Solutions Manual for Digital Control Engineering 3rd Edition by M. Sami Fadali, Antonio Visioli

, Chapter 2 Solutions
2.1 Derive the discrete-time model of Example 2.1 from the solution of the system differential equation
with initial time kT and final time(k+1)T.

The volumetric fluid balance gives the analog mathematical model

d h h qi
 
dt  C
where = R C is the fluid time constant for the tank. The solution of this equation is

1 t
h(t )  e (t t0 ) / h(t 0 )  t e (t ) / q i ()d
C 0




Let qi be constant over each sampling period T, i.e. qi(t) = qi(k) = constant, for t in the interval
[kT, (k+1)T). Then

(i) Let t0 = kT, t = (k + 1)T

(ii) Simplify the integral as follows with : (k  1)T 
1 ( k 1)T [( k 1)T ] / 
e qi (kT )d
C kT

C kT

1 ( k 1)T [( k 1)T ] / 
e dqi (kT )  d: d 
 T,  kT

C T

1 0 / 
e (d) qi (kT )  
0,
  (k  1)T


 1 e T /  qi (kT )
C

We thus reduce the differential equation to the difference equation


h(k  1)  e T / h(k )  R 1 e T /  q i (k ) 
2.2 For each of the following equation, determine the order of the equation then test it for
(i) Linearity. (ii) Time-invariance. (iii) Homogeneousness.

(a) y(k+2) = y(k+1) y(k) + u(k)
(b) y(k+3) + 2 y(k) = 0
(c) y(k+4) + y(k-1) = u(k)
(d) y(k+5) = y(k+4) + u(k+1)  u(k)
(e) y(k+2) = y(k) u(k)

The results are summarized below

Problem Order Linear Time-invariant Homogeneous
(a) 2 No Yes No
(b) 3 Yes Yes Yes
(c) 5 Yes Yes No
(d) 5 Yes Yes No
(e) 2 No Yes No



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,2.3 Find the transforms of the following sequences using Definition 2.1
(a) {0, 1, 2, 4, 0, 0,...} (b) {0, 0, 0, 1, 1, 1, 0, 0, 0,...}
(c) {0, 20.5 , 1, 20.5 , 0, 0, 0, ... }


From Definition 2.1, {u0, u1 , u2 , ... , uk , ... } transforms to U ( z )   uk z k . Hence:
k 0



(a) Z 0,1,2,4,0,0,... z 1  2 z 2  4 z 3 (b) Z 0,0,0,1,1,1,0,0,... z 3  z 4  z 5

(c) Z 0,2 0.5 ,1,2 0.5 ,0,0,... 2 0.5 z 1  z 2  2 0.5 z 3
2.4 Obtain closed forms of the transforms of Problem 2.3 using the table of z-transforms and the time
delay property.

Each sequence can be written in terms of transforms of standard functions

(a) {0, 1, 2, 4,0,0,...} = {0, 1, 2, 4, 8, 16,...}  {0, 0, 0, 0, 8, 16,...}={f(k)}{g(k)}

2 k 1 ,
 k 0 8  2 k 4 , k  4

where f (k )   g(k )  
0,
 k0 0,
 k4

z 8z z 3 8
Z 0,1,2,4,0,0,... z 1 z 4  3
z 2 z 2 z ( z 2)

(b) {0, 0, 0, 1, 1, 1, 0, 0,...} = {0, 0, 0, 1, 1, 1, 1, 1,...}  {0, 0, 0, 0, 0, 0, 1, 1, 1, 1,...}
= {f(k)} {g(k)}

1,
 k 3 1,
 k 6
where f (k )   g(k )  
0,
 k3 0,
 k6

z z z 3 1
Z 0,0,0,1,1,1,0,0,... z 3 z 6  5
z 1 z 1 z ( z 1)

(c) {0,2-0.5,1,2-0.5,0,0,...} = {0,2-0.5,1,2-0.5,0,-2-0.5,-1,-2-0.5,0,...}+ {0,0,0,0,2-0.5,1,2-0.5,0,-2-0.5,-1,-2-0.5,0,...}

= {f(k)} + {g(k)}

sin( k 4) ,
 k 0 sin( k  4) ,
 k 4
where f (k )   g(k )  
0,
 k0 0,
 k4


Z 0,20.5 ,1,20.5 ,0,0,0,... sin( 4) z
z 4 sin( 4) z

20.5 z 4 1  
z 2 2 cos( 4) z  1 
z 2 2 cos( 4) z  1 z 3 z 2 20.5 z  1 

2.5 Prove the linearity and time delay properties of the z-transform from basic principles.

To prove linearity, we must prove homogeneity and additivity using Definition 2.1,

(i) Homogeneity: Z f (k )  Z f (k )


2

, 
Z f (0), f (1), f (2),..., f (i),... f (0)  f (1) z 1  f (2) z 2  ...  f (i) z i  ...   f (i) z i
i 0



Z f (0), f (1), f (2),..., f (i),... f (0)  f (1) z 1  f (2) z 2  ...  f (i) z i  ...   f (i) z i
i 0




(ii) Additivity Z f (k )  g(k ) Z f (k ) Z g(k )

Z f (k )  g(k ) Z f (0)  g(0), f (1)  g(1), f (2)  g(2),..., f (i)  g(i),...
 f (0)  g (0)  f (1)  g (1) z 1  f (2)  g (2) z 2  ...  f (i )  g (i ) z i  ...
 
  f (i ) z i   g (i ) z i  Z f (k ) Z g(k )
i 0 i 0

To prove the time delay property, we write the transform of the delayed sequence

Z 0, f (0), f (1), f (2),..., f (i),... f (0) z 1  f (1) z 2  f (2) z 3  ...  f (i) z i 1  ...

 z 1  f (i ) z i  z 1 Z f (k )
i 0


2.6 Use the linearity of the z-transform and the transform of the exponential function to obtain the
transforms of the discrete-time functions.
(a) sin(kT) (b) cos(kT)

e jkT e jkT
(a) sin( k T ) 
2j
1
Z sin (kT )
2j
Z e Z ejkT jkT

1  z z 
 z e jT z e jT 
2j  


1 
 e jT e jT z  sin (T ) z
2  2
2j z e
jT
ejT

z  1  z 2cos(T ) z  1

e jkT  e jkT
(b) cos(kT ) 
2

Z cos(kT ) 1 Z e jkT  Z e jkT 
2
1 z z 
  
2 z e jT z e jT  
 2
1 2 z e
 2
jT
e jT

z  z 2 cos(T ) z

2 z e jT
e jT

z  1  z 2 2cos(T ) z  1

2.7 Use the multiplication by exponential property to obtain the transforms of the discrete-time functions.
(a) ekTsin(kT) (b) ekTcos(kT)




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