Question

The impulse invariant approximation of an analogue transfer func- tion G(s) including a model of the analogue-to-digital converter in the computer using sample period T = 0.1 seconds, is Q(2) 4.535 x 10-3 2 + 0.905 (z - 1)(2 – 0.7408) (i) In your exam booklet, neatly sketch the z-domain root locus di- agram for Q(2) including location of the asymptotes, and any break-away or break-in points on the real axis. Useful formulae are provided on page 2. 6 marks (ii) What gain is required in a negative unity feedback, proportional control configuration to yield a critically damped response ? What are the resulting closed loop poles ? 4 marks (iii) Explain how you would use a digital finite impulse response lead compensator to improve the transient response of the closed loop system. You are not required to design the compensator, but you should state its general transfer function, and sketch a RLD to illustrate how your approach works. 5 marks What is the steady state error of your closed loop design from part (b)(ii) of this question to (i) a unit step input sequence and (ii) a unit ramp input sequence. Explain your reasoning. 3 marks

Answer 1

solution: Given Q(z) = 4.535x10 z-domain root locus: -3 Z+0.905 (2-1)(z-0.7408) poles location : I, O, THE Leros location : -0.905 To find breakpoints: dk dz K = (2-1)(z-0.7408) 4.53 5x103 (z+0.905) -(x- - 1.7408 +1) 4.535x163 (2+0.905) dz (az – 1.7408) (4.535X163) (2+0.905) - (4.535x16?) (z?-.1.7408240.2400) 2 -6 4.535 x 10 [z+0.905)? > (22-9.7408) (7 7890 5) - (22.1.7408740.700) = 0 2 az 2 .8 z - 2.316 - 3 + 1.7408 Z-0.74080 #174082 => z = 0.865; -2.675

RLD 100 865 0.905 10 0.1968 Break In point (-2.675) Break away point (0.865) There is no need of Assyroptotes P-Z=10 [no. of poles Since -no, o of ters] (?:) Let gain be ko characteristic equation with negative unity feedback becomes, z + (4.535x103 k-1.7408)+ (0.7408 +0.905 X4.53670 2 -3 Z + (4.535x10 K-1.7408)Z + (0.7408 + #.104X102K) =0 for criticall damped system, damping ratio, (coso) Comparing the above equation with standard equation of Z-domain and by using 2² +28 Coso e pe zo eqin 0. 2

f = Vo 10.7408 + 4. 104 x 103 k and also 4.535 X 10 k -1.7408 -3 2.267X10 K-0.87014 2 3 from ② f ③, K 0.106, 30402 closed loop poles: o for K = 0.106 C'E=> x²_107403 Z+0.7412=0 poles: 0.996, 0.7H3 o for k= 30402 C.E=> z? +136.132 Z +125.51 = 0 poles: -0,928 -0,928-135.203.
i have given solution for first two parts