BSc - Sample Examination Digital Control Sytem (5-588-) Prof. L. Guzzella Solution Exam Duration: Number of Quetion: Rating: Permitted aid: minute examination time + 5 minute reading time at the beginning of the exam 5 (differently weighted, in total 35 point) It i not required to anwer all quetion to achieve the maximum grade. The number of point i indicated in front of each quetion eparately. heet A4 (4 page) A pocket calculator will be provided. The aitant are not allowed to anwer technical quetion and you are not permitted to ue any electronic device (exception i the pocket calculator provided officially). Important: Ue only thee prepared heet for your olution. You may ak for extra heet if you need more pace.
Page Sample Examination Digital Control Sytem Quetion (General Knowledge) 5 Point You have to develop a controller for an intable plant, which will run on a microproceor allowing for a ampling time of T =.3. The pectra of the plant, of the diturbance and of the meaurement noie are hown in Figure. The plant ha one untable pole at π + =. The reference for the control i r(t) =. P D N untable pole Magnitude (db) - - -3-4 - ω (rad/) Figure : Spectra of the plant (P), the diturbance (D) and the meaurement noie (N) a) ( point) Draw the block diagram of a generic, complete dicrete-time control ytem, including interface, filter, and all relevant ignal. b) ( point) Can you do controller emulation in the cae at hand? Jutify your anwer. c) ( point) I an anti-aliaing filter neceary? Jutify your anwer. d) ( point) Baed on your anwer to the previou quetion, what plant model do you ue to deign your controller?
Sample Examination Digital Control Sytem Page 3 Solution a) The diagram i hown in Figure. d(t) r(t) = AAF ADC μp DAC Plant y(t) n(t) Figure : Dicrete-time control ytem b) It i generally adviable to have the croover frequency of the loop gain centered between the croover frequencie of the diturbance and the noie, which i at approximately 8 rad /. Alo, the control ytem hould be at leat one octave (better: one decade) fater than the fatet untable pole of the plant. Thi requirement alo force a croover frequency of at leat 8 rad /. The ampling frequency i ω = π T rad /, and thu only.6 time fater than the required croover frequency. For emulation to poibly work well, a factor of approximately i required. Therefore, we cannot do emulation. c) Ye. Even if the control ytem ha a teep decent after the croover frequency (and thu the ampled ignal doe not have ubtantial energy at frequencie higher than the Nyquit frequency of roughly.5 rad /), an anti-aliaing filter (AAF) i till needed to attenuate the noie. d) The four combination of emulation/dicrete controller deign and with/without AAF are poible. For emulation, the continuou repreentation of the plant i ued when deigning the (continuou) controller. In the cae of real dicrete-time controller deign, the plant (and poibly the AAF) have to be dicretized firt. All four cae are lited here, the correct one baed on the anwer to b) and c) i the dicrete-time deign with AAF. dicrete-time deign { { } } with AAF ( z )Z L AAF () P () t=kt { { } } no AAF ( z )Z L P () t=kt emulation AAF () e T P () e T P ()
Page 4 Sample Examination Digital Control Sytem Quetion (Controller Emulation) 6 Point For a continuou-time plant model, an engineer obtained the following continuou-time controller: C() = + + α, wherea α R i a tuning factor. a) ( point) The engineer implement the controller on a microproceor uing the Euler backward emulation approach. What i the reulting tranfer function C(z) for a generic ampling time T (T R + )? b) (3 point) What i the range of the tuning factor α to produce an aymptotically table dicrete-time controller uing the Euler backward emulation approach? c) ( point) What i the condition on α to have an aymptotically table continuou-time controller C() and an aymptotically table dicrete-time controller uing the Euler backward emulation approach? Solution a) The Euler backward emulation approach i given by = z T z. Inerting thi equation into the controller C() yield C(z) = z T z + z T z + α = z( + T ) z( + αt ). b) The controller C(z) i aymptotically table if it pole z p fulfill the condition z p <. The pole z p i obtained from the equation z( + αt ) =, i.e. z p = condition above, α mut atify < + αt <. Thi inequality contraint lead to the olution α > and α < T. +αt. From the c) For C() being aymptotically table, it pole p = α mut lie in the left half of the complex plane, i.e. Re{ p } < α >. Together with the reult from b), the condition on α i that α >.
Sample Examination Digital Control Sytem Page 5 Quetion 3 (Plant Dicretization and Analyi) 8 Point a) (3 point) Derive the dicrete-time tranfer function of the following continuou-time plant: P () = + 3 + Ue a zero-order hold element and aume the ampling time T to be known. b) (3 point) Determine the tate-pace repreentation of the above ytem. Introduce a tate vector x and find the matrice F, G, C, D decribing the tate-pace repreentation of the dicretized ytem: x k+ = F x k + Gu k () y k = Cx k + Du k (3) c) ( point) Ae the tability of the dicrete-time ytem. () Solution 3 a) Dicretizing a continuou-time tranfer function uing zero-order hold a ampling method can be done by evaluating the following formula: P (z) = z { } } Z {L z t=kt P () (4) Partial fraction decompoition i required in order to apply the tranformation table: P () = ( + )( + ) = A + B + + C + A( + )( + ) + B( + ) + C( + ) = ( + )( + ) = (A + B + C) + (3A + B + C) + A ( + )( + ) Comparing coefficient of the numerator obviouly yield A =.5 and olving the remaining two equation for B and C reult in B =, C =.5. By inerting thee value we obtain a tranfer function in a form uitable for direct tranformation:.5 P () = + +.5 + Uing the tranformation table reult in: (.5 P (z) = z z z z z z +.5 z e T ) z e T In order to implify the notation for further calculation, the abbreviation a = e T i introduced: P (z) = z ( z.5 z z z ) z a +.5 z z a () (5) (6) (7) (8) (9)
Page 6 Sample Examination Digital Control Sytem Expanding and rearranging finally yield: P (z) =.5(z a)(z a ) (z )(z a ) +.5(z )(z a) (z a)(z a ) = z(.5a a +.5) + (.5a 3 a +.5a) z z(a + a) + a 3 = Y (z) U(z) () () b) Baed on the dicrete-time tranfer function derived in the foregoing quetion, the input/output ytem decription i readily obtained by applying the hift property of the Z tranformation: Y (z) (z z(a + a) + a 3 ) = U(z) (z(.5a a +.5) + (.5a 3 a +.5a)) (3) y k+ (a + a)y k+ + a 3 y k = (.5a a +.5)u k+ + (.5a 3 a +.5a)u k. (4) Solving for y k+ and hifting one timetep backward yield: y k+ = (a + a)y k a 3 y k + (.5a a +.5)u k + (.5a 3 a +.5a)u k. (5) Rearranging the element in equation (5) in a chronological order, one get y k+ = (a + a)y k + (.5a a +.5)u k a 3 y k + (.5a 3 a +.5a)u }{{ k }. (6) =:ξ k By looking at equation (6), the ytem can be decribed uing two tate ν k = y k and ξ k a deignated (which contain all information from the previou tep), yielding: ν k+ = (a + a)ν k + (.5a a +.5)u k + ξ k (7) ξ k+ = a 3 ν k + (.5a 3 a +.5a)u k, (8) Thi i equivalent to the following tate pace repreentation: [ (a x k+ = ] ( + a).5a a 3 x k + ) a +.5.5a 3 a u +.5a k }{{}}{{} (9) F G y k = ( ) x k + }{{}}{{} u k. D C () c) The pole of the ytem determine whether it i table or not. There are three way to calculate the pole of the dicrete-time ytem: i) Eigenvalue of matrix F, i.e. olve det(π I F ) = ; i Zero of the denominator of P (z) calculated above (mot direct way); Calculate the pole of the continuou-time plant (e.g. from it tranfer function P ()) and ue π z i = e π i T to obtain the pole of the dicrete-time ytem. All three way lead to π = a = e T and π = a = e T. Since T >, it can be concluded that π, < and hence the ytem i aymptotically table.
Sample Examination Digital Control Sytem Page 7 Quetion 4 (Anti Aliaing Filter and Signal Sampling) 6 Point a) Signal Sampling without Anti-Aliaing Filter Given a ignal x(t) which i corrupted by noie n(t): y(t) = x(t) + n(t) x(t) = in(ω t) + in(ω t) n(t) =.5 in(ω n t) with: ω = rad, ω = 3 rad and ω n = 7 rad i) ( point) Draw the amplitude pectrum Y (ω) of y(t). (Ue the firt prepared frame on the next page) ( point) Now aume that the ignal i ampled with the ampling time T = π 4. Calculate the ampling frequency a well a the Nyquit frequency. Draw vertical line to indicate both frequencie. Sketch the amplitude pectrum of the ampled ignal Ȳ (ω). Indicate where the amplitude pectrum of the original ignal Y (ω) i ditorted by the aliaing of the noie. (Ue the econd prepared frame on the next page) b) Signal Sampling with Anti-Aliaing Filter To avoid aliaing, the following idealized filter F i placed before the ampling F (ω).5 ω F ω[ rad ] The filter parameter are α > R and ω F > R. ωf + α i) ( point) Now apply the filter F to the ignal y(t). Ue the filter parameter α =.5 rad and ω F = 4 rad. Draw the magnitude of the filter F (ω) and the amplitude pectrum Y F (ω) of the filtered ignal y F (t). (Ue the third prepared frame on the next page) ( point) Now aume that the filtered ignal y F (t) from quetion c) i) i alo ampled with ampling time T = π 4. Sketch the amplitude pectrum of the filtered and ampled ignal ȲF (ω). (Ue the fourth prepared frame on the next page) i ( Point) You want to ue the filter decribed in b) i) for another unknown ignal. What i the maximum ampling time you have to ue in order to uppre aliaing completely?
Page 8 Sample Examination Digital Control Sytem.5 Prepared frame for quetion b) i) Y (ω).5.5 3 4 5 6 7 8 9.5 Prepared frame for quetion b) Ȳ (ω).5.5 3 4 5 6 7 8 9.5 Prepared frame for quetion c) i) F(ω), YF (ω).5.5 3 4 5 6 7 8 9.5 Prepared frame for quetion c) ȲF (ω).5.5 3 4 5 6 7 8 9 ω[ rad ]
Sample Examination Digital Control Sytem Page 9 Solution 4 a) i) ( point) The frequencie ω i and amplitude A i can be read directly from the ignal definition: ω = rad, ω = 3 rad and ω n = 7 rad A =, A = and A n =.5 ( point) The ampling frequency ω S can be calulated by ω S = π T = 8rad and the Nyquit frequency ω N i ω N = ω S = 4 rad The ampling ha the following effect on the ampled ignal: The ignal y(t) get uperimpoed with infinitely many copy of the ignal y n (t), with each of their amplitude pectra hifted by πn T repectively. Thi create the ignal component at the frequencie ω = {, 5, 6} rad rad. The alia of the noie at ω = < ω N lie in the frequency range of interet and thereby ditort the ignal. b) i) ( point) The filter uppree all ignal component above 6 rad. The noie i uppreed, wherea the two ignal component at frequencie lower than the Nyquit frequency are left untouched. i ( point) Now, that there are no more component above the Nyquit frequency (noie) in the filtered ignal, no aliae appear for frequencie lower than the Nyquit frequency. In the frequency range of interet the ignal i not ditorted anymore. ( Point) The filter cut off all ignal content above a frequency of 6 rad. Thi i the lower limit for the Nyquit frequency ω N : ω N,min = ω S,min = 6 rad T max = π ω S,min = π ω N,min = π 6
Page Sample Examination Digital Control Sytem.5 Solution for quetion b) i) Y (ω).5.5 3 4 5 6 7 8 9.5 Solution for quetion b) Nyquit Limit Sampling Frequency Ȳ (ω).5 Alia of noie.5 3 4 5 6 7 8 9.5 Solution for quetion c) i) F(ω), YF (ω).5.5 3 4 5 6 7 8 9 ȲF (ω).5.5.5 Solution for quetion c) Nyquit Limit Sampling Frequency 3 4 5 6 7 8 9 ω[ rad ]
Sample Examination Digital Control Sytem Page Quetion 5 (Controller Synthei) Point You have to deign a dicrete controller C(z) for the plant P (z) and the given feedback tructure in figure 3. reference C(z) P (z) output Figure 3: Feedback Control P (z) = αz + β z + γz + δ, with α, β, γ, δ R, and T = a) Root-Locu Deign: Aume the controller i C(z) = k p, with k p R + i) ( Point) Where do the pole of the tranfer function of the cloed-loop control ytem T (z) = L(z) +L(z) converge to for k p = and for k p? ( Point) Aume α =.5, β =.45, γ =., δ =. Figure 4 (next page) how the Nyquit diagram of the plant P (z). Doe the Nyquit diagram enure tability of T (z) for k p =? Give reaon for your anwer. b) Plant Inverion: i) ( Point) Aume you want to control P (z) uch that the tranfer function T (z) become: T (z) = A z + Bz +., with A, B R Find a pair of numerical value for A and B uch that all pole of T (z) are inide the nice pole region and the teady tate gain of T (z) i one..5.4.3.. -. -. ω T = π Nice Pole Region ω T = π -.3 -.4 δ =.5 -.5..4.6.8 ( Point) Determine the controller C(z) which realize T (z) for the plant P (z) with the pole in the nice pole region and the teady tate gain of T (z) equal to one. Doe C(z) contain an open integrator? c) MIMO-tuning: Aume now, that a dicrete-time plant of the following form i given: x k+ = F x k + Gu k, y k = Cx k
Page Sample Examination Digital Control Sytem.5 Imaginary Axi.5 -.5 ω = P ( z = e j(π ɛ)) p = {, } P ( z = e jɛ) - -.5-3.5-3 -.5 - -.5 - -.5 Real Axi Figure 4: Nyquit Diagram of the plant P (z) An engineer deigned a dicrete-time tate-feedback-controller (controller gain K) with an oberver (oberver gain H) to control thi ytem. To find the gain K and H, he olved the LQR problem for {F,G,Q,ρ I} and {F T,C T,GG T,μ I}, repectively. He propoed three different deign which only differ in ρ and μ. The pole of thee deign are given in the plot below, with: ( * = F, o = F GK, + = F HC). Deign a) Deign b) Deign c).5.5.5 Im Im Im -.5 -.5 -.5 - - Re - - Re - - Re Note: Pleae anwer the following quetion with a,b,c or none and explain why. i) ( point) Which of the deign ha the lowet tate-feedback-controller? ( point) Which deign ha the lowet tate-oberver? Solution 5 a) i) The pole of the cloed-loop ytem T (z) are z p,, = (γ + k pα) ± (γ + k p α) 4(δ + k p β)
Sample Examination Digital Control Sytem Page 3 k p = : The pole of the cloed-loop ytem T (z) are equal to the pole of the plant P (z): k p = z p,, = γ ± γ 4δ k p : The pole of the cloed-loop ytem T (z) have to fulfill the following equation: z p,, + γz p,, + δ + k p (αz p,, + β) = For k p the pole of T (z) move to the zero of the plant P (z): k p z p, ign(α), z p, β α P (z) =... z + γz + δ pole: z p,, = γ ± γ = {, γ} = {,.} P (z) ha one untable pole and the gain P (e jϕ ) for ϕ [ɛ, π ɛ] encloe the critical point - one time ccw. According to the Nyquit criterion the cloed-loop ytem with k p = i table. b) i) Steady tate unity gain condition: A = + B +. () Pole: p, = B± B.4 make a imple choice and chooe B =.4 to obtain a double real pole at.4 =.36 which i clearly inide the nice pole region and calculate A =.4+. =.4657. C(z) = T (z) P (z) T (z) = z + γz + δ A αz + β z + Bz +. A an open integrator mean a pole at z =. Looking at the denominator we obtain the condition: (αz + β) (z + Bz +. A) z= = α + αb +.α αa + β + βb +.β βa = α ( + B +. A) + β ( + B +. A) = And from equation follow that + B +. A =. The controller C(z) therefore contain an open integrator. c) i) b) i the lowet tate-feedback-controller becaue thi deign ha it eigenvalue of F GK in the lowet location (compared to the other deign) in the complex plane. c) i the lowet tate-oberver becaue thi deign ha it eigenvalue of F HC in the lowet location (compared to the other deign) in the complex plane.