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Question 4 A motor modelled by 8 G(S) S + 4 is controlled by an analogue compensator 2s + 1 D(S) = K S + 3 where the gain K is a positive constant. The control system is to be implemented digitally with a sampling period T = 0.2 seconds. (a) Show that the corresponding z-transfer function of the DC motor preceded by a zero order hold (ZOH) is given by 1.10 Ga(z) = Z-0.45 [6 marks] (b) Use the pole-zero mapping method to show that a digital approximation to the analogue compensator D(s) is given by Da(z) = K 1.5(z - 0.90) z - 0.55 [6 marks] (c) The digital implementation of the control system is given in Figure 4.1. Find the closed loop transfer function Y (2)/R(z). Then, use the Jury stability test to determine the range of K such that the closed loop system is stable. Further, determine the steady state output in response to the unit step signal. [8 marks] 1 R(2) Y(z) 1 1-e-ST е 8 Da(z) T = 0.2 + S S +4 Compensator ZOH Motor Ga(z) Figure 4.1

Table of s-transform and z-transform f(t) F(s) F(z) f(KT) 1 Z 1. u(t) u(KT) s Z-1 2. t 1 SP Tz (z - 1) КТ dn 3. 3. t" . n! s +1 Z lim(-1)" da" z-e (KT)" a 0 at 1 Z 4. e-at akt S + a 2-ear n! 5. t"e at dh Z (-1)" da" z-e a (KT)"e -akT (s + a)+1 0 6. sin at z sinoT 22 - 2z cos oT +1 sin okt s? +62 7. cos ot sto S s? + 02 z(z-COS OT) z? - 2z cos oT + 1 COS OKT 0 8. e at sin cot ze at sinoT COS OT + Z? - 2ze -al -2aT e -akt sin akt (S + a)? + 02 S +a z? - ze al cos oT z- 2ze COS OT+e 9. e -at cos out at 2a7 e -akt cos OKT (s + a)? + m2

Laplace transform theorems z-transform theorems = z[af (t)]=aF(z) 2. + 1. L[f(t)]=F(s) = [<f(t)e-Stdt L[kf(t))=kF(s) 3. L[f, (t)+f;(t)] = F,(s)+F_(s) 4. Leatf(t)]=F(s+a) z f(t)+f(t)}=F,(2)+F2(z) zle*f(t)) = F( ez) z[f(t - nT)]=z "F(z) = dF(z) 5. L[f(t - T)]= e-STF(s) =e z[tf(t)) = = -Tz dz 6. [flat) = f(0) = lim(1 - z ')F(z) 2-1 a 7. df L dt SF(s)-f(0-) f(0+) = lim F(z) 8. L déf dt? = s-F(s) - sf (0-) - f(0-) 9. dhf L dth =s"F(s) - Es»-«pk-1(04) k k=1 10. 4] [sótce L F(s) f(t)dt = S 11. f(0) = lim SF(s) 5-10 Final Value Theorem 12. f(0+)= lim sF(s) 5-

Time Response Frequency Response 02 G(S)= s? +250,5 +0 M 1 25/1-5² (5/+3) x 100 %OS = e 0), = 0, V1-252 <= -- In%OS/100) + In? (%OS/100) BW = 0, = 0,V(1-252)+1454 + 72 1 26 Te Om = tan OV T-25² +81 +45° 4 T = ζω, -1 Amex = tan 1-B 2/B = sin-1 1- B 1+ B 1 Steady-State Error Фmax TB ( )= estep ()= = 1 1+ lim G(S) 1 G.(jQmax) = 5-0 K, = lim G(s) $ 0 Digital Control e(o)= ere (0) = ramp 1 lim SG(S) S-0 e* (-) = lim(1 - z ')E(S) 2-1 K, = lim SG(s) K = lim G(z) 5-0 2-1 1 1 elmo)= e parabola (0) = lims G(s) 2-1 K. = lim(2 – 1)(2) K = lim(2 – 13+G(2) K. = lim sạG(s) 1 5-0

Stability Testing Routh Array For the continuous characteristic equation a,s" + an-45"-1 +.....+as+ao = 0 the Routh Array is given by s an an-2 an-4 an-6. an an-2 an-1 an-3 an-jan-2-anan-3 an-1 an-1 an an-4 an-1 an-5_an-jan-4-a,an-5 an-1 an-1 an-3 an-5 an-7... C = = C2 = = an-1 where SN-2 sh-3 C1 C2 C3. dy d2... : Jan-1 an-3 C1 C2 C1 d1 Can-3--an-102 C1 d2 = Jan 1 an 5 C1 C3 C Can-5-an-1C3 C1 : : . st so 11 kt etc. etc. Jury Array For the discrete time characteristic equation F(z)= anz" + an-12n-1 + ... +212 + 20 Conditions for stability are (0) [F(z)]2-7>0 (ii) [F(z)]z=-1 <0 in odd) > 0 (n even) ( (iii) 120|<lan and within the Jury array below 160116–11. col>Icn-2, etc. ao a az an an-1 an-2 bo ba b2 bn-1 bn-2 bn-3 Со C1 C2 n-2 n-3 n-4 an-2 an-1 an a2 a1 ao bn-2 bn-1 61 Cn-2 bo lao an-k b = an ak K = 0, 1, ..., n-1 bo bn-t-il Ci bn-1 bi i = 0, 1, n-2, etc. Со Brut