Introduction To Z Transform

Introduction To Z Transform:

Z transform of a signal provides a valuable technique for analysis and design of the discrete time signal and discrete time LTI system.

Z transform of a discrete time signal has both imaginary and real part. The plot of imaginary part versus real part is called as the z plane. The poles and zeros of the discrete time signals are plotted in the complex z plane. Pole-zero plot is the main characteristics of the discrete time signals. Using pole zero plot we can check the stability of the system which we will see in the upcoming posts.

Advantages Of The Z Transform:

Following are some of the main advantages of the Z transform:

• We can simplify the solution of differential equation using z transform.
• By the use of z transform we can completely characterize given discrete time signals and LTI systems.
• The stability of the LTI system can be determined using z transform.
• Mathematical calculations can be reduced by using z transform. For example convolution operation is transformed into simple multiplication operation.
• Z Transforms:

There are two types of z transform:

1. Single sided Z transform.

1. Double sided Z transform.

The relation between x(n) and X(Z) is denoted as follows:

Where X(Z) is the Z transform of the signal x(n). The arrow is bidirectional which indicates that we can obtain x(n) from X(Z) also, which is called as inverse Z transform.

x(n) and X(Z) is called as Z transform pair.

Region Of Convergence (ROC):

“Region of convergence is defined as a set of all values of Z for which X(Z) has finite value. It is very important to denote ROC every time when we find Z transform.”

Significance Of ROC:

ROC will decide whether the given system is stable or unstable.
ROC is also useful for determining the type of sequence. i.e. whether the system is causal or non-causal, finite or infinite.

Long transmission lines molding

A power transmission line with its effective length of around 250 Kms or above is referred to as a long transmission line. Calculations related to circuit parameters (ABCD parameters) of such a power transmission is not that simple, as was the case for a short transmission line or medium transmission line.The reason being that, the effective circuit length in this case is much higher than what it was for the former models (long and medium line) and, thus ruling out the approximations considered there like.
long transmission line

1. Ignoring the shunt admittance of the network, like in a small transmission line model.
2. Considering the circuit impedance and admittance to be lumped and concentrated at a point as was the case for the medium line model.

Rather, for all practical reasons we should consider the circuit impedance and admittance to be distributed over the entire circuit length as shown in the figure below.
The calculations of circuit parameters for this reason is going to be slightly more rigorous as we will see here. For accurate modeling to determine circuit parameters let us consider the circuit of the long transmission line as shown in the diagram below.
long transmission line model
Here a line of length l > 250km is supplied with a sending end voltage and current of VS and IS respectively, where as the VR and IR are the values of voltage and current obtained from the receiving end. Lets us now consider an element of infinitely small length Δx at a distance x from the receiving end as shown in the figure where.


V = value of voltage just before entering the element Δx.
I = value of current just before entering the element Δx.
V+ΔV = voltage leaving the element Δx.
I+ΔI = current leaving the element Δx.
ΔV = voltage drop across element Δx.
zΔx = series impedence of element Δx
yΔx = shunt admittance of element Δx
Where, Z = z l and Y = y l are the values of total impedance and admittance of the long transmission line.
Therefore, the voltage drop across the infinitely small element Δx is given by

Now to determine the current ΔI, we apply KCL to node A.
ΔI = (V+ΔV)yΔx = V yΔx + ΔV yΔx
Since the term ΔV yΔx is the product of 2 infinitely small values, we can ignore it for the sake of easier calculation.
Therefore, we can write dI ⁄ dx = V y -----------------(2)
Now derivating both sides of eq (1) w.r.t x,
d2 V ⁄ d x2 = z dI ⁄ dx
Now substituting dI ⁄ dx = V y from equation (2)
d2 V ⁄ d x2 = zyV
or d2 V ⁄ d x2 − zyV = 0 ------------(3)
The solution of the above second order differential equation is given by.
V = A1 ex√yz + A2 e−x√yz --------------(4)
Derivating equation (4) w.r.to x.
dV/dx = √(yz) A1 ex√yz − √(yz)A2
e−x√yz ------------(5)
Now comparing equation (1) with equation (5)

Now to go further let us define the characteristic impedance Zc and propagation constant δ of a long transmission line as
Zc = √(z/y) Ω
δ = √(yz)
Then the voltage and current equation can be expressed in terms of characteristic impedance and propagation constant as
V = A1 eδx + A2 e−δx -----------(7)
I = A1/ Zc eδx + A2 / Zc e−δx ---------------(8)
Now at x=0, V= VR and I= Ir. Substituting these conditions to equation (7) and (8) respectively.
VR = A1 + A2 ---------------(9)
IR = A1/ Zc + A2 / Zc ---------------(10)
Solving equation (9) and (10),
We get values of A1 and A2 as,
A1 = (VR + ZCIR) ⁄ 2
And A1 = (VR − ZCIR) ⁄ 2
Now applying another extreme condition at x=l, we have V = VS and I = IS.
Now to determine VS and IS we substitute x by l and put the values of A1 and
A2 in equation (7) and (8) we get
VS = (VR + ZC IR)eδl ⁄ 2 + (VR −
ZC IR)e−δl/2 --------------(11)
IS = (VR ⁄  ZC + IR)eδl/2 − (VR / ZC − IR)e−δl/2--------------- (12)
By trigonometric and exponential operators we know
sinh δl = (eδl − e−δl) ⁄ 2
And cosh δl = (eδl + e−δl) ⁄ 2
Therefore, equation(11) and (12) can be re-written as
VS = VRcosh δl + ZC IR sinh δl
IS = (VR sinh δl)/ZC + IRcosh δl
Thus comparing with the general circuit parameters equation, we get the ABCD parameters of a long transmission line as,
A = cosh δl
B = ZC sinh δl
C = sinh δl ⁄ ZC
D = cosh δl


Book

• Power System Analysis by Grainger and W.D. Stevenson.
• Power System Analysis and Design by J. Duncan Glover,Mulukutla S. Sarma and Thomas J. Overbye.
• Power System Analysis by Hadi Sadat,
• Principles of Power Systems by Mehta.

   
Author        Bilal masood
lecture       09
Type         ppt
Availability Available online
For read this book online or Download click the below link

Long Transmission line Model

A major section of power system engineering deals in the transmission of electrical power from one particular place (eg. generating station) to another like substations or distribution units with maximum efficiency. So its of substantial importance for power system engineers to be thorough with its mathematical modeling. Thus the entire transmission system can be simplified to a two port network for the sake of easier calculations.The circuit of a 2 port network is shown in the diagram below. As the name suggests, a 2 port network consists of an input port PQ and an output port RS. In any 4 terminal network, (i.e. linear, passive, bilateral network) the input voltage and input current can be expressed in terms of output voltage and output current. Each port has 2 terminals to connect itself to the external circuit. Thus it is essentially a 2 port or a 4 terminal circuit, having two port network

Given to the input port P Q.

Given to the output port R S.
As shown in the diagram below.
Now the ABCD parameters or the transmission line parameters provide the link between the supply and receiving end voltages and currents, considering the circuit elements to be linear in nature.

Thus the relation between the sending and receiving end specifications are given using ABCD parameters by the equations below.

Now in order to determine the ABCD parameters of transmission line let us impose the required circuit conditions in different cases.

ABCD Parameters, When Receiving End is Open Circuited

two port network
The receiving end is open circuited meaning receiving end current IR  = 0.
Applying this condition to equation (1) we get,

Thus its implies that on applying open circuit condition to ABCD parameters, we get parameter A as the ratio of sending end voltage to the open circuit receiving end voltage. Since dimension wise A is a ratio of voltage to voltage, A is a dimension less parameter.
Applying the same open circuit condition i.e  IR  = 0 to equation (2)

Thus its implies that on applying open circuit condition to ABCD parameters of transmission line, we get parameter C as the ratio of sending end current to the open circuit receiving end voltage. Since dimension wise C is a ratio of current to voltage, its unit is mho.
Thus C is the open circuit conductance and is given by
 C = IS ⁄  VR  mho.
ABCD Parameters, When Receiving End is Short Circuited

two port network
Receiving end is short circuited meaning receiving end voltage VR  = 0
Applying this condition to equation (1) we get,

Thus its implies that on applying short circuit condition to ABCD parameters, we get parameter B as the ratio of sending end voltage to the short circuit receiving end current. Since dimension wise B is a ratio of voltage to current, its unit is Ω. Thus B is the short circuit resistance and is given by
 B = VS ⁄ IR  Ω.
Applying the same short circuit condition i.e  VR  = 0 to equation (2) we get


Thus its implies that on applying short circuit condition to ABCD parameters, we get parameter D as the ratio of sending end current to the short circuit receiving end current. Since dimension wise D is a ratio of current to current, it’s a dimension less parameter.
∴ The ABCD parameters of transmission line can be tabulated as:-
Parameter Specification Unit
A = VS / VR Voltage ratio Unit less
B = VS / IR Short circuit resistance Ω
C = IS / VR Open circuit conductance mho
D = IS / IR Current ratio Unit less

Book

• Power System Analysis by Grainger and W.D. Stevenson
• Power System Analysis and Design by J. Duncan Glover,Mulukutla S. Sarma and Thomas J. Overbye
• Power System Analysis by Hadi Sadat
• Principles of Power Systems by Mehta

     
Author        Bilal masood
lecture       08
Type         ppt
Availability Available online
For read this book online or Download click the below link

Short and Medium Transmission Line

The transmission line having its effective length more than 80 km but less than 250 km is generally referred to as a medium transmission line. Due to the line length being considerably high, admittance Y of the network does play a role in calculating the effective circuit parameters, unlike in the case of short transmission lines. For this reason the modeling of a  medium length transmission line is done using lumped shunt admittance along with the lumped impedance in series to the circuit.These lumped parameters of a medium length transmission line can be represented using three different models, namely-

1. Nominal Π representation.

1. Nominal  T representation.

1. End Condenser Method

Let’s now go into the detailed discussion of these above mentioned models.

Nominal Π Representation of a Medium Transmission Line

In case of a nominal Π representation, the lumped series impedance is placed at the middle of the circuit where as the shunt admittances are at the ends. As we can see from the diagram of the Π network below, the total lumped shunt admittance is divided into 2 equal halves, and each half with value Y ⁄ 2 is placed at both the sending and the receiving end while the entire circuit impedance is between the two. The shape of the circuit so formed resembles that of a symbol Π, and for this reason it is known as the nominal Π representation of a medium transmission line. It is mainly used for determining the general circuit parameters and performing load flow analysis.
medium transmission line

Nominal T Representation of a Medium Transmission Line

In the nominal T model of a medium transmission line the lumped shunt admittance is placed in the middle, while the net series impedance is divided into two equal halves and and placed on either side of the shunt admittance. The circuit so formed resembles the symbol of a capital T, and hence is known as the nominal T network of a medium length transmission line .

End Condenser Method

In this method, the capacitance of the line is limped or concentrated at the receiving or load end. This method of localizing the line capacitance at the load end overestimates the effects of capacitance.

Book

• Power System Analysis by Grainger and W.D. Stevenson

• Power System Analysis and Design by J. Duncan Glover,Mulukutla S. Sarma and Thomas J. Overbye

• Power System Analysis by Hadi Sadat

• Principles of Power Systems by Mehta

     
Author        Bilal masood
lecture       07
Type         ppt
Availability Available online
For read this book online or Download click the below link

Exact Equivalent Circuit of transformer

Exact Equivalent Circuit transformer

Resistances and reactances of transformer, which are described above, can be imagined separately from the windings (as shown in the figure below). Hence, the function of windings, thereafter, will only be the transforming the voltage.

The no load current I0 is divided into, pure inductance X0 (taking magnetizing components Iμ) and non induction resistance R0 (taking working component Iw) which are connected into parallel across the primary. The value of E1 can be obtained by subtracting I1Z1 from V1. The value of R0 and X0 can be calculated as, R0 = E1 / Iw and X0 = E1 / Iμ.

But, using this equivalent circuit does not simplifies the calculations. To make calculations simpler, it is preferable to transfer current, voltage and impedance either to primary side or to the secondary side. In that case, we would have to work with only one winding which is more convenient.

From the voltage transformation ratio, it is clear that,
E1 / E2 = N1 / N2 = K

Now, lets refer the parameters of secondary side to primary.
Z2 can be referred to primary as Z2'
where, Z2' = (N1/N2)2Z2 = K2Z2.   ............where K= N1/N2.
that is, R2'+jX2' = K2(R2+jX2)
equating real and imaginary parts,
R2' =  K2R2 and X2' = K2X2 .
And V2' = KV2

Now, as the values of winding resistance and leakage reactance are so small that, V1 and E1 can be assumed to be equal. Therefore, the exciting current drawn by the parallel combination of  R0 and X0 would not affect significantly, if we move it to the input terminals as shown in the figure below.


Now, let R1 + R2' = R'eq  and X1 + X2' = X'eq

Approximate Equivalent Circuit Of Transformer

If only voltage regulation is to be calculated, then even the whole excitation branch (parallel combination of R0 and X0) can be neglected.

Book

• Power System Analysis by Grainger and W.D. Stevenson

• Power System Analysis and Design by J. Duncan Glover,Mulukutla S. Sarma and Thomas J. Overbye.

• Power System Analysis by Hadi Sadat.

• Principles of Power Systems by Mehta.

     
Author        Bilal masood
lecture       06
Type         ppt
Availability Available online
For read this book online or Download click the below link

Synchronous Machines

Synchronous Machines

Like other electrical machines, synchronous machines can be operated as either generators or motors. We will concentrate on synchronous generators, then adapt the theory for synchronous machines.
The prinicples of operation of synchronous generators are relatively simple. They operate like the simple 3-phase generator discussed in the fundamentals section of the course.

Principles of operation

• Create a magnetic field on the rotor of an electric machine
• Apply an external driving force
• As the rotor rotates, a voltage is induced in windings on the stator
• The frequency of the induced voltage will be synchronised with the speed of rotation

ns=120fep

ns=120fep

Synchronous machine construction

A synchronous machine has two mechanical parts: a rotor and a stator. There are also two electrical parts to the machine: a field source and an armature winding. These basic fundamentals of an electric machine are like those for a DC machine, with one significant difference: The field source of a synchronous machine is on the rotor, the armature winding of a synchronous machine is on the stator. Like DC machines, the field source creates a magnetic field, the armature winding has a voltage induced in it by the field. Also like DC machines, the field can be produced using either a field winding or by using permanent magnets. PM (permanent magnet) machines are common in small sizes, whilst large machines are usually made with field windings. (There are some exceptions to this rule, e.g. multi-MW PM motors are built for ship propulsion and direct drive wind turbine generators)
In this part of the course, we will concentrate on wound-field synchronous machines.
Terminology

• The field winding is on the rotor
• The armature winding is on the stator
• There is sometimes a damper or armortisseur winding on the rotor
• The external driving force (e.g. steam or hydro turbine, diesel generator, jet engine) is called the prime mover

Book

• Power System Analysis by Grainger and W.D. Stevenson
• Power System Analysis and Design by J. Duncan Glover,Mulukutla S. Sarma and Thomas J. Overbye.
• Power System Analysis by Hadi Sadat.
• Principles of Power Systems by Mehta.
•         

Author        Bilal masood
lecture       05
Type         ppt
Availability Available online
For read this book online or Download click the below link

Generator and Transformer Models

In a generator, motion of a conductor in a magnetic field induces an EMF. In a transformer, it is the changing field that induces an EMF in a fixed conductor.

Schematic generator

Discussion: Generators

The structure of a simple generator is essentially the same as a motor. The difference is that now mechanical energy is converted into electrical energy. The electrical current to a load is via a commutator for an ac generator or slip rings if ac is required.

Basic ideas can be understood by thinking about a coil rotating in a uniform magnetic field.

Plane of coil at an angle to the magnetic field
Consider a coil of area A with N turns of wire rotating at a constant angular velocity w in a uniform magnetic flux density B. As the coil rotates, it cuts through the lines of flux. Another way to express this is to say that the flux linking the coil is changing.

At what point is the rate of flux-cutting greatest? (When it is horizontal in the diagram above; when it is vertical, the rate of flux cutting is instantaneously zero.)

Rate of flux cutting = induced EMF = BANwcos wt

with a maximum value, Eo = BANw when the coil is parallel to the field.

Demonstrations: A motor in reverse

Show that a motor can operate in reverse, as a generator. One starting point is simply to attach a weight to a small motor and to drop the weight. The motor works in reverse as a generator; the induced EMF can be monitored with a meter

 Practical transformers

Discuss reasons for energy losses in real transformers. These are readily identified as:

ohmic heating of the coils
eddy current heating of the core
hysteresis effects which heat the core
magnetic flux escaping
But even with these it is not unusual to find efficiencies of 95% and higher. Large transformers used in power transmission may be as much as 99.5% efficient.

Where electronics are being used, low voltage ac supplies are usually required so step-down transformers will be an essential part of the power supply. The output from a transformer is ac, so there will have to be some form of rectification (with diodes) and smoothing (with capacitors).

Transformer on the grid

A second widespread use is within the 'Grid' that supplies electricity to the consumer. The connection from a power station to the consumer involves a long length of wire and often, high currents. For a given section of the grid, the resistance, R is fixed and the rate of heating generated in the wire will be I2 R; this energy is wasted. To minimise this energy loss, the current should be as small as possible. To deliver a particular power (VI), a smaller current can be achieved by using as high a voltage as possible. The grid is designed so that transformers are used to step up the voltage at the power station before transmission. Step down transformers reduce the voltage in stages to the level required by industrial and domestic consumers.

Book

• Power System Analysis by Grainger and W.D. Stevenson

• Power System Analysis and Design by J. Duncan Glover,Mulukutla S. Sarma and Thomas J. Overbye.

• Power System Analysis by Hadi Sadat.

• Principles of Power Systems by Mehta.

     
Author        Bilal masood
lecture       04
Type         ppt
Availability Available online
For read this book online or Download click the below link