Substitute Exam-EE647-O,Academic Period 2021-1 Solutionary Jean Estrada Roque Facultad de Ingeniería Eléctrica y Electrónica Universidad Nacional de Ingeniería Lima, Perú [email protected] AbstractโThe purpose of this work is to solve the problems proposed in the substitute exam proposed by engineer Daniel Carbonel, we will also complement this with simulations that will make the concepts clearer. To pass the first and second filter KeywordsโControl theory, root locus, Routh-Hurwitz, controller, Bode diagram. Then: I. PROPOSED EXERCISES 1. The following closed-loop system is shown, you are asked to determine the value or values that ๐ can take, so that the system is considered stable. (Time: 15 min) ๐พ + 1 > 0 โ ๐พ > โ1 ๐พ>0 ๐พ>0 Now that we have passed the first and second filters, let's do the Routh-Hurwitz arrangement ๐ 5 1 2 ๐พ+1 ๐ 4 1 1 ๐พ 3 ๐1 ๐2 ๐ 2 ๐1 ๐2 ๐ 1 ๐1 ๐ 0 ๐1 ๐ Fig. 1. Control system 1 Solution of problem 1 First, we must find the open loop transfer function, for that we open loop and calculate ๐บ๐ป(๐ ): We can quickly see that: ๐2 = ๐1 = ๐พ Then it would only be necessary to find: ๐1, ๐2 , ๐1 , ๐1 Process to find ๐๐ โถ Fig. 2. Control system 2 1 1 2 |=1 ๐1 = โ . | 1 1 1 Process to find ๐๐ โถ Then: ๐บ๐ป(๐ ) = ๐พ(๐ + 1) ๐ ((๐ + 1)(๐ 3 + 1) + 2๐ 2 ) In advance we know that ๐พ > 0 because it is open-loop gain. Remember 1 + ๐บ๐ป(๐ ) = 0 gives you the characteristic closed-loop equation. 1+ ๐พ(๐ + 1) =0 ๐ ((๐ + 1)(๐ 3 + 1) + 2๐ 2 ) 1 1 ๐พ+1 |=1 ๐2 = โ . | ๐พ 1 1 Process to find ๐๐ โถ ๐1 = โ 1 1 1 1 | = โ .| | = 0 โ ๐1 = ๐ > 0 ๐2 1 1 1 Process to find ๐ ๐ โถ ๐1 = โ ๐ 5 + ๐ 4 + 2๐ 3 + ๐ 2 + (๐พ + 1)๐ + ๐พ = 0 1 1 .| ๐1 ๐1 1 ๐1 .| ๐1 ๐1 1 ๐ โ ๐พ โ๐พ ๐2 | = โ . |1 1 | = โ ๐2 ๐ ๐ ๐พ ๐ ๐ Now let's analyze the first column: Using another online simulator [1], we obtain: ๐ 5 1 (+) 2 ๐พ+1 4 1 (+) 1 ๐พ ๐ 3 1 (+) 1 2 ๐ (+) ๐พ ๐ ๐ ๐ 1 ๐ 0 โ๐พ (โ) ๐ ๐พ (+) We notice that there are two sign changes, so I conclude: โ๐ฒ โ ๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐ But how do I certify or verify this? We will simulate in Scilab root locus of ๐บ๐ป(๐ ) using Scilab console. Fig. 5. Root Locus 2 Then we see that for every value of ๐ฒ the system will always have poles in the right half plane, and this represents instability โ๐ฒ. 2. There is a ๐ฎ๐(๐) plant that is a second order system, without zeros, whose settling time (for a step input) is ๐ก๐ = 1 4 ๐ ๐๐, whose factor damping is and unity static gain.It is โ10 desired to control said plant, for which a control system is designed (as shown in the figure below), where the feedback ๐ฏ(๐) is a block represented only by a zero located at -2, with static gain 0.5 and the ๐ฎ๐(๐) block is a โtype 2โ second order system, without zeros and with gain static 3. Fig. 3. Scilab console 1 Note that only 3 command lines were used, this tool is very useful to certify our calculations, then we get: Fig. 6. Control system 3 With the above information, you are asked to sketch the Geometric Place of the Roots for the system presented, indicating in the drawing the location of the parameters that you determined for your respective sketch. (Time: 25 min) Solution of problem 2 Process to find ๐ฎ๐(๐) โถ Fig. 4. Root Locus 1 ๐บ๐(๐ ) = ๐พ๐ค๐2 ๐ 2 + 2๐๐ค๐ ๐ + ๐ค๐2 ๐ก๐ = 4 โ ๐= 1 โ10 4 = 4 โ ๐๐ค๐ = 1 ๐๐ค๐ So ๐บ๐ป(๐ ) is: ๐บ๐ป(๐ ) = ๐บ๐ (๐ ). ๐บ๐(๐ ). ๐ป(๐ ) โ ๐ค๐ = โ10 ๐บ๐ป(๐ ) = 3 10 (๐ + 2) . . ๐ 2 ๐ 2 + 2๐ + 10 4 ๐พ๐ ๐ = 1 โ ๐บ๐(0) = 1 โ ๐พ = 1 ๐บ๐ป(๐ ) = Then: ๐บ๐(๐ ) = ๐ 2 10 + 2๐ + 10 We make 30 4 30(๐ + 2) + 2๐ + 10) 4๐ 2 (๐ 2 equal to ๐พ to find the locus of the roots Process to find ๐ฏ(๐) โถ ๐บ๐ป(๐ ) = ๐พ(๐ + 2) ๐ 2 (๐ 2 + 2๐ + 10) ๐ป(๐ ) = ๐พ(๐ + 2) Step 1: Number of branches ๐พ๐ ๐ = 1 1 1 1 โ ๐บ๐(0) = โ ๐พ. 2 = โ ๐พ = 2 2 2 4 Then: #๐๐๐๐๐โ๐๐ = #๐๐๐๐๐ ๐๐ ๐บ๐ป(๐ ) = 4 Step 2: Existence of locus on the real axis (๐ + 2) ๐ป(๐ ) = 4 We first identify all the poles and zeros of ๐บ๐ป(๐ ) to locate them in the complex plane and to be able to determine whether or not there is a locus on the real axis: Process to find ๐ฎ๐ (๐) โถ ๐บ๐ (๐ ) = ๐พ ๐ 2 ๐ = โ2 โ ๐๐๐๐ ๐ง๐๐๐ ๐ = 0 โ ๐๐๐๐ ๐๐๐๐ ๐๐ ๐๐ข๐๐ก๐๐๐๐๐๐๐ก๐ฆ 2 ๐ = โ1 ± 3๐ โ ๐๐๐๐๐ข๐๐๐ก๐๐ ๐๐๐๐๐๐๐ฅ ๐๐๐๐๐ ๐พ๐ ๐ = 3 โ ๐บ๐(0) = 3 โ ๐พ = 3 Then: ๐บ๐ (๐ ) = 3 ๐ 2 We must find the open loop transfer function, for that we open loop and calculate ๐บ๐ป(๐ ): Fig. 7. Control system 4 Fig. 8. Root Locus 3 Step 3: Asymptotes Step 4: Break or bifurcation points (๐๐ ) a) Number of asymptotes As GH (s) has the following form: ๐บ๐ป(๐ ) = ๐พ #๐๐ ๐ฆ๐๐๐ก๐๐ก๐๐ = #๐๐๐๐๐ ๐๐ ๐บ๐ป(๐ ) โ #๐ง๐๐๐๐ ๐๐ ๐บ๐ป(๐ ) #๐๐ ๐ฆ๐๐๐ก๐๐ก๐๐ = 4 โ 1 = 3 ๐(๐ ) ๐ท(๐ ) From: 1 + ๐บ๐ป(๐ ) = 0 โ 1 + ๐พ b) Convergence point (๐๐ ) โ ๐๐๐๐๐ ๐๐ ๐บ๐ป(๐ ) โ โ ๐ง๐๐๐๐ ๐๐ ๐บ๐ป(๐ ) ๐๐ = #๐๐๐๐๐ ๐๐ ๐บ๐ป(๐ ) โ #๐ง๐๐๐๐ ๐๐ ๐บ๐ป(๐ ) (โ1 + 3๐ โ 1 โ 3๐ + 0 + 0) โ (โ2) ๐๐ = 4โ1 ๐๐ = 0 ๐พ= ๐(๐ ) =0 ๐ท(๐ ) โ๐ท(๐ ) ๐(๐ ) To find break or bifurcation points: ๐๐พ = 0 โ ๐ = โ3.03 ๐๐๐ ๐ = 0 ๐๐ We discard s = 0, then: ๐๐ = โ3.03 c) Angle with the real axis (๐ฝ) Step 5: Intersection with the imaginary axis From: (2๐พ + 1)180° ๐ฝ= #๐๐๐๐๐ ๐๐ ๐บ๐ป(๐ ) โ #๐ง๐๐๐๐ ๐๐ ๐บ๐ป(๐ ) 1 + ๐บ๐ป(๐ ) = 0 โ 1 + ๐พ(๐ + 2) =0 + 2๐ + 10) ๐ 2 (๐ 2 ๐ 4 + 2๐ 3 + 10๐ 2 + ๐พ๐ + 2๐พ = 0 Where ๐พ = 0, 1, 2, โฆ , #๐๐ ๐ฆ๐๐๐ก๐๐ก๐๐ โ 1 Since #๐๐ ๐ฆ๐๐๐ก๐๐ก๐๐ = 3 (2๐พ + 1)180° = (2๐พ + 1)60° 3 Then ๐พ = 0, 1, 2 ๐ฝ= We make ๐ = ๐๐ค , and we obtain: (๐ค 4 โ 10๐ค 2 + 2๐พ) + (๐พ๐ค โ 2๐ค 3 )๐ = 0 = 0 + 0๐ We obtain the following solutions: ๐ฝ0 = 60° ๐ค = โโ6 ๐๐๐ ๐พ = โ12 ๐ฝ1 = 180° ๐ค = 0 ๐๐๐ ๐พ = 0 ๐ฝ2 = 300° ๐ค = โ6 ๐๐๐ ๐พ = 12 We stay with: Remember that the 180° asymptote does not exist, so we discard it. ๐ค = โ6 ๐๐๐ ๐พ = 12 Then intersect at: ๐ = ๐๐ค = ๐โ6 = ๐2.449 โ ๐2.45 Fig. 9. Root Locus 4 Fig. 10. Root Locus 5 Step 6: Departure Angle (๐ฝ๐ ) Step 7: Root locus sketch Due to the presence of complex conjugated poles in GH (s), the departure angle is calculated, taking a test pole and making all of them converge to it. We proceed by sketching according to all the parameters obtained: Fig. 13. Root Locus 8 Fig. 11. Root Locus 6 A cleaner graph would be the following: 3 โ 1 = tanโ1 ( ) = 71.56° 1 3 ๐1 = 180 โ tanโ1( ) = 108.435 ° 1 ๐2 = 108.435° ๐3 = 90° โ โ ๐๐๐๐๐ = ๐1 + ๐2 + ๐3 = 306.87° โ โ ๐ง๐๐๐๐ = โ 1 = 71.56° ๐๐ = 180° โ (โ โ ๐๐๐๐๐ โ โ โ ๐ง๐๐๐๐ ) Fig. 14. Root Locus 9 ๐๐ = โ55.31 But how do I certify or verify this? We will simulate in Scilab the root locus of ๐บ๐ป(๐ ) using Scilab console. Fig. 12. Root Locus 7 Fig. 15. Scilab console 2 Note that only 3 command lines were used, this tool is very useful to certify our calculations, then we get: Fig. 18. Control system 5 Solution of problem 3 Let's find ๐บ1(๐ ) ๐พ ๐ ๐พ๐ ๐ = 10 โ ๐บ๐(0) = ๐พ โ ๐พ = 10 ๐บ1(๐ ) = Then: ๐บ1(๐ ) = 10 ๐ Then ๐บ๐ป(๐ ): Fig. 16. Root Locus 10 Using another online simulator [1], we obtain: ๐บ๐ป(๐ ) = 10 10(๐ + 0.1) 100(๐ + 0.1) . = ๐ ๐ 2 + 2๐ + 2 ๐ (๐ 2 + 2๐ + 2) We make 100 equal to ๐พ: ๐บ๐ป(๐ ) = ๐พ(๐ + 0.1) ๐ (๐ 2 + 2๐ + 2) We make ๐ = ๐๐ค: ๐บ๐ป(๐๐ค) = ๐พ(๐๐ค + 0.1) ๐๐ค((๐๐ค)2 + 2๐๐ค + 2) Using complex notation: ๐บ๐ป (๐๐ค) = โ๐พ(๐ค 2โ1.8) ๐ค 4 +4 โ 1.9๐พ(๐ค 2+0.105263) ๐ค(๐ค 4+4) j a) Since ๐. ๐ = 50 ๐น + 180° = 50° โ ๐น = โ130° So, I add 180 ° to relocate the angle (F) in the correct quadrant, to apply the arctangent, so we get 50 °. Fig. 17. Root Locus 11 3. There is a system like the one shown, where ๐บ1(๐ ) is a firstorder block of type "1", without zeros and with static gain 10, 10(๐ +0.1) in addition: ๐บ๐(๐ ) = ๐ 2+2๐ +2 . โ1.9๐พ(๐ค 2 + 0.105263) ๐ค(๐ค 4 + 4) ) = 50° tanโ1 ( โ๐พ(๐ค 2 โ 1.8) ๐ค4 + 4 Then, we obtain: ๐ค = โ0.6825 You are asked: a) Calculate the value of K if you want the Phase Margin to be 50º. b) What is the value of the gain margin and what can you say about the stability of the closed-loop system? (Time: 20 min) ๐ค = โ0.1032 ๐ค = 2.38 We stay with: ๐ค = 2.38 Evaluating ๐ค = 2.38 in the magnitude of | ๐บ๐ป (๐๐ค)| , it would have to give us the value of 1 | ๐พ(๐2.38 + 0.1) |=1 ๐2.38((๐2.38)2 + 2๐2.38 + 2) 0.16๐พ = 1 โ ๐พ = 6.25 How do we certify? We use Scilab to check if our calculation is right or wrong. We see that we obtain a phase margin of 27.38 °, which does not say that our calculation of ๐พ = 6.25 is not correct, so what do we do? We can go testing various values of ๐พ using the simulator and precisely for the value of ๐พ = 2.609 we obtain the desired phase margin equal to approximately 50 °, obviously we do this here with the simulator, the idea is that we obtain it theoretically, so I kindly ask the teacher to upload the theoretical solution to this problem, for now this experimental way of finding ๐พ has been placed since I do not know how to obtain that value theoretically, however our simulators can help us for this type of case. Fig. 19. Scilab console 3 Fig. 21. Scilab console 4 Let's look at the gain and phase margins on the magnitude and phase bode diagram for ๐พ = 6.25 We see that we already have the desired phase margin equal to 50 ° with ๐พ = 2.609.Let's look at the gain and phase margins on the magnitude and phase bode diagram for ๐พ = 2.609 Fig. 20. Bode plot 1 Fig. 22. Bode plot 2 Then , the asked value is ๐ฒ = ๐. ๐๐๐. b) Now they ask us to determine the gain margin, now we set the phase of ๐บ๐ป (๐ ) to -180, this means that the imaginary part will be zero, so let's set the imaginary part of ๐บ๐ป (๐ ) to zero: Fig. 24. Root Locus Tool 2 โ1.9๐พ(๐ค 2 + 0.105263) =0 ๐ค(๐ค 4 + 4) ๐ค 2 + 0.105263 = 0 We see that there will be no real and positive solutions for ๐ค, what does this mean? It means that the phase Bode diagram never touches โ180°, so the gain margin is infinite ๐ฎ. ๐ด = โ Fig. 25. Root Locus Tool 3 What do we conclude about the closed loop system? Since ๐. ๐ = 50° and ๐ฎ. ๐ด = โ , then the system will be stable. OBSERVATIONS The use of the following online tool [1] is recommended as it will allow you to check the work done here, it is freely accessible for educational purposes. This allows you to graph the locus of the roots and modify the transfer function and choose different view options. Fig. 23. Root Locus Tool 1 REFERENCES [1] โDrawing the root locus.โ [Online]. Available: https://lpsa.swarthmore.edu/Root_Locus/RLDraw.html. [Accessed: 15-Aug-2021].
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