Electrical Engineering Control Systems Homework Solution - Analysis

This document presents a detailed solution to a control systems homework assignment. It begins by analyzing the linearity of systems, distinguishing between linear and non-linear behaviors. The solution proceeds to identify the equilibrium point and linearize a system, followed by deriving the transfer function. The unit impulse and step responses are calculated, providing a comprehensive understanding of system behavior. Additionally, the solution includes MATLAB code for simulating and visualizing the system's response, accompanied by plots. Finally, a comparison is made between the linear and non-linear system responses, highlighting their similarities and differences. The assignment concludes with references to relevant research articles, offering further context and supporting the analysis.

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

Solution f)
Figure 1
t = 0:0.1:2
u = t;
plot(t,u)
hold on
t = 2:0.1:4
u = 2
plot(t,u)
hold on
t = 4:0.1:8
u = 0.5*(8-t)
plot(t,u)
xlabel(‘time’)
ylabel(‘Amplitude’)

g)Matlab Code:
syms y(x)
Dy = diff(y);
ode = diff(y,x,2)+6*diff(y)+5*y == exp(-2*x)
cond1 = y(0) == 0;
cond2 = Dy(0) == 0;
conds = [cond1 cond2 ];
ySol(x) = dsolve(ode,conds);
ySol = simplify(ySol)
display(ySol)
x=0:16
plot (x,ySol(x))
xlabel(‘time’)
ylabel(‘Amplitude’)

Figure 2
COMPARISON:
The comparison shows that the linear and the non linear systems give approximately the same
results in the proximity of the equilibrium point. However, the response is somewhat
different at the points away from the equilibrium points.
References:
[1] Spiegler, Virginia LM, et al. "A technique to develop simplified and linearised models of
complex dynamic supply chain systems." European Journal of Operational Research 251.3
(2016): 888-903.
[2] Zhen, Chong, et al. "Feedback linearisation of nonlinear vibration problems: A new
formulation by the method of receptances." Mechanical Systems and Signal Processing 98
(2018): 1056-1068.