ENGINEERING 6951 AUTOMATIC CONTROL ENGINEERING FINAL EXAM FALL 2011 MARKS IN SQUARE [] BRACKETS INSTRUCTIONS NO NOTES OR TEXTS ALLOWED
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1 NAME: JOE CROW ENGINEERING 6951 AUTOMATIC CONTROL ENGINEERING FINAL EXAM FALL 2011 MARKS IN SQUARE [] BRACKETS INSTRUCTIONS NO NOTES OR TEXTS ALLOWED NO CALCULATORS ALLOWED GIVE CONCISE ANSWERS ASK NO QUESTIONS
2 SYSTEM DESCRIPTION Rollers are used to control the thickness of material sheets. A hydraulic actuator forces rollers onto the sheet. The thickness sensor is located downstream of the rollers and this introduces a transport lag. Here we use the Pade approximant to approximate this lag. The governing equations for the system are: SENSOR + T/2 dp/dt + P = - T/2 dr/dt + R THICKNESS ERROR E = C P CONTROL SIGNAL Q = K E OIL FLOW TO ACTUATOR M = A Q VELOCITY OF ACTUATOR B V = M + N SHEET THICKNESS X dr/dt = V where R is the actual thickness of the sheet at the rollers, P is the thickness of the sheet at the sensor, C is the command thickness, T is the time lag, V is the actuator velocity, M is a control flow to the actuator, N is a disturbance flow due to leakage, Q is the control signal, E is the error signal, T K A B X are constants. T=2 A=1 B=1 X=1
3 Sketch an overall block diagram for the system. [10] The transfer functions are: + T/2 S P + P = - T/2 S R + R P/R = [1 - T/2 S] / [1 + T/2 S] E = C P Q = K E M = A Q B V = M + N V/[M + N] = 1/B X S R = V R/V = 1/[XS]
4 Derive equations for the system Ziegler Nichols gains. [Hint: Differentiate sensor equation wrt time.] [10] Differentiation gives + T/2 d 2 P/dt 2 + dp/dt = - T/2 d 2 R/dt 2 + dr/dt X dr/dt = V X d 2 R/dt 2 = dv/dt V = [A K (C-P) + N] / B dv/dt = [A K (dc/dt-dp/dt) + dn/dt] / B Substitution into the sensor equation gives + T/2 d 2 P/dt 2 + dp/dt = - T/2 [A K (dc/dt-dp/dt) + dn/dt] / [B X] + [A K (C-P) + N] / [B X] Assume that the system is borderline with P: P = P o + P Sin[ωt] Assume that the inputs are constants:
5 C = C o N = N o Substitution into the sensor equation gives - T/2 ω 2 P Sin[ωt] + ω P Cos[ωt] = + T/2 [A K] ω P Cos[ωt] / [B X] + [[A K] [C o -P o ] + N o ] / [B X] - [A K] P Sin[ωt] / [B X] This is of the form i Sin[ωt] + j Cos[ωt] + k = 0 Mathematics requires that i = 0 j = 0 k = 0 K = [T/2]/A ω 2 = [AK]/[T/2] P o = C o + N o /[AK] Plugging in numbers gives K = 1 ω = 1 P o = C o + N o
6 Develop equations that would allow the behavior of the system to be predicted step by step in time. For this, use an exact representation of the transport lag. [10] The exact representation of the transport lag is P(t) = R(t-T) When this is used, the only ODE is X dr/dt = V An application of time stepping to this gives R NEW = R OLD + t [V OLD /X] The algebraic equations are E OLD = C OLD P OLD Q OLD = K E OLD M OLD = A Q OLD V OLD = [M OLD + N OLD ]/B
7 Determine the characteristic equation for the system. [10] The GH function is: [ A * K ] * [ 1 - T/2 S ] B * [ X S ] * [ 1 + T/2 S ] K * [ 1 - S ] S * [ 1 + S ] This is of the form GH = N/D The characteristic equation is: N + D = 0 K * [1 - S] + S * [1 + S] = 0 S 2 + [1 K] S + K = 0
8 Use the Routh Hurwitz criteria to determine the borderline gain of the system. [5] Is the system stable when K is half the borderline gain? [5] For stable operation of a system, all coefficients in its characteristic equation must be positive. In addition, certain tests functions must be positive. A quadratic characteristic equation has the form a S 2 + b S + c = 0 It has no test functions. For stable operation, each of its coefficients must be positive: a>0 b>0 c>0. For thickness control, this implies K > 0 [1 - K] > 0 This gives the borderline gain K=1. When K is half the borderline gain K, [1-K] is 1/2 which is greater than zero so the system is stable with this gain.
9 Sketch the Nyquist plot when K is half the borderline gain. [10] What are the system stability margins? [5] Explain the significance of GH equal to minus one. [2.5] What function would model the transport lag exactly? [2.5] The GH function is [ A * K ] * [ 1 - T/2 S ] B * [ X S ] * [ 1 + T/2 S ] K * [ 1 - S ] S * [ 1 + S ] Along the imaginary axis in the S plane S=jω K * [ 1 - ωj ] ωj * [ 1 + ωj ] K * ( 1 ωj ) * ( 1 ωj ) ωj * ( 1 + ωj )* ( 1 ωj ) K * ( 1-2ωj - ω 2 ) ωj * ( 1 + ω 2 )
10 As ω approaches zero, the GH function reduces to K * ( 1 ) ωj * ( 1 ) which tends to minus infinity j. As ω approaches infinity the GH function reduces to K * ( -ω 2 ) -K ωj * ( +ω 2 ) ωj which tends to plus zero j. A real axis cross over occurs when ω 2 this case, the GH function becomes is equal to one. In 0.5 * ( - 2ωj ) 0.5 * ( - 2j ) ωj * ( 1 + ω 2 ) j * ( 2 ) which is equal to minus 0.5.
11 The GH plot is shown on the next page. Inspection of the plot shows that the net clockwise rotations is zero. Inspection of the GH function shows that the number of unstable poles is zero. The number of unstable zeros is: N = N Z N p N Z = N + N p This gives N Z equal to zero. So the system is stable. The gain margin is one over the magnitude of GH where it crosses the negative real axis. Here it is 2. The phase margin is the angle to where the GH plot crosses a unit circle centered on the origin. One could get this by plotting more points on the GH plot. A GH plot is basically a polar open loop frequency response plot. When GH is equal to minus one, a command sine wave produces a response which has the same magnitude as the command but is 180 o out of phase. If the command was suddenly removed and the loop was suddenly closed, the negative of the response would take the place of the command and keep the system oscillating. If the gain was bigger than K, the command would produce a response bigger than itself. When this takes over, it would produce growing or unstable oscillations. If the gain was smaller than K, the command would produce a response smaller than itself. When this takes over, it would produce decaying or stable oscillations. The function e -TS models transport lags exactly.
12
13 Use Root Locus concept to check the borderline gain. [10] The GH function is: K * [ 1 - S ] S * [ 1 + S ] Nyquist suggests ω=1. In this case the angles are: = -180 The magnitudes are: [K * 2] / [1 * 2] = 1 K = 1
14 Determine the amplitude and the period of the limit cycle generated when the system is controlled by an ideal relay controller with DF=1/E o. [5] Is the limit cycle stable? [2.5] Is the system practically stable? [2.5]. For this problem, assume that R is in millimeters. [BONUS: For relay with deadband: What is the critical deadband?] At a limit cycle the DF is equal to the borderline gain: DF = K = 1/E o E o = 1/K = 1 The limit cycle frequency is ω=1. So the period is T o = 2 /ω = 2 This is a system which is stable when K is below the borderline gain K. So the limit cycle is stable. If the sheet being rolled was around 1mm thick, the system would be practically unstable. If the sheet was 10cm thick, it might be practically stable. Relay with deadband controller: DF theory shows that the critical deadband is half the limit cycle amplitude.
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16 Write a short m code that would predict the behavior of the system step by step in time. For this, use an exact representation of the transport lag. [5] Add statements to the code that would mimic loop rate phenomena. [5] A=1;B=1;X=1;K=0.5; ROLD=0.0; POLD=0.0; COLD=1.0; NOLD=0.0; NIT=10000; MIT=50; PIT=10; EOLD=COLD-POLD; JIT=0; QOLD=K*EOLD; DELT=0.01; for IT=1:NIT JIT=JIT+1; if(jit==1) if(it>mit) POLD=R(IT-MIT); EOLD=COLD POLD; end;end; if(jit==pit) QOLD=K*EOLD; JIT=0;end; MOLD=A*QOLD; VOLD=[MOLD+NOLD]/B; RNEW=ROLD+DELT*[VOLD/X]; ROLD=RNEW;R(IT)=RNEW; T(IT)=IT*DELT; end plot(t,r)
17 NAME: ENGINEERING 6951 AUTOMATIC CONTROL ENGINEERING QUIZ #1 The governing equations for control of a rocket are: PLANT X d 2 R/dt 2 + Y dr/dt + Z R = T + M DRIVE A dt/dt = Q PID Q = K P E + K I Edτ + K D de/dt ERROR E = C R where R is the actual orientation of the rocket, C is the command orientation, T is a control torque, M is a disturbance torque, Q is the control signal, E is the error signal, X Y Z A are constants and K P K I K D are gains.
18 Derive equations for the rocket Ziegler Nichols gains. [40] The Ziegler Nichols gains for a system are based on its borderline proportional gain K P and its borderline period T P. We manipulate the governing equations to leave only the output R and the two inputs C and M. The plant equation gives T = X d 2 R/dt 2 + Y dr/dt + Z R - M dt/dt = X d 3 R/dt 3 + Y d 2 R/dt 2 + Z dr/dt - dm/dt Substitution into the drive equation gives A (X d 3 R/dt 3 + Y d 2 R/dt 2 + Z dr/dt - dm/dt) = K P (C-R) During borderline stable operation R = R o + R Sin[ t] C = C o M = M o Substitution into the modified drive equation gives - AX 3 R Cos[ t] - AY 2 R Sin[ t] + AZ R Cos[ t] = K P C o - K P R o - K P R Sin[ t]
19 This is of the form i Sin[ t] + j Cos[ t] + k = 0 Mathematics requires that i=0 j=0 k=0: - AY 2 + K P = 0 - AX 3 + AZ = 0 K P C o - K P R o = 0 This gives 2 = Z/X K P = AY 2 = AYZ/X = [Z/X] T P = 2 / R o = C o Substitution then gives the ZN gains K P = 0.6*K P K I = K P /T I K D = K P *T D T I = T P /2 T D = T P /8
20 Develop a NEW from OLD template that would allow the motion of the rocket to be predicted step by step in time. [40] For time stepping, we need to convert given ODEs into a set of first order ODEs. For this, we let dr/dt = U With this the plant equation becomes X du/dt + Y U + Z R = T + M du/dt = (T + M Y U Z R) / X The drive equation can be written as dt/dt = Q/A
21 Application of simple time stepping gives R NEW = R OLD + t * U OLD U NEW = U OLD + t * (T OLD + M OLD Y U OLD Z R OLD ) / X T NEW = T OLD + t * (Q OLD ) / A Q OLD = K P E OLD + K I E OLD t + K D E OLD / t E OLD = C OLD R OLD We insert OLD values into the right hand side of the above OLD equations to get OLD values for the NEW from OLD equations. These give NEW values at the end of a step. These NEW values become OLD values for the next step. The rectangle rule is used to get the integral of the error and a backward difference is used to get the error rate.
22 Sketch an overall block diagram for the rocket. [10] Reduce the overall block down to the standard GH form with C as input and R as output. [5] Reduce it down to the standard GH form with D as input and R as output. [5] Laplace transformation gives X S 2 R + Y S R + Z R = T + M A S T = Q Q = K P E + K I /S E + K D S E E = C R The transfer functions are R/(T+M) = PLANT = 1 /(X S 2 + Y S + Z R) T/Q = DRIVE = 1/(A S) Q/E = PID = (K P + K I /S + K D S) The overall block diagram is
23 The command standard GH form is The disturbance standard GH form is
24 NAME: ENGINEERING 6951 AUTOMATIC CONTROL ENGINEERING QUIZ #2 The governing equations for control of a rocket are: PLANT J d 2 R/dt 2 - I R = T + M FEEDBACK P = X dr/dt + Y R SIGNAL IN E = C P SIGNAL OUT Q = K E DRIVE T = Z Q where R is the actual orientation of the rocket, C is the command orientation, T is a control torque, M is a disturbance torque, P is a feedback signal, Q is the signal out of the controller, E is the signal into the controller, J I X Y Z are constants and K is the controller gain. J = 10 I = 40 X = 1 Y = 1 Z = 10
25 Sketch the overall block diagram of the system. [10] Determine the overall transfer function of the system. [5] Determine the characteristic equation of the system. [5] Laplace Transformation gives [ J S 2 I ] R = T + M P = [ X S + Y ] R E = C P Q = K E T = Z Q The overall block diagram is: For the command case, the overall transfer function is: R/C = K Z / [ J S 2 + KZX S + KZY I ]
26 The characteristic equation is: J S 2 + KZX S + KZY I = 0 For the disturbance case, the overall transfer function is: R/M = 1 / [ J S 2 + KZX S + KZY I ] The characteristic equation is: J S 2 + KZX S + KZY I = 0 For the command case, the GH function is: K Z [ X S + Y ] / [ J S 2 I ] This gives the characteristic equation: J S 2 + KZX S + KZY I = 0 For the disturbance case, the GH function is: K Z [ X S + Y ] / [ J S 2 I ] This gives the characteristic equation: J S 2 + KZX S + KZY I = 0
27 Use the Routh Hurwitz criteria to determine the borderline gain of the system. [10] Is the system stable when K is double the borderline gain? [5] The characteristic equation is of the form: m S 2 + c S + k = 0 For stability all coefficients must be positive: m > 0 c > 0 k > 0 Setting k to zero gives K Z Y I = 0 K = I/[Z*Y] = 40/[10*1] = 4 When K is double the borderline K, k is 40. So all coefficients are positive and the system is stable.
28 Use Root Locus concept to check the borderline gain. [20] The GH function is: K Z [ X S + Y ] / [ J S 2 I ] [K*Z*X]/[J] [S+Y/X]/[S 2 -I/J] [K*10*1]/[10] [S+1]/[S 2-4] K [S+1]/[[S-2][S+2]] This has two poles at +2 and -2 and one zero at -1. The angle requirement shows that ω is zero when all angles add up to 180 o. So the imaginary axis crossover is on the real axis. The magnitude requirement gives: K [1]/[[2][2]] = 1 K = 4
29 Sketch the Nyquist plot for the case where K is double the borderline gain. [20] Interpret the plot. [10] Does the system have stability margins? [5] The GH function is: K Z [ X S + Y ] / [ J S 2 I ] With K equal to double the borderline gain, this becomes: 8 * 10 (S + 1) / (10 S 2 40) = 8 (S+1) / (S 2 4) Setting S equal to jω gives 8 (jω+1) / (-ω 2-4) When ω is equal to zero, GH is equal to -2. When ω is equal to, GH is equal to 0 - j. There are no real or imaginary axis crossovers. All of the plot is in the 3 rd quadrant.
30 The GH function has 1 unstable pole so N P is 1. The GH plot shows one counterclockwise rotation of the GH vector so N is -1. Substitution into N Z = N + N P shows N Z is zero. So the system is stable. This system is not a normal system so it does not have gain or phase margins.
31 Determine the amplitude and the period of the limit cycle generated when the system is controlled by an ideal relay controller with DF=0.4/E o. [5] Is the limit cycle stable? [2.5] Is the system practically stable? [2.5] A limit cycle occurs when DF = K. This means This gives DF = K = 0.4/E o E o = 0.4/K = 0.4/4 = 0.1 (6 degrees) The oscillation frequency ω is 0 so the period T is. The DF versus E o plot shows the limit cycle is unstable. Small oscillations decay away from it while large oscillations grow away from it. If E o is in radians the limit cycle would probably be practically unstable.
32 BONUS QUESTION [5+2] Describe how you would get the response of the system to a unit impulse command. [BASIC: 5] Give a rough numerical answer to this question. [EXTRA: 2] The transfer function is R/C = K Z / [ J S 2 + KZX S + KZY I ] For a unit impulse command, C is 1 and R becomes R = K Z / [ J S 2 + KZX S + KZY I ] R = 80 / [ 10 S S ] R = 8 / [ S S + 4 ] R = A / [S-a] + B / [S-b] R = A e +at + B e +bt ) By inspection, a rough approximation to R is R = 8 / [ S + 7.5] [S + 0.5] R = A / [S + 7.5] + B / [S + 0.5] A = 8 / [ ] = -8/7 B = 8 / [ ] = +8/7 R = [-8/7] / [S + 7.5] + [8/7] / [S + 0.5] R = 8/7 (e -0.5t - e -7.5t )
ENGINEERING 6951 AUTOMATIC CONTROL ENGINEERING FINAL EXAM FALL 2011 MARKS IN SQUARE [] BRACKETS INSTRUCTIONS NO NOTES OR TEXTS ALLOWED
NAME: JOE CROW ENGINEERING 6951 AUTOMATIC CONTROL ENGINEERING FINAL EXAM FALL 2011 MARKS IN SQUARE [] BRACKETS INSTRUCTIONS NO NOTES OR TEXTS ALLOWED NO CALCULATORS ALLOWED GIVE CONCISE ANSWERS ASK NO
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