Control Systems: Signal Analysis, System Response, and Discrete Time

This assignment delves into the analysis of control systems, focusing on signal processing, system responses, and discrete-time systems. It begins with a review of signals and systems, utilizing Laplace transforms to determine time series responses and employing MATLAB for visualization. The impact of varying damping coefficients on system behavior, particularly in the frequency domain, is examined using Bode diagrams and sinusoidal inputs. The assignment then transitions to discrete-time systems, deriving state-space representations and Z-transforms. Finally, it explores least squares solutions for system identification, creating datasets and analyzing parameter estimates. The assignment uses MATLAB extensively to simulate system behavior and validate analytical results. Desklib provides this solved assignment and other resources for students.

Control
systems
Student name

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 ( 1−2ζ 2 ) ( ω
ωn )2
+ ( ω
ωn )4
]
To obtain the phase,
2

atan 2 [−2ζ ω
ωn
,1−
( ω
ωn )2
]
When the value of w=1.0 rad/sec,
−10 log [1−2 (1−2 ( 0.005 )2 ) ( 1
2 )2
+( 1
2 )4
]
Magnitude=3.9992
Phase=−178.8542
When the value of w=10.0 rad/sec,
−10 log [1−2 (1−2 ( 0.005 )2 ) ( 10
2 )2
+( 10
2 )4
]
Magnitude=0.04
Phase=−179.8854
When the input is sinusoidal, the steady state output for several cycles is obtained using
MATLAB,
3

0 2 4 6 8 10 12 14 16 18 20
time(sec)
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
x(t)
Time Response
x(t)
Ut=sin(10*t);
Xt1=(1000/707)-(1.41443)*exp(-2.8282*t).*Ut;
figure(2)
plot(t,Xt1,'r-.')
grid on
title('Time Response ')
xlabel('time(sec)')
ylabel('x(t)')
legend('x(t)')
Comment
The time domain captures the different input values unlike the frequency domain. The rise time
is different such that the sinusoidal input is different from a constant input. The sinusoidal input
oscillates shortly before it settles down.
QUESTION 2
REVIEW OF DISCRETE TIME SYSTEMS
The dynamic system for the differential equation considering the initial conditions is zero,
4

´x ( t ) +2 ζ ωn ´x ( t ) +ωn
2 x ( t ) =ωn
2 u ( t )
The state space matrix form of the equation,
´x ( t ) = Ax ( t ) +Bu ( t )
y ( t ) =cx ( t ) +Du ( t )
Expressing the equation in transfer function,
s2 X ( s ) +2 ζ ωn sX ( s ) +ωn
2 X ( s ) =ωn
2 U ( s )
X ( s )
U ( s ) = ωn
2
s2 +2 ζ ωn s+ωn
2
The state space matrix form is given as a collection of 4 variables,
A=
[−2 ζ ωn −ωn
2
u ( t ) 0 ]
B= [u (t )
0 ]
C= [ 0 ωn
2 ]
D=0
The discrete time state space equation is given as,
X ( s )
U ( s ) = 4
s2 +2.8284 s+ 4
Assuming there is a zero order hold, ZOH, on the input signal u(t),
A=
[ −2.8284 −4.0
1 0 ]
B= [1
0 ]
C= [ 0 4 ]
D=0
Obtaining the Z transform of the state space equations,
5

zX ( z ) =zx ( 0 ) = Ad X ( z ) + Bd U ( z )
X ( z )= [ z In −Ad ]−1
{ zx ( 0 ) + Bd U ( z ) }
[ z I n− Ad ] −1 z= [ −1
z Ad ]
−1
I n+ Ad z−1 +Ad
2 z−2 +…+ Ad
i z−i+ …
Converting back to time domain and finding the second order difference equation based on the
sampling rate,
[ z I n− Ad ] −1 z=I n+ Ad z−1 + Ad
2 z−2 +…+ Ad
i z−i+…
Z−1
{ [ z In− Ad ]−1 z }= { In , Ad , Ad
2 ,… , Ad
i ,… }
{ Ad
k } k=0
∞
The MATLAB implementation,
disp('Converting to Z transform')
[Z,P,K] = tf2zp(num1,den1);
% Returning the solution to time domain
[nm,den]=zp2tf(Z,P,K)
QUESTION 3
LEAST SQUARES SOLUTIONS
y ( k )=1.5 y ( k −1 )+0.7 y ( k−2 ) =u ( k−1 ) +0.5u ( k −2 ) +e ( k )
Creating datasets,
function [a0,a1]=leastSquareSolutions(x,y)
x=x(:);
y=y(:);
X=[x,ones(numel(x),1)];
a = (X'*X)\(X'*y);
a0=a(2);
a1=a(1);
end
The standard deviation of the parameter estimates is given as,
std=σ2 ( squared variance)
a0=1.72 , a1=2.64
6

The tests are carried out as,
x= 1:5;
y= [4, 7.1, 10.1, 12.5, 14.5];
[a0,a1]=linear_regression(x,y);
plot(x,y,'r*'); %points
hold on;
plot(x,a0+a1*x,'b'); %line
grid on
0 2 4 6 8 10 12 14 16 18 20
time(sec)
0
5
10
15
x(t)
Time Response
x(t)
data1
data2
7