The document discusses the Z-transform, which is a tool for analyzing and solving linear time-invariant difference equations. It defines the Z-transform, provides examples of common sequences and their corresponding Z-transforms, and discusses properties such as the region of convergence. Key topics covered include the difference between difference and differential equations, properties of linear time-invariant systems, and mapping between the s-plane and z-plane.
Z Transform And Inverse Z Transform - Signal And SystemsMr. RahüL YøGi
The z-transform is the most general concept for the transformation of discrete-time series.
The Laplace transform is the more general concept for the transformation of continuous time processes.
For example, the Laplace transform allows you to transform a differential equation, and its corresponding initial and boundary value problems, into a space in which the equation can be solved by ordinary algebra.
The switching of spaces to transform calculus problems into algebraic operations on transforms is called operational calculus. The Laplace and z transforms are the most important methods for this purpose.
z-Transform is for the analysis and synthesis of discrete-time control systems.The z transform in discrete-time systems play a similar role as the Laplace transform in continuous-time systems
Chapter 7 Controls Systems Analysis and Design by the frequency response analysis . From the book (Ogata Modern Control Engineering 5th).
7-1 introduction.
7-2 Bode diagrams.
Z Transform And Inverse Z Transform - Signal And SystemsMr. RahüL YøGi
The z-transform is the most general concept for the transformation of discrete-time series.
The Laplace transform is the more general concept for the transformation of continuous time processes.
For example, the Laplace transform allows you to transform a differential equation, and its corresponding initial and boundary value problems, into a space in which the equation can be solved by ordinary algebra.
The switching of spaces to transform calculus problems into algebraic operations on transforms is called operational calculus. The Laplace and z transforms are the most important methods for this purpose.
z-Transform is for the analysis and synthesis of discrete-time control systems.The z transform in discrete-time systems play a similar role as the Laplace transform in continuous-time systems
Chapter 7 Controls Systems Analysis and Design by the frequency response analysis . From the book (Ogata Modern Control Engineering 5th).
7-1 introduction.
7-2 Bode diagrams.
Overlap Add, Overlap Save(digital signal processing)Gourab Ghosh
In DSP to solve a convolution of a long duration sequence there are two popular methods. Overlap Add, Overlap Save. In this presentation i've discussed about both.
- Gourab Ghosh
The z-Transform is often time more convenient to use
Definition:
Compare to DTFT definition:
z is a complex variable that can be represented as z=r ej
Substituting z=ej will reduce the z-transform to DTFT
Region of Convergence for a discrete time signal x[n] is defined as a continuous region in z plane where the Z-Transform converges.
The roots of the equation P(z) = 0 correspond to the ’zeros’ of X(z)
The roots of the equation Q(z) = 0 correspond to the ’poles’ of X(z)
The RoC of the Z-transform depends on the convergence of the polynomials P(z) and Q(z),
Uses to analysis of digital filters.
Used to simulate the continuous systems.
Analyze the linear discrete system.
Used to finding frequency response.
This presention is about one of the most important concept of control system called State space analysis. Here basic concept of control system and state space are discussed.
It gives how states are representing in various canonical forms and how it it is different from transfer function approach. and finally test the system controllability and observability by kalman's test
Overlap Add, Overlap Save(digital signal processing)Gourab Ghosh
In DSP to solve a convolution of a long duration sequence there are two popular methods. Overlap Add, Overlap Save. In this presentation i've discussed about both.
- Gourab Ghosh
The z-Transform is often time more convenient to use
Definition:
Compare to DTFT definition:
z is a complex variable that can be represented as z=r ej
Substituting z=ej will reduce the z-transform to DTFT
Region of Convergence for a discrete time signal x[n] is defined as a continuous region in z plane where the Z-Transform converges.
The roots of the equation P(z) = 0 correspond to the ’zeros’ of X(z)
The roots of the equation Q(z) = 0 correspond to the ’poles’ of X(z)
The RoC of the Z-transform depends on the convergence of the polynomials P(z) and Q(z),
Uses to analysis of digital filters.
Used to simulate the continuous systems.
Analyze the linear discrete system.
Used to finding frequency response.
This presention is about one of the most important concept of control system called State space analysis. Here basic concept of control system and state space are discussed.
It gives how states are representing in various canonical forms and how it it is different from transfer function approach. and finally test the system controllability and observability by kalman's test
Welcome to WIPAC Monthly the magazine brought to you by the LinkedIn Group Water Industry Process Automation & Control.
In this month's edition, along with this month's industry news to celebrate the 13 years since the group was created we have articles including
A case study of the used of Advanced Process Control at the Wastewater Treatment works at Lleida in Spain
A look back on an article on smart wastewater networks in order to see how the industry has measured up in the interim around the adoption of Digital Transformation in the Water Industry.
Immunizing Image Classifiers Against Localized Adversary Attacksgerogepatton
This paper addresses the vulnerability of deep learning models, particularly convolutional neural networks
(CNN)s, to adversarial attacks and presents a proactive training technique designed to counter them. We
introduce a novel volumization algorithm, which transforms 2D images into 3D volumetric representations.
When combined with 3D convolution and deep curriculum learning optimization (CLO), itsignificantly improves
the immunity of models against localized universal attacks by up to 40%. We evaluate our proposed approach
using contemporary CNN architectures and the modified Canadian Institute for Advanced Research (CIFAR-10
and CIFAR-100) and ImageNet Large Scale Visual Recognition Challenge (ILSVRC12) datasets, showcasing
accuracy improvements over previous techniques. The results indicate that the combination of the volumetric
input and curriculum learning holds significant promise for mitigating adversarial attacks without necessitating
adversary training.
Cosmetic shop management system project report.pdfKamal Acharya
Buying new cosmetic products is difficult. It can even be scary for those who have sensitive skin and are prone to skin trouble. The information needed to alleviate this problem is on the back of each product, but it's thought to interpret those ingredient lists unless you have a background in chemistry.
Instead of buying and hoping for the best, we can use data science to help us predict which products may be good fits for us. It includes various function programs to do the above mentioned tasks.
Data file handling has been effectively used in the program.
The automated cosmetic shop management system should deal with the automation of general workflow and administration process of the shop. The main processes of the system focus on customer's request where the system is able to search the most appropriate products and deliver it to the customers. It should help the employees to quickly identify the list of cosmetic product that have reached the minimum quantity and also keep a track of expired date for each cosmetic product. It should help the employees to find the rack number in which the product is placed.It is also Faster and more efficient way.
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...Dr.Costas Sachpazis
Terzaghi's soil bearing capacity theory, developed by Karl Terzaghi, is a fundamental principle in geotechnical engineering used to determine the bearing capacity of shallow foundations. This theory provides a method to calculate the ultimate bearing capacity of soil, which is the maximum load per unit area that the soil can support without undergoing shear failure. The Calculation HTML Code included.
Water scarcity is the lack of fresh water resources to meet the standard water demand. There are two type of water scarcity. One is physical. The other is economic water scarcity.
Explore the innovative world of trenchless pipe repair with our comprehensive guide, "The Benefits and Techniques of Trenchless Pipe Repair." This document delves into the modern methods of repairing underground pipes without the need for extensive excavation, highlighting the numerous advantages and the latest techniques used in the industry.
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Ideal for homeowners, contractors, engineers, and anyone interested in modern plumbing solutions, this guide provides valuable insights into why trenchless pipe repair is becoming the preferred choice for pipe rehabilitation. Stay informed about the latest advancements and best practices in the field.
2. Agenda
Difference Equation vs Differential Equati
Z-Transform
Examples
Properties
Inverse Z-Transform
Long Division
Partial Fraction
Solution of Difference Equation
Mapping between s-plane to z-plane
2
3. Difference Equation vs Differential Equation
A difference equation expresses the change in some variable as a result
of a finite change in the other variable.
A differential equation expresses the change in some variable as a result
of an infinitesimal change in the other variable.
3
4. Differential Equation
Following figure shows a mass-spring-damper-system. Where y is
position, F is applied force D is damping constant and K is spring
constant.
Rearranging above equation in following form
4
𝐹 𝑡 = 𝑚 𝑦 𝑡 + 𝐷 𝑦 𝑡 + 𝐾𝑦(𝑡)
𝑦 𝑡 =
1
𝑚
𝐹 𝑡 −
𝐷
𝑚
𝑦 𝑡
𝐾
𝑚
𝑦(𝑡)
7. Difference Equations
Difference equations arise in problems where the time is assumed to
have a discrete set of possible values.
Where coefficients 𝑎 𝑛−1, 𝑎 𝑛−2,… and 𝑏 𝑛, 𝑏 𝑛−1,… are constant.
𝑢(𝑘) is forcing function
7
𝑦 𝑘 + 𝑛 + 𝑎 𝑛−1 𝑦 𝑘 + 𝑛 − 1 + ⋯ + 𝑎1 𝑦 𝑘 + 1 + 𝑎0 𝑦 𝑘
= 𝑏 𝑛 𝑢 𝑘 + 𝑛 + 𝑏 𝑛−1 𝑢 𝑘 + 𝑛 − 1 + ⋯ + 𝑏1 𝑢 𝑘 + 1 + 𝑏0 𝑢 𝑘
8. Difference Equations
Example 1: For the given difference equation, determine the (a)
order of the equation. Is the equation (b) linear, (c) time
invariant, or (d) homogeneous?
𝒚 𝒌 + 𝟐 + 𝟎. 𝟖𝒚 𝒌 + 𝟏 + 𝟎. 𝟎𝟕𝒚 𝒌 = 𝒖 𝒌
Solution:
a) The equation is second order.
b) All terms enter the equation linearly
c) All the terms if the equation have constant coefficients.
Therefore the equation is therefore LTI.
d) A forcing function appears in the equation, so it is
nonhomogeneous.
8
9. Z-Transform
Difference equations can be solved using z-transforms which provide a
convenient approach for solving LTI equations.
The z-transform is an important tool in the analysis and design of
discrete-time systems.
It simplifies the solution of discrete-time problems by converting LTI
difference equations to algebraic equations and convolution to
multiplication.
Thus, it plays a role similar to that served by Laplace transforms in
continuous-time problems.
9
10. Z-transform Definition
Given the causal sequence {x(ktT)}, its z-transform is defined as
The variable z−1 in the above equation can be regarded as a time delay
operator.
10
X 𝑧 = 𝑥 0 + 𝑥 𝑇 𝑧−1
+ 𝑥 2𝑇 𝑧−2
+ ⋯ + 𝑥 𝑘𝑇 𝑧−𝑘
𝑋 𝑧 =
𝑘=0
∞
𝑥 𝑘𝑇 𝑧−𝑘
11. Example
Obtain the Z-transform of the sequence
Solution:
11
x 𝑘𝑇 = {1, 1, 3, 2, 0, 4, 0, 0, 0, … }
X 𝑧 = 1 + 𝑧−1
+3 𝑧−2
+ 2 𝑧−3
+ 4𝑧−5
12. Laplace Transform and Z-Transform
Given the sampled impulse train of a signal
𝑥∗ 𝑡 = 𝑥 0 𝛿 𝑡 + 𝑥 𝑇 𝛿 𝑡 − 𝑇 + ⋯ + 𝑥 𝑘𝑇 𝛿(𝑡 − 𝑘𝑇) + ⋯
𝑥∗
𝑡 =
𝑘=0
∞
𝑥 𝑘𝑇 𝛿 𝑡 − 𝑘𝑇
12
13. Laplace Transform and Z-Transform
𝑋∗
𝑆 = 𝑥 0 + 𝑥 𝑇 𝑒−𝑠𝑇
+ 𝑥 2𝑇 𝑒−𝑠2𝑇
+ ⋯ + 𝑥 𝑘𝑇 𝑒−𝑠𝑘𝑇
+ ⋯
𝑋∗ 𝑆 =
𝑘=0
∞
𝑥 𝑘𝑇 𝑒−𝑠𝑘𝑇 =
𝑘=0
∞
𝑥 𝑘𝑇 (𝑒 𝑠𝑇)−𝑘 → (1)
The z-transform is
𝑋 𝑧 =
𝑘=0
∞
𝑥 𝑘𝑇 𝑧−𝑘
→ (2)
Comparing (1) and (2) yields
𝑧 = 𝑒 𝑠𝑇 where 𝑇 is the sample period
13
14. A Note
In general, given a transfer function in s-domain you cannot just replace
𝑠 by s =
𝑙𝑛𝑧
𝑇
(from 𝑧 = 𝑒 𝑠𝑇
) to get its z-domain transfer function.
The reason is that 𝑧 = 𝑒 𝑠𝑇 is true with the sampled signal.
14
16. z-transform of the Unit impulse
Let x(kT) = δ(kT)
Then
16
otherwise0
0kfor1
(kT)
1(kT)
00
k
k
k
k
zzkTxzX
17. z-transform of the a Shifted Unit impulse
Let x(kT) = δ(kT-qT)
Then
17
otherwise0
qkfor1
qT)-(kT
q
k
k
k
k
zzzkTxzX
00
qT)-(kT)(
18. z-transform of a Unit-Step Function
Let x(kT) = u(kT)
Then
18
0kfor0
0kfor1
(kT)
u
11
1
(kT))( 1
000
z
z
z
zzuzkTxzX
k
k
k
k
k
k
19. z-transform of Sample Exponential
Let x(kT) = ak u(kT)
Then
19
0kfor0
0kfor
(kT)
k
k a
ua
000
(kT))(
k
kk
k
kk
k
k
zazuazkTxzX
az
z
az
azzX
k
k
1
0
1
1
1
)(
20. z-transform of kaku[k]
Let x(kT) = k ak u(kT)
Then
20
0kfor0
0kfor
(kT)
k
k ka
uka
000
k(kT))(
k
kk
k
kk
k
k
zazukazkTxzX
221
0
1
)()1(
1
)(k
az
z
az
azzX
k
k
21. z-transform of Sinusoids
Let x(kT) = (cos ΩkT) u(kT)
Then
Similarly it can be shown:
21
0kfor0
0kfor)cos(
(n))cos(
kT
un
2
)k(cos
kTjkTj
ee
T
)}()({
2
1
u(k)})k({cos kTuekTueZT kTjkTj
TjTj
ez
z
ez
z
kTkT
2
1
)}u()({cos
1)(cos2
)(cos
)(cos21
)(cos1
)}u()({cos 2
2
21
1
Tzz
Tzz
zTz
Tz
kTkT
1)(cos2
)(s
)(cos21
)(s
)}u()({sin 221
1
Tzz
Tinz
zTz
Tinz
kTkT
29. Z-Transform Properties: Convolution
Convolution in time domain is multiplication in z-domain
Example: Let’s calculate the convolution of
Multiplications of z-transforms is
29
zXzXkTxkTx Z
2121
kTukTkTuakTx kT
21 xand
1121
11
1
zaz
zXzXzY
zXkTxzXkTx ZZ
2211 )(&)(
30. Z-Transform Properties: Initial and Final Value Theorems
Initial Value Theorem:
The value of x(n) as k → 0 is given by:
Final Value Theorem:
The value of x(n) as k → is given by:
30
)(limlim)0(
0
zXkTxx
zk
)]()1[(limlim)(
1
zXzkTxx
zk
32. 32
The Inverse Z-Transform
Formal inverse z-transform is based on a Cauchy integral
Less formal ways sufficient most of the time
1) Direct or Long Division Method
2) Partial fraction expansion and Look-up Table
3) Inversion Integral Method (Residue-theorem)
dzzzX
j
zXZkTx k
C
11
2
1
33. 33
Inverse Z-Transform: Power Series Expansion
Using Long Division to expand X(z) as a series
Write the inverse transform as the sequence
0
21
...20
k
k
zkTxzTxzTxxzX
...},2,,0{ TxTxxkTx
35. Inverse Z-Transform: Power Series Expansion
Example: Find x(n) for n = 0, 1, 2, 3, 4, when X(z) is given by:
Solution:
First, rewrite X(z) as a ratio of polynomial in 𝑧−1
, as follows:
Dividing the numerator by the denominator, we have:
35
2.01
510
zz
z
zX
21
21
2.02.11
510
zz
zz
zX
4321
68.184.181710
zzzz
21
2.02.11
zz 21
510
zz
32
217
zz
432
4.34.2017
zzz
43
4.34.18
zz
543
68.308.224.18
zzz
54
68.368.18
zz
654
736.3416.2268.18
zzz
321
21210
zzz Therefore,
• x(0) = 0,
• x(1) = 10,
• x(2) = 17,
• x(3) = 18.4
• x(4) = 18.68
36. Inverse Z-Transform: Partial Fraction Expansion
Assume that a given z-transform can be expressed as
Apply partial fractional expansion
First term exist only if M>N
Br is obtained by long division
Second term represents all first order poles
Third term represents an order s pole
There will be a similar term for every high-order pole
Each term can be inverse transformed by inspection
36
N
0k
k
k
M
0k
k
k
za
zb
zX
s
1m
m1
i
m
N
ik,1k
1
k
k
NM
0r
r
r
zd1
C
zd1
A
zBzX
37. Inverse Z-Transform: Partial Fraction Expansion
Coefficients are given as
Easier to understand with examples
37
s
m
m
i
m
N
ikk k
k
NM
r
r
r
zd
C
zd
A
zBzX
1
1
,1
1
0 11
kdzkk zXzdA
1
1
1
1
1
!
1
idw
s
ims
ms
ms
i
m wXwd
dw
d
dms
C
38. Example:
Find x(kT) if X(z) is given by:
Solution: We first expand X(z)/z into partial fractions as follows :
Then we obtain
then
38
2.01
10
zz
z
zX
2.0
5.12
1
5.12
2.01
10
zzzzz
zX
2.01
5.12
z
z
z
z
zX
)()2.0(5.12)(5.12 kTukTukTx k
)(11
kTZ
kTu
z
z
Z
1
1
kTua
az
z
Z n
1
39. Another Solution
In this example, if X(z), rather than X(z)/z, is expanded into partial
fractions, then we obtain:
However, by use of the shifting theorem we find :
Then
39
2.0
5.2
1
5.12
2.01
10
zzzz
z
zX
2.0
5.2
1
5.12 11
z
z
z
z
z
zzX
)(2.015.12)()2.0(
2.0
5.2
)(5.12)(
)()2.0(5.2)(5.12)( 1
TkTuTkTuTkTukTx
TkTuTkTukTx
kk
k
..
..
48.124
4.123
122
101
00
x
x
x
x
x
TnkTxzXzZ o
no
1
40. Example:
Find x(kT) if X(z) is given by:
Solution: Expand X(z)/z into partial fraction as follows:
Then:
By referring to tables we find that x(kT) is given by:
and therefore,
x(0) = 0 -1 +3 = 2
x(1) = 18 – 2 + 3= 19
40
12
2
2
3
zz
zz
zX
1
3
2
1
2
9
12
12
22
2
zzzzz
z
z
zX
1
3
22
9
2
z
z
z
z
z
z
zX
)(3)()2()()2(5.4 kTukTukTrkTx kk
kTuk
k
kk
kTr
00
0
)(
kTr
z
z
Z
2
1
1
kTra
az
az
Z k
2
1
44. Mapping between s-plane to z-plane
Where 𝑠 = 𝜎 + 𝑗𝜔 for real number 𝜎 and real number 𝜔.
Then 𝑧 in polar coordinates is given by
44
𝑧 = 𝑒(𝜎+𝑗𝜔)𝑇
𝑧 = 𝑒 𝜎𝑇
𝑒 𝑗𝜔𝑇
∠𝑧 = 𝜔𝑇𝑧 = 𝑒 𝜎𝑇
𝑧 = 𝑒 𝑠𝑇
45. Mapping between s-plane to z-plane
We will discuss following cases to map given points on s-plane to z-
plane.
Case-1: Real pole in s-plane (𝑠 = 𝜎) [on the left hand-side]
Case-2: Imaginary Pole in s-plane (𝑠 = 𝑗𝜔)
Case-3: Complex Poles (𝑠 = 𝜎 + 𝑗𝜔)
45𝑠 − 𝑝𝑙𝑎𝑛𝑒 𝑧 − 𝑝𝑙𝑎𝑛𝑒
46. Mapping between s-plane to z-plane
Case-1: Real pole in s-plane (𝑠 = 𝜎)
When 𝑠 = 0
When 𝑠 = −∞
46
∠𝑧 = 𝜔𝑇𝑧 = 𝑒 𝜎 𝑇
𝑧 = 𝑒0𝑇 = 1 ∠𝑧 = 0𝑇 = 0
𝑠 = 0
𝑠 − 𝑝𝑙𝑎𝑛𝑒 𝑧 − 𝑝𝑙𝑎𝑛𝑒
1
𝑧 = 𝑒−∞𝑇 = 1 ∠𝑧 = 0𝑇 = 0
47. Mapping between s-plane to z-plane
Case-2: Imaginary pole in s-plane (𝑠 = ±𝑗𝜔)
When 𝒔 = 𝒋𝝎
47
∠𝑧 = 𝜔𝑇𝑧 = 𝑒 𝜎𝑇
𝒛 = 𝒆 𝟎𝑻 = 𝟏 ∠𝒛 = 𝝎𝑻
𝑠 = 𝑗𝜔
𝑠 − 𝑝𝑙𝑎𝑛𝑒 𝑧 − 𝑝𝑙𝑎𝑛𝑒
1
−1
−1
1
𝜔𝑇