# Lecture 14 - Using the MATLAB Control System Toolbox and Simulink Friday, February 8, 2013

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1 Today s Objectives ENGR 105: Feedback Control Design Winter 2013 Lecture 14 - Using the MATLAB Control System Toolbox and Simulink Friday, February 8, introduce the MATLAB Control System Toolbox 2. introduce Simulink 3. solve problems in class using the toolbox Reference: FPE Appendix C, MATLAB website Note: To use your computer for this lecture, you will need: MATLAB, Simulink, and the Control System Toolbox. All of these come with the Student Edition of MATLAB. 1 Control System Toolbox The Control System Toolbox is a collection of algorithms, written mostly as m-files, that implements common control system design, analysis, and modeling techniques. Convenient graphical user interfaces (GUIs) simplify typical control engineering tasks. The command help control gives a list of the different functions. The most commonly used functions are presented below. You are encouraged to look through the help files for other functions and options that you might find useful in future. 1.1 Transfer Functions Control systems can be modeled as transfer functions, in zero-pole-gain form, or state-space form (take ENGR 205 to learn the latter). Transfer function models are created using the function tf(numerator,denominator), where numerator and denominator are two vectors containing the coefficients of the polynomials in the numerator and denominator of the transfer function. >> num = [1 2 0]; >> den = [2 2 1]; >> sys = tf(num,den) Transfer function: s^2 + 2 s s^2 + 2 s + 1 Some text/examples in this document Mathworks. Some materials also obtained from other universities, including UC Berkeley, USC, CMU, and UMich. 1

2 Zero-Pole-Gain forms of transfer functions, ie, G(s) = K(s z 1)(s z 2 )... (s p 1 )(s p 2 )... can be specified directly as zpk(z,p,k) where z and p are the vectors of zero and pole values in the complex plane (z= [ z1 z2...] and p = [p1 p2...]), and K is the gain. The equivalent to the above system is: >> K = 0.5; >> z = [0-2]; >> p = [ *i *i]; >> zpk(z,p,k) Zero/pole/gain: 0.5 s (s+2) (s^2 + s + 0.5) The above models can be used in conjunction with the operators +, - and * to generate new models. The command sys1 * sys2 produces a series interconnection from input through sys2 and sys1 (in that order) to the output. sys1 ± sys2 represent parallel interconnections. Control system models can also be converted from one form to the other. The functions presented above (tf(), zpk()) are overloaded to perform arbitrary system conversions. For example, if sys1 is a state-space model, we can generate an equivalent transfer function model sys2 by issuing the command >> sys2 = tf(sys1); The function feedback(sys1,sys2) will help you compute a closed-loop transfer function for system consisting of the negative feedback interconnection of model objects sys1 (in the forward path) and sys2 (in the feedback path). 1.2 Time Domain Analysis Time domain analysis in the form of time response of a system to a specified input can be obtained using the following commands: step(sys) plots the step response of a system sys. Also, check out stepinfo for information about common performance metrics. impulse(sys) plots the impulse response of sys. lsim(sys,input_vector,time_vector,initial_state_vector) plots the response of sys to an arbitrary input. The elements of input_vector define the values of the input at times corresponding to the elements of time_vector. The initial state vector can be specified for state-space models only (so, not used in the course). pidtool(sys,type) launches the PID Tuner GUI and designs a controller of type type (which can take on values such as p, pi, and pid ) for plant sys. damp and/or roots (a standard MATLAB function) can also be useful in determining the poles of a system. 2

5 4 Examples and MATLAB Challenge Solutions 4.1 Introductory Material Sample code for the Transfer Function section is in lecture14_code.m. Simulink model for notch filter example is in notch_filer.mdl. But don t show the students the completed model, rather, lead them through its creation. After you drag the elements and use interconnections to create the Simulink model, take the following steps: 1. For the transfer function, assign the numerator coefficients [1 0 (100*2*pi)^2] and the denominator coefficients [1 2*1*100*2*pi (100*2*pi)^2] 2. Change the sine wave parameters so that the the frequency is 1*2*pi for the 1 Hz example 3. Change the simulation configuration parameters to have a stop time of Run the simulation and look at the scope for 1Hz. You can see that the input signal (yellow) feeds right through to the output (magenta) 5. Change the sine wave parameters so that the the frequency is 100*2*pi for the 100 Hz example. You can see that the input signal is completely attenuated the output is zero. The output also looks choppy because the simulation time step is too low. 6. Change the sine wave parameters so that the the frequency is 1000*2*pi for the 1000 Hz example. There are many ways to deal with the time step issue, but here you can just change the stop time to Run the simulation and you can see that the input signal feeds through again (albeit a bit delayed). 4.2 Challenge 1 Prompt: You have a plant G(s) = s+2. You controller is of the form D(s) = K. Find the value s 2 3s of K that stabilizes this system using a simulation-based approach (i.e., test different values of K and see what happens). Note that we do not want a marginally stable system. We suggest that you use a unit step reference input. Answer: K > 3 Solution A: Control System Toolbox (command-line) solution is in challenge1_cst_solution.m. Solution B: Simulink solution is in challenge1_simulink.mdl (image below). Note that it isn t necessary to feed the step input to the scope, I just like to. Step 10 Gain s+2 s 2 3s Transfer Fcn Scope 5

6 4.3 Challenge 2 Prompt: Here is the equation of motion for a damped pendulum: θ + c θ ml + g Tc l sin θ =. The ml 2 input is T c and the output is θ. Choose l = 2.5, m = 0.75, c = 0.15, and g = 9.8. Say we place this plant in a negative feedback system with PD controller (with gains K p = 100 and K d = 10) in the forward path. Consider the response of the system to a step of 45 degrees. To what value does the output converge? Answer: About 38 degrees for the linearized case, and a little more for the nonlinear case. These are very approximate numbers, read off the plots. The point is that the nonlinear version responds a little differently. Solution A: Control System Toolbox (command-line) solution is in challenge2_cst_solution.m. Solution B: Simulink solutions are in challenge2_simulink_linear.mdl and challenge2_simulink_nonlinear.mdl (images below). Note that Simulink won t let you do a pure PD controller, since the order of the denominator is less than the numerator. So students can implement a lead controller with a fast pole. (Considering the description of the lead compensator in Lecture 12, they could make a = 100.) Also, note that my solution assumes that l, m, c, and g have already been defined in the workspace. Step 10s s+1 Transfer Fcn1 1/m/l^2 s 2+c/m/l.s+g/l Transfer Fcn Scope Step 10s s+1 Transfer Fcn1 1/m/l^2 Gain Scope 1 s 1 s t g/l Integrator Integrator1 Sine Wave Function Gain2 Gain1 c/m/l 6

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