Question

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B2. The characteristic equation of a system is given below. Determine and explain the stability of the system 4s8s +2+10+3 = 0 B3. For a unity feedback system with the plant G(8) 500 (8 + 20%(2 + 4s +10) find the steady-state error for the following test inputs: a. 150(1) (i.e. 15 x unit step input) b. 151(1) (i.e. 15 x unit ramp input) 151 () (i.e. 15 x unit ramp input) (6 marks) (6 marks) (8 marks) c.

Answer 1

B2 Given, characteristic Equation 4844853+25+105+3=0. The Routh table shown below. S4 4 2 3 18+ oso 8067 10 882-1084 8 -3 3 18 -3710-843 3 sº 3 18*3-0 18 3. In Routh table, two sign changes occorre Then system is unstable. 2-poles lies in Right Hand side. system is unstable. Gives, Soo GES) (S+20)(574SHO) a) r(t) = 150(+). The steady state ensor is given by ess- lim S. R(1) ltas so R25) = 15 15 ess= lim so SI+GC)

ess=lim sro 15 1+ 500 (5426(575+) ess ess 1 + 500 0+20) (0+0HO) 15 1+ 500 20 XT ess o 15 1+ 2 ess 4.285 b). V(t) = 15+ ult) Steady state The ensor is given by ess - lins. RCS) sto 1+6CS) RCS) = 15152 - ess= lem sto I+ 500 6+20)+45 +0) ess - 1+ 500 20 XTO 15 ess ess ОЮ - ess infinity

c) V(+) =15+UCH) RCS = 15x2 R(S) = 30 83 ess = lim S. S to R(S) 1+G) ess llim x 30 C ܘܠܠ 500 1+ 5+20) S+45+10) ess lin 30 30 sto S2 1 + БОО S+2075745710) ess 11 30 10) 178.5 essa co