# Blog

# Control system-QA56

- March 28, 2018
- Posted by:
- Category: Control system QA

**DC Motor Position: PID Controller Design Using the Root Locus Approach**

**1.Physical Setup:** A common actuator in control systems is the DC motor. It directly provides rotary motion and, coupled with wheels or drums and cables, can provide translational motion. The electric equivalent circuit of the armature and the free-body diagram of the rotor are shown in the following figure. Figure

**System Parameters:** For this problem, we will assume the following values for the physical parameters. These values were derived by experiment from an actual motor in Carnegie Mellon’s undergraduate controls lab.

(J) moment of inertia of the rotor 3.2284e − 6 kg.m2

(b) motor viscous friction constant 3.5077e − 6 N.m.s

(Kb) electromotive force constant 0.0274 V/rad/sec

(Kt) motor torque constant 0.0274 N.m/Amp

(R) electric resistance 4 Ω

(L) electric inductance 2.75e − 6 H

**System Equations:** In general, the torque generated by a DC motor is proportional to the armature current and the strength of the magnetic field. In this problem we will assume that the magnetic field is constant and, therefore, that the motor torque is proportional to only the armature current i by a constant factor Kt as shown in the equation below. This is referred to as an armature-controlled motor.

T = Kti

The back emf, e, is proportional to the angular velocity of the shaft by a constant factor Kb.

e = Kb ˙θ

In SI units, the motor torque and back emf constants are equal, that is, Kt = Kb; therefore, we will use K to represent both the motor torque constant and the back emf constant.

From the figure above, we can derive the following governing equations based on Newton’s 2nd law and Kirchhoff’s voltage law

J ¨θ + b ˙θ = Ki

L di dt + Ri = V − K ˙θ

• By taking the Laplace transform of the above equations, obtain the transfer function of the open-loop system, defined as, G(s) := θ(s) V (s) . Here, the input voltage V (s) is assumed to be the control input for the open-loop system, and the output is taken as position of the motor shaft θ(s).

• Sketch the root locus of the open-loop system using the seven steps covered in your notes. You must do all the required math and support your answer.

• Sketch the root locus of the open-loop system in MATLAB and compare it with your plot.

• Using the root locus approach, design a PID controller for the unity closed-loop control system with the desired input θd (i.e., desired shaft position ) and the actual output θ (i.e., actual shaft position) to achieve the following requirements:

(1) Settling time < 0.04 seconds,
(2) Overshoot < 16%, and
(3) No steady-state error, even in the presence of a step disturbance input.
(You must design the PID controller based on the techniques covered in the root locus approach chapter. Do not use MATLAB pidtuner, sisotool, lqr, or similar packages.)
• Using MATLAB/Simulink, simulate the linear closed-loop system’s response for a step input 0.1 (rad) in the presence of a step-like external disturbance with the magnitude 1 (N.m) over the time interval [0, 10] (s).
• Using final value theorem, compute the final value of the tracking error, i.e., limt→∞ e(t), for a ramp input in the absence of the external disturbance. Repeat this computation for an acceleration input. Check your results with MATLAB/Simulink.
• Using MATLAB, sketch the Bode plots for the original open-loop system G(s) and the modified open-loop system C(s)G(s), where C(s) represents the controller transfer function. Compute the gain and phase margins for the modified open-loop system using MATLAB. What is the maximum actuator delay time to make the system marginally stable?

**Inverted Pendulum: PID Controller Design Using the Root Locus Approach**

**2.Physical Setup:** The system in this problem consists of an inverted pendulum mounted to a motorized cart. The inverted pendulum system is an example commonly found in control system textbooks and research literature. Its popularity derives in part from the fact that it is unstable without control, that is, the pendulum will simply fall over if the cart isn’t moved to balance it. Additionally, the dynamics of the system are nonlinear. The objective of the control system is to balance the inverted pendulum by applying a force to the cart that the pendulum is attached to. A real-world example that relates directly to this inverted pendulum system is the attitude control of a booster rocket at takeoff.

In this case we will consider a two-dimensional problem where the pendulum is constrained to move in the vertical plane shown in the figure below. For this system, the control input is the force F that moves the cart horizontally and the output is the angular position of the pendulum θ.

**System Parameters:** For this problem, let’s assume the following quantities:

(M) mass of the cart 0.5 kg

(m) mass of the pendulum 0.2 kg

(b) coefficient of friction for cart 0.1 N/m/sec

(l) length to pendulum center of mass 0.3 m

(I) mass moment of inertia of the pendulum 0.006 kg.m2

(F) force applied to the cart

(x) cart position coordinate

(θ) pendulum angle from vertical (down)

**System Equations:** Below are the free-body diagrams of the two elements of the inverted pendulum system. Summing the forces in the free-body diagram of the cart in the horizontal direction, you get the following equation of motion.

Mx¨ + bx˙ + N = F

Note that you can also sum the forces in the vertical direction for the cart, but no useful information would be gained.

Summing the forces in the free-body diagram of the pendulum in the horizontal direction, you get the following expression for the reaction force N.

N = mx¨ + ml¨θ cos θ − ml ˙θ 2 sin θ

If you substitute this equation into the first equation, you get one of the two governing equations for this system.

(M + m)¨x + bx˙ + ml¨θ cos θ − ml ˙θ 2 sin θ = F

To get the second equation of motion for this system, sum the forces perpendicular to the pendulum. Solving the system along this axis greatly simplifies the mathematics. You should get the following equation.

P sin θ + N cos θ − mg sin θ = ml¨θ + mx¨ cos θ

To get rid of the P and N terms in the equation above, sum the moments about the centroid of the pendulum to get the following equation.

−Plsin θ − Nl cos θ = I ¨θ

Combining these last two expressions, you get the second governing equation. (I + ml2 ) ¨θ + mglsin θ = −mlx¨ cos θ

Since the analysis and control design techniques we will be employing in this problem apply only to linear systems, this set of equations needs to be linearized. Specifically, we will linearize the equations about the vertically upward equilibrium position, θ = π, and will assume that the system stays within a small neighborhood of this equilibrium. This assumption should be reasonably valid since under control we desire that the pendulum not deviate more than 20 degrees from the vertically upward position. Let φ represent the deviation of the pendulum’s position from equilibrium, that is, θ = π + φ. Again presuming a small deviation (φ) from equilibrium, we can use the following small angle approximations of the nonlinear functions in our system equations:

cos θ = cos(π + φ) ≈ −1

sin θ = sin(π + φ) ≈ −φ

˙θ 2 = φ˙2 ≈ 0

After substituting the above approximations into our nonlinear governing equations, we arrive at the two linearized equations of motion. Note u has been substituted for the input F. (I + ml2 )φ¨ − mglφ = mlx¨ (14)

(M + m)¨x + bx˙ − mlφ¨ = u

• By taking the Laplace transform of the above linearized equations, obtain the transfer function of the open-loop system, defined as, G(s) := φ(s) u(s) .

Here, the input voltage u(s) is assumed to be the control input for the open-loop system, and the output is taken as position of the pendulum φ(s).

• Sketch the root locus of the open-loop system using the seven steps covered in your notes. You must do all the required math and support your answer.

• Sketch the root locus of the open-loop system in MATLAB and compare it with your plot.

• Using the root locus approach, design a PID controller for the unity closed-loop control system with the desired input φd (i.e., desired pendulum position) and the actual output φ (i.e., actual pendulum position) to achieve the following requirements:

(1) Settling time < 5 seconds, and
(2) Pendulum angle φ never more than 0.05 radians from the vertical.
(You must design the PID controller based on the techniques covered in the root locus approach chapter. Do not use MATLAB pidtuner, sisotool, lqr or other available packages.)
• Using MATLAB, simulate the linear closed-loop system’s response for a zero step input 0 (rad) in the presence of an impulse-like external disturbance over the time interval [0, 10] (s).
• Using MATLAB, sketch the Bode plots for the original open-loop system G(s) and the modified open-loop system C(s)G(s), where C(s) represents the controller transfer function. Compute the gain and phase margins for the modified open-loop system using MATLAB. What is the maximum actuator delay time to make the system marginally stable?

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**Bicycle Dynamics and Control: PID Controller Design Using the Root Locus Approach**

3.The bicycle is an ingenious device for recreation and transportation, which has evolved over a long period of time. It is a very effective vehicle that is extremely maneuverable. Feedback is essential for understanding how the bicycle really works. In the bicycle there is no explicit control system with sensing and actuation, instead control is accomplished by clever mechanical design of the front fork which creates a feedback that under certain conditions stabilizes the bicycle.

A detailed model of the bicycle is quite complicated. We will derive a simplified model that captures many relevant balancing properties of the bicycle. To understand how a bicycle works it is necessary to consider the system consisting of the bicycle and the rider. The rider can influence the bicycle in two ways by exerting a torque on the handle bar and by leaning. We will neglect the lean and consider the rider as a rigid body, firmly attached to the bicycle. A schematic picture of the bicycle is shown

To describe the dynamics we must account for the tilt of the bicycle. We introduce a coordinate system fixed to the bicycle with the x-axis through the contact points of the wheels with the ground, the y-axis horizontal and the z-axis vertical, as shown in Fig. 4. Let m = 100 (kg) be the mass of the bicycle and the rider, J = 25 (kg.m2 ) the moment of inertia of the bicycle and the rider with respect to the x-axis. Furthermore let l = 0.5 (m) be the distance from the x-axis to the center of mass of bicycle and rider, θ the tilt angle and F the component of the force acting on rider and the bicycle.

A momentum balance around the x-axis gives

J d 2 θ dt2 = mglsin(θ) + Fl cos(θ),

where g = 9.81 (m/s2 ). The force F has two components, a centripetal force and an inertia force due to the acceleration of the coordinate system. The force can be determined from kinematic relations, see Fig. 4. To describe these we introduce the steering angle β, and the forward velocity V0 = 9 (m/s). Furthermore the distance between the contact point of the front and rear wheel is b = 1 (m) and the distance between the contact point of the rear wheel and the projection of the center of mass of bicycle and rider is a = 0.6 (m). To simplify the equations it is assumed that the angles β and θ are so small that sines and tangent are equal to the angle and cosine is equal to the one. Viewed from the top as shown in Fig. 4 the bicycle has its center of rotation at a distance b/β from the rear wheel. The centripetal force is

Fc = mV 2 0 b β.

The y-component of the velocity of the center of mass is

Vy = V0α = aV0 b β,

Where a is the distance from the contact point of the back wheel to the projection of the center of mass. The inertial force due to the acceleration of the coordinate system is thus

Fi = amV0 b dβ dt .

Inserting the total force F = Fc + Fi into

we find that the bicycle can be described by

J d 2 θ dt2 = mglθ + amV0l b dβ dt + mV 2 0 l b β.

• By taking the Laplace transform of (20), obtain the transfer function of the open-loop system, defined as, G(s) := θ(s) β(s) .

Here, β(s) is assumed to be the control input for the open-loop system, and the output is taken as θ(s).

• Sketch the root locus of the open-loop system using the seven steps covered in your notes. You must do all the required math and support your answer.

• Sketch the root locus of the open-loop system in MATLAB and compare it with your plot.

• Using the root locus approach, design a PID controller for the unity closed-loop control system with the desired input θd (i.e., desired position ) and the actual output θ (i.e., actual position) to achieve the following requirements:

(1) Settling time < 0.5 seconds,
(2) Overshoot < 10%, and
(3) No steady-state error, even in the presence of a step disturbance input.
(You must design the PID controller based on the techniques covered in the root locus approach chapter. Do not use MATLAB pidtuner, sisotool, lqr, or similar packages.)
• Using MATLAB, simulate the linear closed-loop system’s response for a zero step input 0.0 (rad) in the presence of an impulse-like external disturbance over the time interval [0, 10] (s).
• Using MATLAB, sketch the Bode plots for the original open-loop system G(s) and the modified open-loop system C(s)G(s), where C(s) represents the controller transfer function. Compute the gain and phase margins for the modified open-loop system using MATLAB. What is the maximum actuator delay time to make the system marginally stable?
• Using the PID controller designed in the above steps, check the performance of the closed-loop system in the presence of an impulse-like external disturbance over the time interval [0, 10] (s) for ±%10, ±%20 and ±%30 change in the forward velocity V0. What do you conclude? Is your controller capable of working at different speeds?
Product code:Control system-QA56
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