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Nov 15, 2013 (4 years and 6 months ago)

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VUT
Vaal University of Technology

DIGITAL CONTROL SYSTEMS IV

MODULE CODE: EIDBS4

STUDY PROGRAM: UNIT 1 AND UNIT 2

EIDBS4 Chapter 1: Sampled Data Systems Page 1-1
1. SAMPLED DATA SYSTEMS
1.1 Sampling
In a digital control system a digital computer is used to play the role of the controller.
This immediately necessitates that certain analog signals must be digitised. A typical
digital controller is shown in Figure 1-1.

The sampling process implies that the analog signals be sampled every T seconds and
converted into binary numbers. To ensure a good duplication of the analog signal, two
aspects are important.
Firstly unwanted high frequencies in the analog signal should be filtered out and the
sampling rate should, theoretically, be twice the remaining highest frequency
contained in the analog signal. This is known as Shannon's sampling theorem.
Secondly the number of binary bits that are used to digitise the samples, should
For the reconstruction process, a zero order hold circuit is typically used. The raw
signal delivered by the zero order hold circuit still contains high frequencies as a
consequence of the spectrum repetition inherent to the original sampling process, and
these should be cancelled by a low pass filter.
A simplified illustration of the sampling and reconstruction process is given in
Figure 1-2. The signal x(t) is sampled and immediately reconstructed to produce the
signal x
r
(t).

Actual
Output
Plant
Desired
output
Transducer
Computer
Analog to digital
converter
Digital to analog
converter
x(t)
x
r
(t)
T sek
5T 4T 3T 2T
T
0
Sample and hold
t
5T4T 3T 2T
T
0
t
Figure 1-1
Figure 1-2

EIDBS4 Chapter 1: Sampled Data Systems Page 1-2
1.2 Difference equations
Just as differential equations are used to describe the relationship between the output
and input of continuous systems, difference equations are used to describe the input-
output relationship of discrete systems. An everyday example of a discrete process that
can be described by a difference equation, is the following:
An amount of 10 rand is borrowed from the bank. The outstanding amount is
augmented by 10% interest at the end of each month while a payment of x rand is
made also at the end of each month to pay down the loan. The following iterative
process leads to a difference equation that describes the relationship between the
outstanding amount y and the instalment x.
y(0) = 10
y(1) = y(0) + 0.1y(0) – x(0)
y(2) = y(1) + 0.1y(1) – x(1)
y(k+1) = y(k) + 0.1y(k) – x(k)
y(k+1) = 1.1y(k) – x(k)

1.3 Elementary digital signals
1.3.1 Step function
The step function u(k) is defined as follows:
u(k) = 0, k < 0
u(k) = 1 k  0

1.3.2 Impulse function
The impulse function is defined as follows:
(k) = 1 k = 0
(k) = 0, k  0

1.4 Z transform
1.4.1 Definition of the z-transform
The z transform of a digital signal is defined as:
X(z) = x(0) + x(1)z
-1
+ x(2)z
-2
+ x(3)z
-3
+................

X(z) x(k)z
k
k 0

Equation 1-1
u(k)
1
k=0 k
(k)
1
k=0 k

EIDBS4 Chapter 1: Sampled Data Systems Page 1-3
Example 1-1
Determine the sum of the series: 1/2 + 1/4 + 1/8 + 1/16 + ...........
Let S = 1/2 + 1/4 + 1/8 + 1/16 + ...........
1/2S = 1/4 + 1/8 + 1/16 + .........
S - 1/2S = 1/2
1/2S = 1/2
S = 1

Example 1-2
Determine the z transform of the function x(k) = u(kT) .
X(z) = 1 + 1z
-1
+ 1z
-2
+ 1z
-3
+........
X(z) = 1 + 1/z + 1/z
2
+ 1/z
3
+.......
(1/z)X(z) = 1/z + 1/z
2
+ 1/z
3
+.....
X(z) - (1/z)X(z) = 1
X(z) =
z
z
1

Example 1-3
Determine the z transform of the function x(k) = (kT):
X(z) = 1 + 0z
-1
+ 0z
-2
+.....
X(z) = 1 + 0 + 0 + …..
X(z) = 1

Example 1-4
Determine the z transform of the function x(k) = a
k
, for k  0.
X(z) = a
0
+ a
1
z
-1
+ a
2
z
-2
+ a
3
z
-3
+ .........
X(z) = 1 + (a/z) + (a/z)
2
+ (a/z)
3
+ ......
(a/z)X(z) = (a/z) + (a/z)
2
+ (a/z)
3
+......
X(z) - (a/z)X(z) = 1
X(z)(1 - a/z) = 1
X(z) = 1/(1 – a/z)
X(z) =
z
z
a

Example 1-5
Determine the z transform of the function x(k) = sin(k), for k  0.
X(z) =
Z
e
jk
e
jk
j2
 

= (1/j2)[Z{e

jk
} – Z{e

j
}]
= (1/j2)[(z/(z – e

j
) – z/( z – e

j
)]
= (z/j2)[(z – e

j
– z + e

j
)/(z – e

j
)(z – e

j
)]
= (z/j2)[(e
j
– e
-j
)/(z – e

j
)(z – e

j
)]
= z[(e

j
– e

j
)/j2]/(z – e

j
)(z – e

j
)
= z sin/(z – e

j
)(z – e

j
)
x(k)
1
k
k
x(k)
1
1
k
k
x(k)
x(k)
1

EIDBS4 Chapter 1: Sampled Data Systems Page 1-4

X(z) =
z(sin )
(z e
j
)(z e
j
)

 
 

=
z(sin )
z
2
ze
j
- ze
- j
1

 
 
=
z(sin )
z
2
z e
j
+e
- j
1

 
 2 2[( )/]

X(z) =
z(sin )
z
2
2z(cos ) 1

 

1.5 Important z transforms
Important z transforms are given in Table 1-1. The corresponding continuous time
functions and associated Laplace transforms are also indicated.
Notes on using Table 1-1:
1. For all time sequences x(k), assume k  0
2. The second entry into the table, x(k) = 1, corresponds to x(k) = u(kT).
3. The number sequence generated by functions such as x(k) = (kT) or
x(k) = u(kT), is independent of the sampling period T and the z transform
of such functions does not involve the sampling period T.
4. The number sequence generated by a functions such as x(k) = kT is clearly
dependent on the sampling period and the value of T is needed to resolve X(z).
5. The z transform of a sequence such as x(k) = k, may be found from the table
by looking up the z transform of x(k) = kT and setting T = 1.
6. Some very useful transforms are given in the lower window of the table.

Table 1-1

X(s)=
x(t)e
st
dt

x(t) x(k) = x(t)|
t=kT

X(z) =

x(k) z
k
k
0

1

(t)

(kT)
1
1
s

1 1
z
z
1

1
s
2

t kT
Tz
(z 1)
2

2
s
3

t
2

(kT)
2

3
1)(z
1)+z(z
2
T

1
s
a

e
at

e
akT

ⵡ

1
(s a)
2

te
at

kTe
akT

ⵡ
⡺ e
ⵡ-
)
2

EIDBS4 Chapter 1: Sampled Data Systems Page 1-5

a
s(s a)

at
e1

akT
e1

)

eㄩ⡺⡺
)

e稨


s
2 2

sin

t sin

kT
zsin T
z (2cos T) z 1
2

 

(s +a)
2 2

e
at
sin t

e
akT
sin kT

ze
- aT
sin T
z (2e
aT
cos T) z e
2a
T
2



s
s
2 2

cos t cos kT
z
2
zcos T
z (2cos T) z 1
2

 

s +a
(s +a)
2 2


e
a
t
cos t

e
akT
cos kT

z
2
- ze
- aT
cos T
z (2e
aT
cos T) z e
2a
T
2



e
-sT
X(s) x(t-T) x[(k-1)T] z
-1
X(z)
x[(k-2)T] z
-2
X(z)
x[(k+1)T] zX(z) - zx(0)
x[(k+2)T] z
2
X(z) - z
2
x(0) - zx(1)
X(s+a)
e
at
x(t)

e
akT
x(kT)

ㄭe
⵳
s

⠱ 
1

s

a
k

z/(z-a)

2ab
k
cos(k ) 
(a z
z b
(a z
z b


 


 


1.6 Properties of the z transform
1. Z{x(k) + y(k)} = X(z) + Y(z)
2. Z{ax(k)} = aX(z)
3.

Z{x(k - n)} = z
-n
X(z)
4.

Z{x(k + n)} = z
n
X(z) - z
n
x(0) - z
n-1
x(1) - ....... - zx(n - 1)
5.

Z{e
-akT
x(k)} = X(ze
aT
)
6.

Z{x(k)y(k)} = X(z)Y(z)
7.

lim x(k) lim X(z)
k 0 z  

8.

lim x(k) lim (1- z
-1
)X(z)
k z  

1
if (1-z
-1
)X(z) has no poles on or outside |z|=1.

Example 1-6
Determine the z transform of x(k) = 3u(k) + (0.5)
k
u(k).

EIDBS4 Chapter 1: Sampled Data Systems Page 1-6
X(z) = Z{3u(k) + (0.5)
k
u(k)} = Z{3u(k)} + Z{(0.5)
k
u(k)} = 3Z{u(k)} + z/(z - 0.5)
= 3[z/(z - 1)] + z/(z - 0.5) = 3z/(z - 1) + z/(z - 0.5)
= [3z(z - 0.5)+z(z - 1)]/(z - 1)(z - 0.5) = z(4z -2.5)/(z
2
-1.5z +0.5).

Example 1-7
Determine the z transform of a) x(k) = u(k - 3) and b) x(k) = u(k + 3).
a) X(z) = Z{u(k - 3)} = z
-3
Z{u(k)} = z
-3
[z/(z - 1)] = z
-2
/(z - 1).
b) X(z) = Z{u(k + 3)} = z
3
{Z[u(k)] - [x(0) + x(1)z
-1
+ x(2)z
-2
]}
= z
3
{z/(z-1) - [1 + z
-1
+ z
-2
]} = z
4
/(z - 1) - z
3
- z
2
- z

Example 1-8
Find the final value of y(kT) if Y(z)=0.792z
2
/(z-1)(z
2
-0.416z+0.208).
lim y(kT) lim (1- z
-1
)Y(z)
k z 

1
=
lim
z-1
z
0.792z
2
(z-1)(z
2
0.416z +0.208)
z 

1

= li
m

z 1
[0.792z/(z
2
-0.416z+0.208)] = 0.792/(1-0.416+0.208)=1

1.7 Inverse z transform

Method of partial fractions

1. This method assumes that the function on which this method is applied, is a proper
fraction, that is the order of the numerator is smaller than the order of the
denominator.
2. If the inverse transform of X(z) must be found, then it is preferred that X(z)/z is
expanded.

Three possible situations may arise and these will be illustrated with the following
three examples.

Case 1 - different real factors

Example 1-9

Determine the inverse z transform of X(z) = 1/(z - 1)(z - 0.5)

X(z)/z = 1/z(z-1)(z-0.5)
= A/z +B/(z-1) + C/(z-0.5)
A = 1/(0-1)(0-0.5) = 2
B = 1/(1-0.5) = 2
C = 1/0.5(0.5-1) = -4
X(z)/z = 2/z + 2/(z-1) -4/(z-0.5)
X(z) = 2 +2z/(z-1) - 4z/(z-0.5)
x(k) = 2(k) + 2u(k) - 4(0.5)
k
u(k)

EIDBS4 Chapter 1: Sampled Data Systems Page 1-7
Case 2 - complex factors

Example 1-10

Determine y(k) if x(k) = u(k) for the difference equation:
y(k) = x(k) + y(k-1) - 0.5y(k-2)

Z{y(k)} = Z{x(k) + y(k-1) - 0.5y(k-2)}
Z{y(k)} = Z{x(k)} + Z{y(k-1)} - 0.5Z{y(k-2)}
Y(z) = X(z) + z
-1
Y(z) - 0.5z
-2
Y(z)
With x(k) = u(k) is X(z) = z/(z - 1)
Y(z) = z/(z - 1) + Y(z)/z - 0.5Y(z)/z
2

z
2
Y(z) = z
3
/(z - 1) + zY(z) - 0.5Y(z)
Y(z)(z
2
- z + 0.5) = z
3
/(z - 1)
Y(z) = z
3
/(z-1)(z
2
-z+0.5)
Y(z)/z = z
2
/(z-1)(z
2
-z+0.5)
z
2
-z+0.5 = 0  z = 0.5  j0.5 = 0.70710.7854
r

z
2
-z+0.5 = (z-0.70710.7854
r
)(z-0.7071-0.7854
r
)
Y(z)/z = z
2
/(z-1)(z-0.70710.7854
r
)(z-0.7071-0.7854
r
)
= A/(z-1) + B/(z-0.70710.7854
r
) + B
*
/(z-0.7071-0.7854
r
)
A = z
2
/(z
2
-z+0.5)|
z=1
= 2
B = z
2
/(z-1)(z-0.7071-0.7854
r
)|
z=0.70710.7854r

= (0.70710.7854
r
)
2
/(0.70710.7854
r
-1)(0.70710.7854
r
-0.7071-0.7854
r
)
= 0.51.5708
r
/(0.70712.356
r
)(11.5708
r
) = 0.7071-2.356
r

and B
*
= 0.70712.356
r
.
Y(z)/z = 2/(z-1)+0.7071-2.356/(z-0.70710.7854)
+0.70712.356/(z-0.7071-0.7854)
Y(z)=2z/(z-1)+0.7071-2.356z/(z-0.70710.7854)
+0.70712.356z/(z-0.7071-0.7854)
y(k) = 2 + 20.7071(0.7071)
k
cos(0.7854k - 2.356)

using
(
a
z
z b
(
a
z
z b
2ab
k
cos(k )

 

 
 

  (Table 1-1)

Case 3 - repeated factors

Example 1-11

Determine x(k) if X(z) = z/(z-0.5)(z-1)
2

X(z)/z = 1/(z-0.5)(z-1)
2

= A/(z-0.5) + B/(z-1)
2
+ C/(z-1)
A = 1/(z-1)
2
|
z=0.5
= 4
B = 1/(z-0.5)|
z=1
= 2
C = d/dz[1/(z-0.5)]|
z=1
= d/dz[(z-0.5)
-1
]|
z=1
= [-(z-0.5)
-2
]|
z=1

= [-1/(z-0.5)
2
]|
z=1
= -4
X(z) = 4z/(z-0.5) + 2z/(z-1)
2
- 4z/(z-1)

EIDBS4 Chapter 1: Sampled Data Systems Page 1-8
x(k) = 4(0.5)
k
+ 2k - 4

Exercise

1-1
Determine the z transform of the following functions:
a) f(k) = ke
-3k
b) f(k) = ksin2k c) f(k) = e
-2k
sin3k

Ans: a) F(z) = ze
-3
/(z-e
-3
)
2
from table 1-1 with T = 1
b) F(z)=1/j2Z{ke
j2k
-ke
-j2k
}=1/j2[ze
j2
/(z-e
j2
)
2
-ze
-j2
/(z-e
-j2
)
2
]
c) F(z) = ze
-2
sin3/(z
2
- 2ze
-2
cos3 + e
-4
) from table 1-1 with T=1

1-2
Find the z transform of the following sequence:
f(k)=1 for k even (0,2,4...) and f(k)=-1 for k odd (1,3,5..).

Ans: F(z) = 1 - z
-1
+ z
-2
- z
-3
+ z
-4
- z
-5
+ .....= z/(z+1)

1-3
Determine the z transform of the following functions:
a) F(s) = 1/(s+1)(s+2) b) F(s) = 1/s
2
(s+1)
c) F(s) = 10/s(s+5)
2
d) F(s) = 5/s(s
2
+2)

Ans: a) F(s) = 1/(s+1)(s+2) = 1/(s+1) - 1/(s+2)
F(z) = z/(z-e
-1
) - z/(z-e
-2
) = z/(z-0.3679) - z/(z-0.1353)
b) F(s) = 1/s
2
(s+1) = A/s
2
+ B/s + C/(s+1)
A = 1 B = d/ds[1/(s+1)]|
s=0
= [-1/(s+1)
2
]|
s=0
= -1 C = 1
F(s) = 1/s
2
- 1/s + 1/(s+1)
F(z) = Tz/(z-1)
2
- z/(z-1) + z/(z-e
-T
)
c) F(s) = A/s + B/(s+5)
2
+ C/(s+5) = 0.4/s - 2/(s+5)
2
- 0.4/(s+5)
F(z) = 0.4z/(z-1) - 2Tze
-5T
/(z-e
-5T
)
2
- 0.4z/(z-e
-5T
)
d) F(s)=5/s(s+j2)(s-j2)=A/s+B/(s+j2)+B
*
/(s-j2)
=2.5/s-1.25/(s+j2)-1.25/(s-j2)=2.5/s-1.252s/(s
2
+2)
=2.5/s-2.5s/[s
2
+(2)
2
]
F(z) = 2.5z/(z-1) - 2.5(z
2
-zcos2T)/(z
2
-2zcos2T+1)

1-4
Determine the inverse z transform, f(k), of the following functions:
a) F(z)=10z/(z-1)(z-0.2) b) F(z)=z/(z-1)(z
2
+z+1)
c) F(z)=z/(z-1)(z-0.85) d) F(z)=10/(z-1)(z-0.5)

Ans a) F(z)/z = 10/(z-1)(z-0.2) = 12.5/(z-1) - 12.5/(z-0.2)
F(z) = 12.5z/(z-1) - 12.5z/(z-0.2)
f(k) = 12.5 - 12.5(0.2)
k
k  0
b) F(z)/z=1/(z-1)(z
2
+z+1)=1/(z-1)(z-1120)(z-1-120)
=A/(z-1) +B/(z-1120) +B
*
/(z-1-120)
= 0.3333/(z-1)+0.3333120/(z-1120) + 0.3333-120/(z-1-120)
F(z)=0.3333z/(z-1)+0.3333120z/(z-1120)+0.3333-120z/(z+1-120)

EIDBS4 Chapter 1: Sampled Data Systems Page 1-9
f(k) = 0.3333 + 20.3333(1)
k
cos(k120+120)
=0.3333 + 0.6667cos(k120+120) k  0
c) F(z)/z = 1/(z-1)(z-0.85) = A/(z-1)+B/(z-0.85)=6.667/(z-1)-6.667/(z-0.85)
F(z) = 6.667z/(z-1)-6.667z/(z-0.85)
f(k) = 6.667 - 6.667(0.85)
k
……k=0, 1, 2, ....
d) F(z)/z=10/z(z-1)(z-0.5)=A/z+B/(z-1)+C/(z-0.5)=20/z+20/(z-1)-40/(z-0.5)
F(z) = 20 + 20z/(z-1)-40z/(z-0.5)
f(k) = 20(k) +20 - 40(0.5)
k

1-5
Solve the following difference equations with the aid of the z transform.
a) x(k+2) - x(k+1) + 0.1x(k) = u(k), x(0)=x(1)=0.
b) x(k+2) - x(k) = 0, x(0) = 1, x(1) = 0.

Ans a) Z{x(k+2) - x(k+1) + 0.1x(k)} = Z{u(k)}
[z
2
X(z)-z
2
x(0)-zx(1)] - [zX(z) - zx(0)] + 0.1X(z) = z/(z-1)
X(z)(z
2
- z +0.1] = z/(z-1)
X(z) = z/(z-1)(z-0.8873)(z-0.1127)
X(z)/z = 1/(z-1)(z-0.8873)(z-0.1127)=A/(z-1)+B/(z-0.8873)+C/(z-0.1127)
= 10/(z-1) - 11.46/(z-0.8873) + 1.455/(z-0.1127)
X(z) = 10z/(z-1) - 11.46z/(z-0.8873) + 1.455z/(z-0.1127)
x(k) = 10 - 11.46(0.8873)
k
+ 1.455(0.1127)
k

b) Z{ x(k+2) - x(k)} = 0
z
2
X(z) - z
2
x(0) - zx(1) - X(z) = 0
z
2
X(z)-z
2
-X(z)=0
X(z)(z
2
- 1) = z
2

X(z)/z=z/(z+1)(z-1)=A/(z+1)+B/(z-1)
= 0.5/(z+1) + 0.5/(z-1)
X(z) = 0.5z/(z+1) +0.5z/(z-1)
x(k) = 0.5(-1)
k
+0.5, note that z/(z+1) = z/[z-(-1)] and use Z
-1
{z/(z-a)}=a
k

1-6
A car is bought with a loan of P
o
rand. The interest is r percent per month and
the loan is paid down at u rand per month over a period of N months.
a) Write down a difference equation for the outstanding amount p(k).
b) Solve for p(k) by means of the z transform.
c) Solve for the monthly instalment u, if the bond must be settled in N months.
d) Take P
0
as R15000, r=0.01(1 percent per month) en N=48 months. Find
the monthly payment.

Ans: a) p(0) = P
o

p(1) = p(0) + rp(0) - u
p(2) = p(1) + rp(1) - u
p(k+1) = p(k) + rp(k) - u
p(k+1) = (1 + r)p(k) - u

EIDBS4 Chapter 1: Sampled Data Systems Page 1-10
b) Z{p(k+1)} = Z{(1 + r)p(k) - u}
zP(z) - zp(0) = (1 + r)P(z) - uz/(z-1)
P(z)[z - (1+r)] = zP
o
- uz/(z-1) where p(0) = initial loan = P
0

= z[P
0
- u/(z-1)]
= z[P
0
(z-1) - u]/(z-1)
P(z) =
z[P
o
(z 1) u]
(z 1)[z (1 r)]

  

P(z)
z
=
P
o
(z 1) u
(z 1)[z (1 r)]

  

=
A
z 1
B
z (1 r)

 

where A =
P
o
(z 1) u
z (1 r)
z 1

 

= u/r
and B =
P
o
(z 1) u
z 1
z 1
r

= (rP
0
-u)/r

P(z)
z
=
(u/r)
z 1
(rP
o
- u)
/
r
z (1 r)

 

P(z) =
u
r
z
z 1
rP
o
u
r
z
z (1 r)

 

p(k) = (u/r) + [(rP
o
-u)/r](1+r)
k

c) p(N) = (u/r) + [(rP
o
-u)/r](1+r)
N
= 0

u + (rP
o
-u)(1+r)
N
= 0

u[1 - (1+r)
N
] = -rP
o
(1+r)
N

u = -rP
o
(1+r)
N
/[1 - (1+r)
N
]
= rP
o
(1+r)
N
/[(1+r)
N
-1]

d) u = 0.01

15000(1.01)
48
/[(1.01)
48
- 1] = R395

EIDBS4 Chapter 2: Transfer Functions Page 2-1
2. TRANSFER FUNCTIONS
2.1 Transfer function of a zero order hold circuit
Consider again the simplified presentation of a zero order hold system of Figure 1-2,
repeated here in Figure 2-1.

Figure 2-1

The reconstructed version x
r
(t) of the original analog signal x(t), can be expressed as:

x
r
(t) = x(0)[u(t)

u(t

T)]

+

x(T)[u(t

T)

u(t

2T)]

+

x(2T)[u(t

2T)

u(t

3T)]

+

...

where u(t) =

 0 for t , 1
0 <for t , 0
.

X
r
(s) =
.... +
s
sT3
e
s
2sT-
e
x(2T)+
s
sT2
e
s
sT-
e
x(T)+
s
sT
e
s
1
x(0)

=
........ +
s
sT
e1
sT2
(2T)e x+
s
sT
e1
sT
(T)e x+
s
sT
e1
(0)x

=

s
sT
e1
] ......... +
2sT
(2T)e x+
sT
(T)e x+ (0)[x



X
r
(s) = X*(s)

s
sT
e1

where X*(s)
=

....+
2sT
(2T)ex+
sT
(T)ex+(0)x

Equation 2-1

x(t) x
r
(t)
T sek
5T
4T 3T 2T
T
0
Sample and hold
t
5T4T3T 2T
T
0
t

EIDBS4 Chapter 2: Transfer Functions Page 2-2
The sampling process could thus be considered as consisting out of two separate
processes as shown in Figure 2-2.

Figure 2-2

Producing X*(s) from X(s), could be viewed as ideal sampling, because when we take
the inverse Laplace transform of X*(s) to find x*(t) from Equation 2-1:
x*(t) = L
-1
{X*(s)}.
x*(t) = x(0)(t) + x(T)(t

T) + x(2T)(t

2T) + ..……...

Although the impulse sequence x*(t), visualized in Figure 2-3, is never physically
present in the A/D or D/A circuitry, it does accurately represent and model the
generation of a number sequence from a sampled continuous time function. The z
transform of x*(t) is in fact identical to the z transform of the corresponding time
function x(kT) that we considered in the previous chapter.

Figure 2-3

The second block in Figure 2-2 must represent the zero order hold process and we
come to the conclusion that the transfer function of the zero order hold circuit, is given
by:

G
H
(s) =
s
sT
e1

Equation 2-2

It is also important to note that by comparing Equation 1-1;
X(z) = x(0)+x(T)z
-1
+x(2T)z
-2
+....
with Equation 2-1; X
*
(s) =x(0)+x(T)e
-sT
+x(2T)e
-2sT
+... ,
that: X(z) = X
*
(s),
if:
z = e
sT

Equation 2-3

1 e
sT
s

X
r
(s)
x
r
(t)
X
*
(s)X(s)
x
*
(t) x(t)
T
Hold action
Ideal sampling
x*(t)
x(2T)
x(T)
x(3T)
x(4T)
x(0)
x(5T)
5T4T3T
2T
T
0
t

EIDBS4 Chapter 2: Transfer Functions Page 2-3
2.2 Pulse transfer function

Figure 2-4

In Figure 2-4, a system G(s) is shown which is driven by an ideally sampled version
x

(t) of a continuous signal x(t). Although the response y(t) could be determined from
the relationship:
Y(s) = G(s)X

(s),
we

are

really

interested

in

the

values

of

y(t)

at

the

sampling

instants.

For

that

purpose

we

introduce

a

perfect

sampler

at

the

output,

in

synchronism

with

the

input

sampler

and

it

can be shown that the relationship between the output and input is then given by:

Y(z) = G(z)X(z)

Equation 2-4

where G(z) = Z{G(s)}

Equation 2-5

The output y(kT) may now be calculated from Y(z). The z domain transfer function
G(z), calculated by taking the z transform of the system's Laplace transfer function, as
per Equation 2-5, is called the pulse transfer function of the system.

2.3 Transfer function of systems in cascade
When two systems are connected in cascade, it is important to know whether sampling
takes place between the two systems.

a)

b)

Figure 2-5

Y(z)
X(z)
Y(z) X(z)
T
G
1
(s)
G
2
(s)
T
T
G
1
(s)
G
2
(s)
T
T
x(t) y(t)
T
G(s)
T
x

(t) y

(t)

EIDBS4 Chapter 2: Transfer Functions Page 2-4
For the system in Figure 2-5 a) the transfer function G(z) = Y(z)/X(z), is given by:

G(z) = Z{G
1
(s)}

Z{G
2
(s)}
Equation 2-6

= G
1
(z)

G
2
(z).

For the system in Figure 2-5 b) the transfer function G(z) = Y(z)/X(z), is given by:

G(z) = Z{G
1
(s)G
2
(s)}
Equation 2-7

= G
1
G
2
(z).

It is important to understand the meaning of the notation G
1
G
2
(z) and also to realise
that G
1
G
2
(z)

G
1
(z)

G
2
(z).

Example 2-1

Determine the transfer function G(z) = Y(z)/X(z) for the system shown in Figure 2-6.

Figure 2-6

From Equation 2-6: G(z) = Z{G
1
(s)}

Z{G
2
(s)}
= Z{1/s}

Z{1/(s+1)}
= [z/(z-1)]

[z/(z-e
-T
)]
= z
2
/(z-1)(z-e
-T
)

Example 2-2

Determine the transfer function of the system in Figure 2-7.

Figure 2-7

From Equation 2-7: G(z) = Z{G
1
(s)G
2
(s)}
= Z{[1/s]

[1/(s+1)]}
= Z{1/s(s+1)}
= z(1-e
-T
)/(z-1)(z-e
-T
) from Table 1-1.
Y(z)
G
2
(s)G
1
(s)
X(z)
T
1/s
1/(s+1)
T
Y(z) X(z)
G
2
(s)G
1
(s)
T
1/s
1/(s+1)
T
T

EIDBS4 Chapter 2: Transfer Functions Page 2-5
2.4 Feedback systems
Consider the unity feedback system in Figure 2-8.

Figure 2-8

The overall transfer function T(z) = C(z)/R(z), will now be derived.
C(z) = G(z)E(z)........................……………............(1)
and E(z) = R(z) - C(z)……..........……………................(2)
(2) in (1): C(z) = G(z)[R(z) - C(z)]

C(z) = G(z)R(z) - G(z)C(z)

C(z) + G(z)C(z) = G(z)R(z)
:C(z)[1 + G(z)] = G(z)R(z)

C(z)
R(z)
= T(z)
G(z)
1 G(z)

Equation 2-8

Consider now the more general feedback system shown in Figure 2-9.

Figure 2-9

C(z) = G(z)E(z)........................……………............(1)
and E(z) = R(z) - GH(z)E(z)........……………................(2)
From (1): E(z) = C(z)/G(z)..............……………......................(3)
(3) in (2): C(z)/G(z) = R(z) - GH(z)C(z)/G(z)

C(z) = R(z)G(z) - GH(z)C(z)

C(z) + GH(z)C(z) = G(z)R(z)

C(z)[1 + GH(z)] = G(z)R(z)

C(z)
R(z)
= T(z)
G(z)
1 GH(z)

Equation 2-9
E(z) C(z) R(z)
T
G(s)
H(s)
T
E(z) C(z) R(z)
T
G(s)
T

EIDBS4 Chapter 2: Transfer Functions Page 2-6
Example 2-3

Determine the transfer function T(z) = C(z)/R(z) for the control system in Figure 2-10.

Figure 2-10

G(z) = Z{G(s)}
= Z{1/s)}
= z/(z-1)
GH(z) = Z{G(s)H(s)}
= Z{[1/s]

[1/(s+1)]}
= Z{1/s(s+1)}
=
z(1 e
T
)
(z 1)(z e
T
)

 

From Equation 2-9: T(z) =

G(z)
1 GH(z)

=
z
z 1
1
z(1 e
T
)
(z 1)(z e
T
)

 

=
z
z 1
(z -1)(z e
T
) +z(1 e
T
)
(z 1)(z e
T
)

 

=
z
z 1
z
2
- z - ze
T
+e
T
+z ze
T
(z 1)(z e
T
)

 

 

=
z
z 1
(z 1)(z e
T
)
z
2
- z- ze
T
+e
T
+z ze
T

 

 

=
z(z e
T
)
z
2
- 2ze
T
+e
T

 

C(z) R(z)
T
G(s)=1/s
T
H(s)=1/(s+1)

EIDBS4 Chapter 2: Transfer Functions Page 2-7
2.5 Transfer function of systems preceded by a zero order hold device

In Figure 2-11 a plant with transfer function P(s), is shown with a zero order hold
device G
H
(s) at the input. From Equation 2-2, the transfer function of a zero order hold
device is given by G
H
(s) = (1-e
-sT
)/s. The transfer function G(z) = Y(z)/X(z) of this
configuration will now be obtained.

Figure 2-11

G(z) =
Z
1 e
sT
s
P(s)

= Z{P(s)/s - (e
-sT
/s)P(s)}
= Z{P(s)/s} - Z{e
-sT
[P(s)/s]}
= Z{P(s)/s} - z
-1
Z{P(s)/s} {e
-sT
X(s)

††††††††††††††††‽⁚筐⡳⤯獽⠱⵺

††††††††††

s
P(s)
Z)
1
z(1=G(z)

Equation

2-10

Example 2-4

Refer to the system in Figure 2-12.
a) Determine the transfer function G(z) = Y(z)/X(z).
b) Determine y(k) if x(k)=u(kT)

Figure 2-12

a) G(z) = (1-z
-1
)Z{P(s)/s}
= (1-z
-1
)Z{(1/s)/s}
= (1-z
-1
)Z{1/s
2
}
= (1-z
-1
)Tz/(z-1)
2
Table 1-1

Now 1-z
-1
=
1
1
z
z 1
z
 

G
H
(s) =
1 e
sT
s

P(s)
Y(z)
X(z)
G
H
(s) =
1 e
sT
s

Y(z)
X(z)
T
T
T
T
P(s) =
1
s

EIDBS4 Chapter 2: Transfer Functions Page 2-8
G(z) =
z 1
z
Tz
(z 1)
2

= T/(z-1)

b) Y(z) = G(z)X(z)
= T/(z-1)z/(z-1)
= Tz/(z-1)
2

y(k) = kT
{typical integrator (1/s) response}

Example 2-5

Refer to the system in Figure 2-13 and assume that T=ln2 sec.
a) Determine the transfer function G(z) = Y(z)/X(z).
b) Determine y(k), if x(k)=u(kT).
c) Calculate the final value of y(k) with the aid of the final value theorem (property 8).

Figure 2-13

a) G(z) = (1-z
-1
)Z{P(s)/s}
= (1-z
-1
)Z{1/s(s+1)}
= (1-z
-1
)z(1-e
-T
)/(z-1)(z-e
-T
) table 1-1.
But T=ln2  e
T
=2  e
-T
=1/2
G(z) = [(z-1)/z][z(1-0.5)/(z-1)(z-0.5)]
= 0.5/(z-0.5)

b) Y(z) = G(z)X(z)
= 0.5/(z-0.5)z/(z-1)
Y(z)/z = 0.5/(z-1)(z-0.5)
= A/(z-1) + B(z-0.5)
where A=1 and B=-1
Y(z)/z = 1/(z-1) - 1/(z-0.5)
Y(z) = z/(z-1) - z/(z-0.5)
y(k) = 1 - (0.5)
k

c)
lim y(k) lim(1- z
-1
)Y(z)
k z  

1

=
li
m
[(z-1)
/
z][0.5z
/
(z-1)(z - 0.5)]
z 1

=
li
m
[0.5
/
(z - 0.5)]
z 1

= 1
G
H
(s) =
1 e
sT
s

Y(z)
X(z)
P(s) =
1
s +1
T
T
k
43 2
y(k)
1
1/2
1
0
k
3 2
y(k)
3T
2T
T
1
0

EIDBS4 Chapter 2: Transfer Functions Page 2-9
Example 2-6

Refer to the control system in Figure 2-14. Determine the output c(k) if r(k) = u(kT)
and T=1/2 sec.

Figure 2-14

G(z) = (1-z
-1
)Z{P(s)/s}
= (1-z
-1
)Z{1/s
2
}
= (1-z
-1
)

[Tz/(z-1)
2
]
= [(z-1)/z]

[Tz/(z-1)
2
]
= T/(z-1)

According to Equation 2-8 for a unity feedback system:
T(z) =
G(z)
1 G(z)

= [T/(z-1)]/[1+T/(z-1)]
= [T/(z-1)]/[(z-1+T)/(z-1)]
= T/(z-1+T)
With T=0.5: T(z) = 0.5/(z-0.5)

Now T(z) = C(z)/R(z)

C(z) = T(z)R(z)
With r(k) = u(kT), R(z) = z/(z-1)

C(z) = [0.5/(z-0.5)]

[z/(z-1)]
= 0.5z/(z-0.5)(z-1)

C(z)/z = 0.5/(z-0.5)(z-1)
= -1/(z-0.5) + 1/(z-1)

C(z) = -z/(z-0.5) + z/(z-1)

c(k) = -(0.5)
k
+ 1

c(k)
1
k
¾

½

0
4
3
2

1
Zero order hold
G
H
(s)
C(z) R(z)
P(s)=
1
s

T
G
(
z
)
T

EIDBS4 Chapter 2: Transfer Functions Page 2-10
Exercise

2-1
The output of a linear time-invariant discrete system is given by the following time
sequence:
c(k) = 1 - e
-2kT
, k = 0, 1, 2, 3, .....,
when a signal is applied to the input of the system that is described by:
r(k) = 1 for all k

1.
Find the transfer function T(z) = C(z)/R(z).
Ans: C(z) = Z{c(k)} = Z{1 - e
-2kT
}
= z/(z-1) - z/(z-e
-2T
) = [z(z-e
-2T
) - z(z-1)]/(z-1)(z-e
-2T
)
= z(1-e
-2T
)/(z-1)(z-e
-2T
)
and R(z) = Z{1} = z/(z-1)

T(z) = C(z)/R(z) = [z(1-e
-2T
)/(z-1)(z-e
-2T
)]

(z-1)/z
= (1-e
-2T
)/(z-e
-2T
)

2-2
Determine C(z)/R(z) for the systems in Figure P 2-1 a) - f). Take T=0.5 sec.

a)

Figure P 2-1 a)

b)

Figure P 2-1 b)

c)

Figure P 2-1 c)

d)

Figure P 2-1 d)

c(t)
r(t)

1
s +1

T

10
s +2

c(t)
r(t)
c(t)
r(t)

1
s +1

T

10
s+2

1
s(s + 2)
T
T
c(t)
r(t)
T

5
s(s +2)
G
H
(s)=
1 e
sT
s

EIDBS4 Chapter 2: Transfer Functions Page 2-11
e)

Figure P 2-1 e)

f)

Figure P 2-1 f)

Ans:
a) T(z) = Z{1/s(s+2)} = (1/2)Z{2/s(s+2)} = 0.5z(1-e
-1
)/(z-1)(z-e
-1
)
= 0.316z/(z-1)(z-0.3679)

b) T(z) = Z{[1/(s+1)]

[10/(s+2)]} = Z{10/(s+1)(s+2)}
= Z{10/(s+1) - 10/(s+2)} = 10z/(z-e
-0..5
) - 10z/(z-e
-1
)
= [10z(z-e
-1
) - 10z(z-e
-0.5
)]/(z-e
-0.5
)(z-e
-1
) = 10z(e
-0.5
- e
-1
)/(z-e
-0.5
)(z-e
-1
)
= 2.387z/(z-0.6065)(z-0.3679)
c) T(z) = Z{1/(s+1)}

Z{10/(s+2)} = z/(z-e
-0.5
)

10z/(z-e
-1
) = 10z
2
/(z-e
-0.5
)(z-e
-1
)
= 10z
2
/(z-0.6065)(z-0.3679)

d) T(z) = (1-z
-1
)Z{(1/s)

5/s(s+2)} = [(z-1)/z]Z{5/s
2
(s+2)} = [(z-1)/z]Z{5/s
2
(s+2)}
= [(z-1)/z]Z{A/s
2
+ B/s + C/(s+2)} = [(z-1)/z]Z{2.5/s
2
- 1.25/s + 1.25/(s+2)}
= [(z-1)/z][2.5

0.5z/(z-1)
2
- 1.25z/(z-1) + 1.25z/(z-e
-1
)]
= [(z-1)/z]

{[1.25z(z-e
-1
) - 1.25z(z-1)(z-e
-1
) + 1.25z(z-1)
2
]/(z-1)
2
(z-e
-1
)}
= [(z-1)/z]

{1.25z

[z-e
-1
- (z
2
- z - ze
-1
+ e
-1
)+(z
2
- 2z + 1)]/(z-1)
2
(z-e
-1
)}
= 1.25z

[(z-1)/z]

{[z - e
-1
- z
2
+ z + ze
-1
- e
-1
+ z
2
- 2z + 1]/(z-1)
2
(z-e
-1
)}
= 1.25z

[(z-1)/z]

{[ze
-1
- 2e
-1
+ 1]/(z-1)
2
(z-e
-1
)}
= 1.25(ze
-1
- 2e
-1
+ 1)/(z - 1)(z - e
-1
)}
= (0.4598z + 0.3303)/(z - 1)(z - 0.3679)
e) G(z) = (1-z
-1
)Z{(1/s)

5/s(s+2)} = (0.4598z + 0.3303)/(z - 1)(z - 0.3679)
Prob 2-2d
T(z) = G(z)/[1+G(z)] unity feedback system Equation 2-8
= (0.4598z+0.3303)/(z-1)(z-0.3679)/[1+(0.4598z+0.3303)/(z-1)(z-0.3679)]
= (0.4598z+0.3303)/(z-1)(z-0.3679)/
{[(z-1)(z-0.3679)+(0.4598z+0.3303)]/(z-1)(z-0.3679)}
= (0.4598z + 0.3303)/[(z - 1)(z - 0.3679) + (0.4598z + 0.3303)]
= (0.4598z + 0.3303)/(z
2
- 1.3679z + 0.3679 + 0.4598z + 0.3303)]
G
H
(s)=
1 e
sT
s

c(t)
r(t)

5
s(s +2)

T
G
H
(s)=
1 e
sT
s

c(t)
r(t)

5
s(s +1)(s +2)

T

EIDBS4 Chapter 2: Transfer Functions Page 2-12
= (0.4598z + 0.3303)/(z
2
- 0.9081z + 0.6982)
f) G(z) = (1-z
-1
)Z{(1/s)

[5/s(s+1)(s+2)]} = (1-z
-1
)Z{5/s
2
(s+1)(s+2)}
= (1-z
-1
)Z{A/s
2
+ B/s + C/(s+1) + D/(s+2)}
A = 5/(s+1)(s+2)|
s=0
= 2.5
B = d/ds{5/(s+1)(s+2)}|
s=0
= 5d/ds[(s
2
+3s+2)
-1
]|
s=0

= -5[(s
2
+3s+2)
-2
(2s+3)]|
s=0
= -5[(2s+3)/(s
2
+3s+2)
2
]|
s=0
= -5[3/4]
= -3.75
C = 5/s
2
(s+2)|
s=-1
= 5
D = 5/s
2
(s+1)|
s=-2
= -1.25

G(z) = (1-z
-1
)Z{2.5/s
2
-3.75/s + 5/(s+1) - 1.25/(s+2)}
= [(z-1)/z]

[1.25z/(z-1)
2
- 3.75z/(z-1) + 5z/(z-e
-0.5
) - 1.25z/(z-e
-1
)]
= [(z-1)/z]

1.25z

[(z-e
-0.5
)(z-e
-1
)-3(z-1)(z-e
-0.5
)(z-e
-1
)+4(z-1)
2
(z-e
-1
)- (z-1)
2
(z-e
-0.5
)]/
(z-1)
2
(z-e
-0.5
)(z-e
-1
)
= 1.25[z
2
-e
-0.5
z-e
-1
z+e
-1..5
-3(z-1)(z
2
-e
-0.5
z-e
-1
z+e
-1..5
)
+4(z
2
-2z+1)(z-e
-1
)-(z
2
-2z+1)(z-e
-0.5
)]/(z-1)(z-e
-0.5
)(z-e
-1
)
= 1.25[z
2
-e
-0.5
z-e
-1
z+e
-1..5
-3(z
3
-e
-0.5
z
2
-e
-1
z
2
+e
-1..5
z-z
2
+e
-0.5
z+e
-1
z-e
-1..5
)
+4(z
3
-2z
2
+z-e
-1
z
2
+2e
-1
z-e
-1
)-(z
3
-2z
2
+z-e
-0.5
z
2
+2e
-0.5
z-e
-0.5
)]/
(z-1)(z-e
-0.5
)(z-e
-1
)
= 1.25[z
2
-e
-0.5
z-e
-1
z+e
-1..5
-3z
3
+3e
-0.5
z
2
+3e
-1
z
2
-3e
-1..5
z+3z
2
-3e
-0.5
z-3e
-1
z+3e
-1..5
)
+4z
3
-8z
2
+4z-4e
-1
z
2
+8e
-1
z-4e
-1
-z
3
+2z
2
-z+e
-0.5
z
2
-2e
-0.5
z+e
-0.5
)]/
(z-1)(z-e
-0.5
)(z-e
-1
)
= 1.25[(1+3e
-0.5
+3e
-1
+3-8-4e
-1
+2+e
-0.5
)z
2

+ (-e
-0.5
-e
-1
-3e
-1.5
-3e
-0.5
-3e
-1
+4+8e
-1
-1-2e
-0.5
)z
+ (e
-1..5
+3e
-1..5
-4e
-1
+e
-0.5
)]/(z-1)(z-e
-0.5
)(z-e
-1
)
= 1.25[(-2 +4e
-0.5
-e
-1
)z
2
+ (3 - 6e
-0.5
+ 4e
-1
- 3e
-1.5
)z
+ (4e
-1..5
- 4e
-1
+ e
-0.5
)]/(z-1)(z-e
-0.5
)(z-e
-1
)
= 1.25(0.05824z
2
+ 0.1629z + 0.02753)/(z-1)(z-e
-0.5
)(z-e
-1
)
= (0.0728z
2
+ 0.2036z + 0.03441)/(z-1)(z-0.6065)(z-0.3679)

T(z) = G(z)/[1 + G(z)]
= (0.0728z
2
+ 0.2036z + 0.03441)/(z-1)(z-0.6065)(z-0.3679)/
[1 + (0.0728z
2
+ 0.2036z + 0.03441)/(z-1)(z-0.6065)(z-0.3679)]
= (0.0728z
2
+ 0.2036z + 0.03441)/(z-1)(z-0.6065)(z-0.3679)/
[(z-1)(z-0.6065)(z-0.3679)+(0.0728z
2
+ 0.2036z + 0.03441)]/
(z-1)(z-0.6065)(z-0.3679)
= (0.0728z
2
+ 0.2036z + 0.03441)/
[(z-1)(z-0.6065)(z-0.3679)+(0.0728z
2
+ 0.2036z + 0.03441)]
= (0.0728z
2
+ 0.2036z + 0.03441)/
[(z-1)(z
2
-0.9744z+0.2231)+(0.0728z
2
+0.2036z+0.03441)]
= (0.0728z
2
+ 0.2036z + 0.03441)/
(z
3
-0.9744z
2
+0.2231z-z
2
+0.9744z-0.2231+0.0728z
2
+0.2036z+0.03441)
= (0.0728z
2
+ 0.2036z + 0.03441)/(z
3
- 1.902z
2
+ 1.401z - 0.1887)

EIDBS4 Chapter 3: Time Domain Analysis Page 3-1
3. TIME DOMAIN ANALYSIS
3.1 Transient response of a second order continuous-time control system
It is necessary at this stage to consider the response of the prototype second order
continuous-time control system in Figure 3-1.

Figure 3-1

The overall transfer function is given by:
T(s) =
C(s)
R(s)
=
G(s)
1 G(s)
.

T(s) =

 
n
n
2
s
2
+2 s +
2
.
Equation 3-1

The character of the natural response of the system depends on the roots of the
characteristic equation (denominator of T). For the case when 
n
> , the roots will be
two conjugate complex numbers p and p

, as shown in Figure 3-2.

Figure 3-2

G(s) =

n
2
s(s +2 )
C(s)
p
p

Im s

n
Complex s plane
 = cos
Re s

d

R(s)

EIDBS4 Chapter 3: Time Domain Analysis Page 3-2
It is clear from Figure 3-2 that the parameters , 
d
, 
n
,  and  are interrelated and
two equations that will be useful, are given below.

n
=

 
2
d
2

Equation 3-2

and
 = cos =

n
.
Equation 3-3

The parameters are defined as follows:

: the damping factor of the system, representing the real part of the roots

d
: the damped frequency - the imaginary part of the roots

n
: the natural undamped frequency of the system, equal to the magnitude of the roots

: the damping ratio

For this case where the characteristic equation has two complex roots as illustrated in
Figure 3-2, the system will exhibit a damped sinusoidal natural response and the
system's step response, demonstrates this clearly. We may find the step response by
substituting r(t) = u(t), that is R(s) = 1/s, into Equation 3-1:
C(s) =

 
n
n
2
s(s
2
+2 s +
2
)
,
and by means of the inverse Laplace transform:

c(t) = 1 –
e
t
sin
sin(
d
t )

 .
Equation 3-4

A typical step response is shown in Figure 3-3 with the overshoot D, the rise time t
r

and the settling time t
s
indicated.

Figure 3-3

c(t)
D
1
t
r

t
s

t

EIDBS4 Chapter 3: Time Domain Analysis Page 3-3
z
s
e
-T

d

d
T
-
j Im s j Im z
Re z
Re s
The maximum overshoot may be found by setting the derivative of c(t) equal to zero.
From this follows the time when the first maximum occurs:
t
max
=

 
n
sin
,
and the corresponding maximum value of c(t) is,
c(t)|
max
= 1 + e
-cot
.
The overshoot D is thus given by D = e
-cot
, or in terms of

and

d
,

D =
e
d



Equation 3-5

The rise time can be found by setting c(t) = 1, which occurs, from Equation 3-4, when
sin(

t+

) = 0 for the first time, that is

t +

=

or,

t
r
=

tan
1
d
d

.
Equation 3-6

The settling time is normally defined as the time it takes for the decaying function e
-t

(and the corresponding oscillations) to settle down to 2% of the initial value. This
occurs, from equation 3-4, when e
-t
= 0.02 or -

t = ln(0.02)

-4.

t
s
=

4

Equation 3-7

3.2 Mapping between the s plane and the z plane
The fundamental relation between the s plane and the z plane is specified by Equation
2-3:
z = e
sT
.
If a point on the s plane is given by:
s = -

+ j

d
,
then the corresponding point z in the z plane will be given by:
z =
)Tj(-
e
d



z = e
-T

d
T
.
Equation 3-8

EIDBS4 Chapter 3: Time Domain Analysis Page 3-4
3.3 Time response of discrete data control systems
Example 3-1

Consider the system in Figure 3-4 and estimate the overshoot D, the rise time t
r
and the
settling time t
s
.

Figure 3-4

For this system G(z) = (1 - z
-1
)Z
1
s
1
s(s 1)

= (1 - z
-1
)Z
1
s
2
(s 1)

= (1-z
-1
)Z
1
s
2
1
s
1
(s 1)
 

=
z 1
z
Tz
(z-1)
2
z
z-1
z
(z- e
T
)

 

=
z 1
z
z[T(z e
T
) (z 1)(z e
T
) (z 1)
2
]
(z 1)
2
(z e
T
)

  

 
 

=
Tz Te
T
(z
2
z e
T
z +e
T
) z
2
2z 1
(z 1)(z e
T
)

  

  
 

=
(T-1+e
T
)z +(1-e
T
-Te
T
)
(z 1)(z e
T
)

 

T(z) =
G(z)
1 G(z)

=
(T-1+e
T
)z +(1- e
T
- Te
T
)
(z 1)(z e
T
)
(T-1+e
T
)z +(1- e
T
- Te
T
)
(z 1)(z e
T
)

 

  
 

1

=
(
T-1+e
T
)z +(1- e
T
- Te
T
)
(z 1)(z e
T
) +(T-1+e
T
)z +(1- e
T
- Te
T
)

 
   

G
H
(s) =
1 e
sT
s

c(t)
r(t)
G(z)

1
s(s +1)

T=0.5

EIDBS4 Chapter 3: Time Domain Analysis Page 3-5
=
(
T-1+e
T
)z +(1- e
T
- Te
T
)
z
2
z e
T
z +e
T
+(T-1+e
T
)z +(1- e
T
- Te
T
)

 
    

T(z) =
(T-1+e
T
)z +(1- e
T
- Te
T
)
z
2
(T 2)z +(1- Te
T
)

 

Equation 3-9

Analogously to continuous systems, the denominator of T(z) is called the
characteristic equation of the system. The roots of the characteristic equation, also
called the
poles
of T(z), completely determines the transient behaviour of the system.
It is always possible to distinguish between two components of a system's response.
One component is associated with the driving force, r(t) in this case, and the natural
component which is governed by the poles of the system.
The roots of the characteristic equation can also be determined without calculating the
total transfer function first. For a unity feedback system the appropriate equation will
be:

G(z) + 1 = 0

Equation 3-10

and for the general feedback system:

GH(z) + 1 = 0

Equation 3-11

For example, for the system under consideration:
G(z) + 1 =
(T-1+e
T
)z +(1-e
T
-Te
T
)
(z 1)(z e
T
)

 

+ 1 = 0
(T - 1 + e
-T
)z + (1 - e
-T
-Te
-T
) + (z - 1)(z - e
-T
) = 0
z
2
+ (T -2)z + (1 - Te
-T
) = 0,
which is the desired equation according to the transfer function Equation 3-9.

Now with T = 0.5 sec. the transfer function becomes:

T(z) =
0.1065z 0.0902
z
2
z +0.6967

15.
.
Equation 3-12

The characteristic equation is:

z
2
- 1.5z + 0.6967 = 0 ,

with roots (poles):

z = 0.83470.4543
and 0.8347-0.4543
.

T(z) becomes zero when z = -0.8469 and this value is called a zero of T(z). If a zero is
between -1 and 0, it has a small effect on the step response while a zero moving
towards z = +1 greatly increases the overshoot. We should therefore be able to specify
a continuous second order system that behaves similarly to the present discrete second
order system and apply Equations 3-5 to 3-7 to estimate the overshoot, the rise time
and the settling time.

EIDBS4 Chapter 3: Time Domain Analysis Page 3-6
From Equation 3-8:

e
-T
= |z| = 0.8347
 = 0.3614,

and 
d
T = z = 0.4543

d
= 0.9086.

The overshoot for the corresponding continuous system is from Equation 3-5:
D = e
-/d

= e
-(0.3614)/0.9086

= 0.2866 or 28.66%.

The rise time using Equation 3-6:
t
r
= [-tan
-1
(
d
/)]/
d

= [ - tan
-1
(0.9086/0.3614)]/0.9086
= 2.145 sec.

Finally the settling time from Equation 3-7:
t
s
= 4/
= 4/0.3614
= 11.07 sec

We could confirm these findings by applying a step input to the system:
C(z) = T(z)R(z)
=
0.1065z 0.0902
z
2
z +0.6967
z
z 1

15.
.
c(k) may now be calculated in the normal way with the inverse Z transform. This will
be left as an exercise.

With MATLAB a graph of c(k) can easily be created with the following statements:
a=[0.1065 0.0902]
b=[1 -1.5 0.6967]
dstep(a,b,100)

The result is shown in Figure 3-5. From the graph the overshoot, rise time and settling
time may now be determined.
The overshoot is approximately 0.3 or 30% - estimated value 28.66%.
The rise time from the graph is 5 samples corresponding to 50.5 = 2.5 sec - estimated
value is 2.145 sec.
The settling time is difficult to read off from the graph but it should be approximately
25 sample times which is 250.5 = 12.5 sec. - estimated at 11.07 sec.

EIDBS4 Chapter 3: Time Domain Analysis Page 3-7

Figure 3-5

3.4 Relationship between the roots of the characteristic equation and
transient response
For real roots z = a in the characteristic equation, the transient response is of the form
a
k
. This is illustrated in Figure 3-6.

Figure 3-6

EIDBS4 Chapter 3: Time Domain Analysis Page 3-8
In the case of complex roots, z = b and b = 1, the transient response is of the form
cos(k), illustrated in Figure 3-7.

Figure 3-7

In the case of complex roots of the form, z=b and b  1, the transient response is
of the form b
k
cos(k), illustrated in Figure 3-8.

Figure 3-8
=T=/2
=T=3/4
=T=/4
b > 1
b < 1

EIDBS4 Chapter 3: Time Domain Analysis Page 3-9
It is very clear that if any roots of the characteristic equation falls outside the unit
circle on the z plane, the transient response does not decay but rather continue to grow.
We therefore come to the conclusion that
a stable discrete system can not have
any poles outside the unit circle
.

Consider the system in Figure 3-9.

Figure 3-9

E(z) = R(z) – GH(z)E(z)
E(z)[1 + GH(z)] = R(z)
E(z) =

GH(z)1
R(z)

The final error signal e
ss
is therefore given by (property 8):
e
ss
= lim
z1

GH(z)1
)R(z)z-(1
1-

For a step input r(t) = Ru(t), R(z) =
1)(z
Rz

, and
e
ss
= lim
z1

1)(z
Rz
GH(z)]z[1
1)-(z

= lim
z1

GH(z)1
R

=

 GH(z)1
R
1z
lim

The position error constant is now defined as :

E
P
=
lim
z 1 
GH(z)
,
so that
e
ss
=
P
E1
R

H(s)
C(s)
R(s)
E(s)
T
G(s)

EIDBS4 Chapter 3: Time Domain Analysis Page 3-10
In the case of a ramp function r(t) = Rtu(t), R(z) =
2
1)(z
RTz

, and
e
ss
= lim
z 1

2
1)(z
RTz
GH(z)]z[1
1)-(z

= lim
z 1

GH(z)]1)[1-(z
RT

= lim
z 1

 GH(z)])1(1)-[(z
RT
z

= lim
z 1

 GH(z))1(
RT
z

=
 
1)GH(z)(z
R
1z
lim
T
1

The velocity error constant E
v
is now defined as:

E
v
=
 
1)GH(z)(z
1z
lim
T
1

e
ss
=
v
E
R
.

For the quadratic function r(t) =
½
Rt
2
u(t) the acceleration error constant E
a
is defined
as:

E
a
=

GH(z)
2
1)(z
1z
lim
2
T
1

e
ss
=
a
E
R
.

The derivation of E
a
is left as an exercise.

Example 3-2
Determine the position, velocity and acceleration steady state errors for the system in
Figure 3-10, for r(t)=u(t), r(t)=tu(t) and r(t)=
½
t
2
u(t), respectively.

Figure 3-10
C(s)
R(s)
T=0.1
G(s)=
0.5)s(s
1

ZOH
G(z)

EIDBS4 Chapter 3: Time Domain Analysis Page 3-11
G(z) = (1 - z
-1
)Z

)5.0s(s
1
s
1

= (1 - z
-1
)Z

 )5.0(s
2
s
1

= (1-z
-1
)Z


)5.0(s
4
s
4
2
s
2

=



)
T5.0
e-(z
4z
1-z
4z
2
1)-(z
2Tz
z
1z

=



)
05.0
e-(z
4z
1-z
4z
2
1)-(z
0.2z
z
1z

=
)
05.0
e(z
2
1)(z
]
2
1)(z4)
05.0
e1)(z(z4)
05.0
ez[0.2(z
z
1z






=
)
05.0
e1)(z(z
4z8
2
z4)
05.0
e+z
05.0
ez
2
(z4
05.0
e2.00.2z






=
)
05.0
e1)(z(z
)
05.0
0.2e-
05.0
4e-(4+8)z-
05.0
4e+4(0.2



=
)95123.01)(z(z
0.0048364+0.0049177z


E
P
=
1z
lim

G(z) = 
e
ss
=
P
E1
R

=

1
1
= 0

E
v
=
 
1)G(z)(z
1z
lim
T
1

=
1z
lim
0.1
1

)95123.01)(z(z
0.0048364+0.0049177z
1
1z


=
1z
lim
0.1
1

)95123.0(z
0.0048364+0.0049177z

= 100.2 = 2
e
ss
=
v
E
R
.= ½ = 0.5

E
a
=

G(z)
2
1)(z
1z
lim
2
T
1

=
1z
lim
0.01
1

)95123.01)(z(z
0.0048364+0.0049177z
1
2
1)(z


EIDBS4 Chapter 3: Time Domain Analysis Page 3-12
=
1z
lim
0.01
1

)95123.0(z
0.0048364)+77z1)(0.00491-(z

= 1000 = 0
e
ss
=
a
E
R
.= 1/0 = 

Exercise

3-1
This problem refers to the system discussed in example 3-1. Find c(k) for C(z)
given by:
C(z) =
0.1065z 0.0902
z
2
z +0.6967
z
z 1

15.
.
Tabulate c(k) for k = 0 to 30 and compare with Figure 3-5.

3-2
The block diagram of a discrete control system is given in Figure P 3-1.
a) Determine the transfer function T(z) of the system if the sampling period
T = 0.1 sec.
b) Calculate the unit step response c(k) for k = 0 to 20.

Figure P 3-1

Ans:
G(z) = (1-z
-1
)Z{(1/s)5/s(s+2)} = [(z-1)/z]Z{5/s
2
(s+2)}
= [(z-1)/z]Z{5/s
2
(s+2)} = [(z-1)/z]Z{A/s
2
+ B/s + C/(s+2)}
= [(z-1)/z]Z{2.5/s
2
- 1.25/s + 1.25/(s+2)}
= [(z-1)/z][0.25z/(z-1)
2
- 1.25z/(z-1) + 1.25z/(z-e
-0.2
)]
= [(z-1)/z]0.25z{[(z-e
-0.2
) - 5(z-1)(z-e
-0.2
) + 5(z-1)
2
]/(z-1)
2
(z-e
-0.2
)}
= 0.25[z-e
-0.2
- 5(z
2
- z - ze
-0.2
+ e
-0.2
)+5(z
2
- 2z + 1)]/(z-1)(z-e
-0.2
)}
= 0.25[z - e
-0.2
- 5z
2
+ 5z + 5ze
-2
- 5e
-0.2
+ 5z
2
- 10z + 5]/(z-1)(z-e
-0.2
)}
= 0.25[-4z + 5ze
-0.2
- 6e
-0.2
+ 5]/(z-1)(z-e
-0.2
)}
= 0.25[z(-4 + 5e
-0.2
) - 6e
-0.2
+ 5)/(z - 1)(z - e
-0.2
)}
= 0.25(0.09365z + 0.08762)/(z - 1)(z - 0.8187)
= (0.02341z + 0.02191)/(z - 1)(z - 0.8187)
T(z) = G(z)/[1 + G(z)]
= [(0.02341z + 0.02191)/(z - 1)(z - 0.8187)]/
[1+(0.02341z + 0.02191)/(z - 1)(z - 0.8187)]
ZOH
c(t)
r(t)

2)+s(s
5

T

EIDBS4 Chapter 3: Time Domain Analysis Page 3-13
= [(0.02341z + 0.02191)/(z - 1)(z - 0.8187)]/
[(z - 1)(z - 0.8187)+(0.02341z + 0.02191)/(z - 1)(z - 0.8187)]
= (0.02341z + 0.02191)/(z
2
- 1.8187z + 0.8187 + 0.02341z + 0.02191)
= (0.02341z + 0.02191)/(z
2
- 1.795z + 0.8406)
b) C(z) = (0.02341z + 0.02191)/(z
2
- 1.795z + 0.8406)R(z)
= (0.02341z + 0.02191)/(z
2
- 1.795z + 0.8406)z/(z-1)
= z(0.02341z + 0.02191)/(z-1)(z
2
- 1.795z + 0.8406)
{z
2
– 1.795z + 0.8406 = (z – 0.91680.2058
r
)(z – 0.9168– 0.2058
r
)}

C(z)/z = (0.02341z + 0.02191)/(z – 1)(z – 0.91680.2058
r
)(z – 0.9168– 0.2058
r
)
= A/(z – 1) + B/(z – 0.91680.2058
r
) + B
*
/(z – 0.9168– 0.2058
r
)
A = [(0.02341z + 0.02191)/z
2
– 1.795z + 0.8406)]|
z = 1
= 0.9939
B = [(0.02341z + 0.02191)/(z – 1)(z-0.9168– 0.2058
r
)]|
z = 0.91680.2058

=0.53912.743
r

C(z)/z = 0.9939/(z-1) + 0.53912.743
r
/(z – 0.91680.2058
r
)
+ 0.5391– 2.743
r
/(z – 0.9168– 0.2058
r
)
C(z) = 0.9939z/(z-1) + 0.53912.743
r
z/(z – 0.91680.2058
r
)
+ 0.5391– 2.743
r
z/(z – 0.9168– 0.2058
r
)

c(k) = 0.9939 + 20.5391(0.9168)
k
cos(0.2058k + 2.743)
= 0.9939 + 1.078(0.9168)
k
cos(0.2058k + 2.743)

In the table below the values of c(k) were calculated with Excel for k = 0 to 59.

k 0 1 2 3 4 5 6 7 8 9
y(k) 4E-04 0.024 0.088 0.183 0.3 0.43 0.565 0.698 0.823 0.936
k 10 11 12 13 14 15 16 17 18 19
y(k) 1.034 1.114 1.176 1.22 1.247 1.257 1.254 1.24 1.217 1.187
k 20 21 22 23 24 25 26 27 28 29
y(k) 1.153 1.117 1.082 1.048 1.017 0.99 0.967 0.949 0.937 0.928
k 30 31 32 33 34 35 36 37 38 39
y(k) 0.924 0.924 0.927 0.933 0.94 0.949 0.959 0.968 0.977 0.986
k 40 41 42 43 44 45 46 47 48 49
y(k) 0.993 1 1.005 1.008 1.011 1.012 1.013 1.012 1.011 1.009
k 50 51 52 53 54 55 56 57 58 59
y(k) 1.006 1.004 1.001 0.999 0.997 0.995 0.993 0.991 0.99 0.99

EIDBS4 Chapter 3: Time Domain Analysis Page 3-14
With the aid of MATLAB for T(z) = (0.02341z + 0.02191)/(z
2
- 1.795z + 0.8406):

a=[0 0.02341 0.02191]
b=[1 -1.795 0.8406]
dstep(a,b,100)

EIDBS4 Chapter 4: Stability Analysis Page 4-1
4. STABILITY ANALYSIS
4.1 The Jury stability test
A discrete control system is stable if all the roots of the characteristic equation fall
inside the unity circle on the z plane. The Jury test may be used to confirm stability.
Assume the characteristic equation of a discrete system is given by:
Q(z) = a
n
z
n
+ a
n-1
z
n-1
+ ......+ a
1
z + a
0
= 0, a
n
>0.

We now form the following structure:

a
0
a
1
..... a
n-1
a
n

a
n
a
n-1
..... a
1
a
0

b
0
b
1
..... b
n-1

b
n-1
b
n-2
..... b
0

..... ..... ..... .....
l
0
l
1
l
2
l
3

l
3
l
2
l
1
l
0

m
0
m
1
m
2

where b
k
=
a a
a a
o
n k
n
k

, c
k
=
b b
b b
o
n k
n
k

, etc.
The necessary and sufficient conditions for the roots of Q(z), not to fall outside or on
the unit circle, are::
Q(1) > 0
Q(-1) > 0 if n is even or Q(-1) < 0 if n is uneven
|a
0
| < a
n

|b
0
| > |b
n-1
|
|c
0
| > |c
n-2
|
.
|l
0
| > |l
3
|
|m
0
| > |m
2
|

4.2 Application to digital control systems

Example 4-1

The characteristic equation of a system is given by:

Q(z) = z
3
+ z
2
+ 0.5z + 0.25.

Use the Jury test to investigate the stability of the system.

EIDBS4 Chapter 4: Stability Analysis Page 4-2
Q(1) = 1 + 1 + 0.5 + 0.25 = 2.75 > 0
Q(-1) = -1 +1 - 0.5 + 0.25 = -0.25 < 0 for n=3 (uneven)
|a
0
| = 0.25 < a
3
= 1
Form the Jury table:

0.25 0.5 1 1
1 1 0.5 0.25
-0.9375 -0.875 -0.25
(.25

.25 – 1

1) (.25

.5 - 1

1) (.25

1 - 1

.5)

|m
0
| = 0.9375 > |m
2
| = 0.25.
The system is therefore unconditionally stable.

Note:
Stability may also be tested with Matlab:
a=[1 1 0.5 0.25]
roots(a)

|z
1
| = 0.7718
-0.1141 + 0.5576i

|z
2
| = 0.5692
-0.1141 - 0.5576i

|z
2
*
| = 0.5692

Example 4-2

Use the Jury test to determine values for K that will ensure stability for the system
below. Assume T = ln2 sec.

Figure 4-1

B(z) = K
and A(z) = Z
1 e
s
.
1
(s +1)(s 2)
sT

1 e
sT
s

C(z)
R(z)

1
(s +1)(s +2)

T
K
A(z)
ZOH
G(z)
B(z)
T

EIDBS4 Chapter 4: Stability Analysis Page 4-3
= (1 - z
-1
)Z
1
s(s 1)(s +2)

A(z) = (1 - z
-1
)Z
0.5 1 0.5
s s 1 s 2

A(z) =
z 1
z
z
(z 1)
z
z e
0.5z
z- e
0.5
T - 2T

For T=ln2 sek:
A(z) =
z 1
z
z
(z 1)
z
z
0.5z
z- 0.25
0.5

05.

G(z) = B(z)A(z) = K
z 1
z
z
(z 1)
z
z
0.5z
z- 0.25
0.5

05.

= K
z 1
z
z[0.5(z 0.5)(z ) - (z 1)(z ) 0.5(z-1)(z )]
(z 1)(z )(z 0.25)

  

0 25 0 25 0 5
0 5
...
.

= K
0.5(z 0.5)(z ) (z 1)(z 0.25) 0.5(z-1)(z )
(z )(z )
  

 
025 05
05 025
..
..

= K
0.5z z +0.0625 (z z +0.25) z - 0.75z +0.25
(z )(z 0.25)
2 2 2
   
 
0375 075 05
05
...
.

= K
0.125z +0.0625
(z )(z 0.25) 05.

=
0.125K(z +0.5
(z )(z )
)
.. 05 025

The characteristic equation may be determined by calculating T(z).
T(z) = G(z)/[1 + G(z)]
=
0.125K(z +0.5
(z )(z )
0.125K(z +0.5
(z )(z )
)
..
)
.. 

 

05 025
1
05 025

=
0.125K(z +0.5
(z )(z )
(z -0.5)(z -0.25) +0.125K(z +0.5
(z )(z )
)
..
)
..   

05 025 05 025

=
0.125K(z +0.5
(z- 0.5)(z- 0.25) +0.125K(z +0.5
)
)

=
0.125K(z +0.5
z
2
- 0.75z +0.125+0.125Kz +0.0625
K
)

Q(z) = z
2
+ (0.125K - 0.75)z + (0.0625K + 0.125)

or by means of :
1 + G(z) = 0

1 +
0.125K(z +0.5
(z )(z )
)
..
 
05 025
= 0

(z-0.5)(z-0.25) + 0.125K(z+0.5) = 0

z
2
- 0.75z + 0.125 +0.125Kz + 0.0625K = 0

z
2
+ (0.125K - 0.75)z + (0.0625K + 0.125) = 0

Q(z) = z
2
+ (0.125K - 0.75)z + (0.0625K + 0.125)

EIDBS4 Chapter 4: Stability Analysis Page 4-4
Jury test:

Q(1) = 1 + 0.125K - 0.75 + 0.0625K + 0.125 > 0

0.1875K > -0.375

K > -2
Q(-1) = 1 - 0.125K +0.75 + 0.0625K + 0.125 > 0 for n even (2)

-0.0625K > -1.875

0.0625K < 1.875

K < 30
and |0.0625K + 0.125| < 1 (|a
0
| < a
n
)

0.0625K + 0.125 < 1 and 0.0625K + 0.125 > -1

K < 14 and K > -18
Because this is a second order system (n=2), the Jury table is not involved.
We therefore conclude that for stability: -2 < K < 14.

Exercise

4-1
A digital control system is described by the following difference equation:
x(k+1) = (0.368 - 0.632K)x(k) + Kr(k),
where r(k) is the input and y(k) is the output. Calculate values for K so that the
system is stable.

Answ:
zX(z) = (0.368 - 0.632K)X(z) + KR(z) assume x(0) = 0

X(z)[z - 0.368 + 0.632K] = KR(z)

X(z)/R(z) = K/(z - 0.368 + 0.632K)

The characteristic equation is therefore: