j\omega_0)\right| \approx -40 \text{~dB/dec~} \\ \implies \angle G(j\omega_0) \approx -180 \text{~deg}\),
\(\text{Slope of~} 20\log_{10}\left|G(j\omega_0)\right| \approx -60 \text{~dB/dec~} \\ \implies \angle G(j\omega_0) \approx -270 \text{~deg}\).
Of course, these values are approximate and require that the slope around \(\omega_0\) persist for approximately a decade before and after \(\omega_0\).
While we do not directly use Equation 17.1 in practice, it has clear implications for feedback design. In particular, the takeaway message is that to have a good phase margin, the slope of the Bode magnitude plot at the crossover frequency \(\omega_{co}\) cannot be too steep, and that it needs to be reasonably shallow for roughly a decade before and after crossover. As a rule of thumb, a phase margin of \(PM=60\) degrees requires that the slope at crossover is between \(-20\) dB/dec and \(-40\) dB/dec. In other words, the ideal loop shape will need to look something like Fig. 17.10. It is also important to understand that the Bode phase-gain theorem requires that there is adequate frequency separation between the frequency content of signals that we are tracking or rejecting and the frequency content of the sensor noise that is to be attenuated.
Chapter 17 interactive figure
Adjust the loop-shape parameters and use the checklist to connect the ideal Bode picture in Figure 17.10 to the design requirement it represents.
Low-frequency gainDesign goal: make the loop gain large at low frequency.
Why it mattersLarge low-frequency loop gain makes the sensitivity function small where commands and disturbances usually have most of their energy.
How to check or fix itCheck that |PC| is well above 0 dB at low frequency. Increase low-frequency gain with integral action or lag, but watch the crossover frequency.
Current checklist status
Example Problem
3.1.20For the single-link robot arm, use the bode and
margin commands (from Python or MATLAB) to find the phase
and gain margin for the closed-loop system under PID control. On the
same graph, plot the open-loop Bode plot and the closed-loop Bode plot.
What is the bandwidth of the closed-loop system, and how does this
relate to the crossover frequency? Use the gains found in
HW 3.1.13.
Design Study A guided exercise
Use the open-loop Bode plot to identify crossover, gain margin, phase margin, and bandwidth.
Use this plot to identify gain crossover, phase margin, gain margin, and closed-loop bandwidth.
Answer: Gain crossover occurs where the open-loop magnitude is \(0\) dB, or magnitude \(1\).
Answer: Phase margin is the additional lag needed to reach \(-180^\circ\) at gain crossover.
Answer: Closed-loop bandwidth is usually near the open-loop crossover and can be slightly larger for this design.
The complete solution includes margin-plot code. Before opening it, outline how you would compute the open-loop and closed-loop transfer functions, measure margins, and annotate the Bode plot.
This prompt is ungraded. It is intended to connect the margin measurements to the implementation code in the complete solution.
Python code to generate the plots from this problem is included below. However, similar MATLAB code is available on the book website (see http://controlbook.byu.edu).
# 3rd-Party
import control as cnt
import matplotlib.pyplot as plt
# local (controlbook)
import case_studies.A_arm.params as P
from case_studies.A_arm.pid_controller import ArmControllerPID
use_dB = False
plant = cnt.tf(P.tf_num, P.tf_den)
pid = ArmControllerPID()
kp, ki, kd, sigma = pid.kp, pid.ki, pid.kd, pid.sigma
controller = cnt.tf([kd + kp * sigma, kp + ki * sigma, ki], [sigma, 1, 0])
# open-loop system
sys_ol = cnt.series(controller, plant)
# closed-loop system
sys_cl = cnt.feedback(sys_ol, 1)
if __name__ == "__main__":
# You could call bode() once with multiple systems, but it draws margins
# for all of them. Here we only draw them for the open-loop system.
# Calling bode() multiple times reuses the same axes, but removes the title
# and legends, so we manually create the figures and axes to add them later.
fig, ax = plt.subplots(2, sharex=True)
cnt.bode(sys_ol, ax=ax, dB=use_dB, display_margins=True, label="open-loop")
cnt.bode(sys_cl, ax=ax, dB=use_dB, grid=False, label="closed-loop")
# cross-over frequency ≈ closed-loop bandwidth
bandwidth_cl = cnt.bandwidth(sys_cl)
print(f"Closed-loop bandwidth: {bandwidth_cl:.2f} rad/s")
# Margin info is on the plot, so we don't really need to do this, but this
# is how you can calculate it without plotting.
gm, pm, wcg, wcp = cnt.margin(sys_ol)
print(f"Crossover frequency: {wcp:.2f} rad/s")
# draw bandwidth line
ax[0].axvline(x=bandwidth_cl, color="red", linestyle=":", label="bandwidth")
# manually add annotations after running bode() multiple times
ax[0].legend()
fig.suptitle("HW 17 Bode Plots: Robot Arm")
fig.tight_layout()
# keep figure open until closed by user
plt.show()The transfer function for the plant is defined in Line 14. The
transfer function for the PID controller is \[\begin{aligned}
C_{PID}(s) &= k_P + \frac{k_I}{s} + \frac{k_D s}{\sigma s + 1}
= \frac{s(\sigma s+1)k_P + (\sigma s + 1)k_I + k_D
s^2}{s(\sigma s + 1)} \\
&= \frac{(k_D+\sigma k_P)s^2 + (k_P+\sigma k_I)s +
k_I}{s(\sigma s + 1)},
\end{aligned}\] and is defined in Line 15. In Python the
margin command returns information about the phase margin,
gain margin, and frequencies where those are calculated (like the
crossover frequency). The bode command can be used to make
plots of transfer functions (and will show the margin information if
sent the “margin=True” argument). The result of the Python code is shown
in Fig. 17.11.
As seen in Fig. 17.11 the bandwidth for PID control is approximately \(18\) rad/sec, which is slightly larger than the crossover frequency of \(10.8\) rad/sec. The larger bandwidth is due to the small phase margin of \(PM=49.0\) degrees.
Example Problem
3.2.20For this problem, use the gains found in HW 3.2.13.
For the inner loop of the inverted pendulum, use the Python or
MATLAB bode and margin commands to find the
phase and gain margin for the inner loop system under PD control. On the
same graph, plot the open-loop Bode plot and the closed-loop Bode plot.
What is the bandwidth of the inner loop, and how does this relate to the
crossover frequency?
For the outer loop of the inverted pendulum, use the
bode and margin commands to find the phase and
gain margin for the outer loop system under PID control. Plot the open
and closed-loop Bode plots for the outer loop on the same plot as the
open and closed-loop Bode plots for the inner loop. What is the
bandwidth of the outer loop, and how does this relate to the crossover
frequency?
What is the bandwidth separation between the inner (fast) loop, and the outer (slow) loop? For this design, is successive loop closure justified?
Design Study B guided exercise
Compare inner-loop and outer-loop margins and bandwidths.
Use margin plots to compare crossover, phase margin, gain margin, and bandwidth.
Answer: The inner-loop bandwidth is approximately 19 rad/s.
Answer: The outer-loop bandwidth is approximately 1.3 rad/s.
Answer: The bandwidths are separated by about a decade, justifying successive loop closure.
The complete solution computes separate inner-loop and outer-loop margin plots. Before opening it, outline each component separately.
These prompts are ungraded. They are intended to connect the derivation to the implementation code in the complete solution.
second-order The Python code that can be used to generate the plots for this solution is shown below.
# 3rd-party
import control as cnt
import matplotlib.pyplot as plt
# local (controlbook)
import case_studies.B_pendulum.params as P
from case_studies.B_pendulum.pid_controller import CartPendulumControllerPID
use_dB = True
plant_inner = cnt.tf(P.tf_inner_num, P.tf_inner_den)
plant_outer = cnt.tf(P.tf_outer_num, P.tf_outer_den)
pid = CartPendulumControllerPID()
kp_theta, kd_theta, sigma = pid.kp_theta, pid.kd_theta, pid.sigma
controller_inner = cnt.tf([kd_theta + kp_theta * sigma, kp_theta], [sigma, 1])
kp_z, ki_z, kd_z = pid.kp_z, pid.ki_z, pid.kd_z
controller_outer = cnt.tf(
[kd_z + kp_z * sigma, kp_z + ki_z * sigma, ki_z], [sigma, 1, 0]
)
# open-loop system
sys_inner_ol = cnt.series(controller_inner, plant_inner)
sys_outer_ol = cnt.series(controller_outer, plant_outer)
# closed-loop system
sys_inner_cl = cnt.feedback(sys_inner_ol, 1)
sys_outer_cl = cnt.feedback(sys_outer_ol, 1)
if __name__ == "__main__":
# part a)
# You could call bode() once with multiple systems, but it draws margins
# for all of them. Here we only draw them for the open-loop system.
# Calling bode() multiple times reuses the same axes, but removes the title
# and legends, so we manually create the figures and axes to add them later.
fig, ax = plt.subplots(2, sharex=True)
cnt.bode(sys_inner_ol, ax=ax, dB=use_dB, display_margins=True, label="open-loop")
cnt.bode(sys_inner_cl, ax=ax, dB=use_dB, grid=False, label="closed-loop")
# cross-over frequency ≈ closed-loop bandwidth
bandwidth_inner_cl = cnt.bandwidth(sys_inner_cl)
print(f"\nClosed-loop bandwidth (inner): {bandwidth_inner_cl:.2f} rad/s")
# Margin info is on the plot, so we don't really need to do this, but this
# is how you can calculate it without plotting.
gm, pm, wcg, wcp = cnt.margin(sys_inner_ol)
print(f"Crossover frequency (inner): {wcp:.2f} rad/s")
# draw bandwidth line
ax[0].axvline(x=bandwidth_inner_cl, color="red", linestyle=":", label="bandwidth")
# manually add annotations after running bode() multiple times
ax[0].legend()
fig.suptitle("HW 17 Bode Plots: Pendulum on a Cart - Inner Loop")
fig.tight_layout()
# keep figure open until closed by user
plt.show()
# part b)
fig, ax = plt.subplots(2, sharex=True)
cnt.bode(sys_outer_ol, ax=ax, dB=use_dB, display_margins=True, label="open-loop")
cnt.bode(sys_outer_cl, ax=ax, dB=use_dB, grid=False, label="closed-loop")
bandwidth_outer_cl = cnt.bandwidth(sys_outer_cl)
print(f"\nClosed-loop bandwidth (outer): {bandwidth_outer_cl:.2f} rad/s")
gm, pm, wcg, wcp = cnt.margin(sys_outer_ol)
print(f"Crossover frequency (outer): {wcp:.2f} rad/s")
ax[0].axvline(x=bandwidth_outer_cl, color="red", linestyle=":", label="bandwidth")
ax[0].legend()
fig.suptitle("HW 17 Bode Plots: Pendulum on a Cart - Outer Loop")
fig.tight_layout()
plt.show()
# part c)
bandwidth_separation = bandwidth_inner_cl / bandwidth_outer_cl
print(f"\n{bandwidth_separation = :.2f}")The transfer functions for the inner and outer loop plants and
controller are defined in Lines 14–19. For this problem, we first
calculate the gain and phase margins (see Lines 26–35). We do this so
that we can then plot the open-loop and closed-loop Bode plots with
stability margin information shown on the plots (see Lines 38–56). If
all we want is the open-loop Bode plot with stability margins, we can
simply pass the argument “margins=True” to the bode
function. The results of this code are shown in Fig. 17.12. We follow a
similar process for the outer loop with results shown in
Fig. 17.13.
As seen from Fig. 17.12, the bandwidth of the inner loop is approximately \(19\) rad/sec, which is slightly larger than the crossover frequency of \(10\) rad/sec. Similarly, Fig. 17.13 indicates that the bandwidth of the outer loop is approximately \(1.3\) rad/sec, which is slightly larger than the crossover frequency of \(1.0\) rad/sec with a phase margin of \(PM=72\) degrees (the plot says -108, but this is off by \(180\), so that \(PM = 180-108\)).
For a second-order system, making the step response 10 times faster implies that \(t_r\) is divided by 10, or equivalently that \(\omega_n\) is multiplied by 10. A similar principle holds for high order systems. To be 10 times faster, the closed-loop bandwidth of the inner loop should be a decade higher on the Bode plot than the closed-loop bandwidth of the outer loop. The bandwidth separation between the inner and outer loop is about one decade justifying the successive loop closure design approach.
Example Problem
3.3.20For this problem, use the gains found in HW 3.3.13.
For the inner loop of the satellite attitude controller, use the
bode and margin commands (from MATLAB or
Python) to find the phase and gain margin for the inner loop system
under PD control. On the same graph, plot the open-loop Bode plot and
the closed-loop Bode plot. What is the bandwidth of the closed-loop
system, and how does this relate to the crossover frequency?
For the outer loop of the satellite attitude controller, use the
bode and margin commands to find the phase and
gain margin for the outer loop system under PID control. Plot the open
and closed-loop Bode plots for the outer loop on the same plot as the
open and closed-loop Bode plots for the inner loop. What is the
bandwidth of the closed-loop system, and how does this relate to the
crossover frequency?
What is the bandwidth separation between the inner (fast) loop, and the outer (slow) loop? For this design, is successive loop closure justified?
Design Study C guided exercise
Compare the inner and outer loop crossover frequencies, phase margins, bandwidths, and bandwidth separation.
Answer: The inner-loop bandwidth is about 11 rad/s with phase margin about 56 degrees.
Answer: The outer-loop bandwidth is about 1 rad/s, giving close to a decade of separation.
Answer: Yes, the bandwidths are close to a decade apart and the closed inner loop is approximately one over the outer-loop bandwidth.
Python code to generate the plots in this solution is included below.
# 3rd-party
import control as cnt
import matplotlib.pyplot as plt
# local (controlbook)
import case_studies.C_satellite.params as P
from case_studies.C_satellite.pid_controller import SatelliteControllerPID
use_dB = True
plant_inner = cnt.tf(P.tf_inner_num, P.tf_inner_den)
plant_outer = cnt.tf(P.tf_outer_num, P.tf_outer_den)
pid = SatelliteControllerPID()
kp_theta, kd_theta, sigma = pid.kp_theta, pid.kd_theta, pid.sigma
controller_inner = cnt.tf([kd_theta + kp_theta * sigma, kp_theta], [sigma, 1])
kp_phi, ki_phi, kd_phi = pid.kp_phi, pid.ki_phi, pid.kd_phi
controller_outer = cnt.tf(
[kd_phi + kp_phi * sigma, kp_phi + ki_phi * sigma, ki_phi], [sigma, 1, 0]
)
# open-loop system
sys_inner_ol = cnt.series(controller_inner, plant_inner)
sys_outer_ol = cnt.series(controller_outer, plant_outer)
# closed-loop system
sys_inner_cl = cnt.feedback(sys_inner_ol, 1)
sys_outer_cl = cnt.feedback(sys_outer_ol, 1)
if __name__ == "__main__":
# part a)
# You could call bode() once with multiple systems, but it draws margins
# for all of them. Here we only draw them for the open-loop system.
# Calling bode() multiple times reuses the same axes, but removes the title
# and legends, so we manually create the figures and axes to add them later.
fig, ax = plt.subplots(2, sharex=True)
cnt.bode(sys_inner_ol, ax=ax, dB=use_dB, display_margins=True, label="open-loop")
cnt.bode(sys_inner_cl, ax=ax, dB=use_dB, grid=False, label="closed-loop")
# cross-over frequency ≈ closed-loop bandwidth
bandwidth_inner_cl = cnt.bandwidth(sys_inner_cl)
print(f"\nClosed-loop bandwidth (inner): {bandwidth_inner_cl:.2f} rad/s")
# Margin info is on the plot, so we don't really need to do this, but this
# is how you can calculate it without plotting.
gm, pm, wcg, wcp = cnt.margin(sys_inner_ol)
print(f"Crossover frequency (inner): {wcp:.2f} rad/s")
# draw bandwidth line
ax[0].axvline(x=bandwidth_inner_cl, color="red", linestyle=":", label="bandwidth")
# manually add annotations after running bode() multiple times
ax[0].legend()
fig.suptitle("HW 17 Bode Plots: Satellite - Inner Loop")
fig.tight_layout()
# keep figure open until closed by user
plt.show()
# part b)
fig, ax = plt.subplots(2, sharex=True)
cnt.bode(sys_outer_ol, ax=ax, dB=use_dB, display_margins=True, label="open-loop")
cnt.bode(sys_outer_cl, ax=ax, dB=use_dB, grid=False, label="closed-loop")
bandwidth_outer_cl = cnt.bandwidth(sys_outer_cl)
print(f"\nClosed-loop bandwidth (outer): {bandwidth_outer_cl:.2f} rad/s")
gm, pm, wcg, wcp = cnt.margin(sys_outer_ol)
print(f"Crossover frequency (outer): {wcp:.2f} rad/s")
ax[0].axvline(x=bandwidth_outer_cl, color="red", linestyle=":", label="bandwidth")
ax[0].legend()
fig.suptitle("HW 17 Bode Plots: Satellite - Outer Loop")
fig.tight_layout()
plt.show()
# part c)
bandwidth_separation = bandwidth_inner_cl / bandwidth_outer_cl
print(f"\n{bandwidth_separation = :.2f}")Similar to the last two case studies, the transfer functions for the inner and outer loop plants are calculated first. Then we find the stability margins and generate Bode plots for the inner loop (see Fig. 17.14). We follow the same procedure for the outer loop plots shown in Fig. 17.15.
As shown in Fig. 17.14, the bandwidth of the inner loop is approximately \(11\) rad/sec, which is slightly larger than the crossover frequency of \(7\) rad/sec, with a phase margin of \(PM=56\) degrees. Similarly, Fig. 17.15 indicates that the bandwidth of the outer loop is approximately the crossover frequency of \(1\) rad/sec. The bandwidth separation between the inner and outer loops is close to a decade, and the magnitude of the inner-loop closed-loop frequency response is approximately one at frequencies well above the crossover frequency of the outer loop.
Objective: estimate gain crossover, phase margin, phase crossover, and gain margin from sampled Bode data.
Evaluation criteria: the lab checks interpolation at crossover points and compares the computed margins to a reference result.
Phase margin is measured at:
Gain margin is measured relative to:
If the phase at gain crossover is -135 degrees, enter the phase margin in degrees.
Answer: 45 degrees because the phase is 45 degrees above -180 degrees.
At phase crossover, the magnitude is -8 dB. Enter the gain margin in dB.
Answer: 8 dB because the gain could increase by 8 dB before reaching 0 dB at phase crossover.
If the Bode plot has no gain crossover, what should you say about phase margin?
Answer: It is undefined or does not exist because there is no gain-crossover frequency where it can be measured.
At the gain-crossover frequency, the phase is -140 degrees. What is the phase margin?
At phase crossover, the magnitude is -12 dB. What is the gain margin?
What commonly happens when loop gain is increased in a system whose phase becomes more negative at higher frequency?
If the Bode plot does not show a phase crossover in the plotted frequency range, what should you conclude about gain margin from that plot alone?