Difference Equation From State Space Canonical Form of Discrete System

Question 1
The linear system is given as
X ( t ) = AX ( t ) +Bu( t )
But
u ( t )=u(kT )
Hence
X ( t )= AX ( t ) +Bu ( kT ) for kT ≤ t< ( k +1 ) T
If
A=
[1 0
0 1 ], B= [1
0 ]
From state space canonical form of a discrete system, the system is expressed in the
form of
[ x (k +1)T
x1 (k +1) T ]= [−a1 − a2
1 0 ] [ x (kT )
x1 (kT ) ]+[b
0 ]u (kT )
Where
x1 ( k ) T =x (k − 1) T
Substituting the variables in the question, we have
[ x (k +1)T
x1 (k +1) T ]= [−a1 − a2
1 0 ] [ x (kT )
x1 (kT ) ]+[b
0 ]u (kT )
[ x1 (k +1) T
x (k +1)T ]= [ 1 0
0 1 ] [ x1 ( kT )
x (kT ) ] + [ b
0 ] u(kT )
x1 ( k+ 1 ) T =x1 (kT )+ bu(kT )
x ( k +1 ) T =x ( kT )
Which gives the difference equation shown below
x ( k +1 ) T − x (k −1)T =u (kT )
Comparing with the desired difference equation
X [ ( k +1 ) T ] =Φ ( T ) X ( kT ) + φ ( T ) u(kT )
From which
Φ ( T )= [0
1 ]
φ ( T ) = [ 1
0 ]

Question 2
From Real-translation, time advance property of the Z transform,
Z { x ( k +n ) T }=Zn
[ X ( z ) −∑
k=0
n −1
x (kT )Z− k
]
From the difference equation provided
c ( k + 4 ) +2 c ( k +2 ) − c ( k +1 ) +c ( k )=5 u(k )
Taking Z transform on both sides and obtaining the transfer equation, we assume 0
initial conditions since no condition was provided
Z4 C ( Z ) +2 Z2 C ( Z ) − ZC ( Z ) +C ( Z )=5 U (Z)
C ( z )
U ( z) = 5
Z 4 +2 Z2+ Z +1
Using MATLAB, the system’s state space is obtained from the MATLAB script
below
z=tf('z')
NUM = [0 0 0 0 5 ]; % NUM and DEN should be same length
DEN = [1 0 2 1 1];
[A,B,C,D] = tf2ss(NUM,DEN)
z =
z
Sample time: unspecified
Discrete-time transfer function.
A =
0 -2 -1 -1
1 0 0 0
0 1 0 0

0 0 1 0
B = 1
0
0
0
C = 0 0 0 5
D = 0