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
state space modeling of electrical systemMirza Baig
Introduction
As systems become more complex, representing them with differential equations or transfer functions becomes cumbersome. This is even more true if the system has multiple inputs and outputs. This document introduces the state space method which largely alleviates this problem. The state space representation of a system replaces an nth order differential equation with a single first order matrix differential equation. The state space representation of a system is given by two equations :
The first equation is called the state equation, the second equation is called the output equation. For an nth order system (i.e., it can be represented by an nth order differential equation) with r inputs and m outputs the size of each of the matrices is as follows:
Several features:The state equation has a single first order derivative of the state vector on the left, and the state vector, q(t), and the input u(t) on the right. There are no derivatives on the right hand side.The output equation has the output on the left, and the state vector, q(t), and the input u(t) on the right. There are no derivatives on the right hand side.
q is nx1 (n rows by 1 column)q is called the state vector, it is a function of timeA is nxn; A is the state matrix, a constantB is nxr; B is the input matrix, a constant u is rx1; u is the input, a function of time C is mxn; C is the output matrix, a constant D is mxr; D is the direct transition matrix, a constant y is mx1; y is the output, a function of time
Derivation of of State Space Model (Electrical)
To develop a state space system for an electrical system, they choosing the voltage across capacitors, and current through inductors as state variables. Recall that
so if we can write equations for the voltage across an inductor, it becomes a state equation when we divide by the inductance (i.e., if we have an equation for einductor and divide by L, it becomes an equation for diinductor/dt which is one of our state variable). Likewise if we can write an equation for the current through the capacitor and divide by the capacitance it becomes a state equation for ecapacitor
There are three energy storage elements, so we expect three state equations. Try choosing i1, i2 and e1 as state variables. Now we want equations for their derivatives. The voltage across the inductor L2 is e1 (which is one of our state variables)so our first state variable equation is
This equation has our input (ia) and two state variable (iL2 and iL1) and the current through the capacitor. So from this we can get our second state equation
Our third, and final, state equation we get by writing an equation for the voltage across L1 (which is e2) in terms of our other state variables
references:
http://lpsa.swarthmore.edu/Representations/SysRepSS.html
https://en.wikipedia.org/wiki/State-space_representation
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
State space analysis, eign values and eign vectorsShilpa Shukla
State space analysis concept, state space model to transfer function model in first and second companion forms jordan canonical forms, Concept of eign values eign vector and its physical meaning,characteristic equation derivation is presented from the control system subject area.
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
state space modeling of electrical systemMirza Baig
Introduction
As systems become more complex, representing them with differential equations or transfer functions becomes cumbersome. This is even more true if the system has multiple inputs and outputs. This document introduces the state space method which largely alleviates this problem. The state space representation of a system replaces an nth order differential equation with a single first order matrix differential equation. The state space representation of a system is given by two equations :
The first equation is called the state equation, the second equation is called the output equation. For an nth order system (i.e., it can be represented by an nth order differential equation) with r inputs and m outputs the size of each of the matrices is as follows:
Several features:The state equation has a single first order derivative of the state vector on the left, and the state vector, q(t), and the input u(t) on the right. There are no derivatives on the right hand side.The output equation has the output on the left, and the state vector, q(t), and the input u(t) on the right. There are no derivatives on the right hand side.
q is nx1 (n rows by 1 column)q is called the state vector, it is a function of timeA is nxn; A is the state matrix, a constantB is nxr; B is the input matrix, a constant u is rx1; u is the input, a function of time C is mxn; C is the output matrix, a constant D is mxr; D is the direct transition matrix, a constant y is mx1; y is the output, a function of time
Derivation of of State Space Model (Electrical)
To develop a state space system for an electrical system, they choosing the voltage across capacitors, and current through inductors as state variables. Recall that
so if we can write equations for the voltage across an inductor, it becomes a state equation when we divide by the inductance (i.e., if we have an equation for einductor and divide by L, it becomes an equation for diinductor/dt which is one of our state variable). Likewise if we can write an equation for the current through the capacitor and divide by the capacitance it becomes a state equation for ecapacitor
There are three energy storage elements, so we expect three state equations. Try choosing i1, i2 and e1 as state variables. Now we want equations for their derivatives. The voltage across the inductor L2 is e1 (which is one of our state variables)so our first state variable equation is
This equation has our input (ia) and two state variable (iL2 and iL1) and the current through the capacitor. So from this we can get our second state equation
Our third, and final, state equation we get by writing an equation for the voltage across L1 (which is e2) in terms of our other state variables
references:
http://lpsa.swarthmore.edu/Representations/SysRepSS.html
https://en.wikipedia.org/wiki/State-space_representation
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
State space analysis, eign values and eign vectorsShilpa Shukla
State space analysis concept, state space model to transfer function model in first and second companion forms jordan canonical forms, Concept of eign values eign vector and its physical meaning,characteristic equation derivation is presented from the control system subject area.
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Realization of transfer function into state variable models is needed even if the control system design based on frequency-domain design method.
In these cases the need arises for the purpose of transient response simulation.
But there is not much software for the numerical inversion of Laplace transform.
So one ways is to convert transfer function of the system to state variable description and numerically integrating the resulting differential equations rather than attempting to compute the inverse Laplace transform by numerical method.
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state space representation,State Space Model Controllability and Observabilit...Waqas Afzal
State Variables of a Dynamical System
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Why State space approach
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Realization of transfer function into state variable models is needed even if the control system design based on frequency-domain design method.
In these cases the need arises for the purpose of transient response simulation.
But there is not much software for the numerical inversion of Laplace transform.
So one ways is to convert transfer function of the system to state variable description and numerically integrating the resulting differential equations rather than attempting to compute the inverse Laplace transform by numerical method.
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www.wsautter.com
and
http://www.youtube.com/results?search_query=wnsautter&aq=f
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About
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
Technical Specifications
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
Key Features
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface
• Compatible with MAFI CCR system
• Copatiable with IDM8000 CCR
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
Application
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
2. Lecture Outline
• Introduction
• Difference Equations
• Review of Z-Transform
• Inverse Z-transform
• Relations between s-plane and z-plane
• Solution of difference Equations
2
3. Introduction
• Digital control offers distinct advantages over analog
control that explain its popularity.
• Accuracy: Digital signals are more accurate than their
analogue counterparts.
• Implementation Errors: Implementation errors are
negligible.
• Flexibility: Modification of a digital controller is possible
without complete replacement.
• Speed: Digital computers may yield superior
performance at very fast speeds
• Cost: Digital controllers are more economical than
analogue controllers. 3
7. • A difference equation expresses the change in
some variable as a result of a finite change in
another variable.
• A differential equation expresses the change in
some variable as a result of an infinitesimal
change in another variable.
Difference Equation vs Differential Equation
7
8. 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
8
𝐹 𝑡 = 𝑚 𝑦 𝑡 + 𝐷 𝑦 𝑡 + 𝐾𝑦(𝑡)
𝑦 𝑡 =
1
𝑚
𝐹 𝑡 −
𝐷
𝑚
𝑦 𝑡
𝐾
𝑚
𝑦(𝑡)
9. Difference Equations
• Difference equations arise in problems where the
independent variable, usually 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
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
9
10. Difference Equations
• Example-1: For each of the following difference equations,
determine the (a) order of the equation. Is the equation (b)
linear, (c) time invariant, or (d) homogeneous?
1. 𝑦 𝑘 + 2 + 0.8𝑦 𝑘 + 1 + 0.07𝑦 𝑘 = 𝑢 𝑘
2. 𝑦 𝑘 + 4 + sin(0.4𝑘)𝑦 𝑘 + 1 + 0.3𝑦 𝑘 = 0
3. 𝑦 𝑘 + 1 = −0.1𝑦2
𝑘
10
11. Difference Equations
• Example-1: For each of the following difference equations,
determine the (a) order of the equation. Is the equation (b)
linear, (c) time invariant, or (d) homogeneous?
1. 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.
11
12. Difference Equations
• Example-1: For each of the following difference equations,
determine the (a) order of the equation. Is the equation (b)
linear, (c) time invariant, or (d) homogeneous?
2. 𝑦 𝑘 + 4 + sin(0.4𝑘)𝑦 𝑘 + 1 + 0.3𝑦 𝑘 = 0
Solution:
a) The equation is 4th order.
b) All terms are linear
c) The second coefficient is time dependent
d) There is no forcing function therefore the equation is
homogeneous.
12
13. Difference Equations
• Example-1: For each of the following difference equations,
determine the (a) order of the equation. Is the equation (b)
linear, (c) time invariant, or (d) homogeneous?
3. 𝑦 𝑘 + 1 = −0.1𝑦2
𝑘
Solution:
a) The equation is 1st order.
b) Nonlinear
c) Time invariant
d) Homogeneous
13
14. Z-Transform
• Difference equations can be solved using classical methods
analogous to those available for differential equations.
• Alternatively, z-transforms 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. 14
15. Z-Transform
• Given the causal sequence {u0, u1, u2, …, uk}, its z-
transform is defined as
• The variable z −1 in the above equation can be
regarded as a time delay operator.
𝑈 𝑧 = 𝑢 𝑜 + 𝑢1 𝑧−1 + 𝑢2 𝑧−2 + 𝑢 𝑘 𝑧−𝑘
𝑈 𝑧 =
𝑘=0
∞
𝑢 𝑘 𝑧−𝑘
15
17. Relation between Laplace Transform and Z-Transform
• The Laplace Transform of above equation is
• And the Z-transform of 𝑢∗
𝑡 is given as
• Comparing (A) and (B) yields
𝑢∗ 𝑡 = 𝑢 𝑜 𝛿 𝑡 + 𝑢1 𝛿 𝑡 − 𝑇 + 𝑢2 𝛿 𝑡 − 2𝑇 + ⋯ + 𝑢 𝑘 𝛿 𝑡 − 𝑘𝑇
𝑈∗
𝑠 = 𝑢 𝑜 + 𝑢1 𝑒−𝑠𝑇
+ 𝑢2 𝑒−2𝑠𝑇
+ ⋯ + 𝑢 𝑘 𝑒−𝑘𝑠𝑇
𝑈∗ 𝑠 =
𝑘=0
∞
𝑢 𝑘 𝑒−𝑘𝑠𝑇 (𝐴)
𝑧 = 𝑒 𝑠𝑇
17
𝑈 𝑧 =
𝑘=0
∞
𝑢 𝑘 𝑧−𝑘
(𝐵)
18. Conformal Mapping between s-plane to z-plane
• Where 𝑠 = 𝜎 + 𝑗𝜔.
• Then 𝑧 in polar coordinates is given by
𝑧 = 𝑒(𝜎+𝑗𝜔)𝑇
18
𝑧 = 𝑒 𝜎 𝑇 𝑒 𝑗𝜔 𝑇
∠𝑧 = 𝜔𝑇𝑧 = 𝑒 𝜎 𝑇
𝑧 = 𝑒 𝑠𝑇
19. Conformal 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)
– Case-2: Imaginary Pole in s-plane (𝑠 = 𝑗𝜔)
– Case-3:Complex Poles (𝑠 = 𝜎 + 𝑗𝜔)
19
𝑠 − 𝑝𝑙𝑎𝑛𝑒 𝑧 − 𝑝𝑙𝑎𝑛𝑒
20. Conformal Mapping between s-plane to z-plane
• Case-1: Real pole in s-plane (𝑠 = 𝜎)
• We know
• Therefore
20
∠𝑧 = 𝜔𝑇𝑧 = 𝑒 𝜎 𝑇
𝑧 = 𝑒 𝜎 𝑇 ∠𝑧 = 0
21. Conformal Mapping between s-plane to z-plane
When 𝑠 = 0
21
∠𝑧 = 𝜔𝑇𝑧 = 𝑒 𝜎 𝑇
𝑧 = 𝑒0 𝑇 = 1
∠𝑧 = 0𝑇 = 0
𝑠 = 0
𝑠 − 𝑝𝑙𝑎𝑛𝑒 𝑧 − 𝑝𝑙𝑎𝑛𝑒
1
Case-1: Real pole in s-plane (𝑠 = 𝜎)
22. Conformal Mapping between s-plane to z-plane
When 𝑠 = −∞
22
∠𝑧 = 𝜔𝑇𝑧 = 𝑒 𝜎 𝑇
𝑧 = 𝑒−∞ 𝑇
= 0
∠𝑧 = 0
−∞
𝑠 − 𝑝𝑙𝑎𝑛𝑒 𝑧 − 𝑝𝑙𝑎𝑛𝑒
0
Case-1: Real pole in s-plane (𝑠 = 𝜎)
23. Conformal Mapping between s-plane to z-plane
Consider 𝑠 = −𝑎
23
∠𝑧 = 𝜔𝑇𝑧 = 𝑒 𝜎 𝑇
𝑧 = 𝑒−𝑎 𝑇
∠𝑧 = 0
−𝑎
𝑠 − 𝑝𝑙𝑎𝑛𝑒 𝑧 − 𝑝𝑙𝑎𝑛𝑒
1
Case-1: Real pole in s-plane (𝑠 = 𝜎)
0
24. Conformal Mapping between s-plane to z-plane
• Case-2: Imaginary pole in s-plane (𝑠 = ±𝑗𝜔)
• We know
• Therefore
24
∠𝑧 = 𝜔𝑇𝑧 = 𝑒 𝜎 𝑇
𝑧 = 1 ∠𝑧 = ±𝜔𝑇
28. Conformal Mapping between s-plane to z-plane
• Anything in the Alias/Overlay region in the S-Plane will be
overlaid on the Z-Plane along with the contents of the strip
between ±𝑗
𝜋
𝑇
.
28
29. Conformal Mapping between s-plane to z-plane
• In order to avoid aliasing, there must be nothing in this region, i.e. there
must be no signals present with radian frequencies higher than w p/T,
or cyclic frequencies higher than f = 1/2T.
• Stated another way, the sampling frequency must be at least twice the
highest frequency present (Nyquist rate).
29
44. Exercise
• Find the z-transform of following causal sequences.
44
𝑓 𝑘 =
4, 𝑘 = 2,3, …
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Solution: The given sequence is a sampled step starting at k-2 rather than
k=0 (i.e. it is delayed by two sampling periods). Using the delay property,
we have
𝐹 𝑧 = 𝒵{4 × 1 𝑘 − 2 }
𝐹 𝑧 = 4𝑧−2
𝒵{1 𝑘 − 2 }
𝐹 𝑧 = 4𝑧−2
𝑧
𝑧 − 1
=
4
𝑧(𝑧 − 1)
45. Inverse Z-transform
1. Long Division: We first use long division to
obtain as many terms as desired of the z-
transform expansion.
2. Partial Fraction: This method is almost identical
to that used in inverting Laplace transforms.
However, because most z-functions have the
term z in their numerator, it is often convenient
to expand F(z)/z rather than F(z).
45
46. Inverse Z-transform
• Example-4: Obtain the inverse z-transform of
the function
• Solution
• 1. Long Division
𝐹 𝑧 =
𝑧 + 1
𝑧2 + 0.2𝑧 + 0.1
46
51. Inverse Z-transform
• Example-6: Obtain the inverse z-transform of
the function
• Solution
• 2. Partial Fractions
𝐹 𝑧 =
𝑧
(𝑧 + 0.1)(𝑧 + 0.2)(𝑧 + 0.3)
51
52. Inverse Z-transform
• Example-6: Obtain the inverse z-transform of the
function
• Solution
• The most convenient method to obtain the partial fraction expansion
of a function with simple real roots is the method of residues.
• The residue of a complex function F(z) at a simple pole zi is given by
• This is the partial fraction coefficient of the ith term of the expansion
𝐹 𝑧 =
𝑧
(𝑧 + 0.1)(𝑧 + 0.2)(𝑧 + 0.3)
52
𝐴𝑖 = 𝑧 − 𝑧𝑖 𝐹(𝑧)
𝑧→𝑧 𝑖
F z =
𝑖=1
𝑛
𝐴𝑖
𝑧 − 𝑧𝑖
53. Z-transform solution of Difference Equations
• By a process analogous to Laplace transform solution of
differential equations, one can easily solve linear
difference equations.
1. The equations are first transformed to the z-domain (i.e., both
the right- and left-hand side of the equation are z-
transformed) .
2. Then the variable of interest is solved.
3. Finally inverse z-transformed is computed.
53