Control Systems Signals and Systems Review

Control
systems

QUESTION 1
REVIEW OF SIGNALS AND SYSTEMS
́x (t ) +2 ζ wn ́x ( t ) =wn
2 U ( t )
ζ =0.707
wn=2
́x ( t ) +2 ( 2 ) ( 0.707 ) ́x ( t ) =4 U ( t )
́x (t ) +2.828 ́x ( t )=4 U ( t )
Performing Laplace transform considering zero initial conditions,
S2 X ( s ) +2.828 SX ( s ) =4 U ( s )
X ( s )
U ( s ) = 4
s2 +2.828 s
Performing an inverse Laplace transform to determine the time series.
Using the partial solutions,
L−1
{ 4
s2 +2.828 s }
4
s2+ 2.828 s = 1000
707 s − 1.41443
s +2.828
L−1
{1000
707 s − 1.41443
s+ 2.828 }
L−1
{1000
707 s }−L−1
{ 1.41443
s+ 2.828 }
x (t )= 1000
707 −1.41443 e−2.828 t
Matlab implementation to plot the time response,
t=0:0.1:20;
Xt=(1000/707)-(1.41443)*exp(-2.8282*t);
figure(1)
plot(t,Xt,'b-.')
1

grid on
title('Time Response ')
xlabel('time(sec)')
ylabel('x(t)')
legend('x(t)')
0 2 4 6 8 10 12 14 16 18 20
time(sec)
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
x(t)
Time Response
x(t)
Part C
Changing the value of ζ =0.005, the bode diagram for the system is given as,
́x ( t ) +2 ( 2 ) ( 0.005 ) ́x ( t ) =4 U ( t )
X ( s )
U ( s ) = 4
s2 +0.02 s
To obtain the magnitude,
−10 log [1−2 ( 1−2 ζ2 ) ( ω
ωn )2
+ ( ω
ωn )4
]
To obtain the phase,
2