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AE4610

Welcome

AE4610 - Dynamics and Control Laboratory

Welcome to AE4610 - Dynamics and Control Laboratory!

Here you will find the lab manuals for all experiments that will take place this semester. Navigate to the appropriate experiment on the left-hand menu. You can export each experiment as a PDF if you want a copy for your records, or to print as a hard copy. Also, bookmark this page on your smartphone or tablet as well as your PC/laptop as a handy guide during your lab session.

Experiments

Lab 0: MATLAB & SIMULINK

The objective of this lab is to get familiar with MATLAB/SIMULINK in general.

Important Note: Georgia Tech honor code applies to this and all other experiments in this course. Ask the TAs or the instructor for help if you have any questions on these problems.

MATLAB & SIMULINK

It is critical that you read following two pdf files to know how to use MATLAB and SIMULINK to perform well in this course.

Appendix

The following material contains detailed information on how to use MATLAB for control that you might find useful throughout the semester.

Lab 1: Rotary Servo Base

Welcome to the first hands-on experiment!

A. Integration (Week 1)

1. QUANSER Rotary Servo Base

The Quanser Rotary Servo Base Unit rotary servo plant, shown below, consists of a DC motor that is encased in a solid aluminum frame and equipped with a planetary gearbox. The motor has its own internal gearbox that drives external gears. The Rotary Servo Base Unit is equipped with two sensors: a potentiometer and an encoder. The potentiometer and the encoder sensors measure the angular position of the load gear using different methods.

Component

Top plate

Encoder

Bottom plate

Ball-bearing block

Posts

Motor

Motor pinion gear: 72-teeth (low-gear)

Motor Gearbox

Load gear: 72 teeth (low-gear)

Motor connector

Potentiometer gear

Encoder connector

Anti-backlash springs

Potentiometer connector

Load shaft (i.e. output shaft)

Bar inertial load

Motor pinion gear: 24-teeth (high-gear)

Disc inertial load

Load gear: 120-teeth (high-gear)

Thumb screws

Potentiometer

2. Components & Sensors

a) DC Motor

The Rotary Servo Base Unit incorporates a Faulhaber Coreless DC Motor model 2338S006 and is shown in Figure 1c with ID #9. This is a high-efficiency, low-inductance motor that can obtain a much faster response than a conventional DC motor.

b) Potentiometer

A rotary potentiometer, or pot, is a manually controlled variable resistor. See the example shown below. It typically consists of an exposed shaft, three terminals (A, W, and B), an encased internal resistive element shaped in a circular pattern, and a sliding contact known as a wiper. By rotating the shaft, the internal wiper makes contact with the resistive element at different positions, causing a change in resistance when measured between the centre terminal (W) and either of the side terminals (A or B). The total resistance of the potentiometer can be measured by clamping a multimeter to terminals A and B.

c) Encoders

An incremental optical encoder is a relative angular displacement sensor that measures angular displacement relative to a previously known position. Unlike an absolute encoder, an incremental encoder does not retain its position information upon power loss. An incremental encoder outputs a series of pulses that correlate to the relative change in angular position. Encoders are commonly used to measure the angular displacement of rotating load shafts. The information extracted from an incremental encoder can also be used to derive instantaneous rotational velocities.

An incremental optical encoder typically consists of a coded disk, an LED, and two photo sensors. The disk is coded with an alternating light and dark radial pattern causing it to act as a shutter. As shown schematically above, the light emitted by the LED is interrupted by the coding as the disk rotates around its axis. The two photo sensors (A and B) positioned behind the coded disk sense the light emitted by the LED. The process results in A and B signals, or pulses, in four distinct states: (1) A off, B on; (2) A off, B off; (3) A on, B off; (4) A on, B on. Encoders that output A and B signals are often referred to as quadrature encoders since the signals are separated in phase by 90◦. The resolution of an encoder corresponds to the number of light or dark patterns on the disk and is given in terms of pulses per revolution, or PPR. In order to make encoder measurements, you need to connect the encoder to a counter to count the A and B signals. Then use a decoder algorithm to determine the number of counts and direction of rotation. Three decoding algorithms are used: X1, X2, X4.

X1 Decoder: When an X1 decoder is used, only the rising or falling edge of signal A is counted as the shaft rotates. When signal A leads signal B, the counter is incremented on the rising edge of signal A. When signal B leads signal A, the counter is decremented on the falling edge of signal A. Using an X1 decoder, a 1,024 PPR encoder will result in a total of 1,024 counts for every rotation of the encoder shaft. X2 Decoder: When an X2 decoder is used, both the rising and falling edges of signal A are counted as the shaft rotates. When signal A leads signal B, the counter is incremented on both the rising and falling edge of signal A. When signal B leads signal A, the counter is decremented on both the rising and falling edges of signal A. Using an X2 decoder, a 1,024 PPR encoder will generate a total of 2,048 counts for every rotation of the encoder shaft. X4 Decoder: When an X4 algorithm is used, both the rising and falling edges of both signals A and B are counted. Depending on which signal leads, the counter will either increment or decrement. An X4 decoder generates four times the number of counts generated by an X1 decoder resulting in the highest resolution among the three types of decoders. Using an X4 decoder, a 1,024 PPR encoder will generate a total of 4,096 counts for every rotation of the encoder shaft.

The angular resolution of an encoder depends on the encoder’s pulses per revolution (PPR) and the decoding algorithm used:

where N = 1, 2, or 4 corresponds to X1, X2, and X4 decoders respectively

3. Velocity Estimation

While encoders are typically used for measuring angular displacement, they can also measure rotational speeds. The velocity can be found by taking the difference between consecutive angle measurements

4. Filtering the Velocity Signal

In practice, determining rotational speeds by means of derivation of the discontinuous encoder output often results in signal noise. The ripple in the velocity signal due to sampling can be reduced by filtering. Applying a first-order low-pass filter to the measured velocity signal

Rearranging the filter transfer function above,

we can represent the filtering using the differential equation

5. Encoder Counter Wrapping

Experiment

Follow the steps and build the SIMULINK model below:

Follow these steps to read the motor velocity:

First, download and open q_servo_vel_direct.slx Simulink model.
Configure the model to apply a 0.4 Hz square wave voltage signal going between 0.5V and 2.5V.
Change the simulation time to 5 seconds from Simulation tab, stop time.
Go to Modelling tab and open modelling setting. Check that sample time is set to 0.002s and check that solver is set to ode1 (euler).
The velocity measurement using a direct derivative will be noisy. After the SIMULINK has run for 5 seconds and stopped, note down the velocity ripple (crest to peak or height within ripple). You can use the Cursor Measurements tool in the Scope for more accurate measurements. Save the speed data (wl) to plot in the report. Name the file such as integration_sampletime0.002s.
Increase the control loop rate of the QUARC controller by decreasing the sampling interval to 0.001 s, i.e. increasing the sampling rate to 1 kHz. Build, Connect and run the Simulink. Once SIMULINK has run for 5 seconds and stopped, note down the velocity ripple and examine the change in the ripple velocity measurement. Save the speed data (wl) to plot in the report. Name the file such as integration_sampletime0.001s.
Run the controller for a minute or two (stop time to inf). Depending on the data acquisition device you are using (e.g. Q2-USB or Q8-USB), you may notice a spike or discontinuity that occurs in the velocity estimation, as shown below (once the discontinuity occurs stop the simulation). Save the speed data (wl) to plot in the report. Name the file such as integration_wrapping.
Save the speed data (wl) to plot in the report. Name the file such as integration_lowpass. Save a screenshot of the Simulink model designed (block diagram).

Results for Report

(A) From Step 7

Response plots from Step 7.

(B) From Step 8

Response plots from Step 8.

(C) From Step 9

Response plots with and without encoder wrapping from Step 9.

(D) From Step 10

Compare the measured velocity ripple from Step 8 (without filter) with the measured velocity ripple from Step 10 (with filter). How much noise does the filter remove?
Completed SIMULINK block diagram with low-pass filter from Step 10.
Response plots from Step 10.

B. Modeling (Week 1)

The objective of this experiment is to find a transfer function that describes the rotary motion of the Rotary Servo Base Unit load shaft. The dynamic model is derived analytically from classical mechanics principles and using experimental methods.

The angular speed of the Rotary Servo Base Unit load shaft with respect to the input motor voltage can be described by the following first-order transfer function:

\frac{\Omega_l(s)}{V_m(s)} = \frac{K}{\tau s+1} \qquad \qquad \qquad \tag{2.1}

1. Modeling Using First-Principles

a) Electrical Equations

The DC motor armature circuit schematic and gear train is illustrated in Figure 7.

$R_m$ is the motor resistance, $L_m$ is the inductance, and $K_m$ is the back-emf constant.

The back-emf (electromotive) voltage $e_b(\mathrm{t})$ depends on the speed of the motor shaft, $\omega_m$ , and the back-emf constant of the motor, $k_m$ . It opposes the current flow. The back emf voltage is given by:

e_b(\mathrm{t}) = k_m\omega_m(\mathrm{t}) \qquad \qquad \qquad \tag{2.2}

Using Kirchoff’s Voltage Law, we can write the following equation:

V_m(\mathrm{t}) - R_mI_m(\mathrm{t}) - L_m \frac{dI_m(\mathrm{t})}{dt} - k_m \omega_m(\mathrm{t}) = 0 \qquad \qquad \qquad \tag{2.3}

Since the motor inductance $L_m$ is much less than its resistance, it can be ignored. Then, the equation becomes:

V_m(\mathrm{t}) - R_mI_m(\mathrm{t})-k_m\omega_m(\mathrm{t}) = 0 \qquad \qquad \qquad \tag{2.4}

Solving for $I_m(\mathrm{t})$ , the motor current can be found as:

I_m(\mathrm{t}) = \frac{V_m(\mathrm{t}) - k_m\omega_m(\mathrm{t})}{R_m} \qquad \qquad \qquad \tag{2.5}

b) Mechanical Equations

In this section, the equation of motion describing the speed of the load shaft, $\omega_l$ , with respect to the applied motor torque, $L_m$ , is developed. Since the Rotary Servo Base Unit is a one degree-of-freedom rotary system, Newton’s Second Law of Motion can be written as:

J \cdot \alpha = \tau \qquad \qquad \qquad \tag{2.6}

where $J$ is the moment of inertia of the body (about its center of mass), $\alpha$ is the angular acceleration of the system, and $\tau$ is the sum of the torques being applied to the body. As illustrated in Figure 7, the Rotary Servo Base Unit gear train along with the viscous friction acting on the motor shaft, $B_m$ , and the load shaft $B_l$ are considered. The load equation of motion is:

J_l \frac{d\omega_l(\mathrm{t})}{dt} + B_l\omega_l(\mathrm{t}) = \tau_l(\mathrm{t}) \qquad \qquad \qquad \tag{2.7}

where $J_l$ is the moment of inertia of the load and $\tau_l$ is the total torque applied on the load. The load inertia includes the inertia from the gear train and from any external loads attached, e.g. disc or bar. The motor shaft equation is expressed as:

J_m \frac{d\omega_l(t)}{dt} + B_m\omega_m(t) +\tau_{ml}= \tau_m(t) \qquad \qquad \qquad \tag{2.8}

where $J_m$ is the motor shaft moment of inertia and $\tau_{ml}$ is the resulting torque acting on the motor shaft from the load torque. The torque at the load shaft from an applied motor torque can be written as:

\tau_l(t) = \eta_gK_g\tau_{ml}(t) \qquad \qquad \qquad \tag{2.9}

K_{gi} = \frac{N_2}{N_1} \qquad \qquad \qquad \tag{2.10}

This is the internal gear box ratio. The motor gear $N_3$ and the load gear $N_4$ are directly meshed together and are visible from the outside. These gears comprise the external gear box which has an associated gear ratio of

K_{ge} = \frac{N_4}{N_3} \qquad \qquad \qquad \tag{2.11}

The gear ratio of the Rotary Servo Base Unit gear train is then given by:

K_g = K_{ge} K_{gi} \qquad \qquad \qquad \tag{2.12}

Thus, the torque seen at the motor shaft through the gears can be expressed as:

\tau_{ml} (t) = \frac{\tau_l(t)}{\eta_gK_g} \qquad \qquad \qquad \tag{2.13}

Intuitively, the motor shaft must rotate $K_g$ times for the output shaft to rotate one revolution.

\theta_m(t) = K_g\theta_l(t) \qquad \qquad \qquad \tag{2.14}

We can find the relationship between the angular speed of the motor shaft, $\omega_m$ , and the angular speed of the load shaft, $\omega_l$ by taking the time derivative:

\omega_m(t) = K_g\omega_l(t) \qquad \qquad \qquad \tag{2.15}

To find the differential equation that describes the motion of the load shaft with respect to an applied motor torque substitute (Eq 2.13), (Eq 2.15) and (Eq 2.7) into (Eq 2.8) to get the following:

J_mK_g \frac{d\omega_l(t)}{dt} + B_mK_g\omega_l(t) +\frac{J_l \frac{d\omega_l(t)}{dt} + B_l\omega_l(t)}{\eta_gK_g}= \tau_m(t) \qquad \qquad \qquad \tag{2.16}

Collecting the coefficients in terms of the load shaft velocity and acceleration gives

(\eta_gK_g^2J_m + J_l) \frac{d\omega_l(t)}{dt} + (\eta_gK_g^2B_m + B_l)\omega_l(t) = \eta_gK_g\tau_m(t) \qquad \qquad \qquad \tag{2.17}

Defining the following terms:

J_{eq} = \eta_gK_g^2J_m+J_l \qquad \qquad \qquad \tag{2.18}

B_{eq} = \eta_gK_g^2B_m+B_l \qquad \qquad \qquad \tag{2.19}

simplifies the equation as:

J_{eq} \frac{d\omega_l(t)}{dt} +B_{eq}\omega_l(t) = \eta_gK_g\tau_m(t) \qquad \qquad \qquad \tag{2.20}

c) Combining the Electrical and Mechanical Equations

In this section the electrical equation and the mechanical equation are brought together to get an expression that represents the load shaft speed in terms of the applied motor voltage.

The motor torque is proportional to the voltage applied and is described as

\tau_m(t) = \eta_mk_tI_m(t) \qquad \qquad \qquad \tag{2.21}

\tau_m(t) = \frac{\eta_mk_t(V_m(t) - k_m\omega_m(t))}{R_m} \qquad \qquad \qquad \tag{2.22}

To express this in terms of $V_m$ and $\omega_l$ , insert the motor-load shaft speed Eq 2.15, into Eq 2.21 to get:

\tau_m(t) = \frac{\eta_mk_t(V_m(t) - k_mK_g\omega_l(t))}{R_m} \qquad \qquad \qquad \tag{2.23}

If we substitute (Eq 2.23) into (Eq 2.20), we get:

J_{eq} \frac{d\omega_l(t)}{dt} +B_{eq}\omega_l(t) = \frac{\eta_gK_g\eta_mk_t(V_m(t) - k_mK_g\omega_l(t))}{R_m} \qquad \qquad \qquad \tag{2.24}

After collecting the terms, the equation becomes

J_{eq} \frac{d\omega_l(t)}{dt} +\left (\frac{k_m\eta_gK_g^2\eta_mk_t}{R_m}+B_{eq}\right) \omega_l(t) = \frac{\eta_gK_g\eta_mk_tV_m(t) }{R_m} \qquad \qquad \qquad \tag{2.25}

This equation can be re-written as:

J_{eq} \frac{d\omega_l(t)}{dt} +\left (B_{eq,v}\right) \omega_l(t) = A_mV_m(t) \qquad \qquad \qquad \tag{2.26}

where the equivalent damping term is given by:

B_{eq,v} = \frac{k_m\eta_gK_g^2\eta_mk_t + B_{eq}R_m}{R_m} \qquad \qquad \qquad \tag{2.27}

and the actuator gain equals

A_m = \frac{\eta_gK_g\eta_mk_t}{R_m} \qquad \qquad \qquad \tag{2.28}

2. Modeling Using Experiments

A linear model of a system can also be determined purely experimentally. The main idea is to experimentally observe how a system reacts to different inputs and change structure and parameters of a model until a reasonable fit is obtained. The inputs can be chosen in many different ways and there are a large variety of methods. Two methods of modeling the Rotary Servo Base Unit are: (1) frequency response and, (2) bump test.

a) Frequency Response

In Figure 8, the response of a typical first-order time-invariant system to a sine wave input is shown. As it can be seen from the figure, the input signal ( $u$ ) is a sine wave with a fixed amplitude and frequency. The resulting output ( $y$ ) is also a sinusoid with the same frequency but with a different amplitude. By varying the frequency of the input sine wave and observing the resulting outputs, a Bode plot of the system can be obtained as shown in Figure 9.

The Bode plot can then be used to find the steady-state gain, i.e. the DC gain, and the time constant of the system. The cuttoff frequency, $\omega_c$ , shown in Figure 9 is defined as the frequency where the gain is 3 dB less than the maximum gain (i.e. the DC gain) in dB. When working in the linear non-decibel range, the 3 dB frequency is defined as the frequency where the gain is $\frac{1}{\sqrt{2}}$ , or about 0.707, of the maximum gain. The cutoff frequency is also known as the bandwidth of the system which represents how fast the system responds to a given input.

The magnitude of the frequency response of the Rotary Servo Base Unit plant transfer function given in equation Eq 2.1 is defined as:

|G_{\omega l,v}(\omega)| = \left |\frac{\Omega_l(\omega j)}{V_m(\omega j)}\right| \qquad \qquad \qquad \tag{2.29}

where $\omega$ is the frequency of the motor input voltage signal $V_m$ . We know that the transfer function of the system has the generic first-order system form given in Eq 2.1. By substituting $s = j \omega$ in this equation, we can find the frequency response of the system as:

\left |\frac{\Omega_l(\omega j)}{V_m(\omega j)}\right| = \frac{K}{\tau \omega j + 1} \qquad \qquad \qquad \tag{2.30}

Then, the magnitude of it equals

|G_{wl,v}(w)| = \frac{K}{\sqrt{1+\tau^2\omega^2}} \qquad \qquad \qquad \tag{2.31}

Let’s call the frequency response model parameters $K_{e,f}$ and $\tau_{e,f}$ to differentiate them from the nominal model parameters, $K$ and $\tau$ , used previously. The steady-state gain or the DC gain (i.e. gain at zero frequency) of the model is:

K_{e,f} = |G_{wl,v}(0)| \qquad \qquad \qquad \tag{2.32}

b) Bump Test / Unit Step Input Test

The bump test is a simple test based on the step response of a stable system. A step input is given to the system and its response is recorded. As an example, consider a system given by the following transfer function:

\frac{Y(s)}{U(s)} = \frac{K}{\tau s + 1} \qquad \qquad \qquad \tag{2.33}

The step response shown in Figure 10 is generated using this transfer function with $K = 5 rad/Vs$ and $\tau = 0.05s$ .

The step input begins at time $t_0$ . The input signal has a minimum value of $u_{min}$ and a maximum value of $u_{max}$ . The resulting output signal is initially at $y_0$ . Once the step is applied, the output tries to follow it and eventually settles at its steady-state value yss. From the output and input signals, the steady-state gain is

K = \frac{\Delta y}{\Delta u} \qquad \qquad \qquad \tag{2.34}

where $\Delta y = y_{ss} - y_0$ and $\Delta u =u_{max} - u_{min}$ . In order to find the model time constant, $\tau$ , we can first calculate where the output is supposed to be at the time constant from:

y(t_1) = 0.632(y_{ss}-y_0) + y_0 \qquad \qquad \qquad \tag{2.35}

Then, we can read the time $t_1$ that corresponds to $y(t_1)$ from the response data in Figure 10. From the figure we can see that the time $t_1$ is equal to:

t_1 = t_0 + \tau \qquad \qquad \qquad \tag{2.36}

From this, the model time constant can be found as:

\tau = t_1 - t_0 \qquad \qquad \qquad \tag{2.37}

Going back to the Rotary Servo Base Unit system, a step input voltage with a time delay $t_0$ can be expressed as follows in the Laplace domain:

V_m(s) = \frac{A_ve^{(-st_0)}}{s} \qquad \qquad \qquad \tag{2.38}

where $A_v$ is the amplitude of the step and $t_0$ is the step time (i.e. the delay). If we substitute this input into the system transfer function given in Eq 2.1, we get:

\Omega_l(s) = \frac{KA_ve^{(-s t_0)}}{(\tau s + 1)s} \qquad \qquad \qquad \tag{2.39}

We can then find the Rotary Servo Base Unit load speed step response, $\omega_l(t)$ , by taking inverse Laplace of this equation. Here we need to be careful with the time delay $t_0$ and note that the initial condition is

\omega_l(t) = KA_v\left(1-e^{(-\frac{t-t_0}{\tau})}\right) + \omega_l(t_0) \qquad \qquad \qquad \tag{2.40}

Experimental Setup

The q_servo_modeling Simulink diagram shown in Figure 11a will be used to conduct the experiments. The Rotary Servo Base Unit subsystem contains QUARC blocks that interface with the DC motor and sensors of the Rotary Servo Base Unit system. The Rotary Servo Base Unit Model uses a Transfer Fcn block from the SIMULINK library to simulate the Rotary Servo Base Unit system. Thus, both the measured and simulated load shaft speed can be monitored simultaneously given an input voltage. We have implemented a first-order low-pass filter in integration but for modeling second-order low-pass filter has been implemented with cutoff frequency of $2\pi \times35$ rad/s damping ratio of 0.9.

Download the Modeling_files.zip and extract the files to desktop.
Open q_servo_modeling.m SIMULINK file.
Double click on the "Rotary Servo Interface - Speed" subsystem to access the DAQ configuration
Configure DAQ: Double-click on the HIL Initialize block in the SIMULINK diagram and ensure it is configured for the DAQ device that is installed in your system (e.g. Q2-USB).
Open setup_servo_modeling.m file to open the setup script for the q_servo_modeling SIMULINK model.
Run the script. Note: The calculated servo model parameter are default model parameters and do not accurately represent the Rotary Servo Base Unit system.

Frequency Response Experiment

The frequency response of a linear system can be obtained by providing a sine wave input signal to it and recording the resulting output sine wave from it. In this experiment, the input signal is the motor voltage and the output is the motor speed. In this method, we keep the amplitude of the input sine wave constant but vary its frequency. At each frequency setting, we record the amplitude of the output sine wave. The ratio of the output and input amplitudes at a given frequency can then be used to create a Bode magnitude plot. Then, the transfer function for the system can be extracted from this Bode plot.

a) Steady-state gain

First, we need to find the steady-state gain of the system. This requires running the system with a constant input voltage. To create a $2V$ constant input voltage follow these steps:

In the SIMULINK diagram, double-click on the Signal Generator block and ensure the following parameters are set: • Wave form: sine • Amplitude: 1.0 • Frequency: 0.4 • Units: Hertz
Set the Amplitude (V) slider gain to 0.
Set the Offset (V) block to 2.0 V. Note: Execution of steps 1 through 3 generates a step input of 2.0 V magnitude.
Set the Simulation stop time to 5 seconds.
Open the load shaft speed scope, Speed (rad/s), and the motor input voltage scope, Vm (V) .
The Rotary Servo Base Unit unit should begin rotating in one direction. The scopes should be reading something similar to figure below. Note that in the Speed (rad/s) scope, the yellow trace is the measured speed while the blue trace is the simulated speed (generated by Servo Model block).

b) Gain at varying frequencies

In this part of the experiment, we will send an input sine wave at a certain frequency to the system and record the amplitude of the output signal. We will then increment the frequency and repeat the same observation. To create the input sine wave:

In the SIMULINK diagram, double-click on the Signal Generator block and ensure the following parameters are set: • Wave form: sine • Amplitude: 1.0 • Frequency: 1.0 • Units: Hertz
Set the Amplitude (V) slider gain to 2.0 V.
Set the Offset (V) block to 0 V.
Set the Simulation stop time to 5 seconds.
The Rotary Servo Base Unit unit should begin rotating smoothly back and forth and the scopes should be reading a response similar to Figure 13 a) and b).

Step Response Experiment

To create the step input:

Double-click on the Signal Generator block and ensure the following parameters are set: • Wave form: square • Amplitude: 1.0 • Frequency: 0.4 • Units: Hertz
Set the Amplitude (V) gain block to 1.0 V.
Set the Offset (V)gain block to 2.0 V.
Set the Simulation stop time to 5 seconds.
Open the load shaft speed scope, Speed (rad/s), and the motor input voltage cope, Vm (V).
The gears on the Rotary Servo Base Unit should be rotating in the same direction and alternating between low and high speeds. The response in the scopes should be similar to Figure 14 a) and b).
Click the Stop button on the SIMULINK diagram toolbar (or select QUARC | Stop from the menu) to stop the experiment.

Model Validation Experiment

In this experiment, you will use the calculated values of K and $\tau$ from the frequency response experiment, bump test experiment, and nominal values calculated to see if it matches the measured response. To create a step input:

Double-click on the Signal Generator block and ensure the following parameters are set: • Wave form: square • Amplitude: 1.0 • Frequency: 0.4 • Units: Hertz
Set the Amplitude (V) slider gain to 1.0 V.
Set the Offset (V) block to 1.5 V.
Set the Simulation stop time to 5 seconds.
Open the load shaft speed scope, Speed (rad/s), and the motor input voltage scope, V_m (V).
The gears on the Rotary Servo Base Unit should be rotating in the same direction and alternating between low and high speeds and the scopes should be as shown in Figure 15 a) and b). Recall that the yellow trace is the measured load shaft rate and the purple trace is the simulated trace. By default, the steady-state gain and the time constant of the transfer function used in simulation are set to: K = 1 rad/s/V and $\tau$ = 0.1s. These model parameters do not accurately represent the system.
Notice that the model response does not match with the measured response.
Enter the nominal values, $K$ and $\tau$ , that were found in Step 10 in the MATLAB Command Window. Update the parameters and examine how well the simulated response matches the measured one.
Save wl data and name it as modeling_section#_Group#_K#tau#.

If the calculations of the nominal values were done properly, then the model should represent the actual system quite well. Compare the responses you get by using the $K$ and $\tau$ values from the frequency response experiment, bump test experiment (this needs to be done while writing the report), and when using nominal values. Check which of the three responses matches the measured response best, and record this $K$ and $\tau$ as your model validation values. Use this set of values for controller implementation (next week). If there is no clear set of values that matches the measured response, take the average. Note: Changing the model parameters can be done by manually changing the Servo Model Transfer Function block parameters OR by changing the K and tau parameters in the MATLAB Command Window and going to Simulation | Update Diagram in the SIMULINK model.

Tables for Report

Table B.1 Collected frequency response data.

Table B.2 Summary of results for the Rotary Servo Base Unit Modeling laboratory.

Results for Report

(A) Frequency Response Experiment

(B) Step Response Experiment

Plot the maximum load speed response (saved in the MATLAB workspace under the wl variable) from Step 9 in MATLAB.

(C) Model Validation Experiment

Create a MATLAB figure that shows the simulation nominal values response, the step response (generated by using the K and tau values calculated from the bump test experiment), the frequency response (generated by using the K and tau values calculated for frequency response experiment), the response you got from step 8, and the measured response as five plots on the same graph in the report. Provide two reasons why the nominal model might not represent the Rotary Servo Base Unit with better accuracy. Which response matches most closely with the measured response?
From the plots, describe how the gain and time constant affect the system's response.

(D) Additional Results

Explain how well the nominal model, the frequency response model, and the bump-test model represent the Rotary Servo Base Unit system (you may refer to Result D.1).

Appendix

Table A.1 Main Rotary Servo Base Unit Specifications

Table A.2 Rotary Servo Base Unit Gearhead Specifications

C. Control Design (Week 2)

Objective

The objective of this experiment is to design a position control system for the motor to meet a given set of specifications in frequency-domain.

Experiment Notes

Now that we have identified the time constant and the DC gain for the system in the modeling part of this experiment, the next step is to design a position control system in order to convert the given DC motor into a position servo. A classical PID controller will be used to control the position of the motor subject to a given set of specifications, for example, bandwidth, steady-state error requirements, etc.

\displaystyle\frac{\Omega_l(s)}{V_m(s)} = \frac{A}{\tau s+1} \qquad \qquad \qquad \tag{2.1}

let $A = K$ , DC gain of the motor.

Proportional (P) Control

Since $\dot{\theta} = \Omega$ , where $\theta$ is the angular position of the motor, the transfer function from the applied voltage to the motor angular position is

G(s) = \displaystyle\frac{\theta (s)}{V_m (s)} = \frac{A}{s(\tau s+1)} \qquad \qquad \qquad \tag{3.1}

The block diagram of a DC motor with a proportional controller is shown in Fig.1. In this case, the controller transfer function ( $H$ ) is a simple gain $K_P$ . From Fig. 1, the closed-loop transfer function from the commanded angular position $\theta_c$ to the actual angular position $\theta$ is given by

\frac{\theta(s)}{\theta_c(s)} = \frac{AK_p}{\tau s^2 + s + AK_p} = \frac{\cfrac{AK_p}{\tau}}{s^2 + \cfrac{1}{\tau}s + \cfrac{AK_p}{\tau}}

Comparing the above transfer function to the standard form for a second-order system, i.e.,

\frac{\theta(s)}{\theta_c(s)} = \frac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}

we notice that the proportional gain $K_p$ affects the natural frequency (and hence the bandwidth) of the closed-loop system.

Proportional-plus-Derivative (PD) Control

The block diagram of a DC motor with an inner loop angular rate (i.e., derivative) feedback and an outer loop angular position error (proportional) feedback is shown in Fig. 2. From Fig. 2, the closed-loop transfer function from the commanded angular position $\theta_c$ to the actual angular position $\theta$ is given by

\frac{\theta(s)}{\theta_c(s)} = \frac{AK_p}{\tau_m s^2 + (1 + AK_d)s + AK_p} = \frac{\cfrac{AK_p}{\tau_m}}{s^2 + \cfrac{(1 + AK_d)}{\tau_m}s + \cfrac{AK_p}{\tau_m}}

Comparing the above transfer function with the standard form for a second-order system, we notice that the derivative gain $K_d$ affects the damping while the proportional gain $K_p$ affects the natural frequency of the closed-loop system.

Proportional-plus-Integral-plus-Derivative (PID) Control

The block diagram of a DC motor with an inner loop angular rate (i.e., derivative) feedback and an outer loop angular position error plus integral of angular position error (i.e., proportional plus integral) feedback is shown in Fig. 3.

From Fig. 3, the open-loop transfer function $GH$ becomes

GH = \frac{(K_ps+K_i)}{s} \frac{A}{(\tau_ms+1+AK_d)}\frac{1}{s}

indicating that with integral feedback, the type of the system increases from 1 to 2 and hence, results in zero steady-state error to both ramp and step command inputs.

Controller Gain Limits

In control design, a factor to be considered is saturation. This is a nonlinear element and is represented by a saturation block as shown in Figure 4. In a system like the Rotary Servo Base Unit, the computer calculates a numeric control voltage value. This value is then converted into a voltage, $V_{\text{dac}}(t)$ , by the digital-to-analog converter of the data-acquisition device in the computer. The voltage is then amplified by a power amplifier by a factor of $K_a$ . If the amplified voltage, $V_{\text{amp}}(t)$ , is greater than the maximum output voltage of the amplifier or the input voltage limits of the motor (whichever is smaller), then it is saturated (limited) at $V_{\text{max}}$ . Therefore, the input voltage $V_m(t)$ is the effective voltage being applied to the Rotary Servo Base Unit motor.

The limitations of the actuator must be taken into account when designing a controller. For instance, the voltage entering the Rotary Servo Base Unit motor should never exceed V_{\text{max}} = 10.0 V \qquad \qquad \qquad \tag{3.2}

Sampling Time

The most important difference between analog and digital control is that digital systems operate using a clock. The timing of this clock and the number of operations necessary to implement a controller place a limit on how frequently the controller can access sensors, make calculations, and modify the control inputs. (In addition, other elements of the control system themselves may introduce an additional time delay. For this reason, the sampling time $T_s$ of the digital system is an important parameter.) Digital controllers, in general, will have a maximum $T_s$ . Increasing the $T_s$ further will make the closed-loop system unstable.

Design Specifications

The closed-loop DC motor system must meet the following specifications.

Bandwidth frequency of at least 25 rad/s (Bandwidth is an important measure of the frequency range over which the system output follows well the command signal. It is defined as the frequency at which the magnitude ratio is -3 dB in the closed-loop system frequency response plot. Sinusoidal inputs with frequencies less than the bandwidth frequency are tracked ‘reasonably well’ by the system.)
Phase margin of at least 70 deg. (Gain and phase margins are stability margins to accommodate model variations. A flight control system actuator is typically subject to varying load, resulting in model variations. Hence, it is important to design a controller with sufficient stability margins to accommodate such variations and other effects such as wear with usage.)
Zero steady-state error to both step and ramp commands.

Guidelines for saving plots

Save root locus open-loop Bode, closed-loop bode and step response plots as separate figures.
In open-loop or closed-loop Bode plots, you should show Bandwidth frequency and Phase margin values.
In the step response plot, you should show the steady-state value.
Only for controlSystemDesigner or rltool only, you may use the Snipping tool and save the figures in a Word document.
For other Simulink-generated data, you need to use the plot MATLAB command.

Results from each step below must be checked with TAs before proceeding to the next:

Step 5 - Plots and Kp value
Step 6 - Plots and Kp and Kd values
Step 7-9 - Plots and Kp, Kd and Ki values
Step 10 - Closed loop TF
Step 11 - Closed loop poles
Step 12 - Simulink model
Step 13 - One response plot with PD and PID, One error plot with PD and PID.

Procedure

Using the model, i.e., K and tau values, you have obtained from the Modeling - Model Validation, input the transfer function $G(s)$ (Equation 3.1) into MATLAB. Use K and tau from the model validation values.
First consider a proportional controller, i.e., C = Kp. Input C = 1 into MATLAB.
Run controlSystemDesigner(G,C) (you can alternatively run rltool(G,C) for versions R2020b and older). Select ‘New Plot -> New Step’ to launch the step response plot (if it is not automatically launched). In the time response plot, only the ‘closed loop: r to y' (blue plot) should be visible. Always do this in all the labs. Select ‘New Plot -> New Bode’ and choose the appropriate option to launch the closed-loop Bode plot. In the closed-loop Bode response window, select the magnitude plot only and readjust the magnitude plot limits to -3.5 dB and -2.5 dB under properties and limits. The bandwidth frequency (defined above) can be read from the closed-loop Bode plot. It may be useful to add a grid to the bode plot. Note: If the closed-loop Bode plot is not visible after adjusting magnitude plot limits, it indicates that the design specification is not met, which would require gain tuning. This will be performed in Step 5.
If Open-loop Bode Editor is not automatically launched. In Control System Designer or rltool, select ‘Open-loop Bode’ under ‘Tuning Methods -> Bode Editor’. The phase margin (PM) of the system can be read off the open-loop bode plot. Note: The LoopTransfer_C Bode is the open-loop Bode.
Change the controller gain by double-clicking ‘C’ under ‘Controllers and Fixed Blocks’ or by using your cursor on the root locus and check if all the design specifications can be met using a proportional controller, which would be optimal from a design standpoint. If not, at least select a value of $K_p$ such that the bandwidth is more than 25 rad/sec. Save the root locus, open and closed-loop Bode, time response plot and P gain.
Readjust the proportional and derivative gains (if necessary, by modifying the zeros, not the overall gain, $K_i$ ) such that bandwidth and phase margin specs are met.
Save the root locus plot, open and closed-loop system Bode plots, and time response plot as graphs. Also, save your gains (if you readjusted any values in Step 8, you must convert from the Control System Designer or rltool compensator format, back to our conventional gain representation – $K_p$ , $K_d$ , $K_i$ ).
Using the controller you have designed, obtain the closed-loop transfer function and save your result. (Do this in MATLAB using the transfer function variables of $G$ and $C$ , or export the closed-loop transfer function from Control System Designer or rltool(T_r2y), using the ‘Export’.) (May be listed as ‘IOTransfer_r2y’). In MATLAB, change the exported state space into a transfer function using tf(IOTransfer_r2y)
Obtain poles of the closed-loop system. Compute the natural frequency and damping of the dominant poles of the closed-loop system. You can do all this from your exported transfer function using the function damp(). Save your results.
Construct a SIMULINK diagram using Fig. 3 as a guide. Use a transfer function block for the plant (use K and tau from the model validation). Add saturation of max 10V and min -10V block before the Plant transfer function. Set the solver type to Fixed-step with size of 0.002. Include the controller gains from your design. Do not use the transfer function block or the inbuilt PID controller block for the controller gains. Save your Simulink model for further use in Part C of the experiment. Save the Simulink block diagram as a .jpg or .png image.
Run the responses to a unit ramp input first with a PD controller (by setting Ki = 0) and then with a PID controller (Ki = 1% Kp). Run the simulation for about 30 sec. Make comparison plots of ramp responses with PD and PID controllers (on the same plot) to show any differences between the two responses, especially in steady-state response (i.e., for t>>0, also you may zoom in to see the difference clearly). The difference can be more easily spotted by plotting the error variable (i.e. error between the commanded position and the actual position) for each controller. Save the comparison plot and error plot.

You must have your work checked out by one of the TAs before leaving the lab.

Results for Report

Note:

The plots for controller design, i.e., root locus, open-loop Bode, closed-loop Bode, and step response, must be shown as separate figures.
In open-loop or closed-loop Bode plots, you should show Bandwidth frequency and Phase margin values.
In the step response plot, you should show the steady-state value. This can be done by right-clicking on the plot -> "Characteristics" -> "Steady-state".
Only for controlSystemDesigner or rltool plots, you can include the plots as obtained using the Snipping tool.
For other Simulink-generated data, you need to use the plot MATLAB command.

(A) From Step 5

Root locus plot
Open-loop Bode plot
Closed-loop Bode plot
Time response plot
P (proportional) gain

(B) From Step 6

Root locus plot
Open-loop Bode plot
Closed-loop Bode plot
Time response plot
PD (proportional and derivative) gains

(C) From Step 9

Root locus plot
Open-loop Bode plot
Closed-loop Bode plot
Time response plot
PID (proportional, derivative and integral) gain values (mention if the gains are the same values or readjusted values from Step 8)

(D) From Step 10

Closed-loop transfer function

(E) From Step 11

Poles of the closed-loop system
Natural frequency of the closed-loop system dominant poles
Damping of the closed-loop system dominant poles

(F) From Step 12

PID controller SIMULINK block diagram (image)

(G) From Step 13

Plots comparing the ramp responses of PD and PID controllers (on the same figure)
Plots comparing the error variable of PD and PID controllers (on the same figure)

Questions for Report

What is the difference between Fig. 2 and the control structure screenshot that you saved in Step 6? Hint: Focus on where $K_d$ is located in the control loop and what it is multiplied by.
Why is this difference between the two control structures acceptable when dealing with a step input? Use and modify the Simulink model that you built in Step 12 for the two control structure cases (i.e., $K_d$ is multiplied by the derivative of error or speed directly). You may create two separate Simulink files or build two models within the same Simulink file. Run each Simulink with a unit step input and a unit ramp input for 5 seconds. Use the PD gains found in Step 6 (Result B.5) and set Ki = 0. Create plots below to help you explain the question asked above:
1. Step responses (two plots on the same figure)
2. Ramp responses (two plots on the same figure)
3. Derivative of error signal and speed value in one figure for the step input from both models
4. Derivative of error signal and speed value in one figure for the ramp input from both models

D. Controller Implementation & Evaluation (Week 3)

Objectives

The objective of this part of the DC motor experiment is to implement the controller designed in part B and evaluate its performance.

Equipment Required

The following is a list of the required equipment to perform this experiment:

Q8-USB or Q2-USB interface board
MATLAB and SIMULINK software
Servo Base

Controller Implementation

1. Step Response Experiments

Open q_servo_pos_cntrl.slx file. Do not change anything except the PID gains.
Double-click on the Signal Generator block and ensure that a 0.4 Hz square wave input is being applied. Note that the gain blocks outside the Signal Generator block are introduced to generate the square wave with amplitude between 0 V to 0.4 V.
Check that the sampling time is set to 0.002 s (located under “Modeling” -> “Model Settings” -> “Solver Details”).
Turn on the power amplifier.
Open the position scope.
Save your data. No need to Build again unless you change any blocks or configuration settings in the model.
Sampling Time Test: Change the sampling time to 0.04 sec. To set the sampling time, you need to go to the block diagram file, and choose “Modeling” -> “Model Settings” -> “Solver Details”. Set the fixed step to 0.04 sec (it should have been 0.002 previously). Build, connect and Run PD Controller, observe the behaviour and save the data with an appropriate filename (e.g. - PD_sampling_0.04).

2. Ramp Response Experiments

Double-click on the Signal Generator and set the following parameters to generate a triangular wave (i.e. ramp reference):
- Signal Type = triangle
- Amplitude = 1
- Frequency = 0.4 Hz
In the Simulink diagram, set the Amplitude gain block to $\pi/6$ and the Constant offset gain block to $\pi/6$ . This will generate a triangular wave with amplitude between 0 to $60^{\circ}$ .
Change the sampling time back to 0.002 sec (“Modeling” -> “Model Settings” -> “Solver Details”).
PID Controller Tuning: If the PID controller test has not eliminated the steady-state error, tune the gain values ( $K_p$ , $K_d$ and $K_i$ ) until the system ramp response has zero (or nearly zero) steady-state error. Save the data and gain values after the final tuning.

Results and Questions for Report

Note: Some results require simulation response. This would require running a simulation using your SIMULINK model from Part C Control Design using the model validation K and tau values. The command input will be identical to either the square wave signal or triangle signal implemented during the experiment. The attributes of the simulation will be mentioned in the corresponding section.

(A) Step Response

Any simulation results in this subsection will have the following attributes:

Command input: Square wave
- Amplitude: Maximum of 0.4 and minimum of 0
- Frequency: 0.4 Hz
Sampling time: 0.002 s (Go to "Modeling" -> "Model Settings" -> "Solver details" -> "Fixed-step size")

i) P Controller

Plot angle or rotary position, i.e., commanded, experimental from Step 1.9, and simulation responses on one figure. Use the same $K_p$ gain value evaluated in Step 1.4 for the simulation response.
What is the effect of proportional controller gain on closed-loop system behaviour?

ii) PD Controller

Plot angle or rotary position, i.e., commanded, experimental from Step 1.10, and simulation responses on one figure. Use the same $K_p$ and $K_d$ gain values evaluated in Step 1.10 for the simulation response.
What is the effect of derivative controller gain on closed-loop system behaviour?
Compare the PD simulation (that shows zero steady-state error) and experimental (that shows a steady-state error) results to also discuss regarding the dead zone in the DC motor and how it affects the steady-state error.

iii) Sampling Time

Plot angle or rotary position (commanded and experimental) and control input Vm from Step 1.11.
What is the effect of sampling time on closed-loop system behaviour?

(B) Ramp Response

Any simulation results in this subsection will have the following attributes:

Command input: Triangle wave
- Amplitude: Maximum of $\pi/3$ and minimum of 0
- Frequency: 0.4 Hz
Sampling time: 0.002 s (Go to "Modeling" -> "Model Settings" -> "Solver details" -> "Fixed-step size")

i) PD Controller

Plot angle or rotary position, i.e., commanded, experimental from Step 2.3, and simulation responses on one figure. Use the same $K_p$ and $K_d$ gain values evaluated in Step 2.4 for the simulation response.

iii) PID Controller

Plot angle or rotary position, i.e., commanded, experimental from Step 2.5 (or 2.4 if applicable), and simulation responses on one figure. Use the tuned $K_p$ , $K_d$ and $K_i$ gain values evaluated in Step 2.6 for the simulation response.
Mention the final PID gains after tuning.

Lab 2: 3 DOF Gyroscope

A & B. Modeling and Control Design (Week 1)

Objective

The purpose of this experiment is to design a controller that maintains the direction of a gyroscope under base excitation. The controller can also be used to rotate the gyro platform to a desired orientation using the gyroscopic principle.

Introduction

Gyroscopes have become of great practical interest as they are used in control and guidance systems for air, sea, and space vehicles. One of their biggest advantages is that they can generate a large output (orientation change of an entire spacecraft) using a small input (torque input tilting a rotating disk) through torque amplification.

Description of Quanser Gyroscope system

The Quanser 3 DOF Gyroscope system (see Figure 1) can be actuated about all of its frames using the mounted motors while encoders measure the angle about each axis. In addition, the rotor itself is actuated and measured in the same manner.

The gyroscope setup is shown in Figure 2, and its components are provided in a table in Figure 3. The outer rectangular gray frame, the outer red gimbal and the inner blue gimbal are designed such that they can be individually fixed in place upon desire. This allows the users to perform a variety of different experiments using the device.

In this experiment, the gyroscopic effect will be employed to control the angle of the gray frame by applying the control command about the blue gimbal. The red outer gimbal will be fixed. In order to do this, the rotor has to have acquired enough angular momentum (RPM) for the gyroscopic effect to take place. Therefore, a controller is required to control the angular speed of the disk while another is required to control the blue gimbal angle.

In the next sections, we will develop a model of the gyroscope device from first principles in order to obtain a transfer function of the plant, which we will later use in a control scheme.

A. Modeling

The angular position about the outer gray frame (Body A), i.e., the yaw angle $\psi$ , is controlled by changing the angular position, the roll angle $\phi$ , of the blue gimbal (Body C), with the torque generated by motor 2, $\tau_2$ (see Figure 4). The red gimbal/y-axis (Body B) is locked in our experiment.

The rotational dynamics of rigid bodies can be modeled using the Newton-Euler equation, which states that the rate of change of angular momentum of a rigid body about its mass center equals the total external moment applied about its mass center. Using the Newton-Euler equation, one can derive the equations for rotational motion of individual gimbals and the outer frame. The resulting equations will be non-linear. They are linearized with the disk spinning at a constant high speed $(\Omega)$ with the gyroscope system at rest, i.e., roll $(\phi)$ , pitch $(\theta)$ and yaw $(\psi)$ equal zero. In our experiment, we will lock the the rotation of the red gimbal, i.e., $\theta=0$ , for all the time. Further, we will align the disk spin axis along the y-axis (see Figure 4).

The resulting linearized equations of motion when the red gimbal is locked, i.e., $\theta=0$ , and after neglecting the small amount of bearing friction associated with gimbal and frame rotations can be obtained as

J_d\dot{\omega}=\tau_1 \qquad \qquad \tag{1}

J_x\ddot{\phi}=\tau_2+J_d\Omega\dot{\psi}\qquad \qquad \tag{2}

J_z \ddot{\psi}=-J_d\Omega\dot{\phi}\qquad \qquad \tag{3}

where $\omega$ is the change in disc speed from its steady state value of $\Omega$ , $J_d$ is the disc mass moment of inertia about its spin axis, $J_x$ is the mass moment of inertia of the disk and motor 1 plus the blue gimbal and motor 2 about the x-axis (roll axis), $J_z$ is the mass moment of inertia of the entire system, i.e., including the disk, red and blue gimbals, motors 1 and 2 and the outer gray frame about the z-axis.

Equation (1) is for the disc spinning about its own axis. Equation (2) is for the motion about the roll/blue x-axis that is driven by the torque applied from motor 2 and the gyroscopic reaction torque. Equation (3) is for the motion about the yaw/gray frame that is driven exclusively by the gyroscopic reaction torque.

The Quanser 3 DOF Gyroscope is driven by a current-controller amplifier. The relation between the torque outputs of motors 1 and 2 and their respective current inputs are

\tau_1=k_tu_1\qquad \qquad \tag{4}

\tau_2=k_tu_2\qquad \qquad \tag{5}

where $k_t$ is the current-to-torque constant of the motor, $u_1$ is the current applied to motor 1, and $u_2$ is the current applied to motor 2.

The disc is rotated at a fixed, steady-state speed of $\Omega$ . Given this and applying Laplace transforms to Equations (1) to (3) we arrive at the following transfer function models:

\frac{\omega(s)}{\tau_1 (s)}=\frac{1}{J_d s }\qquad \qquad \tag{6}

\frac{Φ(s)}{τ_2 (s)}=\frac{J_z}{J_x J_z s^2+J_d^2 Ω^2}=P_1 (s)\qquad \qquad \tag{7}

\frac{Ψ(s)}{Φ(s)}=\frac{-ΩJ_d}{J_z s}=P_2 (s)\qquad \qquad \tag{8}

A state-space model representation of Equations (2-3) with the state vector as $[\phi \quad \psi \quad \dot{\phi} \quad \dot{\psi}]^T$ can be written as

\frac{d}{dt} \begin{bmatrix} \phi \\ \psi \\ \dot{\phi} \\ \dot{\psi} \\ \end{bmatrix} = \begin{bmatrix} 0& 0& 1& 0 \\ 0& 0& 0& 1 \\ 0& 0& 0& J_d\Omega/J_x \\ 0& 0& -J_d\Omega/J_z& 0 \end{bmatrix}\begin{bmatrix} \phi \\ \psi \\ \dot{\phi} \\ \dot{\psi} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1/J_x \\ 0 \\ \end{bmatrix}\qquad \qquad \tag{9}

The gyroscope system parameters are provided in Table 1:

Parameter Name

Parameter Symbol

Value

Unit

Rotor mass

1.907

Rotor radius

0.0762

MOI of disk about spin axis

0.0055

MOI about x-axis/blue gimbal (disc rotor + blue gimbal)

0.00388552

kgm^2

MOI about y-axis/red gimbal (disc rotor + blue gimbal + red gimbal)

0.02344554

kgm^2

MOI about z-axis/gray gimbal, i.e., disc and blue, red, gray frames

0.06632120

kgm^2

Gimbal motor current-torque constant

0.115

Nm/A

Disc motor current-torque constant

0.0704

Nm/A

Max current

Max torque

0.3450

Disk rotor steady-state speed

78.5398

rad/s

B. Control Design

The objective is to design a controller for regulation and control of the outer frame (gray frame) angular position $\psi$ using the inner gimbal (blue gimbal) torque $\tau_2$ . This mimics the following problem: Given a spacecraft with a momentum gyro, we would like to regulate and control the spacecraft yaw angle $\psi$ using the tilting of a high-speed spinning disk mounted in the spacecraft. In this lab, the rotation about the z-axis, $\psi$ , is controlled by torque applied about the x-axis through motor 2.

Controller Specifications:

Specification 1: All closed-loop poles must be in the left half of the complex plane.

Specification 2: The peak time in response to a $10\degree$ step command must be less than 0.45 seconds.

Specification 3: The 2% settling time to a $10\degree$ step command input must be less than 0.50 seconds.

Specification 4: Control current must not saturate (see Table 1 for max current input).

Specification 5: For the outer loop, $|k_p|<6$ and $|k_d|<0.4$ .

In order to meet the given controller specifications, we will make use of a technique known as successive loop closures. We will first design an inner loop rate feedback, i.e., feedback of $\dot{\phi}$ , to damp the rotation of the inner blue gimbal. Subsequently, an outer loop PD controller is used to control the outer frame (gray frame) rotation $\psi$ . The selected control logic is summarized in Figure 5.

The transfer functions $P_1(s)$ and $P_2(s)$ are previously obtained in the modeling section as Equations (7) and (8). In the following section, we will form the overall transfer function and do controller design.

Controller Design Procedure:

Open the script gyro_sim_week1.mlx and run the first section to initialize symbolic system parameters.
At the second section of the script, generate the transfer functions $P_1(s)$ and $P_2(s)$ using the given parameters. Note that $P_2(s)$ has a negative sign in the numerator.
Obtain the overall transfer function from $r_i$ to $\psi$ , which will be your plant $G(s)$ (see Figure 5) for the outer loop PD controller design. Remember to close the loop for $P_1(s)$ .
Select a nominal value for the inner loop rate feedback gain $k_v$ , for example $k_v=0.2$
Based on the symbolic G function from step 3, generate a numeric transfer function $G(s)$ from $r_i$ to $\psi$ that you will use for the controlSystemDesigner window. This function should not contain any symbols. Note: Check your $G(s)$ with your TA before proceeding.
Initialize C = 1, then open controlSystemDesigner(G,C) in MATLAB.
The control architecture shown in Figure 5 is different from the default architecture that opens in controlSystemDesigner. To account for this change, on the Control System Designer window, select CONTROL SYSTEM > Edit Architecture, and select the following architecture:
Select some arbitrary negative values (within the limits of Specification 5) for both $k_p$ (C1 in Figure 6) and $k_d$ (C2 in Figure 6). Since C2 is a pure derivative gain, you need to add a differentiator in C2, not a real zero.
Tune the values for $k_p$ and $k_d$ as needed to meet the given set of specifications. If needed, readjust the $k_v$ value of the inner loop, which is positive.
Save the controlSystemDesigner window plots, and your final gains $k_p, \, k_d, \, k_v.$ Note: Check your final gains with your TA before proceeding.
Use Figure 5 as a guide to create the SImulink diagram for this experiment. Adjust the step size to be a fixed-step size of 0.002 under model settings.
To the model created in step 10, add the following:
- A scope for the commanded input to the system.
- A saturation block before $P_1(s)$ to limit the torque $\tau_2$ applied to the system. Note: Since we need to satisfy Specification 4 (current saturation), we should ensure torque saturation as torque and current are linearly related (see Equation (5)).
- A scope to the torque applied to the system, $\tau_2$ (after the saturation block).
- A scope between $P_1(s)$ and $P_2(s)$ that measures the gyro angle, $\phi$ .
- A scope at the final output, $\psi$ . Note: Check your final Simulink diagram with your TA before proceeding.
Run the Simulink diagram for a reference angle of $\psi_{ref}=10^o$ . This can be accomplished in one of two ways:
1. Provide a step input of an initial value of 0 and a final value of 10. This will generate a step input of $\psi_{ref}=10^o$ .
2. Provide a square wave of amplitude 5, frequency of 0.333 Hz and an offset of +5. This will generate a square wave of an upper value $\psi_{ref}=10^o$ and a lower value of 0.
Verify if the Simulink diagram output is similar to the controlSystemDesigner step-input plot.
Check if the Simulink response satisfies all the given specifications (not on controlSystemDesigner). If not, further tune your $k_p$ and $k_d$ values (and $k_v$ if required) till all requirements are met. Note: Check your final gains with your TA before proceeding.

To check if your Simulink response meets Specification 2 and Specification 3, you can do one of the following approaches:

Place markers on the plot to manually check for peak time and 2% settling time.
Use stepinfo() command in MATLAB window to obtain the peak time and 2% settling time.

Results and Questions for Report

Modeling

Results

General principle used to obtain equations of motion for the gyroscope.
The three time-domain linearized equations of motion for the experiment.
$P_1(s)$ (both symbolic and numerical forms).
$P_2(s)$ (both symbolic and numerical forms).

Control Design

Results

Brief description of control design objective.
The closed-loop transfer function for $P_1(s)$ using rate feedback.
$G(s)$ , the transfer function from $r_i$ to $\psi$ (both symbolic and numerical forms).
Gain values $k_p, \, k_d, \, k_v$ from controlSystemDesigner (Step 10).
Control System Designer plots (root locus and bode plots for C1 and C2 and step response plot with peak time, settling time and steady-state values indicated).
Simulink model (Step 12).
Simulink response plots (to be generated using MATLAB, not Scope).
Validation of control design specifications being met or not.
Final tuned gain values $k_p, \, k_d, \, k_v$ from Simulink (Step 15).

Questions

How is the default Control System Designer control architecture different from the one used in this experiment? In your explanation, mention how the locations of the PD controller gains in Figure 5 are different from a traditional PD controller.
Briefly explain the need for the inner loop rate ( $\dot{\phi}$ ) feedback.

C. Controller Implementation (Week 2)

Objective

The controller will also enable the gyroscope to track a small step, large step and ramp-input commands.

The q_3dgyro_reaction_torque_m2021.slx diagram shown in Figure 8 is used to implement the controller on the Quanser Gyroscope system. The 3 DOF Gyroscope system shown in Figure 9 contains the Quanser blocks that interface with their hardware.

Part 1: Heading-hold experiment

1. Orient the gyroscope discs as shown:

2. Make sure the pitch axis/red gimbal is locked.

Note: Show your setup to your TA before proceeding. This will be the default orientation for all subsequent experiments.

4. Turn on the amplifier (big box) and verify the DAQ (small box) is connected to a power source. In previous experiments the smaller DAQ boards could be powered by the computer USB connection, but this is no longer the case.

5. Run gyro_sim_data.mlx script in MATLAB 2021 (not 2019) to initialize values.

6. Open q_3dgyro_reaction_torque.m2021.slx in MATLAB 2021 as well.

8. Set the Psi_Ref to 0. This will make the controller hold the 0-degree heading.

9. Set Hardware > Stop Time to 20 seconds.

10. Open all three scopes (Disc Rate, phi, psi) in a separate window.

11. When ready with the provided ruler, press Monitor & Tune (no need to build).

12. Watch the Disc Rate scope. When the angular velocity of the wheel reaches its maximum of ~78 rad/s, push the gray axis using the ruler at different angles/locations. Observe the behavior of the setup and notice the resistance to your disturbance by the gray axis, even though it is stationary.

IMPORTANT: After the experiment stops, use the provided towel to slowly stop the disc from spinning. Be careful since touching the disc while it is rotating may cause injury.

13. Check the saved data for disk rate, phi and psi, labelled as omega, phi, psi respectively. These are automatically saved as .mat files in your working directory. Rename them with "part1_" as the prefix, followed by the variable name. The data is saved as follows:

The variable omega has the rows saving time, commanded speed, and actual speed respectively.
The variable phi saves time, phi response, and phi gimbal current respectively.
The variable psi saves time, reference psi angle, and psi response respectively.

Part 2: Small step response

1. Make sure the disc is fully stopped from the previous experiment.

2. Orient the gyroscope as shown in Figure 10.

3. Set simulation Stop Time to 20 seconds.

4. Set Psi_Ref to 0, apply the change, but keep the window open.

5. Open all three scopes in a separate monitor.

6. Click Monitor & Tune, and watch the Disc Rate scope.

7. When disc rate reaches its maximum, change Psi_Ref (deg) gain to 10 degrees.

8. Observe the gyroscope tracking the desired heading command.

IMPORTANT: After the experiment stops, use the provided towel to slowly stop the disc from spinning. Be careful since touching the disc while it is rotating may cause injury.

9. Refer to Part 1 Step 13 for the saved data format. Rename them with "part2_" as prefix, followed by the variable names as shown in Part 1 step 13.

Part 3: Large step response

1. Make sure the disc is fully stopped from the previous experiment.

2. Orient the gyroscope as shown in Figure 10.

3. Set simulation Stop Time to 20 seconds.

4. Set Psi_Ref (deg) gain to 0, apply the change, but keep the window open.

5. Open all three scopes in a separate monitor.

6. Click Monitor & Tune, and watch the Disc Rate scope.

7. When disc rate reaches its maximum, change Psi_Ref (deg) gain to 40 degrees.

8. Observe the gyroscope tracking the desired heading command.

IMPORTANT: After the experiment stops, use the provided towel to slowly stop the disc from spinning. Be careful since touching the disc while it is rotating may cause injury.

9. Refer to Part 1 Step 13 for the saved data format. Rename them with "part3_" as the prefix, followed by the variable names shown in Part 1 step 13.

Part 4: Ramp input response

While the step response is useful for system characterization, it is seldom used for an actual in-service trajectory because it is excessively harsh (high acceleration and jerk, which is the rate of change of acceleration). A more common trajectory used for tracking applications is a ramp.

1. Make sure the disc is fully stopped from the previous experiment.

2. Orient the gyroscope as shown in Figure 10.

3. Set simulation Stop Time to 20 seconds.

4. Open the Signal Generator block, and select the “sawtooth” waveform

5. Set Psi_Ref (deg) gain to 0, apply the change, but keep the window open.

6. Open all three scopes in a separate monitor.

7. Click Monitor & Tune, and watch the Disc Rate scope.

8. When disc rate reaches its maximum, change Psi_Ref (deg) gain to 20 degrees.

9. Observe the gyroscope tracking the desired heading command.

IMPORTANT: After the experiment stops, use the provided towel to slowly stop the disc from spinning. Be careful since touching the disc while it is rotating may cause injury.

10. Refer to Part 1 Step 13 for the saved data format. Rename them with "part4_" as the prefix, followed by the variable names shown in Part 1 step 13.

Results for Report

Briefly describe the controller implementation experiments.

Part 1: Heading hold

Explain the behavior of the gray axis as you disturb it.

Part 2: Small step input

Plot motor 2 current simulated and experimental responses on the same axes.
Explain any differences between simulated and experimental responses.

Part 3: Large step input

Plot motor 2 current.

Part 4: Ramp input

Plot motor 2 current.

Questions for Report

Which axis of the setup is fixed to the spacecraft?
Using the gyroscopic principle, what changes would you make to the gyroscope system for measuring yaw rate of the spacecraft?

Lab 3: Rotary Flexible Link

A. System Identification (Week 1)

Objective

The purpose of this experiment is to understand the existence of vibrations in a rotary flexible link and the resulting mode shapes. Any beam-like structure exhibits vibrations either due to external changing loads or due to reorientation via actuators. The experiment deals with the modeling and identification of modal frequencies and mode shapes of free vibrations for a rotary flexible link. This analysis is particularly useful to understand structural vibrations and modes and how to contain them in real-world applications like aircraft and spacecraft structures as well as robotic links/manipulators.

Equipment Required

Flexible link plant (Quanser FLEXGAGE module)
Power amplifier (Quanser VoltPaq-X2)
Data acquisition board (Quanser Q2-USB)
Rotary servo plant (Quanser SRV02)
MATLAB and SIMULINK

Modeling and Vibration Analysis

This experiment involves system identification and modeling of the flexible link. The objective is find the stiffness and the viscous damping coefficient of the flexible link and conduct a frequency sweep across a range to determine the structural frequencies and mode shapes of the link.

Rotary Flexible Link Model

The main components of the setup are labeled in Figs. 2 and 3 and are listed in Table 1.

Table 1. Setup Components

The FLEXGAGE module consists of the strain gauge, the strain gauge circuitry, and a sensor connector. The flexible link is attached to this module and the strain gauge is fixed at the root of the link. The module is mounted onto the servo motor, which is the actuator for this system. The strain gauge sensor produces an analog signal proportional to the deflection of the link tip.

where the various constants are SRV02 parameters which are mentioned in Table 2.

Table 2. Setup Parameters

Equations of Motion

Part 1: Determination of Link Stiffness and Viscous Damping from Experiment

Part 1: Theory

The characteristic polynomial for a second-order system is:

The damping ratio of this second-order system can be found from its response (underdamped system) using the logarithmic decrement given by

The period of oscillation in a system response can be found using the equation

From this, the damped natural frequency (in rad/s) is

and the natural frequency is

Part 1 Experimental Procedure:

Mount the flexible link onto the calibration bench.
Download and open FlexLink_FreeOsc_Q2_USB.slx. This is the block diagram for this part of the experiment. Change the simulation time to 5-10 seconds.
Open the scope alpha.
Turn on the power supply.
Check you have good data and save the link deflection angle data for the free oscillation using the filename FreeOsc_1 to your folder.
Repeat the steps 7 and 8 two more times for different perturbation locations along the link (for example, around the middle and near the base) or different perturbation angles.

Part 1: Analysis:

Plot the measured angular deflection of the link vs. time for each case. Ensure that the data is centered around 0°, and if it is not, adjust it so that it is. Select peaks that are smooth (focus on the region after initial transients) as shown in the figure below. Ensure that there are at least 4-5 peaks in between the two chosen peaks. Option: Usage offindpeak MATLAB function might be helpful, but not required. If you use this function, verify the peaks identified.
Compute the mass moment of inertia of the link using the equation in Table 2.

Part 2 Theory:

The flexible link can be considered as a thin continuous (uniform) cantilever beam anchored at one end and free at the other end. Using the Euler-Bernoulli beam theory, the equation of motion can be written as

The general solution to Eq. (1.15) can be obtained using separation of variables, as in Eq. (1.17) below.

Equation (1.18) can be rewritten as

The constants in Eq. (1.22) are determined from four boundary conditions, while the constants in Eq. (1.23) are determined from two initial conditions.

For a clamped-free or cantilever beam, the geometric boundary conditions are

and the natural boundary conditions are

Hence, Eq. (1.22) becomes

For a non-trivial solution, the determinant of the above matrix must be 0, i.e.

which gives the following characteristic equation

The above equation simplifies to

There are infinite solutions to this characteristic equation, which are given by

Thus, the mode shapes are

The mode shapes of a uniform cantilever beam are shown in Fig. 7.

Part 2 Experimental Procedure:

In order to determine the modal frequencies and mode shapes, the link is rotated using a sinusoidal input to the link via external means. When the frequency of the input coincides with either the fundamental frequency or higher frequency modes, the corresponding modes will be excited due to resonance and their mode shapes can be visually observed. Hence, the following experiment involves a frequency sweep across a range provided as input to the flexible link via the servo motor to identify the frequencies and observe the corresponding mode shapes.

Mount the flexible link onto the rotary servo base.
Download and open FlexLink_ExciteMode.mdl
Open the scope alpha.
Turn on the power supply.
Ensure that the manual switch is connected to the Chirp signal input. This signal will provide a sinusoidal signal of fixed amplitude, with frequency increasing at a linear rate with time.
Open the Chirp signal command block and make sure that the Initial frequency is 0.1 Hz, the target time is 0.25 s and the Frequency at target time is 0.2 Hz. This will allow the frequency sweep to take place at a reasonable rate and ensure that the relevant frequencies are covered within the span of time.
Save the data using the format Osc_ChirpSignal into your folder.

Part 2 Analysis:

Plot the measured angular deflection of the link vs. time. Using the time domain plot, perform frequency analysis using Fast Fourier Transform (FFT) [refer to the Appendix for the code] or a similar technique to identify the number of dominant frequencies and their magnitudes present in the signal.
Using Eqn. 1.26 and the values of necessary system parameters, calculate the first and second modal frequencies.

Part 3 Observation of Mode Shapes

Part 3 Theory:

Analytical expressions for the mode shapes are given in Eq 1.25.

Part 3: Experimental Procedure

Open the FlexLink_ExciteMode.mdl Simulink file used in part 2.
Open the scope alpha.
Turn on the power supply.
Connect the manual switch to the Sine wave signal input. This signal will provide a sinusoidal signal of fixed amplitude and fixed frequency.
Open the Sine wave signal command block and make sure that the Amplitude is 3 and Phase is 0 rad. Enter the first modal frequency determined from Eq. 1.26 in rad/sec.
Tune the frequency value by increasing or decreasing in steps of 1 rad/sec until the first mode shape is clearly visible (do this while the simulation is running).
Observe the corresponding mode in the link and note down the number of nodes and their locations with respect to length of the link.
Save the data using the file name Osc_Sineinput into your folder. This is the result for the first modal frequency.
Repeat steps 6 to 10 for the second modal frequency determined from Eq. 1.26. Save the results for this as well (second modal frequency).

Part 3: Analysis

Record the number of nodes and their locations for each mode. Determine the locations as a ratio of the link length and compare them with the values between the theoretical and experimental mode shape plots.

Results for Report

(A) From Part 1

Plot of the measured angular deflection of the link vs. time with the chosen peaks marked for each dataset, i.e., three plots corresponding to three perturbation locations. Plot as three subplots in one figure.
Equations for time period of oscillation, damped frequency, undamped frequency, damping ratio, stiffness, and viscous damping coefficient.

(B) From Part 2

Plot of measured angular deflection of the link vs. time
Fast Fourier Transform (FFT) plot of link deflection with dominant frequencies and their magnitudes identified (either marked on the plot or mentioned in writing).
Calculation of mass per unit length of the link
Calculations and values for the theoretical first and second modal frequencies (calculated using Eqn. 1.26).
Compare the first dominant frequency (from FFT) with the natural frequency of the flexible link and state your observations. Explain any similarities or differences observed and the reasons behind them.
Compare the first and second dominant frequencies obtained from the FFT with the corresponding calculated first and second modal frequencies (Result B.4) and explain your observation.

(C) From Part 3

1st and 2nd Mode shapes (theoretical and experimental) are plotted on the same graph with node locations marked. (plot of v(x) vs x/L)
Number of nodes and their locations for each mode (theoretical and experimental). Use Table B.2 to document the values.
Comparison of experimental and theoretical mode shapes/node locations and state the reason(s) for why they differ.

Table B.1 System Identification Parameters

Table B.2 Locations of Nodes

Questions for Report

Compare the damped and undamped frequencies of the link and report your observation. What does this signify with respect to the flexible link damping?

Appendix

The following code performs FFT analysis on the flexible link angle response

B & C. Control Design and Controller Implementation (Week 2)

Objective

The objective is to design a controller that allows rapid reorientation of the link with very little vibrations from the link flexibility. This analysis is a preliminary approach to containing structural vibrations in real-world applications like aircraft and spacecraft structures as well as robotic links/manipulators

B. Control Design

The methodology used is to design the control law assuming full state feedback to meet a given set of specifications. This will be followed by studying the loss of controller performance with servo feedback alone. The method we will use for full state feedback is based on the linear quadratic regulator (LQR) theory, which requires the given system to be controllable.

State Space Model

The equations that describe the motions of the servo and the link with respect to the servo motor voltage are obtained using the lumped mass method and can be written as

the EOM can be rearranged as

The various system parameters can be found in Table 2 of Part A: System Identification.

Equations 3 and 4 can be represented in state-space form

In the output equation, only the position of the servo and link angles are being measured.

As such, the velocities of the servo and link angles can be computed in the digital controller if required, e.g., by taking the derivative and filtering the result though a high-pass filter. Hence, with all the states of the system as outputs, the C and D matrices are

Linear Quadratic Regulator (LQR) Design

Controllability

equals the number of states of the system,

LQR controller

Assuming the time-domain design specifications are

Then one can use the Bryson’s rule to choose the weighting matrices as

In our case,

The reference state is defined as

and the controller is

Controller Specifications:

Controller Design Procedure:

Note: The results from the following steps have to be included in the Analysis section of the report.

>> [K,S,E] = lqr(A,B,Q,R);
Using Figure 1 as a guide, develop a Simulink model of the system.
Make sure to include servo position command input, saturation blocks, and appropriate outputs. Use scopes and To Workspace blocks so that reference, output, and response signals can be tracked. (Make sure to maintain consistency between deg and rad in SIMULINK)
- Check if all the given specifications are met.
- Check your plots with the TA.
- Check if the specifications are met (no need to tune Q and R matrices.)

C. Controller Implementation and Evaluation

In this experiment, the servo position is controlled while minimizing the link deflection using the LQR based control found. Measurements will then be taken to ensure that the specifications are satisfied.

Warning: In closed loop experiments, the link may swing around rather quickly. Stay clear of it during closed loop experiments.

Experiment

Turn the amplifier power on.
Download the q_flex.zip and open the Simulink file q_flex for controller implementation. This is the block diagram for this part of the experiment. Make sure you have K gain variable in the MATLAB workspace. Note: The Smooth Signal Generator block generates a 0.33 Hz square wave (with amplitude of 1) that is passed through a Rate Limiter block to smooth the signal (Rising slew rate of 100 and falling slew rate of -100). The Amplitude (deg) gain block is used to change the desired servo position command. The state-feedback gain K is set in the LQR Control gain block and is read from the MATLAB workspace.
Align the flexible link along the 0 mark on the servo stand.
Double click on the scopes and open them.
Ensure that the manual switch is connected to the full state feedback.
A periodic square wave is given as commanded link rotation to the servo motor, causing the link to move left and right alternately for 10 seconds.
Copy the response data to your group folder. The response data is saved in variables data_theta, data_alpha, and data_vm. DO NOT DELETE or SAVE THE SIMULINK MODEL.
Connect the manual switch to the partial state feedback and repeat steps 4-9 again. Make sure your files name is different from full state feedback.
Turn the power off.

Results & Questions for Report

Control Design

Results

Include the results from the controller design procedure Steps 1-6.

Questions

>> sysclosed = ss((A-B*K), B, [1 0 0 0], 0);

>> bw = bandwidth(sysclosed)

>> bode(sysclosed)

The bode() function produces closed loop system frequency response plot. Include Bode Plots of the closed loop system and list the bandwidth of the system with full and partial state feedback.

Estimate the stability margins (gain and phase margins) of your full state and partial state feedback controllers. Hint: Use 'margin(sysopen)' command in MATLAB where 'sysopen' is the state space model of the open loop system with the controller.

>> sysopen = ss(A,B,K,0)

>> margin(sysopen)

We get bode plot of the open loop system with the controller with margins listed on the plots. Include the Bode plots for the full state and partial state feedback.

Controller Implementation

Results

Plots of servo angle, link deflection angle and input voltage vs time for full state and partial state feedback. Compare the results and explain the differences.
Run the Simulink that you built in control design with the same command input used in the experiment. Compare the simulated response with the experimental response for both full state and partial state feedback and explain any differences between them. Note 1: Generate two sets of figures with one set containing the plots of simulation and experiment for full state feedback and the second set containing the plots of simulation and experiment for partial state feedback. For each feedback case, the simulated response and corresponding experimental response must be on the same graph. Note 2: Average the first few seconds of the experimental data to obtain initial conditions for your state-space block in Simulink.

Questions

Explain why the link deflection angle goes opposite to servo rotation initially.
For a controllable system, can partial state feedback guarantee stability like full state feedback? Briefly explain.

Lab 4: Rotary Inverted Pendulum

A & B. Modeling and Balance Control

Objective

The objectives of this laboratory experiment are as follows:

Obtain the linear state-space representation of the rotary pendulum system.
Design a controller that balances the pendulum in its upright position using Pole Placement.
Simulate the closed-loop system to ensure the given specifications are met.
Implement the balance controller on the Quanser Rotary Pendulum system and evaluate its performance.

Equipment

Rotary pendulum (Quanser Rotary inverted pendulum module)
Rotary servo base (Quanser SRV02)
Power amplifier (Quanser VoltPaq-X2)
DAQ (Quanser Q2-USB)
MATLAB and SIMULINK

A. Modeling

This experiment involves modeling of the rotary inverted pendulum.

Rotary Inverted Pendulum Model

A picture of the rotary servo base with rotary pendulum module is shown in Fig 1. The numbered components in Fig. 1 are listed in Table 1 along with the numerical values of the system parameters in Table 2.

Table 1 Rotary Pendulum components (Figure 1)

Table 2 Main Parameters associated with the Rotary Pendulum module

Model Convention

Nonlinear Equations of Motion

Instead of using classical (Newtonian) mechanics, the Lagrange method is used to find the equations of motion of the system. This systematic method is often used for more complicated systems such as robot manipulators with multiple joints.

Specifically, the equations that describe the motions of the rotary arm and the pendulum with respect to the servo motor voltage, i.e., the dynamics, are obtained using the Euler-Lagrange equation:

With the generalized coordinates defined, the Euler-Lagrange equations for the rotary pendulum system are

The Lagrangian of the system is described by

and acting on the pendulum is

Once expressions for the kinetic and potential energy are obtained and the Lagrangian is found, then the task is to compute various derivatives to get the EOMs. After going through this process, the nonlinear equations of motion for the Rotary Pendulum are:

Linearization

Linearization of a nonlinear function about a selected point is obtained by retaining upto first order term in the Taylor Series expansion of the function about the selected point. For example, linearization of a two variable nonlinear function f(z) where

about the point

can be written as

Linearization of Inverted Pendulum Equations

The nonlinear equations of the inverted pendulum system obtained as Equations (7) and (8) are linearized about the equilibrium point with the pendulum in the upright position, i.e.,

Linearization of Equation (7) about the above equilibrium state gives

Likewise, linearization of Equation (8) about the equilibrium point with the pendulum in the upright position gives

Equations (10) and (11) can be arranged in the matrix form as

Equation (12) can be rewritten in state space form as

Analysis: Modeling

Download the Part 1.zip file and extract the folder contents.
Note down your state-space matrices for your report. Note: You may want to cross-check the state-space matrix with TAs before proceeding to balance control.
Find the open-loop poles of the system. Hint: Use eig(A).

B. Balance Control

Specification

Stability

The stability of a system can be determined from its poles ([1]):

Stable systems have poles only in the left-half of the complex plane.
Unstable systems have at least one pole in the right-half of the complex plane and/or poles of multiplicity greater than 1 on the imaginary axis.
Marginally stable systems have one pole on the imaginary axis and the other poles in the left-half of the complex plane.

The poles are the roots of the system’s characteristic equation. From the state-space, the characteristic equation of the system can be found using

Controllability

Rank Test The system is controllable if the rank of its controllability matrix

equals the number of states in the system, i.e.

Companion Matrix

For a controllable system with nxn system matrix A and nx1 control matrix B. The companion matrices of A and B are

and

Now define W,

Then

and

Pole Placement Theory

Desired Poles

and

Simulation Model with Feedback

The feedback control loop that balances the rotary pendulum is illustrated in Figure 4. The reference state is defined as

When running this on the actual system, the pendulum begins in the hanging, downward position. We only want the balance control to be enabled when the pendulum is brought up around its upright vertical position. The controller is therefore

B.1 Experiment: Designing the Balance Control

In the same rotpen_part1_student.mlx live script, go to the Balance Control section. Enter the chosen zeta and omega_n values.
Find gain K using a predefined Compensator Design MATLAB command K = acker(A,B,DP), which is based on pole-placement design. Note: DP is a row vector of the desired poles found in Step 3.

For sanity check, if you use damping ratio of 0.7 and natural frequency of 4 rad/s, you should get around K = [-12 63 -5.5 7].

B.2 Experiment: Simulating the Balance Control

Run rotpen_part1_student.mlx live script. Ensure the gain K you found is loaded in the workspace (type K matrix in the command window).
Open and run the s_rotpen_bal.mdl for 10 seconds. The responses in the scopes shown in Figure 6 were generated using an arbitrary feedback control gain. Note: When the simulation stops, the last 10 seconds of data is automatically saved in the MATLAB workspace to the variables data_theta, data_alpha, and data_Vm.
Save the data corresponding to the simulated response of the rotary arm, pendulum, and motor input voltage obtained using your obtained gain K. Note: The time is stored in the data_theta(:,1) vector, the desired and measured rotary arm angles are saved in the data_theta(:,2) and data_theta(:,3) arrays. Similarly, the pendulum angle is stored in the data_alpha(:,2) vector, and the control input is in the data_Vm(:,2) structure.

B.3 Experiment: Implementing the Balance Controller

In this section, the state-feedback control that was designed and simulated in the previous sections is run on the actual Rotary Pendulum device.

Experiment Setup

The q_rotpen_bal_student SIMULINK diagram shown in Figure 7 is used to run the state-feedback control on the Quanser Rotary Pendulum system. The Rotary Pendulum Interface subsystem contains QUARC blocks that interface with the DC motor and sensors of the system. The feedback developed in the previous section is implemented using a Simulink Gain block.

Download the Part 2.zip file, extract the folder contents and open the part2setup.m script.
Open the q_rotpen_bal_student SIMULINK diagram.
As shown in Figure 7, the SIMULINK diagram is incomplete. Add the necessary blocks from the Simulink library to implement the balance control.
- You need to add a switch logic to implement Equation 30. Use Multi-port switch with 2 data points and zero-based contiguous. The output from the compare to constant block will be 0 if false and 1 if true. Check your block with TA.
- Ensure that you connect the final signal going into the u(V) terminal of the Rotary Pendulum Interface, which is also connected to the u scope terminal.
Turn ON the power amplifier.
Ensure the pendulum is in the hanging down position, with the rotary arm aligned with the 0 marking, and is motionless.
Once it is running, manually bring up the pendulum to its upright vertical position. You should feel the voltage kick-in when it is within the range where the balance control engages. Once it is balanced, the controller will introduce the ±20 degree rotary arm rotation.
The response should look similar to your simulation. Once you have obtained a response, click on the STOP button to stop the controller (data is saved for the last 10 seconds, so stop SIMULINK around 18-19 seconds once the response looks similar to Figure 8).
CAUTION Be careful, as the pendulum will fall down when the controller is stopped.
Similar to the simulation Simulink model, the response data will be saved to the workspace. Copy and paste into your group's folder. Ensure that the data variables have 10 seconds of data saved.

Results for Report

A) Modeling

Open-loop poles of the system.

B.1) Balance Control Design

B.2) Balance Control Simulation

B.3) Balance Controller Implementation

Questions for Report

A) Modeling

Did you expect the stability of the inverted pendulum to be as what was determined? Justify.

B.1) Balance Control Design

For the questions below, calculations and intermediate steps must be shown.

Appendix

Table A Main Rotary Servo Base Unit Specifications

Reference

[1] Norman S. Nise. Control Systems Engineering. John Wiley & Sons, Inc., 2008.

[2] Bruce Francis. Ece1619 linear systems, 2001.

Aero LQR

Copy of Lab 4: Rotary Inverted Pendulum

B. Modeling (Week 1)

The angular speed of the Rotary Servo Base Unit load shaft with respect to the input motor voltage can be described by the following first-order transfer function:

\frac{\Omega_l(s)}{V_m(s)} = \frac{K}{\tau s+1} \qquad \qquad \qquad \tag{2.1}

where $\Omega_l(\mathrm{s})$ is the Laplace transform of the load shaft speed $\omega_l(\mathrm{t})$ , $V_m(\mathrm{s})$ is the Laplace transform of motor input voltage $v_m(\mathrm{t})$ , $K$ is the steady-state gain, $\tau$ is the time constant, and $\mathrm{s}$ is the Laplace operator. The Rotary Servo Base Unit transfer function model is derived analytically in and its $K$ and $\tau$ parameters are evaluated. These are known as the nominal model parameter values. The model parameters can also be found experimentally. describes how to use the frequency response and bump-test methods to find $K$ and $\tau$ . These methods are useful when the dynamics of a system are not known, for example in a more complex system. After the lab experiments, the experimental model parameters are compared with the nominal values.

1. Modeling Using First-Principles

a) Electrical Equations

The DC motor armature circuit schematic and gear train is illustrated in Figure 7.

$R_m$ is the motor resistance, $L_m$ is the inductance, and $K_m$ is the back-emf constant.

e_b(\mathrm{t}) = k_m\omega_m(\mathrm{t}) \qquad \qquad \qquad \tag{2.2}

Using Kirchoff’s Voltage Law, we can write the following equation:

V_m(\mathrm{t}) - R_mI_m(\mathrm{t}) - L_m \frac{dI_m(\mathrm{t})}{dt} - k_m \omega_m(\mathrm{t}) = 0 \qquad \qquad \qquad \tag{2.3}

Since the motor inductance $L_m$ is much less than its resistance, it can be ignored. Then, the equation becomes:

V_m(\mathrm{t}) - R_mI_m(\mathrm{t})-k_m\omega_m(\mathrm{t}) = 0 \qquad \qquad \qquad \tag{2.4}

Solving for $I_m(\mathrm{t})$ , the motor current can be found as:

I_m(\mathrm{t}) = \frac{V_m(\mathrm{t}) - k_m\omega_m(\mathrm{t})}{R_m} \qquad \qquad \qquad \tag{2.5}

b) Mechanical Equations

J \cdot \alpha = \tau \qquad \qquad \qquad \tag{2.6}

J_l \frac{d\omega_l(\mathrm{t})}{dt} + B_l\omega_l(\mathrm{t}) = \tau_l(\mathrm{t}) \qquad \qquad \qquad \tag{2.7}

J_m \frac{d\omega_l(t)}{dt} + B_m\omega_m(t) +\tau_{ml}= \tau_m(t) \qquad \qquad \qquad \tag{2.8}

\tau_l(t) = \eta_gK_g\tau_{ml}(t) \qquad \qquad \qquad \tag{2.9}

where $K_g$ is the gear ratio and $\eta_g$ is the gearbox efficiency. The planetary gearbox that is directly mounted on the Rotary Servo Base Unit motor (see for more details) is represented by the $N_1$ and $N_2$ gears in Figure 7 and has a gear ratio of

K_{gi} = \frac{N_2}{N_1} \qquad \qquad \qquad \tag{2.10}

K_{ge} = \frac{N_4}{N_3} \qquad \qquad \qquad \tag{2.11}

The gear ratio of the Rotary Servo Base Unit gear train is then given by:

K_g = K_{ge} K_{gi} \qquad \qquad \qquad \tag{2.12}

Thus, the torque seen at the motor shaft through the gears can be expressed as:

\tau_{ml} (t) = \frac{\tau_l(t)}{\eta_gK_g} \qquad \qquad \qquad \tag{2.13}

Intuitively, the motor shaft must rotate $K_g$ times for the output shaft to rotate one revolution.

\theta_m(t) = K_g\theta_l(t) \qquad \qquad \qquad \tag{2.14}

We can find the relationship between the angular speed of the motor shaft, $\omega_m$ , and the angular speed of the load shaft, $\omega_l$ by taking the time derivative:

\omega_m(t) = K_g\omega_l(t) \qquad \qquad \qquad \tag{2.15}

To find the differential equation that describes the motion of the load shaft with respect to an applied motor torque substitute (Eq 2.13), (Eq 2.15) and (Eq 2.7) into (Eq 2.8) to get the following:

J_mK_g \frac{d\omega_l(t)}{dt} + B_mK_g\omega_l(t) +\frac{J_l \frac{d\omega_l(t)}{dt} + B_l\omega_l(t)}{\eta_gK_g}= \tau_m(t) \qquad \qquad \qquad \tag{2.16}

Collecting the coefficients in terms of the load shaft velocity and acceleration gives

(\eta_gK_g^2J_m + J_l) \frac{d\omega_l(t)}{dt} + (\eta_gK_g^2B_m + B_l)\omega_l(t) = \eta_gK_g\tau_m(t) \qquad \qquad \qquad \tag{2.17}

Defining the following terms:

J_{eq} = \eta_gK_g^2J_m+J_l \qquad \qquad \qquad \tag{2.18}

B_{eq} = \eta_gK_g^2B_m+B_l \qquad \qquad \qquad \tag{2.19}

simplifies the equation as:

J_{eq} \frac{d\omega_l(t)}{dt} +B_{eq}\omega_l(t) = \eta_gK_g\tau_m(t) \qquad \qquad \qquad \tag{2.20}

c) Combining the Electrical and Mechanical Equations

In this section the electrical equation and the mechanical equation are brought together to get an expression that represents the load shaft speed in terms of the applied motor voltage.

The motor torque is proportional to the voltage applied and is described as

\tau_m(t) = \eta_mk_tI_m(t) \qquad \qquad \qquad \tag{2.21}

where $K_t$ is the current-torque constant ( $N \cdot m / A$ ), $\eta_m$ is the motor efficiency, and $I_m$ is the armature current. See for more details on the Rotary Servo Base Unit motor specifications. We can express the motor torque with respect to the input voltage $V_m(t)$ and load shaft speed $\omega_l(t)$ by substituting the motor armature current given by Eq 2.5, into the current-torque relationship given in Eq 2.21:

\tau_m(t) = \frac{\eta_mk_t(V_m(t) - k_m\omega_m(t))}{R_m} \qquad \qquad \qquad \tag{2.22}

To express this in terms of $V_m$ and $\omega_l$ , insert the motor-load shaft speed Eq 2.15, into Eq 2.21 to get:

\tau_m(t) = \frac{\eta_mk_t(V_m(t) - k_mK_g\omega_l(t))}{R_m} \qquad \qquad \qquad \tag{2.23}

If we substitute (Eq 2.23) into (Eq 2.20), we get:

J_{eq} \frac{d\omega_l(t)}{dt} +B_{eq}\omega_l(t) = \frac{\eta_gK_g\eta_mk_t(V_m(t) - k_mK_g\omega_l(t))}{R_m} \qquad \qquad \qquad \tag{2.24}

After collecting the terms, the equation becomes

J_{eq} \frac{d\omega_l(t)}{dt} +\left (\frac{k_m\eta_gK_g^2\eta_mk_t}{R_m}+B_{eq}\right) \omega_l(t) = \frac{\eta_gK_g\eta_mk_tV_m(t) }{R_m} \qquad \qquad \qquad \tag{2.25}

This equation can be re-written as:

J_{eq} \frac{d\omega_l(t)}{dt} +\left (B_{eq,v}\right) \omega_l(t) = A_mV_m(t) \qquad \qquad \qquad \tag{2.26}

where the equivalent damping term is given by:

B_{eq,v} = \frac{k_m\eta_gK_g^2\eta_mk_t + B_{eq}R_m}{R_m} \qquad \qquad \qquad \tag{2.27}

and the actuator gain equals

A_m = \frac{\eta_gK_g\eta_mk_t}{R_m} \qquad \qquad \qquad \tag{2.28}

2. Modeling Using Experiments

a) Frequency Response

The magnitude of the frequency response of the Rotary Servo Base Unit plant transfer function given in equation Eq 2.1 is defined as:

|G_{\omega l,v}(\omega)| = \left |\frac{\Omega_l(\omega j)}{V_m(\omega j)}\right| \qquad \qquad \qquad \tag{2.29}

\left |\frac{\Omega_l(\omega j)}{V_m(\omega j)}\right| = \frac{K}{\tau \omega j + 1} \qquad \qquad \qquad \tag{2.30}

Then, the magnitude of it equals

|G_{wl,v}(w)| = \frac{K}{\sqrt{1+\tau^2\omega^2}} \qquad \qquad \qquad \tag{2.31}

K_{e,f} = |G_{wl,v}(0)| \qquad \qquad \qquad \tag{2.32}

b) Bump Test / Unit Step Input Test

\frac{Y(s)}{U(s)} = \frac{K}{\tau s + 1} \qquad \qquad \qquad \tag{2.33}

The step response shown in Figure 10 is generated using this transfer function with $K = 5 rad/Vs$ and $\tau = 0.05s$ .

K = \frac{\Delta y}{\Delta u} \qquad \qquad \qquad \tag{2.34}

y(t_1) = 0.632(y_{ss}-y_0) + y_0 \qquad \qquad \qquad \tag{2.35}

Then, we can read the time $t_1$ that corresponds to $y(t_1)$ from the response data in Figure 10. From the figure we can see that the time $t_1$ is equal to:

t_1 = t_0 + \tau \qquad \qquad \qquad \tag{2.36}

From this, the model time constant can be found as:

\tau = t_1 - t_0 \qquad \qquad \qquad \tag{2.37}

Going back to the Rotary Servo Base Unit system, a step input voltage with a time delay $t_0$ can be expressed as follows in the Laplace domain:

V_m(s) = \frac{A_ve^{(-st_0)}}{s} \qquad \qquad \qquad \tag{2.38}

where $A_v$ is the amplitude of the step and $t_0$ is the step time (i.e. the delay). If we substitute this input into the system transfer function given in Eq 2.1, we get:

\Omega_l(s) = \frac{KA_ve^{(-s t_0)}}{(\tau s + 1)s} \qquad \qquad \qquad \tag{2.39}

\omega_l(t) = KA_v\left(1-e^{(-\frac{t-t_0}{\tau})}\right) + \omega_l(t_0) \qquad \qquad \qquad \tag{2.40}

Experimental Setup

Download the Modeling_files.zip and extract the files to desktop.
Open q_servo_modeling.m SIMULINK file.
Double click on the "Rotary Servo Interface - Speed" subsystem to access the DAQ configuration
Configure DAQ: Double-click on the HIL Initialize block in the SIMULINK diagram and ensure it is configured for the DAQ device that is installed in your system (e.g. Q2-USB).
Open setup_servo_modeling.m file to open the setup script for the q_servo_modeling SIMULINK model.
Run the script. Note: The calculated servo model parameter are default model parameters and do not accurately represent the Rotary Servo Base Unit system.

Frequency Response Experiment

a) Steady-state gain

First, we need to find the steady-state gain of the system. This requires running the system with a constant input voltage. To create a $2V$ constant input voltage follow these steps:

In the SIMULINK diagram, double-click on the Signal Generator block and ensure the following parameters are set: • Wave form: sine • Amplitude: 1.0 • Frequency: 0.4 • Units: Hertz
Set the Amplitude (V) slider gain to 0.
Set the Offset (V) block to 2.0 V. Note: Execution of steps 1 through 3 generates a step input of 2.0 V magnitude.
Set the Simulation stop time to 5 seconds.
Open the load shaft speed scope, Speed (rad/s), and the motor input voltage scope, Vm (V) .
To build the model, click the down arrow on Monitor & Tune under the Hardware tab and then click Build for monitoring . This generates the controller code.
Click the Deploy button under Monitor & Tune, followed by the Connect button and then run SIMULINK by clicking Start .
The Rotary Servo Base Unit unit should begin rotating in one direction. The scopes should be reading something similar to figure below. Note that in the Speed (rad/s) scope, the yellow trace is the measured speed while the blue trace is the simulated speed (generated by Servo Model block).
Measure the speed of the load shaft and enter the measurement in below under the f = 0 Hz row. Since amplitude gain is set to 0, technically no sine input when f = 0 Hz is applied. Note: No need to save wl variable data or scope plot. Hint: The measurement can be done directly from the scope using the Cursor Measurements tool. Alternatively, you can use MATLAB commands max(wl(:,2)) to find the maximum load speed using the saved wl variable. When the controller is stopped, the Speed (rad/s) scope saves data to the MATLAB workspace in the wl parameter. It has the following structure: wl(:,1) is the time vector, wl(:,2) is the measured speed, and wl(:,3) is the simulated speed. Calculate the steady-state gain both in linear and decibel (dB) units, i.e. $20\log_{10}(\text{Gain})$ , as explained in . Enter the resulting numerical value in the f = 0 Hz row of . Also, enter its non-decibel value in under the 'Frequency Response Experiment' section.

b) Gain at varying frequencies

In the SIMULINK diagram, double-click on the Signal Generator block and ensure the following parameters are set: • Wave form: sine • Amplitude: 1.0 • Frequency: 1.0 • Units: Hertz
Set the Amplitude (V) slider gain to 2.0 V.
Set the Offset (V) block to 0 V.
Set the Simulation stop time to 5 seconds.
Click Connect button under Monitor & Tune and then run SIMULINK by clicking Start .
The Rotary Servo Base Unit unit should begin rotating smoothly back and forth and the scopes should be reading a response similar to Figure 13 a) and b).
Measure the maximum positive speed of the load shaft at f = 1.0 Hz input and enter it in below. As before, this measurement can be done directly from the scope using the Cursor Measurements tool or you can use MATLAB commands to find the maximum load speed using the saved wl variable. Calculate the gain of the system (in both linear and dB units) and enter the results in .
Increase the frequency to f = 2.0 Hz by adjusting the frequency parameter in the Signal Generator block. Measure the maximum load speed and calculate the gain. Repeat this step for each of the frequency settings in .

Step Response Experiment

In this method, a step input is given to the Rotary Servo Base Unit and the corresponding load shaft response is recorded. Using the saved response, the model parameters can then be found as discussed in .

To create the step input:

Double-click on the Signal Generator block and ensure the following parameters are set: • Wave form: square • Amplitude: 1.0 • Frequency: 0.4 • Units: Hertz
Set the Amplitude (V) gain block to 1.0 V.
Set the Offset (V)gain block to 2.0 V.
Set the Simulation stop time to 5 seconds.
Open the load shaft speed scope, Speed (rad/s), and the motor input voltage cope, Vm (V).
To build the model, click the down arrow on Monitor & Tune under the Hardware tab and then click Build for monitoring . This generates the controller code.
Click Connect button under Monitor & Tune and then run SIMULINK by clicking Start .
The gears on the Rotary Servo Base Unit should be rotating in the same direction and alternating between low and high speeds. The response in the scopes should be similar to Figure 14 a) and b).
Save wl data and name it as modeling_section#_Group#_step. Note down the magnitude of input voltage as this will be required to calculate the steady-state gain in the section.
Click the Stop button on the SIMULINK diagram toolbar (or select QUARC | Stop from the menu) to stop the experiment.

Model Validation Experiment

Double-click on the Signal Generator block and ensure the following parameters are set: • Wave form: square • Amplitude: 1.0 • Frequency: 0.4 • Units: Hertz
Set the Amplitude (V) slider gain to 1.0 V.
Set the Offset (V) block to 1.5 V.
Set the Simulation stop time to 5 seconds.
Open the load shaft speed scope, Speed (rad/s), and the motor input voltage scope, V_m (V).
To build the model, click the down arrow on Monitor & Tune under the Hardware tab and then click Build for monitoring . This generates the controller code.
Click Connect button under Monitor & Tune and then run SIMULINK by clicking Start .
The gears on the Rotary Servo Base Unit should be rotating in the same direction and alternating between low and high speeds and the scopes should be as shown in Figure 15 a) and b). Recall that the yellow trace is the measured load shaft rate and the purple trace is the simulated trace. By default, the steady-state gain and the time constant of the transfer function used in simulation are set to: K = 1 rad/s/V and $\tau$ = 0.1s. These model parameters do not accurately represent the system.
Notice that the model response does not match with the measured response.
Calculate the nominal values, $K$ and $\tau$ , using Eqs. 2.1, 2.26, 2.27, 2.28 and the specifications provided in and for the high-gear configuration. Note: Eq. 2.26 must be converted to the Laplace domain to obtain the transfer function, which should be simplified to resemble Eq. 2.1. This would yield the formulae to determine $K$ and $\tau$ . You may also refer to the lecture slides.
Enter the nominal values, $K$ and $\tau$ , that were found in Step 10 in the MATLAB Command Window. Update the parameters and examine how well the simulated response matches the measured one.
Save wl data and name it as modeling_section#_Group#_K#tau#.

Tables for Report

Table B.1 Collected frequency response data.

Table B.2 Summary of results for the Rotary Servo Base Unit Modeling laboratory.

Results for Report

(A) Frequency Response Experiment

Using the MATLAB plot command and the data collected in , generate a Bode magnitude plot. Make sure the amplitude and frequency scales are in decibels. When making the Bode plot, ignore the f = 0 Hz entry as the logarithm of 0 is not defined. Label the Bode plot where the gain is reduced by 3 dB from the DC gain and the corresponding frequency as the cutoff frequency. Time constant $\tau_{e,f}$ (units are seconds) of a first-order system is equal to the inverse of its cutoff frequency (units are rad/s). Enter the resulting time constant in . Hint: May use the Data Tips tool to obtain values from the MATLAB Figure. Include the labelled Bode plot along with tabulated values of in the report.

(B) Step Response Experiment

Plot the maximum load speed response (saved in the MATLAB workspace under the wl variable) from Step 9 in MATLAB.
Find the steady-state gain using the measured step response (Result B.1) and enter it in under the 'Step Response Experiment' section. Hint: Use the MATLAB ginput command to measure points off the plot.
Find the time constant from the obtained response (Result B.1) using Eqs. 2.35, 2.36 and 2.37, and enter the value in under the 'Step Response Experiment' section. Show the calculations in the report.

(C) Model Validation Experiment

Create a MATLAB figure that shows the simulation nominal values response, the step response (generated by using the K and tau values calculated from the bump test experiment), the frequency response (generated by using the K and tau values calculated for frequency response experiment), the response you got from step 8, and the measured response as five plots on the same graph in the report. Provide two reasons why the nominal model might not represent the Rotary Servo Base Unit with better accuracy. Which response matches most closely with the measured response?
From the plots, describe how the gain and time constant affect the system's response.

(D) Additional Results

Include the completed .
Explain how well the nominal model, the frequency response model, and the bump-test model represent the Rotary Servo Base Unit system (you may refer to Result D.1).

Appendix

Table A.1 Main Rotary Servo Base Unit Specifications

Table A.2 Rotary Servo Base Unit Gearhead Specifications

Gyroscope

This manual documents the modelling of a gyroscope for which an angular position controller using gimbal torque is developed. Finally, the controller is implemented on the hardware and evaluated.

Objective

Equipment Required

Gyroscope (Electromechanical plant)
Input/Output Electronic ECP Model 750 box
DSP based Controller/Data Acquisition Board which is already installed in a PC
ECP software

Part A: Modeling

The Gyroscope Model

This experiment will be performed using the Model 750 Control Gyroscope. The system, shown in Fig. 3.1, consists of an electromechanical plant and a full complement of control hardware and software. The user interface to the system is via an easy-to-use PC-based environment that supports a broad range of controller specification, trajectory generation, data acquisition and plotting features. A picture of the setup including the DSP card and input/output electronics is shown in Fig. 3.2.

Description of the Gyroscope

The gyroscope consists of a high inertia brass rotor suspended in an assembly with four angular degrees of freedom, as seen in Fig. 3.3. The rotor spin torque is provided by a rare Earth magnet type DC motor (Motor #1), whose angular position is measured by an optical encoder (Encoder #1) with a resolution of 2,000 counts per revolution. The motor drives the rotor through a 10:3 gear reduction ratio, which amplifies both the torque and encoder resolution by this factor. The first transverse gimbal assembly (body C) is driven by another rare Earth magnet motor (Motor #2) to effect motion about Axis #2. The motor drives a 6.1:1 capstan to amplify the torque between the adjoining bodies C and B. A 1000 line encoder (Encoder #2) with 4x interpolation is mounted on the motor to provide feedback of the relative position between bodies C and B with resolution of 24,400 counts per revolution.

The subsequent gimbal assembly, body B, rotates with respect to body A about Axis #3. There is no active torque applied about this axis. A brake, which is actuated via a toggle switch on the Controller box, may be used to lock the relative position between bodies A and B and hence reduce the system's degrees of freedom. The relative angle between A and B is measured by Encoder #3 with a resolution of 16,000 counts per revolution. Finally, body A rotates without actively applied torque relative to the base frame (inertial ground) along Axis #4. The Axis #4 brake is controlled similar to the Axis #3 brake and the relative angle between body A and the base frame is measured by an optical encoder (Encoder #4) with a resolution of 16,000 counts per revolution.

Inertial switches or “g-switches” are installed on bodies A, B, and C to sense any over-speed condition in the gimbal assemblies. The switches are set to actuate at 2.1 g’s. For Axis #2, limit switches and mechanical stops are provided at the safe limit of travel. When any of these normally closed switches sense a high angular rate condition, they open and thereby cause a relay to turn off power to the controller box. When this power is lost, the fail-safe brakes (power-on-to-release type) at Axes #3 and #4 engage. Also, upon loss of power, the windings of Motor #1 and #2 have shorted, thereby causing electromechanical damping. Thus, all axes are actively slowed and stopped whenever an over-speed or over-travel condition is detected.

Metal slip rings are included at each gimbal axis to allow continuous angular motion. These low noise, low friction, slip rings pass all electrical signals including those of the motors, encoder, g-switches, limit switches, and brakes to the control box.

Mathematical Model of the Gyroscope

Basic Equations

The basic gyroscope equation can be written as:

Substituting equations 3.2, 3.3 and 3.4 into Eq. 3.1 results in the following set of nonlinear equations:

Linearized Model

In this section, we proceed to linearize the nonlinear gyroscopic equations (Eq. 3.5) by imposing certain assumptions.

Assumptions

The angle of rotation of the rotor disk D about the gimbal axis 2 is small.

Part B: Controller Design

Control System Overview

The equations defining the numerators and denominators in the block diagram are as follows.

System Parameters

Table 1: Gyroscope Inertia Values

Table 2: Encoder Gains

Table 3: Control Effort Gains

Specifications

All closed-loop poles must be in the left half of the complex plane
The peak time in response to a unit step command input must be less than 0.2 sec.
The 2% settling time to a unit step command input must be less than 0.5 sec.

Procedure

Develop a SIMULINK diagram of the block diagram shown in Fig. 3.6 and save it for use in part C of the experiment. Use two separate transfer functions as shown in Fig 3.6.
Obtain the SIMULINK response to a unit step command using the set of gains you have selected. Assess the SIMULINK step response using stepinfo() command. If the response does not meet the design requirements, then modify the gains until it does. Save the response, and make sure it proves you meet the requirements.
The TAs will need to see: Gains, controlSystemDesigner/rltool plots, SIMULINK diagram, SIMULINK response to a step command (proving peak and settling time requirements are met), and the 3 modified responses.

You must have your work checked out by one of the TAs before leaving the lab to get credit for your work.

Part C: Controller Implementation & Evaluation

Safety Briefing

Safety Note 1: In the event of an emergency, control effort should be immediately discontinued by pressing the red “OFF” button on the front of the control box.

Safety Note 2: Stay clear of and do not touch any part of the mechanism while the rotor is moving, while a trajectory is being executed, or before the active controller has been safety checked.

Safety Note 3: The rotor should never be operated at speeds above 825 RPM. The user should take precautions to assure that this limit is not exceeded. If this accidentally happens, hit "Abort Control" in the ECP software and re-initiate rotor.

Safety Note 4: Never leave the system unattended while the ECP control box is powered on.

Experimental Setup

These steps should be performed primarily by the TA, but the students should follow along to gain a deeper understanding of how the experiment works.

Functionality Test

This test ensures if everything is sound mechanically, electrically, and in the software.

Hardware checks
1. Ensure the rotor cover panel is secure.
2. Switch on the ECP control box and disengage Axis 3 and Axis 4 brakes.
3. Ensure all axes can turn freely with no significant friction felt by hand.
4. Ensure the two Axis 2 limit switches function correctly (with rotor not turning) by checking auto-shutoff occurs when manually pressing each of them. Switch the ECP control box back on.
5. Ensure Axis 3 and Axis 4 inertial switches function correctly (with rotor not turning) by aggressively turning the appropriate axis by hand to initiate auto-shutoff. Switch the ECP control box back on.
Software checks
1. Launch E2Usr32 - Model750, the ECP CMG control software, a shortcut for which can be found on the Desktop. Alternatively, the executable can be found at C:\Program Files (x86)\ECP Systems_MV\mv
2. If the program doesn't launch, make sure it is being executed in Windows XP SP3 compatibility mode
3. Go to Setup > Communications > PMAC 00 - PCIO - Plug and play > Test and check that the dialog confirms successful PMAC connection. If it does not, the PCI card is likely not seated in the PC correctly. Have the TA try wiggling the ribbon cable in the back of the PC to get a good connection and re-try communication check. If nothing works, have a TA contact the Lab Manager.
4. If Apparatus is not listed as "C.M.G. Model 750" go to Utility > Download Controller Personality File and load "gyro20.pmc" from C:/Program Files (x86)/ECP Systems_MV/mv/.
5. Load Configuration file "default.cfg" from C:\Program Files (x86)\ECP Systems_MV\mv.
6. Go to Utility > Reset Controller and check the following values on the main screen have been set:
  1. Encoders 1/2/3/4 Pos: 0 COUNTS
  2. Control Loop Status: OPEN
  3. Motor 1 Status: OK
  4. Motor 2 Status: OK
  5. Servo Time Limit: OK
7. Ensure the 4 encoders are reading correctly by manually turning each axis and watching the corresponding encoder counts value change on the main screen (you may want to zero the encoders prior to this step to see it more clearly). Note that Encoder 1 is the rotor, which is hard to move by hand since it is enclosed, so aggressively rotate Axis 3 to let the rotor inertia create some encoder position change.

Orientation setting

With the ECP control box powered on, all brakes off, and flywheel not turning, roughly orient the gyroscope according to Figure 3.7 (as viewed from the front such that labels are the correct way up).
Apply all brakes and turn off the ECP control box.
Ensure Axis 3 is parallel with the platform base by using a spirit level or phone app, making small adjustments by hand (with the brakes on).
Ensure Axis 2 is perpendicular to the platform base and Axis 3 by using the spirit level app of a phone app, making small adjustments by hand (with the brakes on).
It is not critical where Axis 4 is located since the controller will be doing a relative angle maneuver.

Controller Setup

Configure Data Logging

Go to the Data menu and click on Setup Data Acquisition.
Be sure that the selected variables to log match those below. When finished, hit OK to exit this menu.

Initialize Rotor

After these steps, the rotor will be turning, causing a potential snag hazard. Ensure a safe distance between the gyroscope and all personnel. Never touch any part of the gyroscope.

The motor must never exceed 800 RPM!

Switch on the ECP control box.
Brake all axes so that the accelerating rotor does not cause the axes to move from the initialized positions:
1. Set Axis #3 and Axis #4 on the ECP control box to ON.
2. Set Axis #2 V-Brake on the ECP software main screen to ON.
Click on the Command menu and then on Initialize Rotor Speed.
Being careful not to make a typo, input 400 RPM and click OK. If you do accidentally input a speed higher than intended, click Abort Control on the main screen, go back to the Initialize Rotor Speed input and input the correct RPM.
Once the rotor has reached the desired RPM, release Axis 2 and Axis 4 brakes. Axis 3 should remain braked throughout the entirety of the lab.
Observe gyroscopic precession and nutation in open-loop using a ruler or similar object to perturb the system about Axis #4:
1. Apply a constant torque to Axis #4 by pushing on the outer part of the frame. You should notice that the gyro rotates about Axis #2 at a constant angular rate. Contrast this with a constant angular acceleration that would have occurred if the rotor were not turning. This is known as gyroscopic precession.
2. Apply a momentary torque to Axis #4 by sharply jabbing the outer part of the frame. You should notice that the gyro slightly "wobbles" about Axis #2. This is known as gyroscopic nutation. In the absence of damping from friction in the bearings and air resistance due to the angular motion, this wobble would last indefinitely.
3. Return Axis #2 to its vertical position by torquing Axis #4.

Configure Control Algorithm

Go to the Setup menu and clicking on Control Algorithm.
Click the Load From Disk button and select the default controller file, GyroControl_Gimbal4_PID.alg. The original file can be found at C:\AE4610-ControlsLab\Gyro_Controller but, since this version is read-only, it should be loaded from a locally saved copy on the TA's login profile.
1. Check the values with a TA before entering them.
3. Change values by clicking the Edit Algorithm button, editing the relevant section of code, and exiting the edit algorithm section by selecting Save Changes and Quit under the File menu.
Enable the algorithm by clicking Implement Algorithm.
Click OK, and check that the Control Loop Status now reads CLOSED on the main screen.

Controller Evaluation

With the rotor initialized and the closed-loop controller active, you will now evaluate your calculated gains in three different trajectories.

Trajectory 1: Small Step Input

Select the Command menu and then Trajectory 1.
Select the Step Input and then click the Setup button. Enter the following parameters for this case:
1. Step size = 200 counts.
2. Dwell time = 1000 ms.
3. 1 repetition.
Hit OK, then OK again to leave the setup trajectory menus.
Ensure the rotor is oriented vertically using the wooden ruler.
Ensure all people are at a safe distance from the gyroscope.
Go to the Command menu and click Execute. Make sure that both Normal Data Sampling and Execute Trajectory 1 Only are checked.
Click the Run button. The input trajectory will be run on the gyroscope now. When the box on the screen says Upload Complete, click on the OK button.
Plot and review the data:
1. Go to the Plotting menu and select Setup Plot. Set Command Position 1 and Encoder 4 Position on the left axis and Control Effort 2 on the right axis. Click OK when finished to view the plot.
2. If desired, zooming the plot can be accomplished by going to the Plotting menu and selecting Axis Scaling.
3. Discuss the results, particularly the quality of your controller's response and how it may be improved, with your TAs. Given enough time at the end of the lab, you may return to this trajectory and see how gain changes may improve the controller's response.
Save the data by clicking on Data > Export Raw Data.

Trajectory 2: Large Step Input

Repeat steps 1-8 from Trajectory 1, but this time using the following trajectory settings:
1. Step size = 1000 counts.
2. Dwell time = 1000 ms.
3. 1 repetition.
Save the data by clicking on Data > Export Raw Data.

Trajectory 3: Ramp Input

The step response is useful for system characterization but is seldom used for an actual in-service trajectory because it is excessively harsh (high acceleration and jerk ("rate of change of acceleration")). A more common trajectory used for tracking applications is a ramp.

Repeat steps 1-8 from Trajectory 1, this time using the following trajectory settings:
1. Ramp input (Unidirectional Moves not checked).
2. Distance = 6000 counts.
3. Velocity = 2000 counts/sec.
4. Dwell Time = 1000 ms.
5. 2 repetitions.
Save the data by clicking on Data > Export Raw Data.
In addition to the previous plots, plot Control Effort 2 and Encoder 2 Velocity data. (This should be done in separate plots since the scaling of these two variables is greatly different.) Note the relatively close tracking and rapid accelerations at each end of the constant velocity sections. This would not be possible for the small actuator of Axis #2 acting on the massive assembly in a conventional fashion. By using gyroscopic control actuation and the associated transfer of momentum stored in the rotor, the high authority control is made possible.

Trajectory 1 Re-Run

Only if time permits at the end of your session, go back to Trajectory 1 and see if you can improve the controller's response by manually adjusting the gains using your intuition and knowledge of individual gain effects. This task is intended only to allow you an opportunity to get a real-life feel for the effect of gain changes, it will not be graded.

Analysis & Lab Report

Include an overview of the controller design from Part B.
Using the SIMULINK block diagram you have developed of Fig. 3.6 in Part B, simulate the system response for the same step and ramp commands you have used in Part C. Compare your simulated responses with the corresponding experimental results. What is the rise time and % overshoot in the experiment and simulation responses to the step command used?

C. Position Control (Week 2)

1 Background

1.1 Desired Position Control Response

The output of this system can be written as:

As you can see from this equation, the plant is a second-order system. In fact, when a second-order system is placed in series with a proportional compensator in the feedback loop as in Figure 16, the resulting closed-loop transfer function can be expressed as:

1.1.1 Peak Time and Overshoot

Consider a second-order system as shown in Equation 3.4 subject to a step input given by

This is called the peak time of the system.

In a second-order system, the amount of overshoot depends solely on the damping ratio parameter and it can be calculated using the equation

The peak time depends on both the damping ratio and natural frequency of the system and it can be derived as:

Generally speaking, the damping ratio affects the shape of the response while the natural frequency affects the speed of the response.

1.1.2 Steady State Error

We can find the steady-state error of this system using the final-value theorem:

Applying the final-value theorem gives

When evaluated, the resulting steady-state error due to a step response is

Based on this zero steady-state error for a step input, we can conclude that the Rotary Servo Base Unit is a Type 1 system.

1.1.3 Time-Domain Control Specifications

The desired time-domain specifications for controlling the position of the Rotary Servo Base Unit load shaft are:

and

Thus, when tracking the load shaft reference, the transient response should have a peak time less than or equal to 0.20 seconds, an overshoot less than or equal to 5%, and the steady-state response should have no error.

1.2 PD Controller Design

1.2.1 Closed Loop Transfer Function

The proportional-derivative (PD) compensator to control the position of the Rotary Servo Base Unit has the following structure

This is a variation of the classic PD control where the D-term is in the feedback path as opposed to in the forward path.

From the Plant block in Figure 19 and Equation 3.3, we can write

1.2.2 Controller Gain Limits

The limitations of the actuator must be taken into account when designing a controller. For instance, the voltage entering the Rotary Servo Base Unit motor should never exceed

1.2.3 Ramp Steady State Error Using PD Control

From our previous steady-state analysis, we found that the closed-loop Rotary Servo Base Unit system is a Type 1 system. In this section, we will investigate the steady-state error due to a ramp input when using PD controller.

Given the following ramp setpoint (input)

1.3 PID Controller Design

Adding integral control can help eliminate steady-state error. The proportional-integral-derivative (PID) algorithm to control the position of the Rotary Servo Base Unit is shown in Figure 1.6. The motor voltage will be generated by the PID according to:

This is a variation of the standard PID control with the D-term in the feedback path as opposed to the feedforward path.

From the Plant block in Figure 21 and Equation 3.3, we can write

1.3.1 Ramp Steady-State Error using PID Controller

To find the steady-state error of the Rotary Servo Base Unit for a ramp input under the control of the PID substitute the closed-loop transfer function from Equation 3.31 into Equation 3.11

1.3.2 Integral Gain Design

It takes a certain amount of time for the output response to track the ramp reference with zero steady-state error. This is called the settling time and it is determined by the value used for the integral gain.

In steady-state, the ramp response error is constant. Therefore, to design an integral gain the velocity compensation (the V signal) can be neglected. Thus, we have a PI controller left as:

When in steady-state, the expression can be simplified to

2 Pre-Lab Questions

Do problems 1 - 7

↓↓↓ In Lab Exercise ↓↓↓

The main goal of this laboratory is to explore position control of the Rotary Servo Base Unit load shaft using PD and PID controllers. In this laboratory, you will conduct the following experiments:

Step response with PD controller,
Ramp response with PD controller, and
Ramp response with PID controller.

In each experiment, you will first simulate the closed-loop response of the system. Then, you will implement the controller using the Rotary Servo Base Unit hardware and software to compare the real response to the simulated one.

Position Control MATLAB/SIMULINK Files

3.1 Step Response using PD Controller

3.1.1 Simulation

First, you will simulate the closed-loop response of the Rotary Servo Base Unit with a PD controller to step input. Our goals are to confirm that the desired response specifications in an ideal situation are satisfied and to verify that the motor is not saturated. Then, you will explore the effect of using a high-pass filter, instead of a direct derivative, to create the velocity signal in the controller.

Simulation Setup

Open the setup_servo_pos_cntrl.m file to open the setup script for the position control Simulink models.
Replace the default model parameters (K and tau) with the values calculated from the Modelling section in the setup script.
Run the script, which will generate the model parameters, specifications, and default PID gains. Note: The calculated PID gains are all set to zero by default and they need to be changed over the course of the experiment.

Simulation: Closed-loop Response with PD Controller

Enter the proportional and derivative control gains found in Pre-Lab Question 4 in the Matlab command window as kp and kd.
Set the integral gain to 0 in the Matlab window, denoted as ki.
To generate a step reference, ensure the Signal Generator is set to the following:
- Signal type = square
- Amplitude = 1
- Frequency = 0.4 Hz
Set the Manual Switch such that the velocity of the motor load shaft is fed back directly.
Open the servo position Position (rad) scope and the motor input voltage Vm (V) scope.
Start the simulation. By default, the simulation runs for 5 seconds. The scopes should be displaying responses similar to Figure 23a and Figure 23b. Note that in the theta_l (rad) scope, the yellow trace is the setpoint position while the purple trace is the simulated position (generated by the Rotary Servo Base Unit Model block). This simulation is called the Ideal PD response as it uses the PD compensator with the derivative block.
After each simulation run, each scope automatically saves its response to a variable in the MATLAB workspace. That is, the Position (rad) scope saves its response to the variable called data_pos and the Vm (V) scope saves its data to the data_vm variable. The data_pos variable has the following structure: data_pos(:,1) is the time vector, data_pos(:,2) is the setpoint, and data_pos(:,3) is the simulated angle. For the data_vm variable, data_vm(:,1) is the time and data_vm(:,2) is the simulated input voltage.
Save Vm and data_pos data and name it suitably (e.g. posnctrl_section#_Group#_step). Result: Generate a MATLAB figure showing the Ideal PD position response and the ideal input voltage.

Simulation: Using a Filtered Derivative

When implementing a controller on actual hardware, it is generally not advised to take the direct derivative of a measured signal. Any noise or spikes in the signal becomes amplified and gets multiplied by a gain and fed into the motor which may lead to damage. To remove any high-frequency noise components in the velocity signal, a low-pass filter is placed in series with the derivative, i.e. taking the high-pass filter of the measured signal. However, as with a controller, the filter must also be tuned properly. In addition, the filter has some adverse effects (ex. filter could add some delays to the system).

Set the Manual Switch block to the down position to enable the derivative and filter.
Start the simulation. The response in the scopes should still be similar to Figure 23a and Figure 23b. This simulation is called the Filtered PD response as it uses the PD controller with the high-pass filter block. Result: Generate a MATLAB figure showing the Filtered PD position and input voltage responses.

3.1.2 Experiment

In this experiment, we will control the angular position of the Rotary Servo Base Unit load shaft, i.e. the disc load, using the PD controller. Measurements will then be taken to ensure that the specifications are satisfied.

Experimental Setup

From the extracted files, open q_servo_pos_cntrl Simulink file.
Configure DAQ: Double-click on the HIL Initialize block in the Rotary Servo Interface subsystem (which is located inside the Rotary Servo Interface - Position subsystem) Simulink diagram and ensure it is configured for the DAQ device that is installed in your system (e.g. Q2-USB).
Open setup_servo_pos_cntrl.m file and run the script.

Experiment

Enter the proportional and derivative control gains found in Pre-Lab Question 4.
Set Signal Type in the Signal Generator block to square to generate a step reference.
Set the Amplitude (rad) gain block to 0.4 to generate a step with an amplitude of 45.8 degrees.
Open the load shaft position scope, Position (rad), and the motor input voltage scope, Vm (V).
When a suitable response is obtained, click on the Stop button in the SIMULINK diagram toolbar to stop running the code. As in the s_servo_pos_cntrl SIMULINK diagram, when the controller is stopped each scope automatically saves its response to a variable in the MATLAB workspace. Thus the theta_l (rad) scope saves its response to the data_pos variable and the Vm (V) scope saves its data to the data_vm variable.
Save the data in the MATLAB workspace with a suitable name. Result: Generate a MATLAB figure showing the PD position response and its input voltage.
Click the Stop button on the SIMULINK diagram toolbar to stop the experiment.

3.2 Ramp Response using PD Controller

3.2.1 Simulation

In this simulation, the goal is to verify that the system with the PD controller can meet the zero steady-state error specification without saturating the motor.

Clear the workspace and run the setup_servo_pos_cntrl.m script in MATLAB.
Enter the proportional and derivative control gains found in Pre-Lab Question 4 and set the integral gain to 0.
Set the Signal Generator parameters to the following to generate a triangular reference (which corresponds to a ramp input):
- Signal Type = triangle
- Amplitude = 1
- Frequency = 0.8 Hz
Inside the PID Control subsystem, set the Manual Switch to the down position so that the High-Pass Filter block is used.
Open the load shaft position scope, Position (rad), and the motor input voltage scope, Vm (V).
Start the simulation. The scopes should display responses similar to Figure 26a and Figure 26b.
Save the data in the workspace using a suitable name. Result: Generate a MATLAB figure showing the Ramp PD position response and its corresponding input voltage trace.
Measure the steady-state error. Result: Compare the simulation measurement with the steady-state error calculated in Pre-Lab Question 6.

3.2.2 Experiment

In this experiment, we will control the angular position of the Rotary Servo Base Unit load shaft, i.e. the disc load, using a PD controller. The goal is to examine how well the system can track a triangular (ramp) position input. Measurements will then be taken to ensure that the specifications are satisfied.

Clear the workspace and run the setup_servo_pos_cntrl.m script in MATLAB.
Enter the proportional and derivative control gains found in Pre-Lab Question 4.
Set the Signal Generator parameters to the following to generate a triangular reference (i.e. ramp reference):
- Signal Type = triangle
- Amplitude = 1
- Frequency = 0.8 Hz
Open the load shaft position scope, Position (rad), and the motor input voltage scope, Vm (V).
Save the data in the workspace using a suitable name. Result: Generate a MATLAB figure showing the Ramp PD position response and its corresponding input voltage trace.
Measure the steady-state error. Result: Compare it with the steady-state error calculated in Pre-Lab Question 6.

3.3 Ramp Response with No Steady-State Error

Design an experiment to see if the steady-state error can be eliminated when tracking a ramp input. First, simulate the response, then implement it using the Rotary Servo Base Unit system.

List the independent and dependent variables of your proposed controller. Explain their relationship.
Your proposed controller, like the PD compensator, is a model-based controller. This means that the control gains generated are based on a mathematical representation of the system. Given this, list the assumptions you are making in this control design. State the reasons for your assumptions.
For each case, generate a Matlab figure showing the position response of the system and its corresponding input voltage.
In each case, measure the steady-state error.

4 Results for the Report

Pre-lab calculations.

Week 1

Objective

The objectives of this laboratory experiment are as follows:

Obtain the linear state-space representation of the rotary pendulum plant.
Design a state-feedback control system that balances the pendulum in its upright position using Pole Placement.
Simulate the closed-loop system to ensure the given specifications are met.

Equipment

Rotary pendulum (Quanser Rotary inverted pendulum module)
Rotary servo base (Quanser SRV02)
Power amplifier (Quanser VoltPaq-X2)
DAQ (Quanser Q2-USB)
MATLAB and SIMULINK

Week 1

A. Modeling

This experiment involves modeling of the rotary inverted pendulum.

Rotary Inverted Pendulum Model

Table 1 Rotary Pendulum components (Figure 1)

ID#

Component

Rotary Servo

Thumbscrews

Rotary Arm

Shaft Housing

Shaft

Pendulum T-Fitting

Pendulum Link

Pendulum Encoder Connector

Pendulum Encoder

Table 2 Main Parameters associated with the Rotary Pendulum module

Symbol

Description

Value

Unit

Mass of pendulum

0.127

Total length of pendulum

0.337

Distance from pivot to center of mass

0.156

Pendulum moment of intertia about center of mass

0.0012

Pendulum viscous damping coefficient as seen at the pivot axis

0.0024

Mass of rotary arm with two thumb screws

0.257

Rotary arm length from pivot to tip

0.216

Rotary arm length from pivot to center of mass

0.0619

Rotary arm moment of inertia about its center of mass

9.98 x 10^-4

Rotary arm viscous damping coefficient as seen at the pivot axis

0.0024

Rotary arm moment of inertia about pivot

0.0020

Pendulum encoder resolution

4096

Model Convention

The rotary inverted pendulum model is shown in Figure 2. The rotary arm pivot is attached to the Rotary Servo system and is actuated. The arm has a length of , a moment of inertia of , and its angle, $\bm\theta$ , increases positively when it rotates counterclockwise (CCW). The servo (and thus the arm) should turn in the CCW direction when the control voltage is positive, i.e. > 0.

The pendulum link is connected to the end of the rotary arm. It has a total length of $L_p$ and it center of mass is . The moment of inertia about its center of mass is . The inverted pendulum angle, $\bm\alpha$ , is zero when it is perfectly upright in the vertical position and increases positively when rotated CCW.

Nonlinear Equations of Motion

Specifically, the equations that describe the motions of the rotary arm and the pendulum with respect to the servo motor voltage, i.e., the dynamics, will be obtained using the Euler-Lagrange equation:

The variables $\bm q_i$ are called generalized coordinates. For this system let

where, as shown in Figure 2, $\theta(t)$ is the rotary arm angle and $\alpha(t)$ is the inverted pendulum angle. The corresponding velocities are

With the generalized coordinates defined, the Euler-Lagrange equations for the rotary pendulum system are

The Lagrangian of the system is described by

where $T$ is the total kinetic energy of the system and $V$ is the total potential energy of the system. Thus the Lagrangian is the difference between a system’s kinetic and potential energies.

The generalized forces $Q_i$ are used to describe the nonconservative forces (e.g. friction) applied to a system with respect to the generalized coordinates. In this case, the generalized force acting on the rotary arm is

and acting on the pendulum is

Once the kinetic and potential energy are obtained and the Lagrangian is found, then the task is to compute various derivatives to get the EOMs. After going through this process, the nonlinear equations of motion for the Rotary Pendulum are:

Linearization

z = \begin {bmatrix} z_1 & z_2\end {bmatrix}^T

about the point

z_0 = \begin {bmatrix} a & b\end {bmatrix}^T

can be written as

f_{lin}(z) = f(z_0) + \frac{\partial f(z)}{\partial z_1}\Bigr|_{z = z_0} (z_1 - a)+\frac{\partial f(z)}{\partial z_2}\Bigr|_{z=z_0}(z_2 - b)

Linearization of Inverted Pendulum Equations

The nonlinear equations of the inverted pendulum system obtained as Equations (7) and (8) are linearized about the equilibrium point with the pendulum in the upright position, i.e.,

$\theta_o = 0$ , $\alpha_o = 0$ , $\dot{\theta}_o = 0$ , $\dot{\alpha}_o = 0$ , $\ddot{\theta}_o = 0$ , $\ddot{\alpha}_o = 0$

Linearization of Equation (7) about the above equilibrium state gives

(J_r + m_pL_r^2)\ddot{\theta} - \frac{1}{2}m_pL_pL_r\ddot{\alpha} = kV_m-b\dot{\theta} - B_r\dot{\theta} \quad\quad (10)

where, from Equation(9) the servo motor torque coefficient $k$ is

k = \frac{\eta_gK_g\eta_mk_t}{R_m}

and the back emf coefficient $b$ is

b = \frac{\eta_gK_g^2\eta_mk_tk_m}{R_m}

Likewise, linearization of Equation (8) about the equilibrium point with the pendulum in the upright position gives

-\frac{1}{2}m_pL_pL_r\ddot{\theta} + (J_p + \frac{1}{4}m_pL_p^2)\ddot{\alpha} - \frac{1}{2}m_pL_pg\alpha= -B\dot{\alpha} \quad \quad (11)

where $g$ is the acceleration due to gravity. Note that the negative sign of the gravity torque ( $\alpha$ term) in Equation 11 indicates negative stiffness with the pendulum in the vertically upright position.

Equation (10) and (11) can be arranged in the matrix form as

[M]\begin{bmatrix}\ddot{\theta} \\ \ddot{\alpha} \end{bmatrix} +[D] \begin{bmatrix}\dot{\theta} \\ \dot{\alpha} \end{bmatrix} + [K] \begin{bmatrix}\theta \\ \alpha \end{bmatrix} = \begin{bmatrix}k \\ 0 \end{bmatrix}V_m \quad \quad (12)

Where mass matrix $M$ , damping matrix $D$ , and the stiffness matrix $K$ become

[M] = \begin{bmatrix} (J_r + m_pL_r^2) & -\frac{1}{2}m_pL_pL_r \\ -\frac{1}{2}m_pL_pL_r & (J_p + \frac{1}{4}m_pL_p^2) \end{bmatrix}

[D] = \begin{bmatrix} b+B_r & 0 \\ 0 & B_p \end{bmatrix}

[K] = \begin{bmatrix} 0 & 0 \\ 0 & -\frac{1}{2}m_pL_pg \end{bmatrix}

Defining the state vector $x$ , output vector $y$ and the control input $u$ as

x = \begin{bmatrix} \theta & \alpha & \dot{\theta} & \dot{\alpha}\end{bmatrix}^T , \quad y = \begin{bmatrix} \theta & \alpha \end{bmatrix}^T, \quad u = V_m

Equation (12) can be rewritten in state space form as

\dot{x} = Ax + Bu \quad \quad (13)

y = Cx + Du \quad \quad (14)

where the system state matrix $A$ , control matrix $B$ , output matrix $C$ and the feedthrough of input to the output matrix $D$ become

A = \begin{bmatrix} O_{2\times2} & I_{2\times2} \\ -[M]^{-1}[K] & -[M]^{-1}[D] \end{bmatrix}

B = \begin{bmatrix} O_{2\times1} \\ M^{-1} \begin{bmatrix} k \\0 \end{bmatrix} \end{bmatrix}

C = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix}

D = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

Analysis: Modeling

Download the above zip file and extract it.
Note down your state-space matrices for your report. Note: You may want to cross-check the state-space matrix with TAs before proceeding to balance control.
Find the open-loop poles of the system. Hint: Use eig(A).

B. Balance Control

Specification

The control design and time-response requirements are: Specification 1: Damping ratio: 0.6 < $\zeta$ < 0.8 Specification 2: Natural frequency: 3.5 rad/s < < 4.5 rad/s Specification 3: Maximum pendulum angle deflection: $|\alpha|$ < 15 deg. Specification 4: Maximum control effort / voltage: < 10 V. The necessary closed-loop poles are found from specifications 1 and 2. The pendulum deflection and control effort requirements (i.e. specifications 3 and 4) are to be satisfied when the rotary arm is tracking a $\pm 20$ degree angle square wave.

Stability

The stability of a system can be determined from its poles ([2]):

Stable systems have poles only in the left-hand plane.
Unstable systems have at least one pole in the righthand plane and/or poles of multiplicity greater than 1 on the imaginary axis.
Marginally stable systems have one pole on the imaginary axis and the other poles in the left-hand plane.

The poles are the roots of the system’s characteristic equation. From the state-space, the characteristic equation of the system can be found using where $\mathrm{det}()$ is the determinant function, $s$ is the Laplace operator, and $I$ the identity matrix. These are the eigenvalues of the state-space matrix $A$ .

Controllability

If the control input $u$ of a system can take each state variable, $x_i$ where $i = 1 ... n$ , from an initial state to a final state in finite time then the system is controllable, otherwise it is uncontrollable ([2]).

Rank Test The system is controllable if the rank of its controllability matrix

equals the number of states in the system,

Companion Matrix

If (A,B) are controllable and B is n×1, then A is similar to a companion matrix ([1]). Let the characteristic equation of A be Then the companion matrices of A and B are

and

Define where $T$ is the controllability matrix defined in Equation 16 and Then and

Pole Placement Theory

If $(A,B)$ are controllable, then pole placement can be used to design the controller. Given the control law $u = -Kx$ , the state-space model of Equation (13) becomes

We can generalize the procedure to design a gain $K$ for a controllable $(A,B)$ system as follows:

Step 1 Find the companion matrices $\tilde{A}$ and $\tilde{B}$ . Compute $W = T\tilde{T}^{-1}$ . Step 2 Compute $\tilde{K}$ to assign the poles of $\tilde{A} - \tilde{B}\tilde{K}$ to the desired locations.

Step 3 Find $K = \tilde{K}W^{-1}$ to get the feedback gain for the original system $(A,B)$ . Remark-1: It is important to do the $\tilde{K} \rightarrow K$ conversion. Remember that $(A,B)$ represents the actual system while the companion matrices $\tilde{A}$ and $\tilde{B}$ do not.

Remark-2: The entire control design procedure using the pole placement method can be simply done in MATLAB using the function called 'place' or 'acker'. For a selected desired set of closed loop poles DP, the full state feedback gain matrix $K$ is obtained from

>> K = acker(A,B,DP);

Desired Poles

Simulation Model with Feedback

The feedback control loop that balances the rotary pendulum is illustrated in Figure 4. The reference state is defined as where $\theta_d$ is the desired rotary arm angle. The controller is Note that if $x_d = 0$ then $u = -Kx$ , which is the control used in the pole-placement algorithm.

B.1 Experiment: Designing the Balance Control

Select $\zeta$ and $\omega_n$ such that they satisfy the given design specifications.
In the same rotpen_week1_student.mlx live script, go to the Balance Control section. Enter the chosen zeta and omega_n values.
Determine the locations of the two dominant poles $p_1$ and $p_2$ based on the specifications and enter their values in the MATLAB live script. Ensure that the other poles are placed at p3 = -30 and p4 = -40. Hint: Use Equation 27.
Find gain K using a predefined Compensator Design MATLAB command K = acker(A,B,DP), which is based on pole-placement design. Note: DP is a row vector of the desired poles found in Step 3.

For sanity check, if you use damping ratio of 0.7 and natural frequency of 4 rad/s, you should get around K = [-12 63 -5.5 7].

B.2 Experiment: Simulating the Balance Control

Run rotpen_week1_student.mlx live script. Ensure the gain K you found is loaded in the workspace (type K matrix in the command window).
Open and run the s_rotpen_bal.mdl for 10 seconds. The responses in the scopes shown in Figure 6 were generated using an arbitrary feedback control gain. Note: When the simulation stops, the last 10 seconds of data is automatically saved in the MATLAB workspace to the variables data_theta, data_alpha, and data_Vm.
Save the data corresponding to the simulated response of the rotary arm, pendulum, and motor input voltage obtained using your obtained gain K. Note: The time is stored in the data_theta(:,1) vector, the desired and measured rotary arm angles are saved in the data_theta(:,2) and data_theta(:,3) arrays. Similarly, the pendulum angle is stored in the data_alpha(:,2) vector, and the control input is in the data_Vm(:,2) structure.

Results for Report

A) Modeling

The linear state-space representation of the rotary inverted pendulum system, i.e., $A$ , $B$ , $C$ and $D$ matrices (numerical values).
Open-loop poles of the system.

B.1) Balance Control Design

Chosen $\zeta$ and $\omega_n$ based on design specifications.
Corresponding locations of the two dominant poles $p_1$ and $p_2$ .
Gain vector $K$ .

B.2) Balance Control Simulation

Plots of the commanded position of the rotary arm ( $\theta_c$ ), simulated responses of the rotary arm ( $\theta$ ), pendulum ( $\alpha$ ), and motor input voltage ( $V_m$ ) obtained using your obtained gain K.

Questions for Report

B.1) Balance Control Design

Determine the controllability matrix $T$ of the system. Is the inverted pendulum system controllable? Hint: Use Equation 17.
Using the open-loop poles, find the characteristic equation of $A$ . Hint: The roots of the characteristic equation are the open-loop poles.
Determine the controllability matrix $\tilde{T}$ of the companion system.
Determine the transformation matrix $W$ .
Check if $\tilde{A} = W^{-1}AW$ and $\tilde{B} = W^{-1}B$ with the obtained matrices.