four parameters

PERFOMANCE OF FOUR PARAMETERS, TWO PORT ELECTRICAL NETWORKS IN

TERMS OF Y PARAMETERS

Consider a two port, four parameters electrical networks below

Y-Parameters these are parameters representing Admittance

Admittance is the reciprocal of impedance.

1( tan )

Im ( )Y Admit ce

pedance Z

SI units of Admittance is Siemens(S)

Z-parameters are the parameters representing Impedance in the circuit

The equations representing Y-parameters , are different from Z-parameters, and can be written as

follows

1 11 1 12 2

2 21 1 22 2

sec

sec

or input tion

I Y V Y V

For output tion

I Y V Y V

In this case, each term, of each equation has units of currents, compared to voltage for each term

in Z-parameters equation

Basically, just like in Z-parameters, also there are four Y-parameters to consider, these are

i. Input Admittance parameters (Y11)

Since it is input Y parameters, its formula is obtained from input equation

1 11 1 12 2

11 2

1 11 1 12

1 11 1

111 2

1

sec

, , 0, ,

(0)

,

, , 0

For input tion

I Y V Y V

To get Y let V Then substitute in above

I Y V Y

I Y V

IY when V

V

Remember, To Obtain Input Admittance Y11 , Output voltage V2 =0V (V2 should be short

circuited by replacing it with a wire)

ii. Reverse Transfer Admittance Parameters (Y12)

Since it is input Y parameters, its formula is obtained from input equation again

1 11 1 12 2

12 1

1 11 12 2

1 12 2

112 1

2

sec

, , 0, ,

(0)

,

, , 0

For input tion

I Y V Y V

To get Y let V Then substitute in above

I Y Y V

I Y V

IY when V

V

Remember, To Obtain reverse Transfer Admittance Y12 , Input voltage V1 =0V (V1

should be short circuited by replacing it with a wire)

iii. Forward Transfer Admittance Parameters (Y21)

Since it is output Y parameters, its formula is obtained from Output equation again

2 21 1 22 2

21 2

2 21 1 22

2 21 1

221 2

1

sec

, , 0 ,

(0)

, , 0

For output tion

I Y V Y V

to obtain Y let V V substitute above

I Y V Y

I Y V

IY when V V

V

Remember, To Obtain Forward Transfer Admittance Y21 , Output voltage V2 =0V (V2

should be short circuited by replacing it with a wire)

iv. Output Admittance Parameters (Y22)

Since it is output Y parameters, its formula is obtained from Output equation again

2 21 1 22 2

22 1

2 21 22 2

2 22 2

222 1

2

sec

, , 0 ,

(0)

, , 0

For output tion

I Y V Y V

to obtain Y let V V substitute above

I Y Y V

I Y V

IY when V V

V

Remember, To Obtain Output Admittance Y22 , Input voltage V1 =0V (V1 should be short

circuited by replacing it with a wire)

Example 1

With reference of the figure below, which shows, a four terminal two port network

a. State the defining equation for voltage V1 and V2 in terms of the four Y-

parameters, and the current I1 and I2

b. Derive/ calculate the following Y-parameters for the network

i. Y11

ii. Y12

iii. Y21 iv. Y22

SOLUTION

i. To calculate Input admittance Y11

Remember the formula

111 2

1

, , 0 , , , , ( )I

Y when V V output replaced by short circuit a wireV

The effect of introducing short circuit in the output (A wire), the admittance Y3 will be

ineffective,, because current will pass through a wire, rather than Y3 ., hence, the branch

containing Y3 should be removed, hence the overall circuit will be as shown below

This circuit can be re-drawn nicely as shown below

From the above circuit, Y2 is parallel to Y1, Y1//Y2 , And because admittance is the

opposite of Impedance (Resistance), to find the sum of parallel admittances, just add

, therefore

11 1 2

11

11

0.5 0.25

0.75

Y Y Y

Y S S

Y S

ii. To calculate Reverse transfer Admittance Y12


112 1

2

, , 0 , , , , ( )I

Y when V V input replaced by short circuit a wireV

The effect of introducing short circuit (A wire), in input the admittance Y1 will be



This circuit cab be re-drawn nicely as shown below

Therefore , the circuit above

12 2

12 0.25

Y Y

Y S

iii. To calculate Forward Transfer Admittance (Y21)

This is output parameters


221 2

1

, , 0 , , , , ( )I

Y when V V output replaced by short circuit a wireV

The effect of introducing short circuit (A wire), in Output, the admittance Y3 will be



This circuit cab be re-drawn nicely as shown below

Therefore, from the circuit above

21 2

12 0.25

Y Y

Y S

iv. To calculate Output Admittance (Y22)

This is output parameters


222 1

2

, , 0 , , , , ( )I

Y when V V input replaced by short circuit a wireV

The effect of introducing short circuit (A wire), in input, the admittance Y1 will be



This circuit can be re-drawn nicely as shown below

From the above circuit, Y2 is parallel to Y3, Y2//Y3 , And because admittance is the

opposite of Impedance (Resistance), to find the sum of parallel admittances, just add

admittances

22 2 3

22

22

0.25 0.85

1.1

Y Y Y

Y S S

Y S

Analyzing DC and AC circuit using THEVENIN & NORTON THEOREMS

THEVENIN THEOREMS state that, the current in any branch of the circuit , is that

which would results if an E.M.F equal to the p.d across a break made in the branch ,

were introduced into the branch, all other E.M.F being removed and represented by the

internal resistances of the source

Procedures to be followed when applying Thevenins theorems, when determining

current from any branch in active network

1. Removes Resistance R, from that branch, determine the Open circuit voltage

from that branch and name it E

2. If E.M.Fs are available in the circuit and they dont have internal resistance,

replace them with the short circuit (Wire), if they have internal resistance, replace

them with that internal resistances r

3. Find the total resistance of the circuit without including The resistance R

removed before ,by looking through Open circuit introduced in step 1, and name

it RT ,

4. Replace in the circuit the removed resistance R together with open circuit voltage

obtained, and change letter RT to small r.

5. Then calculate current in the branch using the following circuit

E

IR r

Example

Use Thevenins Theorem to find the current flowing in the 10 resistor for the

circuit shown below

Follows the following steps

1. Removes Resistance R, from that branch, determine the Open circuit voltage

from that branch and name it E

There will be no current flowing in the R3 resistor, therefore R3 branch is

ineffective, Hence R1 and R2 will be effective and they are connected in series,

Find current I1 in the circuit

1

1 2

10 101

2 8 10

EI A

R R

Open circuit voltage will be equal to voltage across R2 resistor; hence voltage

across R2 is given by

1 2 1 8 8

8

V I xR Ax V

V E V

2. If E.M.Fs are available in the circuit and they dont have internal resistance,

replace them with the short circuit (Wire), if they have internal resistance, replace

them with that internal resistances r

The EMF in the circuit provided has no internal resistance, so we replace it with a

wire in the circuit as shown below

3. Find the total resistance of the circuit without including The resistance R

removed before ,by looking through Open circuit introduced in step 1, and name

it RT ,

Therefore to find total resistances RT we can say that

1 2 3

1 23

1 2

/ /

2 85

2 8

6.6

T

T

T

T

R R R R

R xRR R

R R

xR

R

4. Replace in the circuit the removed resistance R together with open circuit voltage

obtained, and change letter RT to small r.

5. Then calculate current in the branch using the following circuit

EI

R r

80.482

10 6.6

VI A

REPREENTATION OF SOURCE OF ELECTRICAL ENERGY

Electrical Energy may be represented in two forms

a. Thevenins Constant Voltage source

Consists of a constant voltage source in series with the internal resistor r

This is called, Thevenins equivalent circuit

b. Constant current source

In this case, the current source is in parallel with the internal resistor r

The above circuit is Current Source or Nortons equivalent circuit

CHANGING BETWEEN CURRENT SOURCE EQUIVALENT

CIRCUIT & THEVENINS EQUIVALENT CIRCUIT

Where to get value of ISC use the following approach

SC

EI

r

Example 1

Find the Thevenins equivalent circuit of the figure below, shown below

between terminal a and b

SOLUTION

Remember that Thevenins equivalent circuit can be drawn as shown

below

Where VTH =Thevenins voltage, can be obtained by applying either Mesh

current analysis or Nodal analysis

RTH=Thevenins resistance, which is total of resistance of the circuit

obtained by short circuiting all voltage source in the circuit, and Open

circuiting all the current source in the circuit

Step One

To calculate RTH (Thevenins resistance), Short circuit all voltage source,

and Open circuit all current source in the circuit, and find equivalent

resistance by looking through terminal a and b

Then over all circuit will be as shown below

Then RTH is given as follows

4 / /12 1

4 121

4 12

3 1 4

TH

TH

TH

R

xR

R

Step two

To calculate Thevenins Voltage VTH

Here you can use either Nodal analysis or mesh current analysis

By Nodal Analysis

Note that Voltage V1 at Node V1 is electrically equal to VTH because they

are in parallel connection

V1 =VTH

But do not consider R=1, because, this resistor will be ineffective due to

open circuit on the terminals a and b, hence no current that will be

flowing in this branch

Assume the direction of current entering the node V1 are as indicated

above

1 2

1 1 11 2

1 1

1 1

1 1

1

1

1

,

32 0, 2 ,

4 12 12

322

4 12

322

4 4 12

8 212 4

1 1( ) 1012 4

110

3

30 30Th

I I I

V V VI I A I

V V

V V

V V

V

V

V V V V

The Thevenins equivalent circuit will be shown as indicated below

Example 2

Obtain Thevenins equivalent circuit between terminals a and b

Ans: VTH =6V, RTH =3

The equivalent circuit will be

NORTONs THEOREM

Nortons theorem state that, a linear two terminal circuit can be replaced

by an equivalent circuit consisting of a current source IN in parallel with a

resistor RN , where IN is the short circuit current through the terminals and

RN is the input or equivalent resistances at the terminals when all

independent sources are turned off

STEPS TO APPLY NORTONs THEOREM

To determine the current flowing in a resistor R of a branch AB of an

active circuit follows the following steps

1. Short circuit the branch AB , that contains Resistor R

2. Determine the short circuit current ISC flowing in the branch

3. If E.M.Fs are available in the circuit and they dont have internal

resistance, replace them with the short circuit (Wire), if they have

internal resistance, replace them with that internal resistances r

Then determine the equivalent resistance of the circuit by looking

through a break made across AB, then this equivalent resistance let

it be r

Then, draw the over-all Nortons equivalent circuit as shown

below

4. Replace the removed resistor R back in the Nortons equivalent

circuit Terminals AB, and determine the current flowing in the this

resistor using current divider theorem

Therefore, by current divider theorems, the current flowing

through R, is given by

( )R SCr

I xIr R

Example

Use Nortons theorems to calculate current through R=10

SOLUTION

1. Short circuit the branch that contains Resistor R=10, and name it

as AB

After introducing, a short circuit (A wire), across Branch containing R=10. Then The current in

the circuit wont be flowing in both R=5, and R=10, instead current will be flowing through

wire, hence, this branch become ineffective, hence should be removed completely from the

circuit, this will produce a totally different circuit as indicated below

2. Determine the short circuit current ISC flowing in the branch

Before calculating value of ISC , again the wire AB, introduced,

also will bypass resistor R=8, since current in the circuit, will

again not be flowing through R=8, again this resistor will not be

effective in the circuit, and must be completely removed from the

circuit, hence, the more simpler circuit produced will be as shown

below

Then, from here we can calculate value of ISC , as shown below

105

2SC

V VI A

R

3. If E.M.Fs are available in the circuit and they dont have internal

resistance, replace them with the short circuit (Wire), if they have

internal resistance, replace them with that internal resistances r

Then determine the equivalent resistance of the circuit by looking

through a break made across short circuit branch AB, and then this

equivalent resistance let it be r

Before Removing EMFs the circuit was as shown below

Introduce the break across branch AB, and replace EMF by a short

circuit (A wire), this will produce the following circuit

To calculate value of r

2 8( )

2 8

xr parallel connection

1.6r

Then, draw the over-all Nortons equivalent circuit as shown

below, using ISC =5A, and r=1.6

4. Replace the removed resistors R=10 and R=5 back in the

Nortons equivalent circuit Terminals AB, and determine the

current flowing in the these resistor using current divider theorem

Since, the replaced resistors, are in series, then same current will

be flowing, therefore, combine them to obtain single resistor, and

the new circuit will be as shown below

Therefore value of current I, will be given by current divider

theorem, which state that

( )

1.65

1.6 15

1.65

16.6

0.482

SC

rI xI

r R

I x A

I x A

I A

Example

National Technician Awards In engineering, JUNE 2014, Question 4

a) Obtain the Thevenins equivalent circuit between terminals p and

q of the circuit for the circuit below

b) When 9V battery is connected across the terminals pq with point p

being positive

i. Calculate current flowing through battery

ii. Briefly explain with reason if the battery in (i) is receiving or dissipating power

SOLUTION

a) Obtain the Thevenins equivalent circuit between terminals p and

q of the circuit for the circuit below

We have to calculate value of RTH and VTH

To find value of RTH replaces all voltage sources with short circuit and

current sources with open circuit, since we have only current sources,

replace them with open circuit only, then calculate equivalent

resistance RTH looking from terminals p and q

Then the over-all circuit will be given as shown below

All resistances will be connected in series connection,

2 6 4 12THR

To calculate Thevenins Voltage VTH apply Nodal Voltage analysis

Consider the nodal point V1, then V1 =VTH because of parallel

connection.

Consider the direction of current as indicated at the node

During analysis dont include R=2 and R=4 in the far right of the

circuit because no current will be flowing through these resistors,

hence they will be ineffective

1 2

1 11 2

1

1

1

1

0, 2.5 , , 4.5

6 6

2.5 4.56

4.5 2.56

26

12 12TH

I I I

V Vwhere I A I I A

V

V

V

V V V V

Then Thevenins equivalent circuit will be

c) When 9V battery is connected across the terminals pq with point p

being positive

i. Calculate current flowing through battery

The circuit will look as shown below

The easiest theorem to apply here is NORTONs Theorem, But before doing that, simplify

the circuit by transforming Current source below to Voltage source

Remember to transform to voltage source use the formula

, 2.5 , 6

2.5 6 15

S

S

V IR I A R

V Ax V

Re-drawing the circuit, we get the new circuit as shown below

Consider the Junction A

Therefore, current will be dividing, assume in the indicated direction

Since R=2, 9V battery, and R=4, are connected in series, then same I1 will be flowing

in this branch, also, this branch will be in parallel with 15V battery voltage branch, hence

the sum of voltage across that branch will be equal to 15V, then use this fact to calculate

value of I1

1 1

1

1

1

1

4( ) 9 2( ) 15

6 9 15

6 15 9

6 6

1

I I

I V

I

I

I A

ii. Briefly explain with reason if the battery in (i) is receiving or dissipating power

By considering the direction of flowing of current in the 9V battery branch, i.e. the battery

is receiving current; definitely the battery will be receiving power instead of dissipating

power

Example

National Technician Awards In engineering. JUNE 2009, Question 3

a) i. determine the Thevenin Equivalent circuit across terminals AB of the circuit shown

figure below

ii. Convert the Thevenin equivalent circuit in i) to Norton equivalent circuit

iii. Calculate the power produced in a 10k resistor connected across the output terminals AB

in circuit ii.

SOLUTION

i. To find Thevenin equivalent circuit

We have to calculate value of RTH and VTH

To find value of RTH replaces all voltage sources with short circuit and

current sources with open circuit, then calculate equivalent resistance

RTH looking from terminals A and B

Therefore RTH will be given as series combination of resistors

RTH =2k+3K=5K

To determine Voltage VTH across terminals AB, find Voltage across

node V1 as shown in the circuit below

From the circuit above, battery 5V will be ineffective, because of open

circuit terminals AB, therefore battery must be removed

Also V1=VTH because of parallel connection, therefore same current I1

=4mA will be flowing through series Resistors

But V1 =VTH =I1 (2k+3K)

V1=VTH =4mAx5K=20V

Therefore Thevenin equivalent circuit will be as indicated below

ii. Convert the Thevenin equivalent circuit in i) to Norton equivalent circuit

The Thevenin circuit is shown below

To convert it to Norton circuit, calculate value of the Norton current using Ohms law

204 , , 5

5

THN

TH

N N

VI

R

VI mA where R K

K

iii. To calculate power produced in R=10K Resistor connected across the output terminals

AB in circuit ii.

If you connect R=10K in the above circuit, the following circuit will be produced

Power In R=10K, is obtained by the following formula

22 2, , ( )

NN

N

RPower I xR where I xI Current divider theorem

R R

2

5 54 4 1.3

5 10 15

K KI x mA x mA mA

K K K

Then from 2

2

2(1.3 ) 10 0.0169

Power I xR

Power mA x K W

Example

National Technician Awards In engineering DEC 2012, Question 5

With Reference to figure 4 which shows Norton equivalent circuit,

i. Convert to its Thevenin equivalent circuit

ii. Calculate current drawn in a load of 2-j10, when connected across terminals a-b

Figure 4

SOLUTION

This is AC circuit, remember complex number application

i. Convert to its Thevenin equivalent circuit

Calculate both values of RTH and VTH

To find RTH ,looking through terminals a-b

To find value of RTH replaces all voltage sources with short circuit and current sources with open

circuit, then calculate equivalent resistance RTH looking from terminals a and b

From above circuit, all impedances are in series, so value of RTH is obtained by summing

up

(3 2 ) (4 1) (7 )ThR j j j

To calculate VTH , looking through terminal a-b again consider the circuit below

The branch containing R=3+2j, will become ineffective, because no current that will be

flowing in this branch due to open circuit terminals a-b, therefore same current will be

flowing through branch containing R=4-j1

But in this case also, V1 =VTH this is a parallel connection

V1=Ix (4-j1)=2.5x(4-j1)=(10-2.5j)V=VTH

Then Thevenin equivalent circuit, will be shown below

ii. Calculate current drawn in a load of 2-j10, when connected across terminals a-b

When you connect load R=2-10, across terminals a-b, you will produce following circuit

Then we will have current I, flowing in the circuit, containing series resistors, therefor

sum of these series resistors is RT = (7+j) +(2-10j) =9-9j

But

(10 2.5 )

(9 9 )T

V j VI

R j

Remember to change to polar form when dividing complex number

00 0

0

0 0

0

10.3 14 10.314 ( 45 )

12.7 45 12.7

10.314 45

12.7

0.811 31

VI

I

I A

Since both impedances are in series, same current I, will be flowing, hence current through

R=2-10j, will also be equal to 00.811 31I A

ATTENUATORS

Attenuator is any device or system that can reduce the power or voltage levels of a signal

while introducing little or no distortion. Attenuators are the opposite of AMPLIFIERS

They are generally two types of attenuators

a) Passive attenuators

b) Active Attenuators

This discussion is going to concentrate only to passive attenuators

TYPES OF PASSIVE ATTENUATORS

a) T Attenuators

b) -Attenuators

APPLICATION OF ATTENUATORS

Can be used to match between Source Impedance and Load Impedances

four parameters

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