Digital Control System: Analysis and Design

Running head: DIGITAL CONTROL SYSTEM
DIGITAL CONTROL SYSTEM
Name of the Student
Name of the University
Author Note

DIGITAL CONTROL SYSTEM1
Question 1:
a.
The discrete linear time invariant system is given below.
x(k+1) = [0.1 0 0.1
0 0.5 0.2
0.2 0 0.4 ]x ( k ) + [ 0.01
0
0.005 ]u(k )
Here, A = [0.1 0 0.1
0 0.5 0.2
0.2 0 0.4 ], B = [ 0.01
0
0.005 ], C = [1 0 0
0 1 0
0 0 1 ]
Hence, the controllability matrix will be,
Con = [B,AB,A^2B]
Now, AB = [0.1 0 0.1
0 0.5 0.2
0.2 0 0.4 ]∗
[ 0.01
0
0.005 ] = [0.0015
0.001
0.004 ]
A^2*B = [0.1 0 0.1
0 0.5 0.2
0.2 0 0.4 ]2
∗
[ 0.01
0
0.005 ] = [0.00055
0.0013
0.0019 ]
Con = [ 0.01 0.0015 0.00055
0 0.001 0.0013
0.005 0.004 0.0019 ]
Now, the rank of controllability of C is 3 or C is a full rank matrix and hence the system is
controllable.
b. Now, the characteristics equation is obtained from the system transfer function which is
given by,
T(s) = C*(sI- A)^(-1)*B. This is computed by the following MATLAB code.

DIGITAL CONTROL SYSTEM2
Hence, the characteristics equation of the system is the denominator of the transfer function.
c.
T(s) = ( 100∗( 100∗s3 – 100∗s2+27∗s – 1 ) )
Now, desired closed loop poles at z = 0.4 + j0.4, z = 0.4 – j0.4 and at z = 0.1.
Hence, desired characteristics equation
Td(s) = (s-(0.4+j0.4))(s-(0.4-j0.4))(s-0.1) = ((s-0.4)^2 + 0.16)(s-0.1) = (s^2 -0.8s +0.32)(s-
0.1)
= s^3 -0.1s^2 -0.8s^2 + 0.08s + 0.32s -0.32
= s^3 -0.9s^2 +0.4s – 0.32-100
d. Now, the gain will be such that the system characteristics equation goes close to the
desired characteristics equation.
Hence,

DIGITAL CONTROL SYSTEM3
1 + K*[100,-100,27] = [1,-0.9,0.4]
This is obtained in MATLAB code:
Question 2:
Given, sampling time Ts = 0.25 sec and transfer function
G ( s )= 0.2 ( s−1 )
( 0.25 s−1 ) ( 0.5 s−1 ) = 0.2 s−0.2
0.125 s2−0.75 s +1
The above transfer function can be discretized by using the following MATLAB code.
MATLAB code:

DIGITAL CONTROL SYSTEM4
0.7715(1−1.289 z−1 )
( 1−2.7183 z−1 ) ( 1−1.6487 z−1 )
= z−k∗B ( z )
A ( z )
b. Now, separating the bad and the good parts of A and B
Ag = ( 1−1.6487 z−1 )
Ab = ( 1−2.7183 z−1 )
Bg = 0.7715
Bb = (1−1.289 z−1 )
K = 1
Tc(z) = Ag T 1 , Rc=Bg R1 , Sc= Ag S1 , Bγ ( z ) =Bb T 1
Here, let assume T 1=1, γ = Φc ( 1 )
Br (1 ) .

DIGITAL CONTROL SYSTEM5
Here, Φc ( z ) = desired closed loop characteristics equation.
Now, for finding the location of the desired poles
P=Pexp ( ± jω )
Now, rise time must be less than 2 secs.
Rise time <= 2s => N <= 2/Ts = 2/0.25
 N<= 8 => N = 8
Hence, ω = π/2N = π/16 = 0.1963
Now, overshoot must be less than 10% or less than 0.1.
P <= 0.1
ω
π = 0.1
0.1963
π = 0.8657 => P <= 0.8657
 P = 0.87
Φc1 ( z ) = ( z−Pexp ( jω ) ) ( z−Pexp ( − jω ) )
= ( z−0.87 exp ( j 0.1963 ) ) ( z−0.87 exp (− j 0.1963 ))
= z^2 -0.87z*exp(− j0.1963 ¿ -0.87z*exp( j0.1963 ) + 0.87^2
= z^2 -0.87z(cos(0.1963) –jsin(0.1963)) – 0.87z(cos(0.1963) + jsin(0.1963)) + 0.7569
= z2−1.7066 z+ 0.7569
= 1 – 1.7066 z−1+ 0.7569 z−2
Now, from the Aryabhatta’s identity:
Ab∗R1 + z (−k ) Bb δ 1=Φc 1 ( z) (1)
Here, δ1=a+ b z−1