8 Design Strategies for Second-Order Systems

8.1 Theory

In the previous section we saw that for second-order systems, PD control could be used to exactly specify the pole location of the closed-loop system. This chapter examines the design problem of how to select the pole locations to achieve desired behavior. In Section 11.1.1 we derive the relationship between a single pole location and the step response of the system. In Section 11.1.2 we show that the time-domain response for a second-order system can be understood in terms of natural frequency and damping ratio. Finally, in Section 11.1.3 we show the effect that a zero has on the step response of a second-order system.

8.1.1 First Order Response

Consider a first-order system given by \[Y(s) = \frac{p}{s+p} U(s),\] where \(p>0\). The pole location of the first-order system is at \(-p\). If \(U(s)=\frac{A}{s}\) is a step of magnitude \(A\), then \[Y(s) = \frac{Ap}{s(s+p)}.\] As shown in Appendix 29, the inverse Laplace transform is given by \[\tag{8.1} y(t) = \begin{cases} A(1-e^{-pt}), & t\geq 0 \\ 0, & \text{otherwise} \end{cases}.\] The response is shown in Fig. 8.1, which shows the response to steps of size \(A=1\), \(A=2\), and \(A=3\), respectively.

If we define the rise time \(t_r\) of the system to be the time after the step when the response reaches 90% of its final value, then we note from Fig. 8.1 that the rise time is independent of the step size \(A\). Also, by differentiating Equation 8.1 at time \(t=0\), it is straightforward to show that the slope of the response at \(t=0\) is equal to \(Ap\). This is also shown graphically in Fig. 8.1.

When \(A=1\), the step response for different pole locations is shown in Fig. 8.2. Pole locations corresponding to \(p=1\), \(p=2\), and \(p=4\) are shown. Note that as the pole location moves further into the left half of the complex plane, the rise time decreases.

An important concept for the steady-state response of a system is the DC gain, which is defined as follows.

The DC gain of a transfer function \(H(s)\) is \[\left.H(s)\right|_{\text{DC gain}} = \lim_{s\to 0} H(s).\]

An important fact is that if the poles of \(H(s)\) are in the open left half-plane, then the response of \(H(s)\) to a step of size \(A\) approaches \(A\left.H(s)\right|_{\text{DC gain}}\) as \(t\to\infty\). This is true for first-order systems, as well as general \(H(s)\). For example, the DC gain of the first-order system \(H(s) = \frac{q}{s+p}\) is \(\frac{q}{p}\), as long as \(p>0\).

8.1.2 Second-Order Response

Suppose that the closed-loop system has a second-order transfer function with two real poles and no zeros: \[Y(s) = \frac{K}{(s+p_1)(s+p_2)}R(s).\] To find the output of the system when the input is a step of magnitude \(A\), we find the inverse Laplace transform of \[Y(s) = \frac{K}{(s+p_1)(s+p_2)}\frac{A}{s}.\] Using partial fraction expansion we get \[Y(s) = \frac{\frac{KA}{p_1p_2}}{s} + \frac{\frac{KA}{(-p_1)(p_2-p_1)}}{s+p_1} + \frac{\frac{KA}{(-p_2)(p_1-p_2)}}{s+p_2}.\] Taking the inverse Laplace transform gives \[y(t) = \begin{cases} \frac{KA}{p_1p_2} + \frac{KA}{(-p_1)(p_2-p_1)}e^{-p_1t} + \frac{KA}{(-p_2)(p_1-p_2)} e^{-p_2t}, & t\geq 0 \\ 0, & \text{otherwise} \end{cases}.\] Note that as \(t\to\infty\), \(y(t)\to \frac{KA}{p_1p_2}\), which is equal to the step size \(A\) times the DC gain of the system.

Now suppose that the closed-loop system has two complex poles and no zeros. We will write a generic second-order system with poles in the left half-plane using the notation \[H(s) = \frac{K\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2},\] where \(K\) is the DC gain of the system (typically \(K=1\)), \(\omega_n>0\) is called the natural frequency, and \(\zeta>0\) is called the damping ratio. The poles are given by the roots of the characteristic polynomial \[\Delta(s) = s^2 + 2\zeta\omega_n s + \omega_n^2=0,\] in other words, the poles are \[\begin{aligned} p_{1,2} &= -\zeta\omega_n \pm \sqrt{(\zeta\omega_n)^2-\omega_n^2} \\ &= -\zeta\omega_n \pm \omega_n\sqrt{\zeta^2-1}. \end{aligned}\] If \(0<\zeta<1\), then the roots are complex and given by \[\tag{8.2} p_{1,2} = -\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^2}.\] Using the Python code

import control as cnt
import matplotlib.pyplot as plt

#start by varying the natural frequency
plt.figure(1), plt.clf()
zeta = 0.707

for wn in [0.5, 1.0, 2.0, 5.0]:
    G = cnt.tf(wn**2, [1, 2*zeta*wn, wn**2])
    t,y = cnt.step_response(G)
    plt.plot(t, y)

plt.plot([0, 20.0] , [1, 1], 'k--')
plt.xlabel('Time (seconds)')
plt.ylabel('Amplitude')
plt.title('Step Response')
plt.legend(('wn=0.5', 'wn=1.0', 'wn=2.0', 'wn=5.0'))

#now we'll see the effect of varying zeta
plt.figure(2), plt.clf()
wn = 1.0

for zeta in [0.2, 0.4, 0.707, 1.0, 2.0]:
    G = cnt.tf(wn**2, [1, 2*zeta*wn, wn**2])
    t,y = cnt.step_response(G)
    plt.plot(t, y)
    
plt.plot([0, 35.0] , [1, 1], 'k--')
plt.xlabel('Time (seconds)')
plt.ylabel('Amplitude')
plt.title('Step Response')
plt.legend(('zeta=0.2', 'zeta=0.4', 'zeta=0.707',
            'zeta=1.0', 'zeta=2.0'))

plt.show() # necessary to make plots appear

we obtain second-order response plots similar to the ones shown in Figures 8.3 and 8.4. Note from Fig. 8.3 that as the natural frequency increases, the rise time decreases. We should note that a similar phenomenon to Fig. 8.1 also occurs, where the rise time is independent of the size of the step. Note from Fig. 8.4 that the damping ratio affects the amount of ringing in the system. For a small damping ratio, e.g., \(\zeta=0.2\), there is a large overshoot and significant ringing in the system. At the other extreme, when \(\zeta\) is larger than one, the poles are real and the response is highly damped. The sweet spot is when \(\zeta=0.707\), where the rise time is small but the overshoot is minimal.

Note from Equation 8.2 that when \(\zeta = 0.707 = \frac{1}{\sqrt{2}}\), the poles are at \[p_{1,2} = -\frac{\omega_n}{\sqrt{2}} \pm j\frac{\omega_n}{\sqrt{2}}.\] In the complex plane, the pole locations are shown in Fig. 8.5, where \(\omega_n\) is the distance from the origin to the poles. When \(\zeta=\frac{1}{\sqrt{2}}\) the angle \(\theta\) equals \(45\) degrees.

In rectangular coordinates, the poles are given by \[\tag{8.3} p_{1,2} = -\sigma \pm j\omega_d,\] as shown in Fig. 8.5. Equations 8.2 and 8.3 give \[\begin{aligned} \sigma &= \zeta \omega_n \\ \omega_d &= \omega_n\sqrt{1-\zeta^2}. \end{aligned}\] From Fig. 8.5 it is also clear that \[\sin\theta = \frac{\sigma}{\omega_n} = \zeta.\] Therefore \[\theta = \sin^{-1} \zeta.\]

Since the pole locations are at \(-\zeta\omega_n \pm \omega_n\sqrt{\zeta^2-1}\), the poles are both real if \(\zeta\geq 1\). When \(\zeta=1\) the poles are both located at \(-\omega_n\). When \(0<\zeta<1\) the poles are imaginary and occur in complex conjugate pairs. As shown in Fig. 8.6, we say that the system is overdamped when \(\zeta>1\), we say that the system is critically damped when \(\zeta=1\), and we say that the system is underdamped when \(0<\zeta<1\).

The response of \(H(s)\) to a step of size \(A\) is the inverse Laplace transform of \[Y(s) = \frac{A\omega_n^2}{s(s^2+2\zeta\omega_n s + \omega_n^2)},\] which, as shown in Appendix 29, when the poles are complex results in \[\tag{8.4} y(t) = A\left[ 1-\frac{1}{\sqrt{1-\zeta^2}}e^{-\sigma t}\cos\left(\omega_d t - \tan^{-1}\left(\frac{\sigma}{\omega_d}\right)\right)\right].\] A plot of this signal is shown in Fig. 8.7.

The rise time is defined to be the time that it takes the output to transition from 10% to 90% of its final value \(A\). A suitable approximation is that the rise time is one half of the peak time, which is the time when \(y(t)\) reaches its first peak. To find the peak time \(t_p\), differentiate \(y(t)\) in Equation 8.4 and then solve for the first instance when the derivative is zero. The result is \[t_p = \frac{\pi}{\omega_d} = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}}.\] Therefore, the rise time is approximately \[t_r \approx \frac{1}{2}t_p = \frac{1}{2}\frac{\pi}{\omega_n\sqrt{1-\zeta^2}}.\] When \(\zeta=0.707\) we get \[\tag{8.5} t_r \approx \frac{2.2}{\omega_n}.\]

Specifications for second-order systems are usually given either in terms of the rise time \(t_r\) and the damping ratio \(\zeta\), or in terms of the natural frequency \(\omega_n\) and the damping ratio \(\zeta\).

8.1.3 Effect of a Zero on the Step Response

In the previous section, we looked at the step response when there were two complex poles and no zeros. In this section we will briefly discuss the effect of zeros on the step response of the system. This is a topic that we could treat at a much deeper level.

A transfer function is called rational if it is the ratio of two polynomials in \(s\). For example, \[H(s) = \frac{s+1}{s^3+3s^2+s+4}\] is a rational transfer function, whereas \[H(s) = \frac{(s+1)e^{-s}}{s^3+3s^2+s+4}\] is not. A rational transfer function can be written as \[H(s) = \frac{N(s)}{D(s)},\] where \(N(s)\) is the numerator polynomial and \(D(s)\) is the denominator polynomial. The zeros of the transfer function are roots of the equation \(N(s)=0\), and the poles of the system are the roots of the equation \(D(s)=0\).

One way to understand the effect of zeros is to consider the effect that zeros have on the partial fraction expansion of a transfer function. For a review of partial fraction expansion, see Appendix 29. First consider a two-pole system with characteristic equation given by \(\Delta(s) = (s+2)(s+3)\). The transfer function with DC gain equal to one and without zeros is given by \[H(s) = \frac{6}{(s+2)(s+3)}.\] The response to a unit step is given by \begin{align} y(t) &= \mathcal{L}^{-1}\left\{\frac{6}{s(s+2)(s+3)}\right\} \notag \\ &= \mathcal{L}^{-1}\left\{ \frac{1}{s} + \frac{-3}{s+2} + \frac{2}{s+3}\right\} \notag \\ &= 1 - 3e^{-2t} + 2e^{-3t}, \quad t\geq 0. \tag{8.6} \end{align} Now consider the step response when a zero has been added at \(z=-2.1\) and the gain adjusted so that the DC gain is still one, i.e., when \[H(s) = \frac{\frac{6}{2.1}(s+2.1)}{(s+2)(s+3)}.\] The response to a unit step is given by \[\begin{aligned} y(t) &= \mathcal{L}^{-1}\left\{\frac{\frac{6}{2.1}(s+2.1)}{s(s+2)(s+3)}\right\} \\ &= \mathcal{L}^{-1}\left\{ \frac{1}{s} + \frac{-0.1429}{s+2} + \frac{-0.8571}{s+3}\right\}\\ &= 1 - 0.1429e^{-2t} - 0.8571e^{-3t}, \quad t\geq 0. \end{aligned}\] Note that \(y(t)\) still settles to one, but the response of the pole at \(-2\) has almost been removed. If, however, the zero is far away from the poles, then the effect is much less. Suppose that the zero is at \(z=-60\) and the gain is adjusted so that the DC gain is still one, i.e., \[H(s) = \frac{\frac{6}{60}(s+60)}{(s+2)(s+3)}.\] The response to a unit step is given by \[\begin{aligned} y(t) &= \mathcal{L}^{-1}\left\{\frac{\frac{6}{60}(s+60)}{s(s+2)(s+3)}\right\} \\ &= \mathcal{L}^{-1}\left\{ \frac{1}{s} + \frac{-2.9}{s+2} + \frac{1.9}{s+3}\right\}\\ &= 1 - 2.9e^{-2t} + 1.9e^{-3t}, \quad t\geq 0. \end{aligned}\] In this case, the response is similar to Equation 8.6.

Another way to understand the effect of zeros on the system is to recall that an \(s\) in the numerator is essentially a differentiator. Therefore \[\begin{aligned} H(s) &= \frac{\omega_n^2(\tau s+1)}{s^2 + 2\zeta\omega_n s + \omega_n^2} \\ &= \tau\frac{\omega_n^2 s}{s^2 + 2\zeta\omega_n s + \omega_n^2} + \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}. \end{aligned}\] Therefore, the step response is given by \[\begin{aligned} y(t) &= \mathcal{L}^{-1}\left\{\tau\frac{\omega_n^2s}{s(s^2 + 2\zeta\omega_n s + \omega_n^2)} + \frac{\omega_n^2}{s(s^2 + 2\zeta\omega_n s + \omega_n^2)}\right\} \\ &= \tau\frac{d}{dt}\mathcal{L}^{-1}\left\{\frac{\omega_n^2}{s(s^2 + 2\zeta\omega_n s + \omega_n^2)}\right\} % + \mathcal{L}^{-1}\left\{\frac{\omega_n^2}{s(s^2 + 2\zeta\omega_n s + \omega_n^2)}\right\}. \end{aligned}\] Therefore, the effect of the zero is to add \(\tau\) times the derivative of the step response to what the step response would have been without the zero. Fig. 8.8 shows the step response for different values of \(\tau\), illustrating that adding a zero has a strong effect on the rise time and damping of the response. Note also that when \(\tau<0\), i.e., when zeros are in the right half-plane, the system response first goes negative before going positive. Right-half-plane zeros are often called nonminimum-phase zeros , and the associated response is often called a nonminimum-phase response. Physical systems that have right-half-plane zeros include the inverted pendulum and aircraft altitude-control systems. For the inverted pendulum, to move the cart to the right from a stopped position, the cart must first move left so that the rod leans right before the cart can move to its desired position. Similarly, an aircraft pitches up to climb. However, when the aircraft pitches up, it loses speed, which causes it to lose lift and altitude. As the aircraft loses altitude, it gains speed and produces the lift necessary to climb.

8.1.4 Successive Loop Closure

The natural frequency \(\omega_n\) and the damping ratio \(\zeta\) are only defined for second-order systems, and therefore, the PD design technique based on \(\omega_n\) and \(\zeta\) are only valid for second-order systems. However, these concepts can be extended to higher-order systems through a technique called successive loop closure, or inner-loop, outer-loop design. When we derived the Laplace transform in Chapter 8, we saw that for many systems, the transfer functions can be arranged in cascade form as shown in Fig. 8.9 where the input to the system is \(u_1\) and the output is \(y_2\), but where \(y_1\) and \(u_2\) are intermediary values that represent the natural physical coupling between internal variables. Many physical systems can be represented by a cascade of systems.

When the system is represented as a cascade, the control strategy can be designed using successive loop closure. The basic idea is to close feedback loops in succession around the open-loop plant dynamics rather than designing a single control system. To illustrate this approach, consider the open-loop system shown in Fig. 8.9. The open-loop dynamics are given by the product of two transfer functions in series: \(P(s)=P_1(s)P_2(s)\). We assume that each of the transfer functions has an output (\(y_1\), \(y_2\)) that can be measured and used for feedback. Typically, each of the transfer functions, \(P_1(s)\) and \(P_2(s)\), is of relatively low order – usually first or second-order. In this case, we are interested in controlling the output \(y_2\). Instead of closing a single feedback loop around \(y_2\), we will close feedback loops around \(y_1\) and \(y_2\) in succession as shown in Fig. 8.10.

We will design the compensators \(C_1(s)\) and \(C_2(s)\) in succession. A necessary condition in the design process is that the inner loop must be much faster than each successive loop. As a rule of thumb, if the rise time of the inner loop is \(t_{r_1}\), the rise time of the next loop should be 5 to 10 times longer, i.e., \(t_{r_2}> W_1 t_{r_1}\), where \(W_1\) is a design parameter, usually on the order of \(5-10\). In part V of this book, we will talk about the frequency response of the system. At that time, we will provide a deeper explanation that can be understood in terms of the bandwidth of each successive loop.

Examining the inner loop shown in Fig. 8.10, the goal is to design a closed-loop system from \(r_1\) to \(y_1\) having a rise time of \(t_{r_1}\). If the rise time of the inner loop is significantly faster than the rise time of the outer loop, and if the DC gain of the inner loop is \(k_{DC_1}\), then relative to the outer loop the inner loop can be effectively modeled as the DC gain. This is depicted schematically in Fig. 8.11. With the inner-loop transfer function modeled as its DC gain, design of the outer loop is simplified because it only includes the plant transfer function \(P_2(s)\) and the compensator \(C_2(s)\) and the DC gain \(k_{DC_1}\).

Because both plant models \(P_1(s)\) and \(P_2(s)\) are first or second-order, conventional PID compensators can be employed. The loopshaping techniques discussed in Part V can also be used for successive loop closure of higher-order systems.

8.1.5 Input Saturation and Limits of Performance

A natural question is to ask if the rise time can be made arbitrarily small. Unfortunately, saturation limits on the actuators place a lower bound on the achievable rise time. All physical systems have saturation limits on their actuators. For example, in the robot arm problem, the torque that can be applied to the arm is limited by the motor that is used to apply that torque. Force and torque will always be limited by physical constraints like current limits. Similarly, for an airplane, the rudder command is limited by how far the rudder can physically move, which is typically on the order of \(\pm 40\) degrees. A natural way to model input constraints is to add a saturation block preceding the physical plant. Fig. 8.12 shows a saturation block in conjunction with a PD control scheme.

Mathematically, the saturation block is given by \[u_{sat} = \begin{cases} u_{\max}, & \text{~if~} u\geq u_{\max} \\ u, & \text{~if~} -u_{\max} \leq u \leq u_{\max} \\ -u_{\max}, & \text{~otherwise} \end{cases},\] where the input to the block is \(u\), the output is \(u_{sat}\), and the saturation limit is \(u_{\max}\). Note that the saturation block is non-linear. Therefore, one design strategy is to select the feedback gains so that the system remains linear by keeping the input out of saturation.

To see how this might be done, consider the second-order system shown in Fig. 8.12 with proportional feedback on the output error and derivative feedback on the output. When the input is not saturated, the closed-loop transfer function is \[Y(s) = \left(\frac{b_0 k_p}{s^2 + (a_1 + b_0 k_D)s + (a_0 + b_0 k_p)}\right) R(s). \tag{8.7}\] We can see that the closed-loop poles of the system are defined by the selection of the control gains \(k_p\) and \(k_D\). Note also that the actuator effort \(u\) can be expressed as \(u=k_p e - k_D\dot{y}\). When \(\dot{y}\) is zero or small, the size of the actuator effort \(u\) is primarily governed by the size of the control error \(e\) and the control gain \(k_p\). If the system is stable, the largest control effort in response to a step input will occur immediately after the step when \(\dot{y}=0\). If the system is to be kept just out of saturation, then immediately after the step we have \(\left|u\right|=u_\text{max} = k_p e_\text{max}\). Rearranging this expression, we find that the proportional control gain can be determined from the maximum anticipated output error and the saturation limits of the actuator as \[k_p = \pm \frac{u_\text{max}}{e_\text{max}}, \tag{8.8}\] where \(u_{\max}\) is the maximum control effort the system can provide, \(e_{\max}\) is the step error that results from a step input of nominal size, and the sign of \(k_p\) is determined by the physics of the system.

Suppose that the desired closed-loop characteristic polynomial is given by \[\Delta_{cl}^d(s) = s^2 + 2\zeta\omega_n s + \omega_n^2,\] then by comparing with the actual closed-loop characteristic polynomial \[\Delta_{cl}(s) = s^2 + (a_1 + b_0 k_D)s + (a_0 + b_0 k_p)\] we have that \[\begin{aligned} \omega_n &= \sqrt{a_0 + b_0 k_p} \leq \sqrt{a_0 + b_0 \frac{u_\text{max}}{e_\text{max}}}, \end{aligned}\] which is an upper limit on the natural frequency of the closed-loop system, ensuring that saturation of the actuator is avoided. The corresponding rise time constraint is \[t_r \geq \frac{2.2}{\omega_n} = \frac{2.2}{\sqrt{a_0 + b_0 \frac{u_\text{max}}{e_\text{max}}}},\] which is the smallest possible rise time for an error step of \(e_{\text{max}}\).

If the input has an equilibrium value, then the saturation constraint used in the above calculation must be modified to account for the equilibrium. For example, suppose that \(u=u_e+\tilde{u}\) where \(u_e\) is the equilibrium value and \(\tilde{u}\) is the output of the PID controller. Also suppose that the input saturation constraint is given by \(u_{\min}\leq u\leq u_{\max}\). The objective is to find a value for \(\tilde{u}_{\max}\) where \(\left|\tilde{u}\right|\leq \tilde{u}_{\max}\) guarantees that \(u_{\min}\leq u\leq u_{\max}\). We have that \[\begin{aligned} & u_{\min}\leq u \leq u_{\max} \\ \implies & u_{\min} \leq u_e + \tilde{u} \leq u_{\max} \\ \implies & u_{\min}-u_e \leq \tilde{u} \leq u_{\max}-u_e. \end{aligned}\] The desired value for \(\tilde{u}_{\max}\) is therefore the minimum (in absolute value) of the right and left-hand sides of this expression. The situation is shown in Fig. 8.13, where it is clear that the value for \(\tilde{u}_{\max}\) can be written as \[\tilde{u}_{\max} = \begin{cases} \left|u_{\max}-u_e\right|, & \text{~if~} u_e \geq \frac{u_{\max}+u_{\min}}{2} \\ \left|u_e - u_{\min}\right|, & \text{~if~} u_e \leq \frac{u_{\max}+u_{\min}}{2}. \end{cases}\]

The takeaway from this discussion is that the rise time is limited by the input saturation constraint and cannot be made arbitrarily small. In practice, the control parameters are tuned so that the closed-loop system has the fastest possible rise time without driving the plant input into saturation.

8.2 Design Study A. Single Link Robot Arm

8.2.1 Example Problem 3.1.11

For the single-link robot arm, do the following:

(a) Suppose that the design requirements are that the rise time is \(t_r \approx 0.8\) seconds, with a damping ratio of \(\zeta=0.707\). Find the desired closed-loop characteristic polynomial \(\Delta_{cl}^d(s)\) and the associated pole locations. Find the proportional and derivative gains \(k_P\) and \(k_D\) to achieve these specifications, and modify the simulation from HW 3.1.8 to verify that the step response satisfies the requirements.

(b) Suppose that the input torque for the robot arm is limited to \(\left|\tau\right|\leq\tau_{\max}=1\) Nm. Modify the simulation to include a saturation on the torque \(\tau\). Using the rise time \(t_r\) and damping ratio \(\zeta\) as tuning parameters, tune the PD control law so that the input just saturates when a step of size \(50\) degrees is placed on \(\tilde{r}_{\theta}\).

# 3rd-party
import numpy as np

# local (controlbook)
import case_studies.A_arm.params as P
from case_studies.common.controller_base import ControllerBase


class ArmControllerPD(ControllerBase):
    def __init__(self, use_feedback_linearization: bool = True):
        # tuning parameters
        p1 = -3.0
        p2 = -4.0

        # system parameters
        b0 = P.tf_num[-1]
        a1, a0 = P.tf_den[-2:]

        # desired characteristic equation parameters
        # des_CE: s^2 + alpha1*s + alpha0 = s^2 + (a1 + b0*kd)s + (a0 + b0*kp)
        des_CE = np.poly([p1, p2])
        alpha1, alpha0 = des_CE[-2:]

        # find gains
        self.kp = (alpha0 - a0) / b0
        self.kd = (alpha1 - a1) / b0
        print(f"{self.kp = :.2f}, {self.kd = :.3f}")

        self.tau_eq = P.tau_eq
        self.use_feedback_linearization = use_feedback_linearization

    def update_with_measurement(self, r, y):
        raise NotImplementedError("This controller only uses state feedback.")

    def update_with_state(self, r, x):
        # unpack references and states
        theta_ref = r[0]
        theta, thetadot = x

        # theta (modified) PD
        error = theta_ref - theta
        tau_tilde = self.kp * error - self.kd * thetadot

        if self.use_feedback_linearization:
            tau_fl = P.m * P.g * P.ell / 2 * np.cos(theta)
            tau = tau_tilde + tau_fl
        else:  # use equilibrium (Jacobian linearization)
            # NOTE: likely not work well (or at all) for large angles
            tau = tau_tilde + self.tau_eq

        u = np.array([tau])
        return u

# 3rd-party
import numpy as np

# local (controlbook)
import case_studies.A_arm.params as P
from case_studies.A_arm.dynamics import ArmDynamics
from case_studies.A_arm.pd_controller import ArmControllerPD
from case_studies.A_arm.visualizer import ArmVisualizer
from case_studies.common.signal_generator import SignalGenerator
from case_studies.common.simulation import run_simulation


arm = ArmDynamics()
controller = ArmControllerPD(use_feedback_linearization=True)
theta_ref = SignalGenerator(amplitude=np.radians(50), frequency=0.05)

time, x_hist, u_hist, r_hist, *_ = run_simulation(
    arm,
    [theta_ref],
    controller,
    controller_input="state",
    t_final=20,
    dt=P.ts,
)

viz = ArmVisualizer(time, x_hist, u_hist, r_hist)
viz.plot()
# viz.animate()

8.3 Design Study B. Inverted Pendulum

8.3.1 Example Problem 3.2.11

For the inverted pendulum, do the following:

(a) Using the principle of successive loop closure, draw a block diagram that uses PD control for both inner loop control and outer loop control. The input to the outer loop controller is the desired cart position \(r_z\) and the output of the controller is the desired pendulum angle \(r_{\theta}\). The input to the inner loop controller is the desired pendulum angle \(r_{\theta}\) and the output is the force \(F\) on the cart.

(b) Focusing on the inner loop, find the PD gains \(k_{P_\theta}\) and \(k_{D_\theta}\) so that the rise time of the inner loop is \(t_{r_\theta}=0.5\) seconds, and the damping ratio is \(\zeta_{\theta}=0.707\).

(c) Find the DC gain \(k_{DC_\theta}\) of the inner loop.

(d) Replace the inner loop by its DC gain and consider the outer loop. Add a low-pass filter that cancels the left-half-plane zero, and then design a PD control strategy to stabilize the position of the cart.

(e) Implement the control design to control on the inverted pendulum. Modify the simulation to include saturation on the force \(F\) to limit the size of the input force on the cart to \(F_{\max}=5\) N. Give the pendulum angle an initial condition of 10 deg and simulate the response of the system for 10 seconds to verify that the cart is able to balance the pendulum. You may also tune rise times or damping to improve performance.

Design Study B guided exercise

Organize the successive-loop pendulum controller

Separate the inner angle loop from the outer cart-position loop.

The inner loop stabilizes pendulum angle before the outer loop commands cart position.

Step 1. Loop signal order

Signal order

Check Step 1

Step 2. Inner-loop design target

Inner controlled variable Rise time

Check Step 2

Step 3. Outer-loop design feature

Added block

Check Step 3

Pseudocode

The complete solution includes a nested PD controller class and a zero-canceling filter class. 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.

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8.4 Design Study C. Satellite Attitude Control

8.4.1 Example Problem 3.3.11

(a) For the simple satellite system, using the principle of successive loop closure, draw a block diagram that uses PD control for the inner and the outer loop control. The input to the outer loop controller is the desired angle of the solar panel \(r_{\phi}\) and the output is the desired angle of the satellite \(r_{\theta}\). The input to the inner loop controller is \(r_{\theta}\) and the output is the torque on the satellite \(\tau\).

(b) Focusing on the inner loop, find the PD gains \(k_{P_\theta}\) and \(k_{D_\theta}\) so that the rise time of the inner loop is \(t_{r_\theta}=1\) second, and the damping ratio is \(\zeta_{\theta}=0.9\).

(c) Find the DC gain \(k_{DC_\theta}\) of the inner loop.

(d) Replacing the inner loop by its DC gain, find the PD gains \(k_{P_\phi}\) and \(k_{D_\phi}\) so that the rise time of the outer loop is \(t_{r_\phi}= 10 t_{r_\theta}\) with damping ratio \(\zeta_{\phi}=0.9\).

(e) Implement the successive loop closure design for the satellite system in simulation where the commanded solar panel angle is given by a square wave with magnitude \(15\) degrees and frequency \(0.015\) Hz.

(f) Suppose that the size of the input torque on the satellite is limited to \(\tau_{\max}=5\) Nm. Modify the simulation to include saturation on the torque \(\tau\). Use rise time as a tuning parameter and tune the PD control law to get the fastest possible response without input saturation when a step of size \(30\) degrees is placed on \(r_\phi\).

Design Study C guided exercise

Plan the satellite successive-loop controller

Use the body loop as the fast inner loop and the flexible-panel loop as the slower outer loop.

Step 1. Signal architecture

Inner controlled variable Outer controlled variable Actuator input

Check Step 1

Step 2. Inner-loop design target

Inner rise time Damping ratio Inner DC gain

Check Step 2

Step 3. Outer-loop separation

Outer rise-time relationship Why not tune too fast?

Check Step 3

Pseudocode

Before opening the implementation code, outline the nested PD controller and saturation check.

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8.5 Coding Lab: PD Controller Implementation

8.6 Knowledge Check

Use these quick checks to test understanding of Chapter 8 concepts before moving on.

8.7 Notes and References