VAAL UNIVERSITY OF TECHNOLOGY FACULTY OF ENGINEERING

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1 VAAL UNIVERSITY OF TECHNOLOGY FACULTY OF ENGINEERING DEPARTMENT: PROCESS CONTROL AND COMPUTER SYSTEMS BACCALAUREUS TECHNOLOGIAE: ENGINEERING ELECTRICAL SUBJECT : DIGITAL CONTROL SYSTEMS IV EIDBS4A ASSESSMENT : UNIT 1 FIRST ASSESSMENT DATE : 24 AUGUST 2017 DURATION : 10H00 11H30 EXAMINER : R FITCHAT MODERATOR : L COETSEE REQUIREMENTS: Pocket calculators may be used INSTRUCTIONS: 1. Neat work is required 2. Number your answers clearly and correctly Full marks = 50 Total = 50 QUESTION PAPER CONSISTS OF: 1 typed page DO NOT TURN THE PAGE BEFORE PERMISSION IS GRANTED

2 Digital Control Systems IV EIDBS4A Unit 1 First Assessment 24 August 2017 Page 1 Question 1 Find the z transform X(z), of the following time functions x(k): a) x(k) = k(k + 1) u(k) (3) b) x(k) = (-1) k u(k) (2) Question 2 The z transform X(z) of a function x(k), is given by: X(z) = z(0.1065z ) (z 1)(z 2 1.5z ) a) Determine an expression for x(k), by finding the inverse z transform of X(z). (7) b) Use the expression obtained in Question 2 a) for x(k), to calculate x(7). (3) Question 3 Determine the pulse z transform, X(z), of the following functions X(s). Assume that the sampling period is 0.5 sec. (T = 0.5). a) X(s) = (s 4) (s + 2)(s 5) (3) b) X(s) = 1 s (s 2 1) Question 4 Solve for x(k) from the following difference equation, using z transform methods: x(k + 2) x(k) = 1 for k 0, and with x(0) = 0 and x(1) = 0. (10) Question 5 a) Find the transfer function, C(z)/R(z), for the system in Figure 1, if the sampling period is 0.1 second (T = 0.1). (15) r(t) T = 0.1 ZOH s(s 5 2) b) Estimate the overshoot D, of the output c(t), if a step signal r(t) = u(t). is applied to the input of the digital control system in Figure 1. (3) c(t) (4) Figure 1 ---ooo000ooo--- Total: 50 Appendix: z transforms X(s) x(t) x(kt) X(z) X(s) x(t) x(kt) X(z) 1 (t) (kt) 1 e -st X(s) x(t-t) x[(k-1)t] z -1 X(z) 1/s 1 1 z/(z-1) x[(k-2)t] z -2 X(z) 1/s 2 t kt Tz/(z-1) 2 x[(k+1)t] zx(z)-zx(0) 2/s 3 t 2 (kt) 2 T 2 z(z+1)/(z-1) 3 x[(k+2)t] z 2 X(z)-z 2 x(0)-zx(1) 1/(s+a) e -at e -akt z/(z-e -at ) a k z/(z-a) 1/(s+a) 2 te -at kte -akt Tze -at /(z-e -at ) 2 ka k az/(z-a) 2 a/s(s+a) 1-e -at 1-e -akt z(1-e -at )/(z-1)(z-e -at ) X(s+a) e -at x(t) e -akt x(kt) X(ze at ) (sa)2 sa (sa)2 e -at sint e -akt sinkt e -at cost e -akt coskt ze -at sin T z 2 -(2e -at cost)ze - 2ab k cos(k+) 2aT z 2 -ze -at cost z 2 -(2e -at cost)ze -2aT 1-e-sT s za za z-b z-b X(s) (1-z -1 X(s) )Z s

3 Faculty: Department: Diploma: Subject: Internal code: VAAL UNIVERSITY OF TECHNOLOGY Engineering Cover Sheet for Memorandum (Unit 1 First Assessment 24 August 2017) Process Control and Computer Systems Baccalaureus Technologiae: Engineering: Electrical Digital Control Systems IV EIDBS4A Nine digit code: Hours: Examiner: Moderator: 1½ R Fitchat L Coetsee Total Marks: 50 Full marks: 50 Signature of Examiner: Date: Signature of Moderator: Date:

4 Digitale Beheerstelsels IV EIDBS4A Eenheid 1 Eerste Evaluasie 24 Augustus 2017 Memorandum Bladsy 1 1. a) x(k) = k(k+1) = k 2 +k X(z) = Z{k 2 }+Z{k} = [z(z+1)/(z 1) 3 ] + z/(z 1) 2 {= 2z 2 (z 1) 3 } (3) [5] b) x(k) = ( 1) k X(z) = z/[z ( 1)] {using a k z/(z-a)} X(z) = z/(z+1) (2) 2. a) X(z)/z = (0.1065z )/(z 1)(z 2 1.5z ) (z 2 1.5z =0z=0.8347± ) = (0.1065z )/(z 1)(z r )(z r ) [2] =1/(z 1) /(z ) /(z ) [3] X(z)=z/(z 1)+( )z/(z )+( )z/(z ) x(k) = u(k) ( k )cos(0.4543k )u(k) = ( k )cos(0.4543k ) k 0 [2] (7) [10] [7] [10] [18] b) x(7) = ( )cos( ) = ( )cos( ) = = (3) 3. a) X(s) = (s+4)/(s+2)(s+5) = /(s+2) /(s+5) [2] X(z) = z/(z e 2T ) z/(z e 5T ) = z/(z e 1 ) z/(z e 2.5 ) = z/(z ) z/(z ) [1] (3) b) X(s)=1/s(s 2 +1)=1/s(s+j)(s j)=1/s 0.5/(s+j) 0.5/(s j)=1/s [0.5(s j)+0.5(s+j)]/(s+j)(s j)=1/s s/(s 2 +1) [3] X(z)=z/(z 1) (z 2 zcost)/(z 2 2zcosT+1)=z/(z 1) (z z)/(z z+1) [1] (4) or X(s)=1/s(s 2 +1)=A/s+(Bs+C)/(s 2 +1)=[A(s 2 +1)+(Bs+C)s]/s(s 2 +1)=[(A+B)s 2 +Cs+A]/s(s 2 +1) A+B=0, C=0 and A=1 A=1, B= 1 and C=0 X(s) = 1/s s/(s 2 +1) 4. x(k+2)+¼x(k)=1 [z 2 X(z) z 2 x(0) zx(1)]+¼x(z) = Z{1} [z 2 X(z) z 2 0 z0]+¼x(z)=z/(z 1) [2] z 2 X(z)+¼X(z)=z/(z 1) X(z)(z 2 +¼)=z/(z 1) X(z)=z/(z 1)(z 2 +¼) X(z)/z=1/(z 1)(z 2 +¼) [1] X(z)/z = 1/(z 1)(z ½/2)(z ½ /2) [1] X(z)/z = 0.8/(z 1) /(z ½/2) /(z ½ /2) [3] X(z) = 0.8z/(z 1) + ( )z/(z ½/2) + ( )z/(z ½ /2) x(k) = (½) k cos(k/ ) = (½) k cos[k(/2) ], k 0 [3] {or: X(z)/z = 1/(z 1)(z 2 +¼) = 0.8/(z-1) 0.8[(z+1)/(z 2 +¼)] X(z) = 0.8z/(z 1) 0.8[(z 2 +z)/(z 2 +¼)] X(z) = 0.8z/(z 1) 0.8[z 2 /(z 2 +¼)] 0.8[z/(z 2 +¼)] = 0.8z/(z 1) 0.8{z 2 /[z 2 +(½) 2 ]} 1.6{½z/[z 2 +(½) 2 ] x(k) = (½) k cos(/2)k 1.6(½) k sin(/2)k (let T = 1, e -at = ½ and T = /2 in table) x(k) = (½) k [ 0.8cos(/2)k 1.6sin(/2)k] = (½) k {1.7889cos[(/2)k ]} x(k) = (½) k cos[(/2)k ] 5. a) G(z) = (1 z -1 )Z{P(s)/s} = (1 z -1 )Z{5/s 2 (s + 2)} = [(z 1)/z]Z{2.5/s /s /(s + 2)} [3] = [(z 1)/z][0.25z/(z 1) z/(z 1) z/(z e 0.2 )] [3] = 0.25/(z 1) (z 1)/(z e 0.2 ) = [0.25(z e 0.2 ) 1.25(z 1)(z e 0.2 ) (z 1) 2 ]/(z 1)(z e 0.2 )] = [0.25z 0.25e (z 2 z ze e 0.2 ) (z 2 2z + 1)]/(z 1)(z e 0.2 )] = [0.25z 0.25e z z ze e z 2 2.5z )]/ (z 1)(z e 0.2 )] = [(1.25e 0.2 1)z + ( e 0.2 )]/(z 1)(z e 0.2 )] = ( z )/(z 1)(z )] [5] = (z )/(z z ) T(z) = G(z)/[1 + G(z)] = [( z )/(z 1)(z )]/ {1 + [( z )/(z 1)(z )]} = ( z )/[(z 1)(z ) + ( z )] = ( z )/(z z z ) = ( z )/(z z ) [4] (15) b) z z = 0 z = ± j = ± [½] e 0.1 = = ln( ) = [½] and d 0.1 = d = [½] D = e -/d = e / = [1½] (26.44%) (3)

5 VAAL UNIVERSITY OF TECHNOLOGY FACULTY OF ENGINEERING DEPARTMENT: PROCESS CONTROL AND COMPUTER SYSTEMS BACCALAUREUS TECHNOLOGIAE: ENGINEERING ELECTRICAL SUBJECT : DIGITAL CONTROL SYSTEMS IV EIDBS4A ASSESSMENT : UNIT 2 FIRST ASSESSMENT DATE : 21 SEPTEMBER 2017 DURATION : 10H00 11H30 EXAMINER : R FITCHAT MODERATOR : L COETSEE REQUIREMENTS: Pocket calculators may be used INSTRUCTIONS: 1. Neat work is required 2. Number your answers clearly and correctly Full marks = 50 Total = 50 QUESTION PAPER CONSISTS OF: 1 typed page DO NOT TURN THE PAGE BEFORE PERMISSION IS GRANTED

6 Digital Control Systems IV EIDBS4A Unit 2 First Assessment 21 September 2017 Page 1 r(t) T = 0.5 K T = 0.5 ZOH s(s 1 1) c(t) Figure 1 Question 1 Refer to the digital control system in Figure 1, with P(s) = 1/s(s+1), ZOH, a proportional controller with gain K and a sampling period of T = 0.5 sec. a) Determine the characteristic equation Q(z), of the system. (10) b) Use the Jury test to determine the marginal values of the gain K, that will ensure stability of the system. (10) Question 2 a) Draw the root locus for the system in Figure 1, for variations in the value of K. (8) b) Use the root locus to determine the marginal value of K for stability. (5) Question 3 Consider the digital control system shown in Figure 2. Design a digital controller B(z), such that the dominant closed loop poles have a damping ratio of 0.5 ( = 0.5) and a settling time of 4 sec. (t s = 4). Use a sampling period of 0.5 second (T = 0.5) for your design. Assume a first order controller device with transfer function of the form: B(z) = K z a z b (17) R(z) T=0.5 Digital controller B(z) T=0.5 Zero order hold. P(s) 1 s(s 1) C(z) Appendix: z transforms ---ooo000ooo--- Total: 50 X(s) x(t) x(kt) X(z) X(s) x(t) x(kt) X(z) 1 (t) (kt) 1 e -st X(s) x(t-t) x[(k-1)t] z -1 X(z) 1/s 1 1 z/(z-1) x[(k-2)t] z -2 X(z) 1/s 2 t kt Tz/(z-1) 2 x[(k+1)t] zx(z)-zx(0) 2/s 3 t 2 (kt) 2 T 2 z(z+1)/(z-1) 3 x[(k+2)t] z 2 X(z)-z 2 x(0)-zx(1) 1/(s+a) e -at e -akt z/(z-e -at ) a k z/(z-a) 1/(s+a) 2 te -at kte -akt Tze -at /(z-e -at ) 2 ka k az/(z-a) 2 a/s(s+a) 1-e -at 1-e -akt z(1-e -at )/(z-1)(z-e -at ) X(s+a) e -at x(t) e -akt x(kt) X(ze at ) (sa)2 sa (sa)2 e -at sint e -akt sinkt e -at cost e -akt coskt Figure 2 ze -at sin T z 2 -(2e -at cost)ze - 2ab k cos(k+) 2aT z 2 -ze -at cost z 2 -(2e -at cost)ze -2aT 1-e-sT s za za z-b z-b X(s) (1-z -1 X(s) )Z s

7 Faculty: Department: Diploma: Subject: Internal code: VAAL UNIVERSITY OF TECHNOLOGY Engineering Cover Sheet for Memorandum (Unit 2 First Assessment 21 September 2017) Process Control and Computer Systems Baccalaureus Technologiae: Engineering: Electrical Digital Control Systems IV EIDBS4A Nine digit code: Hours: Examiner: Moderator: 1½ R Fitchat L Coetsee Total Marks: 50 Full marks: 50 Signature of Examiner: Date: Signature of Moderator: Date:

8 Digitale Beheerstelsels IV EIDBS4A Eenheid 2 Eerste Evaluasie 21 September 2017 Memorandum Bladsy 1 1. a) G(z) = (1 z 1 )Z{(1/s)K/s(s + 1)} = K(1 z -1 )Z{1/s 2 (s + 1)} = K[(z 1)/z]Z{1/s 2 1/s + 1/(s + 1)} [3] = K[(z 1)/z][Tz/(z 1) 2 z/(z 1) + z/(z e T )] [2] = K[0.5/(z 1) 1 + (z 1)/(z )] =K[0.5(z ) (z 1)(z ) + (z 1) 2 ]/[(z 1)(z )] = K[0.5z (z 2 z z ) + (z 2 2z + 1)]/[(z 1)(z )] = K(0.5z z 2 + z z z 2 2z + 1)/(z 1)(z ) = K( z )/(z z ) [3] G(z) + 1 = 0 K( z )/(z z ) + 1 = 0 K( z ) + (z z ) = 0 Q(z) = z 2 + ( K )z + ( K ) [2] (10) b) Q(1) = K K = K so Q(1) > K > 0 K > 0 [2] and Q( 1) = 1 ( K ) + ( K )= K therefore Q( 1) > K > 0 K < [2] also a 0 <a K < K < 1 and K > 1 [20] K < [2] and K > [2] For stability: 0< K < [2] (10) 2. a) G(z)= K( z )/(z 1)(z ) (from Question 1 a) Thus Q(z) = G(z) + 1 = 0 K( z )/(z 1)(z ) + 1 = 0 K( z ) + (z 1)(z ) = 0 Im z K = (z 1)(z )/( z ) = ( z z )/( z ) P z =1 dk/dz=[( 2z )( z ) [13] [17] ( z z )] /( z ) z z z z = z z = z z = 0 [4] {z z = 0} z = [ ±( )]/( ) =[ ± ]/ = ± z = and breakaway points are and (8) b) At P z = 1 and from Q(z) = z 2 + ( K )z + ( K ) ( K ) = K = K = (5) 3. t s = 4/ 4 = 4/ = 1 [1] = / n 0.5 = 1/ n n = 2 [1] n = ( 2 + d 2 ) d = ( n 2 2 ) = ( ) = r/s [1] z = e -T d T = e ( ) = r [2] A(z) = (1 z 1 )Z{(1/s)1/(s + 1)} = (1 z 1 )Z{1/s 2 (s + 1)} = (1 z 1 )Z{1/s 2 1/s + 1/(s + 1)} = [(z 1)/z]{Tz/(z 1) 2 z/(z 1) + z/(z e T )]} = 0.5/(z 1) 1 + (z 1)/(z ) = [0.5(z ) (z 1)(z ) + (z 1) 2 ]/[(z 1)(z )] = [0.5z (z 2 z z ) + (z 2 2z + 1)]/[(z 1)(z )] = (0.5z z 2 + z z z 2 2z + 1)/(z 1)(z ) = ( z )/(z 1)(z ) [4] The characteristic equation of the final system is: B(z)A(z)+1=0 with B(z) = K(z )/(z + b) [K(z )/(z + b)][( z )/(z 1)(z )] + 1 = 0 [( Kz K)/(z + b)(z 1)] + 1 = 0 [3] z 2 z + bz b Kz K = 0 z 2 + ( K + b 1)z + ( K b) = r must be a solution of the equation: K b = K b = (1) and K + b 1 = cos = K + b = (2) From (1) and (2): K = [2] and b = [2] B(z) = (z )/(z ) [1] [1] Amplitude [1] Step Response [1] [1] T(z)=(0.3151z )/ (z z ) Rez Samples (sec)

VAAL UNIVERSITY OF TECHNOLOGY FACULTY OF ENGINEERING

VAAL UNIVERSITY OF TECHNOLOGY FACULTY OF ENGINEERING DEPARTMENT: PROCESS CONTROL AND COMPUTER SYSTEMS BACCALAUREUS TECHNOLOGIAE: ENGINEERING ELECTRICAL SUBJECT : CIRCUIT ANALYSIS IV EICAM4A ASSESSMENT