EE 370L Controls Laboratory. Laboratory Exercise #7 Root Locus. Department of Electrical and Computer Engineering University of Nevada, at Las Vegas

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1 EE 370L Controls Laboratory Laboratory Exercise #7 Root Locus Department of Electrical an Computer Engineering University of Nevaa, at Las Vegas 1. Learning Objectives To emonstrate the concept of error an feeback To emonstrate how to raw an analyze the root locus of a given system. 2. Equipment Usage In this laboratory exercise, you will be expose to the FEEDBACK control system. You will use MATLAB an SIMULINK software to esign a feeback control system. 3. Error an Feeback Open Loop Systems: An open-loop controller, also calle a non-feeback controller, is a type of controller which computes its input into a system using only the current state an its moel of the system. A characteristic of the open-loop controller is that it oes not use feeback to etermine if its input has achieve the esire goal. This means that the system oes not observe the output of the processes that it is controlling. Consequently, a true open-loop system cannot engage in machine learning an also cannot correct any errors that it coul make. It also may not compensate for isturbances in the system. Close Loop Systems: A close-loop or feeback control system is one in which an input forcing function is etermine in part by the system response. The measure response of a physical system is compare with a esire response. The ifference between these two responses initiates actions that will result in the actual response of the system to approach the esire response. This in turn rives the ifference signal towar zero. 1 P a g e

2 Unity Gain Feeback System Response: Transfer function of the system above can be efine as: Assuming G = N/D, transfer function reuces to If terms in numerator an enominator are not comprise, no cancellation of terms exists an poles of transfer function can be restate as the roots of. System stability: For a feeback system to be stable all the poles of transfer function shall be locate at the left half plane. A simple technique to locate stable poles is root locus systems. A root locus corresponing to G traces all the poles of H, as K varies from 0 to +. Thus, root locus is a family of curves of the roots of. A common technique in control system esign is to plot the root locus corresponing to a plant in orer to iscover where the close-loop poles can be place using a gain compensator. A typical root locus is epicte below. p 3 = -2 X X p 2 = -1 O z 1 = -0.5 X p 1 = 0 s = s = P a g e

3 Secon Orer Systems Response: In this Experiment, you are given G( having the form when System is critically ampe when, is over-ampe when an uner-ampe For uner-ampe case, systems response is as follows, where Tr = rise time Tp = peak time Ts = settling time Mp = Peak value Figure 1. Step Response of an Uner-ampe Systems Ieally, we woul like the motor output to follow the input. Therefore, the esign goal is to have Tr an Ts as small value an Mp as close to unity. Explicit formulas can be foun for these parameters. First, note that the unit step response for t > 0 is given by { } [ ] {[ ] } From the above relationship, we can erive the following relationships: 3 P a g e

4 ( ) ( [ ] ) Steay state error to a step response is as follows: More often, we are also intereste in steay state error for ramp response. A typical ramp response is shown below: ess1 Figure 2. Ramp Response of a close loop Systems 4 P a g e

5 Steay state error to a ramp response is as follows: Using final value theorem, we get the following formula: [ ] In aition to time response, we are also intereste in frequency response parameters. Frequency response of a close loop systems is as follows: [ ] Figure below epicts H(jω) (with ω plotte on a logarithmic axi for an uner-ampe systems: Frequency Response where, Figure 3. Frequency Response m b = banwith M m = Peak at resonance From the above relationship of H(jω), we can erive the following relationships: 5 P a g e

6 Given the target specification of Tr, Tp, Ts, Mp, e ss1, ω m, ω b, an M m, the above formulas can be use to etermine whether a choice of K exists such that the specification goals are satisfie an, if so, what range of values of K achieves the specifications. 4. Root Locus Steps for sketching Root Locus: 1. Root locus begins at pole an ens at zero. The number of separate loci is equal to the number of poles. m loci ens at zeros an n-m loci ens on infinities. 2. Root loci are symmetrical with respect to the horizontal real axis. 3. On the real axis, a point is on the root locus if the sum of poles an zeros on its right rie is an o number. 4. Asymptotes centroi is on the real axis as The angle of the asymptotes with the respect to the real axis is 5. Points that locus crosses the imaginary axis can be erive by: a. Routh-Hurwitz criterion. b. Characteristic equation: (For low orer equations only) 6. Breakaway points on the real axis: or 7. The tangents to the loci at the breakaway point are equally space over. 8. The angle of locus eparture from a pole: Matlab Commans Plotting root locus: First set the gain range. Example. K = -50:.2:50; (this creates a vector varying from -50 to 50 with increment of 0.2) Type r = rlocus(num, en,k), where num an en are the plant numerator an enominator. To plot the root locus, type plot(r) 6 P a g e

7 To fin the gain an frequency corresponing to a point on the locus, type k=rlocfin(num,en) In this experiment we will calculate close-loop response from open-loop ata. Before calculating either step response or Boe plots, the close-loop transfer function must be foun by typing the following commans nc=k0*num; c=en+k0*[0 0 num]; where k0 is a specific gain. Note that two extra zeros are pae to ensure that en an [0 0 num] have the same length for aition. The close-loop unit step response can be generate by typing step(nc,c) If you nee more control over the time interval in which the computation is carrie out, efine a vector t an type step(nc,c,t) The magnitue of the close-loop transfer function H(jω) versus ω can be plotte with the commans [mc,pc,wc]=boe(nc,c); semilogx(wc,mc) Root Locus Example: Consier the control system R( C( with Then K P s 2 K(, s 4s 2 s 1 s G (, H ( s 1 s 4 s 2 s 1 s 4 K( G( H( K P k where the parameter k 4K P s s Now raw the Root Locus: 7 P a g e

8 1. There are 3 branches. 2. The real axis between 1<s<0 an 4<s<-2 is on the root locus because these areas are to the left of an o number of poles an zeros. 3. The Root Loci start ( K 0 ) at the poles locate at 4, -1 an 0 then en at the zero locate at -2 as well as (2) zeros at s. 4. For large values of s, the Root Loci are asymptotic to asymptotes with angles, a O 2k 1180 O O , The intersection of the asymptotes lies on the real axis at s a A Breakaway point occurs along the root locus 1 s 0 at a relative maximum value of k. D( ss 1 s 4 Calculate k an tabulate versus s. N( s 2 s k N k D( ss 1 s 4 ( s D( ss 1 s ( s D( ss 1 s ( s D( ss 1 s ( s D( ss 1 s ( s D( ss 1 s ( s k N k N k N k N k N max From the table it can be seen the maximum k is acquire aroun which is the breakaway point. 8 P a g e

9 s s j k j s = Root Locus can be calibrate to fin gain at important points. Here 1.5 K k P The Matlab commans to generate the root locus are: >> Num=[1 2];Den=conv([1 0],conv([1 1],[1 4])); >> GH=tf(Num,Den) Transfer function: s s^3 + 5 s^2 + 4 s >> rlocus(gh) 9 P a g e

10 Imaginary Axis Root Locus System: GH Gain: 1.48 Pole: i Damping: Overshoot (%): 4.43 Frequency (ra/sec): More Root Locus Examples: MatLab Program: num = [1] poles = [0;-1+j;-1-j;-3] en = poly(pole rlocus(num,en) Notes: n=4 poles, m=0 zeros asymptotes at ±45,±135 asymptote ( )/4 = angle of eparture, + arctan(0.5) ) = 180 ( = eg Real Axis 10 P a g e

11 MatLab Program: zeros = [-1+3*j;-1-3*j] num = poly(zero poles = [0;-1+2*j;-1-2*j;-2] en = poly(pole rlocus(num,en) Notes: n = 4 poles, m = 2 zeros asymptotes at ±90 egrees asymptote [( )-(-1-1)]/2 = -1.0 angle of eparture, [( + arctan(2)+(180-arctan(2))+ 90) - (-90+90) ] = 180 = -90 eg Angle of Arrival, a [(90+arctan(3)+(180-arctan(3))+90)-(-( a +90)]=180 a = 90 eg MatLab Program: num = [1] poles=[0;-1+4*j;-1-4*j;-2] en = poly(pole rlocus(num,en) Notes: n = 4 poles, m = 0 zeros asymptotes at ±45,±135 asymptote ( )/4 = -1.0 angle of eparture, ( + arctan(4) + (180-arctan(4)) + 90) = 180 = -90 eg 5. Prelab 1. The loop transfer function of a single-loop negative feeback system is a. Base on the secon-orer moel an equations given above, fin the value of K for critical amping. b. Using formulas given above, fin the range of K for which the close-loop system satisfies the following specifications: 11 P a g e

12 Tr 100ms Tp 120ms Mp 1.1 Ts 150ms ess1.045 ωm 10r/s ωb 50r/s Mm The loop transfer function of a single-loop negative feeback system is a. Analyze the system, o the han calculation an plot root loci of the system step by step. b. Using MATLAB, plot the root loci again an compare will your han sketch. c. Determine the range of feeback gain K such that the close-loop system is stable. For each value of K where instability begins, fin the corresponing frequency of oscillation. 6. Experiment an Postlab 1. The loop transfer function of a single-loop negative feeback system is a. Construct a Simulink moel. You shoul plot the step input an the response of the system on a single scope using the MUX block. b. Verify the system is oscillatory for any K value in the corresponing interval by printing the output of the scope block for this value of K. c. Verify the system is stable for any K value in the corresponing interval by printing the output of the scope block for this value of K.. Suppose we want the final value of our response to be within 0.01% of the final value of our input (step) signal. That is, the steay state error is 0.01%. Further, let us assume the final value is achieve for t 10 secons. Using your Simulink block, etermine for what values of K is the steay state error 0.01% (for t 10 secon? Notice that in esigning a control system we first analyze the stability of our open loop plant. If our open loop plant is unstable, we use feeback to stabilize the system. Then we pick the values for the parameters in our control law so our control objectives are achieve. Satisfying all the esign requirements is the goal of control theory. 2. The loop transfer function of a single-loop negative feeback system is 12 P a g e

13 a. Analyze the system, o the han calculation an plot root loci of the system step by step. b. Using MATLAB, plot the root loci again an compare will your han sketch. c. Fin each value of K where system is critically ampe.. Fin each value of K where instability begins, fin the corresponing frequency of oscillation. 13 P a g e

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