Controls: Equation (C.1)is a system. A is called the state matrix, of

Caribbean Maritime University
Controls 2
December 05, 2019
Dr. Milton T. Richardson
Email: dr.mtrichardson@gmail.com
Mobile: 469-5940

2
Discrete Signals and Z-Transform

5
Frequency domain vs Time domain
 Frequency domain is a term used to describe the analysis of mathematical functions or signals
with respect to frequency.
 (communications point of view) A plane on which signal strength can be represented graphically
as a function of frequency, instead of a function of time.
 control systems) Pertaining to a method of analysis, particularly useful for fixed linear systems
in which one does not deal with functions of time explicitly, but with their Laplace or Fourier
transforms, which are functions of frequency.
 Speaking non-technically, a time domain graph shows how a signal changes over time, whereas
a frequency domain graph shows how much of the signal lies within each given frequency band
over a range of frequencies.

6
Cont:
 A frequency domain representation can also include information on the phase shift that must be
applied to each sinusoid in order to be able to recombine the frequency components to recover
the original time signal.
 The frequency domain relates to the Fourier transform or Fourier series by decomposing a
function into an infinite or finite number of frequencies. This is based on the concept of Fourier
series that any waveform can be expressed as a sum of sinusoids (sometimes infinitely many.)
 In using the Laplace, Z-, or Fourier transforms, the frequency spectrum is complex and
describes the frequency magnitude and phase. In many applications, phase information is not
important. By discarding the phase information it is possible to simplify the information in a
frequency domain representation to generate a frequency spectrum or spectral density. A
spectrum analyser is a device that displays the spectrum.

7/13
 Sampling is the transformation of a continuous signal into a discrete signal
 Widely applied in digital analysis systems
1. Sample the continuous time signal
2. Design and process discrete time signal
3. Convert back to continuous time
What is Discrete Time Sampling?
x(t),
x[n]
t=nT
Discrete
Time sampler
Discrete
time system
Signal
reconstruction
x(t) x[n] y[n] y(t)
T is the sampling
period

8/13
Why is Sampling Important?
 For many systems (e.g. Matlab, …) designing
and processing discrete-time systems is more
efficient and more general compared to
performing continuous-time system design.
 How does Simulink perform continuous-time
system simulation?
 The signals are sampled and the systems are
approximately integrated in discrete time
 Mainly due to the dramatic development of
digital technology resulting in inexpensive,

Sample and Hold Gates
 It is often useful to be able to sample a signal
and then hold its value constant
 this is useful when performing analogue-to-
digital conversion so that the signal does not
change during conversion
 it is also useful when doing digital-to-
analogue conversion to maintain the output
voltage constant between conversions
 This task is performed by a sample and hold

 Sample and hold gate circuits

11
Modeling the Zero-order Hold

13
Ideal Sampling and Zero-order Hold

14
Impulse Response of the Zero-order Hold

 Most sample and hold gates are constructed
using integrated circuits
 Typical devices require a few microseconds to
sample the incoming waveform, which then
decays
(or droops) at a rate of a few millivolts per
millisecond
 High speed devices, such as those used for video
applications, can sample an input in a few
nanoseconds, but may experience a droop of a

16
The Direct Z-Transform
The z-transform of a discrete time signal is defined as the power series
(1)
Where z is a complex variable. For convenience, the z-transform of a signal x[n] is denoted by
X(z) = Z{x[n]}
Since the z-transform is an infinite series, it exists only for those values of z for which this
series converges. The Region of Convergence (ROC) of X(z) is the set of all values of z for
which this series converges.
We illustrate the concepts by some simple examples.






n
n
z
]
n
[
x
)
z
(
X

17
Example 1: Determine the z-transform of
the following signals
(a) x[n] = [1, 2, 5, 7, 0, 1]
Solution: X(z) = 1 + 2z-1
+ 5z-2
+ 7z-3
+ z-5,
ROC: entire z plane except z = 0
(b) y[n] = [1, 2, 5, 7, 0, 1]
Solution: Y(z) = z2
+ 2z + 5 + 7z-1
+ z-3
ROC: entire z-plane except z = 0 and z = .
(c) z[n] = [0, 0, 1, 2, 5, 7, 0, 1]
Solution: z-2
+ 2z-3
+ 5z-4
+ 7z-5
+ z-7,
ROC: all z except z=0

18
(d) p[n] = [n]
Solution: P(z) = 1, ROC: entire z-plane.
(e) q[n] = [n – k], k > 0
Solution: Q(z) = z-k
, entire z-plane except
z=0.
(f) r[n] = [n+k], k > 0
Solution: R(z) = zk
,
ROC: entire z-plane except z = .

19
x[n] = (1/2)n
u[n]
Solution:
ROC: |1/2 z-1
| < 1, or equivalently |z| > 1/2
1
2
1
n
0
n
1
n
n
0
n
n
n
z
2
1
1
1
.......
z
2
1
z
2
1
1
z
2
1
z
2
1
z
]
n
[
x
)
z
(
X












































20
Example 3: Determine the z-transform of the signal x[n] =
an
u[n]
Solution:
 
 
|
a
|
|
z
:|
ROC
az
1
1
.......
az
az
1
az
z
a
)
z
(
X
1
2
1
1
n
0
n
1
n
0
n
n





















21
Properties of z-transform
 Linearity
If x1[n]  X1(z)
and x2[[n]  X2(z)
then
a1x1[n] + a2x2[n]  a1X1(z) + a2X2(z)

22
Example: Determine the z-transform of
the signal x[n] = [3(2n
) – 4(3n
)]u[n]
Solution:
  1
1
n
n
1
n
z
3
1
1
4
z
2
1
1
3
]
3
4
)
2
(
3
[
z
az
1
1
]]
n
[
u
a
[
z












the signal (cosw0n)u[n]
 
 
 
2
0
1
0
1
1
jw
1
jw
0
n
jw
n
jw
0
z
w
cos
z
2
1
w
cos
z
1
z
e
1
1
2
1
z
e
1
1
2
1
]
n
[
u
n
w
cos
z
e
2
1
e
2
1
]
n
[
u
n
w
cos
0
0
0
0




















23
Example: Find the z-transform of a unit step
function. Use time shifting property to find z-
transform of u[n] – u[n-N].
The z-transform of u[n] can be found as
Now the z-transform of u[n]-u[n-N] may be
found as follows:
1
2
1
0
n
n
n
n
z
1
1
.......
z
z
1
z
z
]
n
[
u
]]
n
[
u
[
z

















 

1
N
1
N
1
z
1
z
1
z
1
1
z
z
1
1
]]
N
n
[
u
]
n
[
u
[
z















24
Inverse z-transform
In general, the inverse z-transform may be
found by using any of the following
methods:
 Power series method
 Partial fraction method

25
Power Series Method
2
1
z
5
.
0
z
5
.
1
1
1
)
z
(
X 




By dividing the numerator of X(z) by its
denominator, we obtain the power series
...
z
z
z
z
1
z
z
1
1 4
16
31
3
8
15
2
4
7
1
2
3
2
2
1
1
2
3














 x[n] = [1, 3/2, 7/2, 15/8, 31/16,…. ]

26
Power Series Method
Example 2:Determine the z-transform of
2
1
1
z
z
2
2
z
4
)
z
(
X 






By dividing the numerator of X(z) by its
denominator, we obtain the power series
 x[n] = [2, 1.5, 0.5, 0.25, …..]

Some common Z-transform pairs
Chapter 3: The Z-Transform 27

28
Z-Transform
 Region of Convergence
Here’s what the ROC can look like:
|z|<a b<|z| b<|z|<a all |z|

SEQUENCE TRANSFORM ROC
1

z
 
 
0
m
if
or
0
m
if
0
except
z
All



1

z
1
1
1

 z
1
1
1

 z
m
z
 
 
 
 
m
n
n
u
n
u
n






1
1 z
ALL

 
 
 
 
 
 
     
 
1
:
cos
2
1
cos
1
cos
:
1
1
:
1
:
1
1
1
:
1
1
2
1
0
1
0
0
2
1
1
2
1
1
1
1




































z
ROC
z
z
z
n
u
n
a
z
ROC
az
az
n
u
na
a
z
ROC
az
az
n
u
na
a
z
ROC
az
n
u
a
a
z
ROC
az
n
u
a
Z
Z
n
Z
n
Z
n
Z
n




31
Partial Fraction Method:
Example 1: Find the signal corresponding to
the z-transform
2
1
3
z
z
3
2
z
)
z
(
X 





Solution:
  
5
.
0
z
1
z
z
5
.
0
z
5
.
0
z
5
.
1
z
5
.
0
z
z
3
2
z
)
z
(
X 2
3
2
1
3








 


   5
.
0
z
4
1
z
1
z
1
z
3
5
.
0
z
1
z
z
5
.
0
z
)
z
(
X
2
2










5
.
0
z
z
)
4
(
1
z
z
z
1
3
)
z
(
X







or 1
1
1
z
5
.
0
1
1
4
z
1
1
z
3
)
z
(
X 








  ]
n
[
u
5
.
0
4
]
n
[
u
]
1
n
[
]
n
[
3
]
n
[
x
n









32
Partial Fraction Method:
Example 2: Find the signal corresponding to the z-
transform
  2
1
1
z
2
.
0
1
z
2
.
0
1
1
)
z
(
Y





Solution:
  2
3
2
.
0
z
2
.
0
z
z
)
z
(
Y



    2
2
2
2
.
0
z
1
.
0
2
.
0
z
75
.
0
2
.
0
z
25
.
0
2
.
0
z
2
.
0
z
z
z
)
z
(
Y









 2
2
.
0
z
z
1
.
0
2
.
0
z
z
75
.
0
1
z
z
25
.
0
)
z
(
Y






 2
1
1
2
.
0
1
.
0
1
1
z
2
.
0
1
z
2
.
0
z
2
.
0
1
1
75
.
0
z
2
.
0
1
1
25
.
0










      ]
n
[
u
2
.
0
n
5
.
0
]
n
[
u
2
.
0
75
.
0
]
n
[
u
2
.
0
25
.
0
]
n
[
y
n
n
n






Thank you very much for your attention

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