- 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 1 (113.91 KiB) Viewed 49 times
- 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 2 (73.35 KiB) Viewed 49 times
- 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 3 (80.34 KiB) Viewed 49 times
- 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 4 (63.06 KiB) Viewed 49 times
- 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 5 (63.93 KiB) Viewed 49 times
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) given by 1.5(z - 0.90) Da(z) = K 2 - 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(z)/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] R(2) Y(z) 1-e-ST 8 Da(z) + T = 0.2 S S +4 Compensator ZOH Motor Ga(z)
USEFUL FORMULAE (Symbols have their usual meaning) 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 1 2. t TZ (2-1) КТ S n! z 3. dn lim(-1)" da" (KT)" S +1 a- Z-e ат 1 z 4 4. e* e akt Sta 2-e n! 5. the-ar (s + a)"+1 dn z (-1)" daz-e-T (KT)'e akt 0 6. sin oot z sinoT z? - 2z cos oT + 1 sin okt s? +62 7. cos at S s? +62 z(z - COS OT) z? - 2z cos oT + 1 cos okt 0 8. e sin at aT ze sin oT z? - 2ze -al cos OT + -2aT -akt e sin ckT (s + a)? + m2 9. -at cos at z? - ze Z? - 2ze S + (s + a)? +02 e COST al cos oT +e-2aT e -akt cos okT
Laplace transform theorems z-transform theorems 1. L[f(t)] = F(s) = 1*f(t)e-st dt z[af(t)] = aF(z) 0- 2. = Z[kf(t)] = kF(s) L[f,(t)+f;(t)]=F,(s)+F (s) 4. Le-atf(t)] = F(s+a) 3. = z[f, (t) + f (t)]=F, (z)+F (2) zle-atf(t)]=F(etz) z[f(t - nT)]=z"F(z) = 5. L[f(t-T)]= e STF(s) е z[tf(t)]= = -Tz dF(z) dz 6. L[f(at)) = 2F ( f(0) = lim(1 - z ')F(z) a 7. df L dt SF(S) - f(0-) f(0+) = lim F(z) = - 8. df at² = sạF(s) - sf(0-) - f(0-) n 9. L d'f dth = s"F(s) – Ës*-*pk+(0-) -1 k=1 10. 4[161 f(t)dt mot] - - S 11. f() = lim sF(s) 5-0 Final Value Theorem 12. f(0+) = lim SF(s) $-
Time Response Frequency Response G(S) = s? +250,8+ on me Me 1 25/1-5² x 100 0 = 0 √1-25² (5/17-3*) %OS = e - In(%OS/100) 5 = va? + In? (%OS/100) BW = 0 = 0,V(1-252)+1454 + 2 TT 25 To = O = tan 0, 11-32 ✓-25² +81 +45° + 4 T = ζω, Cao = tan1. 1-B 2/B sin-1 1-B 1+ B Steady-State Error max - 1 TB 1 e() = estep (o)= 1 + lim G(S) 1 3-0 G. (jo max) = VB Ke = lim G(S) 5-0 Digital Control 1 e()= e ramp (0) = limsG(S) 5-0 e* () = lim(1 - z ')E(s) -Z K =lim SG(s) 50 Ke = lim G(z) 21 e() = e parabola 1 (0) = lim s’G(s) T 251 5-0 K, = lim(2 – 1)G(z) K. -7 lim(2 – 16(2) K, = lim s’G(s) 50 2-1
Stability Testing Routh Array For the continuous characteristic equation a,s" + an-sh-1 +..... +as+ao =0 the Routh Array is given by sh an an-2 an-4 an-6... an an | an-1 an-3 an-jan-2-anan-3 an-1 an an-4 Jan-1 an-5 an1 an-1 an-3 an-5 an-7... C1 = C2 = an-1an-4 - anan-5 an-1 an-1 where sn-2 C3. C1 dh C2 d2... Jan 1 an- C1 C2 Jan-1 an 5 C1 C3 C1 Gjan-3-an-102 do C1 d = Can-5-an-103 C1 : : C1 : : s' j1 CO k1 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)]z=1> 0 >0 (n even) (ii) [F(z)],—-1 <o in odd) (iii) |a0|<lan and within the Jury array below 1601>\6n-11. |co|>lcn-2, etc. ao a a2 an an-1 an-2 bo by b2 br-1 n-2 bn-3 Со C1 C2 n-2 n-3 Cn-4 an-2 an-1 an az a ao bn-2 bn-1 61 bo Cn-2 lao ankl bk = an ak K = 0, 1,..., n-1 bo bn-1-i Ci bn-1 bi i = 0, 1, n-2, etc. Со