Electrical Engg. Control System Solution

Electrical Engg.
Control System
Solution 1 )
a)
. y (t) = 3 e-t u(t)
. y1 (t) = 3 e-t u1(t)
. y2 (t) = 3 e-t u2(t)
. a y1 (t) + b y2 (t) = 3a e-t u1(t) + 3b e-t u2(t)
= 3 e-t [ a u1(t) + b u2(t)]
It is a linear system.
b)
It is a non-linear system ( as it contains a product term of y(t) and dy ( t ) / dt ).
Solution 2)
a) Equilibrium Point :
(0,0) represents the equilibrium point.
b) Linearized System :
. d2y/dt2 + 6 dy/dt + 5 y = - e-2t u(t)

c) Transfer Function :
Y(s) / U(s) = 1 / (s+2) (s2+6s+5)
d) Unit Impulse Response :
U(s) = 1
Y(s) = 1 / (s+2) (s2+6s+5) = 1/8 (s+1) -1/3(s+2) + 1/12(s+5)
. y( t ) = 1/8 e-t -1/3 e-2t + 1/12 e-5t
Step Impulse Response :
U(s) = 1/s
Y(s) = 1 / s(s+2) (s2+6s+5) = -1/8 (s+1) +1/6(s+2) - 1/60(s+5) + 1/10s
. y( t ) = -1/8 e-t +1/6 e-2t - 1/60 e-5t + 1/10
e) U (s) = 1/s2 ( 1 – e –st1 + t1/t2-t3 (e-st2-e-st3))
f) t1 = 2, t2 = 4, t3 = 8 and t = 16 s