Lecture – 26 Two – port Network parameters

Lecture – 26 Two – port Network parameters


This is the twenty sixth lecture and we are
continuing our discussion on 2 port network parameters. Today’s topic would be interrelationships
and applications of 2 port parameters. Before I take the interrelationships we would like
to work out a couple of examples on ABCD parameters and 1 of them is the ideal transformer. As I have already discussed, an ideal transformer
has primary as well as secondary inductances going to infinity. The mutual inductance also
goes to infinity, but the ratio of the 2 inductances is finite and this ratio in terms of trans
ratio is equal to what, L1 by L2. Do you know the relation between inductance and number
of terms inductance is proportional to? Square. So, this is equal to n square, square
of the number of terms. That is why the ratio is finite because the ratio the number of
terms is finite. And in addition we indicate the dots and we write the word ideal here.
In addition to inductance is being infinite the coefficient of coupling has to be 1 exactly
1. If the inductances are finite and the coefficient of coupling is 1, we call it a perfect transformer.
A perfect transformer becomes ideal if L1, L2 and M all go to infinity; in such a manner
that the ratio of L1 to L2 is finite. Now, as I have already pointed out an ideal
transformer does not have impedance or admittance matrices. Impedance and admittance matrices
cannot be defined for an ideal transformer because z11 is infinity. So, is y11 y22 of
course, z22 also yes correct and therefore, z and y parameters are not defined. Now, if
I take the voltage current relationships you will see that port h parameters and ABCD parameters
can be defined. The voltage current relationships are that V1 is equal to n times V2 and I1
is equal to minus 1 over n I2, this is the relationship. Now, let me write it down again V1 equals
to n V2 and I1 is equal to minus 1 over n I2, from which the ABCD parameters are obvious.
You see for ABCD V1 is to be identified with AV2 minus BI2 and I1 is to be identified with
CV2 minus DI2 agreed. This is the definition of ABCD parameters and therefore, for the
ideal transformer the ABCD matrix is simply if you compare this with this you see that
A is simply equal to n B is 0 C is 0 and D is. 1 over n, 1 over n and you also notice
that AB minus CD is equal to 1. This has to be obeyed because the ideal transformer
is a reciprocal device all right. If I want the h parameters the definition of h parameters
are that V1 is h11 I1 plus h12 I2. Sir V2 V 2 agreed h12 V2 and I2 the other current is equal to h21 I1 plus h22 V2. So, can you
tell me what is the h matrix now, h matrix if you compare these relationship with this
you see V1 h11 would be 0 h11 is 0 h12 would be equal to n agreed h21.
Minus n. Minus n and h22 would be. 0. 0 and you notice that h12 is indeed equal
to minus h21 which is the condition for reciprocity. So, if you write the voltage current relationship
at the port if you are able to do that a parameter should be obvious. And for an ideal transformer
the only way to describe it is either an h matrix or an ABCD matrix transmission matrix
z and y parameters do not exist. Now, take an ideal transformer and understand
why it is called a transformer if I have an impedance z L here, the voltages are V1 I1
V2 I2 if it is terminated in ZL. And the ratio is n is to 1 these are the dots then V1 is
equal to n V2 I1 is equal to minus 1 over n I2. And therefore, the input impedance V1
by I1 is simply equal to minus n squared V2 by I2 Now, if I transfer this minus sign here then; obviously, V2 by minus I2 is equal to ZL.
And therefore, input impedance Zin is equal to n squared ZL and this is Y it is called
a transformer not only transforms voltages and currents V2 is 1 by n times V1. Current
I2 is equal to minus n times I1 it transforms voltages and current, it also transforms the
impedance. The secondary impedance z L is transformed into n squared times z L when
referred to the primary. For example, if this is an inductance then
if it is the inductance L then the effective inductance looking at the primary shall be
n squared times L. If this is a capacitance C then the effective capacitance looked at
from the primary would be C divided by n squared. If it is a resistance R then it would be n
square times R this is the property of a transform. The other example that I take for ABCD parameters
is I do not know if I have done this the pi network have I done this? ABCD parameters
or did I derive the h parameters h we derived. Let us derive the ABCD parameters V1 I1 V2
I 2. We wish to derive the ABCD parameters our relationships is V1 equal to AV2 minus
BI2 and I1 is equal to CV2 minus DI2 all right. A is equal to V1 by V2 under the condition
I2 equals to 0 and therefore, what I do is I keep this open connect a source here connect
a source here. This is a constraint because I2 equal to 0 I connect, connect a source
here all right this I have explained already. So, what you have to find out is V2 by V1
and then find the reciprocal of this. So, under this condition V2 by V1 V2 by V1 would
be a potential division. YA is ineffective; potential division between YC and YB and you
can easily show that this is YC divided by YC plus YB. In terms of impedances it is ZB divided by ZB plus ZC. In terms of admittances it is
YC divided by YC plus YB and therefore, A is equal to YC plus YB divided by YC. This
is the value of A. To find the parameter B, well I can find another parameter from the same network you notice
that C is equal to I1 divided by V2 with I2 equal to 0. So, I can find out C by connecting
a current source I1 to the network YA YC YB. And then finding out the voltage here with
I2 equal to 0 this is open circuited. Now, this can be very simply solved this current,
this current is equal to I1 multiplied by 1 by YA divided by 1 by YA plus 1 by YB plus
1 by YC is that ok? This is this current, current division between YA and YB in series
with YC. And the voltage V2 the voltage V2 then shall be equal to this current multiplied
by 1 over YB. And therefore, V2 by I1 is equal to V2 by I1 is equal to YC divided by YA YB
plus YB YC plus YC YA. Is that ok? Are the steps all right I have done it by
inspection and therefore, C which is the reciprocal of this C would be equal to YA YB plus YB
YC plus YC YA divided by Y sub C, this is the z parameter. To find the B and D parameters
B and D parameters we have to make to V2 equal to 0. If you recall B is equal to V1 by minus I2
with V2 equal to 0 and D is equal to I1 by minus I2 with V2 equal to 0. This is the definition
and therefore, I make V2 equal to 0, which means that YB goes out of the picture. So,
we have a YA I do not care what this source is; obviously, we require 2 sources: we require
voltage source and the current source or a current source. I do not care what the sources
is all I know is this voltage is V1 and this current is I 1. Then you have an YB and it is short circuited YC short circuited. So, this current must
be I2 all right. The first thing to find out is V1 by minus I2 which is obviously that
should be. YC on the top. On the top it is YC correct this is YC. So, what is V1? V1 by minus I2; obviously, this
is equal to 1 over YC Are the signs all right? V1 appears across YC the sign I2 opposes V1
and therefore, the negative sign is taken care of. And this must therefore, be B As
far as D is concerned D is I1 divided by minus I2. Now, minus I2 is; obviously, equal to
I1 times yes YC divided by. YA Is it not that equal to this? Yes. I just skipped 1 step and therefore, D is equal to I1 by minus I2. So, it is YC plus
YA divided by YC all right, I have found out all the parameters. Let me let me write them down. A B C D D is
equal to YC plus YA divided by YC B is equal to 1 over YC then C is equal to YA YB plus
YB YC plus YC YA divided by YC and A is equal to YC plus YB divided by YC. And you can see
that AD yeah, AD minus BC is equal to 1 AD minus BC; yeah it is exactly equal to 1.
Yes sir. It has to be, there is no other way. Suppose,
in a problem with a reciprocal network three of the parameters are given you can find the
fourth agreed because you have this relationship AD minus BC is equal to 1 all right. We next
go to the relationships between the parameters. Interrelationships the most commonly used parameters are z and y and therefore, we start with z and y and we write this matrix Z matrix as z11 z12 z21 and z22. And the Y matrix as: y11 y12 y21 y22. The defining relations are that the voltage
vector V1 V2. Yeah pardon me. Would you please say loud? Sure all right. The V matrix which is V1 V2 and the I matrix is I1 I2 these are vectors
this is a column vector this is a column vector. And if you recall the defining relation is
that V equal to Z I. The 2 equations that we wrote can be expressed in this form V1
equal to z11 I1 plus z12 I2 and V2 is equals to z21 I1 plus V22 I2. The other equation
is that I equals to Y matrix multiplied by the V vector all right. If I substitute equation
2 in equitation 1 2 in 1 what do I get? I get V equal to z instead of I I write YV,
this is this clear how I wrote this? Would I repeat? What I have is V equal to Z I and
I is Y matrix multiplied by V that is what I wrote here. And therefore, this matrix is
the same as this matrix the pre-multiplying matrix must be an identity matrix. Therefore,
Z Y must be the identity matrix U of dimension 2 by 2 this is 2 by 2 this is 2 by 2. So,
multiplication of 2 by 2 by 2 by 2 which gives 2 by 2. What is the definition of the identity
matrix? Diagonal elements are one of diagonals are
0and similarly Y matrix must be the inverse of the Z matrix. This is the interrelationships between Z and Y matrices provided the inverse exists. And the condition for that is that
Del Y the determinant of the Y matrix must not be equal to 0, and the condition for this
is that the determinant of the z matrix must not be equal to 0. And if I look at if I look
at the expanded version of these inverse relationships. That is z 11 z12 z21 z22 is equal to the inverse of y matrix
if I look at this it is a 2 by 2 matrix very simple. I can write down the relationships
immediately z11 shall be equal to y22 divided by Del y z22 shall be equal to y11 divided
by Del y. Z12 shall be equal to. Minus y21 or 1 2 1 2 there is a transposition
12 by Del y1 must remember this 12. Now, it is not 21 there is a transposition
after taking de 12 by Del there is a transposition. And z21 is equal to minus y21 divided by Del
y. We are lucky that this do not interchange it is easy to remember all right. In a similar
manner if you look at the inverse relationship; obviously, you can write y11 is equal to yes
z22 by Del z y22 is equal z11 by Del z y12 is equal to z12 by Del z not quite.
Minus sign. Minus sign and y21 is equal minus z21 divided
by Del C. Where Del stands for the determinant of the particular matrix for example, Del
y is equal to y11 y22 minus y12 y21. Now, what we have said about conversion of z to
y or y to z. The other 2 matrices that is the h and the ABCD; obviously, they do not
obey inverse relationships because, the variables are different the independent set of parameters
is different. So, 1 has to work out from ABCD from the from no what is it called ab initio.
Ab initio means, going back to the roots for example. I will take only 1 example. Suppose, I want
to convert the h parameters to the ABCD parameters suppose, I want to convert this. Then, what
I do is I write both the relationship that is I write V1 h parameter relates V1 I2 to
V1 I2 to I1 V2. So, h 1 one I1 plus h12 V2 and I2 equal to h21 I1 plus h22 V2. And I
also write the ABCD parameters V1 I1 at the dependent variables and this is AV2 minus
BI2 1 must remember this and I1 equal to CV2 minus DI2 all right. So, what we have to do is express V1 in terms of V2 and I2. In other words, I have to eliminate
I1 from here and this is easily found from here I1 is I2 minus h22 V2 divided by h21.
Which incidentally, also gives me C and D if you compare these 2 do not you see that
I1 has been expressed in terms of I2 and V2. So, what is C? Minus h22 Minus h22 divided by h21. And what is D? Minus 1 by h11 There is a minus sign because there is a minus
sign here all right. Now, if I substitute this if I substitute this relation in the
first 1. Then I get V1 equals to h11 by h21 I2 minus h22 V 2, this is the first term h11
I1 plus h12 V2. And if I look at this relationship and this 1 if I compare these 2 then I get
the following equations for A and B. A is the coefficient of V2 that is a that
would be h12 minus h11 h22 divided by h21. And B, B would be equal to minus h11 h22 divided
by h21. Is it a minus sign? You do not have h22 this is a redundant term. Is there a minus sign? Yes there is. There is now let me write C and D also C was minus h22 by h21 and D was minus 1 over h21.
It appears that only this term does not come with a minus sign, but if you if you simplify
this, the denominator is minus Del h divided by h21 agreed. So, there is a uniformity all
come with a negative sign all of them have a denominator of h21 h21 h21 h21 h21. The
3 parameters BCD have a single term in the numerator h11 h22 1 whereas, A has the total
determinant of the h matrix. In a similar manner, we could go back from
ABCD to h or z parameters to h parameters or y parameter to ABCD parameters. All them
can be done and this exercise at least for some parameters you should perform. This is the table that we get ultimately.
This is the complete table where it does not assume that the network is reciprocal. It
is a general table and you should keep a copy of this ready with you in working out problems
on problems in network theory. Because, you never know where you shall require a conversion
this incorporates T to pi and pi to T T to pi means z parameters to y parameters and
pi to T is y to z parameters. You see for example, the table is read like
this the z matrix and the z matrix. So, this is this gives the matrix z matrix. Is this
visible on the monitor? No. No, what is the problem you tell me to the shift. Is that ok now? Z parameters z and
z that is the 1 1 is simply matrix. If you go from z if you wish to derive z from y the
row is all z parameters and the column is in terms of those parameters. If you want
to find z from y parameters, then you use this relation that is z11 is y22 by Del y
z12 is minus y12 by Del y as we have done already. Or let us say you want to find out
z parameters from ABCD parameters, then z1 z11 you A by C all right z12 is Del t by C
Del t is simply AD minus BC. Which is equal to 1 for reciprocal network, this is a general
table and therefore, they have used Del t. Similarly, z21 is 1 by C and z22 is D by C
and similarly for all other entries in the table. This is a very important table and
very useful table and I would advise that you keep it ready for reference all right.
Any question? From the book yeah course book any network
theory any respectable network theory book would have this table any respectable network
theory book. But, be aware some of the books particularly the Indian authors have many
miss prints and it is better that you take from course book all right. We then have a brief discussion on applications
of the 2 port parameters. Application of the 2 port parameters in finding out network functions
network functions are driving point. And transfer could be impedance could be admittance or
in the case of transfer it could be dimensionless. Now, given a network given a network N and
its 2 port parameters any transfer function can be found out any transfer function. And
we shall have a graduated series of examples to illustrate the applications.
The first 1 that I have is suppose I connect the voltage source here V1 and I want to find
out the voltage output voltage V 2; that is, my transfer function is V2 by V1 all right.
The parameters you can use any set of parameters, but the condition is that I2 equals to 0.
If I2 equals to 0 then you know that V1 would be equal to I1 z11, If I work in terms of
the z parameters and V2 shall be equal to I1 times z21 that is correct because I2 equal
to 0. Therefore, my V2 by V1 is simply z21 by z11. Can I explain you see my condition is this is kept open and therefore, I2 is 0.
Is it the implied condition? Not implied condition there is a given condition
given conditions if I connect something here then; obviously, I2 shall not be equal to
this. But what I want is opens circuit voltage transfer function. If I2 is 0 then my z parameters
give these 2 relations and therefore, I find V2 by V1. Suppose, 1 is per say I do not know
the z parameter I know the y parameter and I do not want to convert fine. What I will
do is I will write the equation for I2 I2 is equal to 0 equal to y21 V1 plus y22 V2.
And therefore, V2 by V1 is equal to minus y21 divided by y22 all right.
I can find out in terms of z parameters or y parameter and since, I know the relationship
between z and any other set of parameters. What I will require is only to look at the
table to be able to convert this for example, in terms of ABCD parameters. All that I do
is substitute for z21 in terms of ABCD substitute for z11 in terms of ABCD or else I go back
to the roots. That is I write the defining equations put I2 equal to 0 and then what
ever V2 by V1 comes I accept it. Is this all right? This is the first example the simplest 1. The second example well it is also very simple
I have this well situation in which the network N is driven by current generator I1. The output
is short circuited I2 and what I want to find out is I2 by I1; this is my transfer function.
If I take z parameters let us say, then in the second equation V2 is equal to 0 is equal
to I1 z21 plus I2 z22. And therefore, I2 by I1 would be equal to minus z21 divided by
z22 agreed, as simple is that from the second equation. Or if I want in terms of the y parameters what I will do is I take the 2 equations I1
shall be V1 Y11 because V2 is 0 and I2 is V1 y21 all right therefore, I2 by I1 is equal
to y21 divided by y 1 no negative sign. Similarly, I can find out from any other set of parameters
I will confine my attention to z and y other parameters you can try for yourself. The third example that I take is slightly
more involved that is a terminated network I have a network N I can connect either source
here. But all that matters is V1 and I1 and I terminate this by means of resistance let
us say R. The function of interest is I2 by V1 if capital R was replaced by a voltage
source V 2. Then what would have I2 by V1 become? What would you like me to try, z parameters or y parameters? Let us find this out then
we will conclude let us use y parameters. Are they easy to use? Well either of them
is easy there is no problem. Why do I use y parameters? Minus V2 by R that is perfectly all right that is what we shall use you see I2 is y21
V1 I want I2 by V1 I2 is y21 V1 plus y22 V2. But what is V2? Minus I2 R. So, it is minus y 22 R I2 therefore, I2 by
V1 is equal to I can write it down y21 divided by 1 plus y22 R all right. Now, I want to
ask you the following question: if R is 0 that is if this is short circuit. Then what
is I2 by V1? It is simply small I2 1 that is the definition. So, a now I am introducing
a notation, you must be careful about this notation when I write a small y it refers
to a transfer admittance of the network all right. Now, I2 by V1 is also a transfer admittance, I2 is the current in the load and V1 is the
voltage at port 1. So, I2 by V1 is also a transfer admittance. How do you distinguish
between the 2? Use a Y and use the subscripts 21. Y21, now you must be able to distinguish
between Y and y, make them quite different do not make them look alike because, then
you might make a mistake. Why you use this you see if R is 0 if this
is short circuited then I2 by V1 is simply small y21. Which means, the parameter of the
network N it has nothing to do with terminations. The parameter y is defined without termination
now with termination I2 by V1 is still the dimension of admittance and it is a current
at 1 port to the voltage at the other port. So, it is transfer admittance all right.
The y21 we call it short circuit transfer admittance that is it belongs to the network
N. Whereas, under terminated condition to distinguish between y21 the short circuit
transfer admittance and the transfer admittance of a terminated network. We use the symbol
Y. The subscripts are still 21 2 the first subscript refers to the port at which the
response is taken. That is, at port 2 and the second subscript refers to the port at
which the excitation is applied this will be our convention. Y21 shall be that we are interested in a current at port 2 due to a voltage at port 1. Similarly,
if I had written z12 this will mean that we are interested in a voltage at port 1 due to. Current at port 2. A current at port 2 is that this will be our convention all right and in the context things
will be absolutely clear. Now, suppose you are fussy and you say no I do not want to
work with y parameters I want to work with z parameters. Well all that I have to do is
to refer to the table the place y21 by what? z parameter . What is the z parameter? z12 by Del z. Plus or minus? Sir minus Minus z12 by Del z and y21 I shall replace it by z11 by Del z and work it out or I can
go back to the rules; I can do that. Let’s take the next example before taking the next example let me point out 1, 1 of the interesting equivalent circuits. You see I told you that as per as z parameter
is concerned if the network is reciprocal and 3 terminal then you can replace the network
by a T network like this. Where, these parameters are z11 minus z12 z22 minus z12 and z12. If
the original network is reciprocal, but not 3 terminal then this represents only a mathematical
equivalent circuit all right not a physical 1. On the other hand, if the network is reciprocal
reciprocity is a must. If the network is reciprocal and the three terminal this represents the
physical equivalent circuit also suppose it is neither. Suppose, the network is not necessarily reciprocal not necessarily reciprocal; that means, it
can be nonreciprocal also. And it is truly four terminal can you draw an equivalent circuit
well this is simplicity itself the drawing of the equivalent circuit. If I write I1 z11plus
I2 z12 and V2 equal to I1 z21 plus I2 z22, then it is common sense that this equivalent circuit describes the network. That is you have a z 1 one a current I1 V1.
So, V1 equal to I1 z11 plus a quantity I2 z12 which is the dimension of voltage. So,
I connect a voltage generator here which is I2 z12; obviously, is that it is very simple
V1 equal to I1 z11 plus this voltage source. Now; obviously, this is not an independent
voltage source it is a controlled source, it is a dependent source, the source depends
on the current at port number 2; which is I2. This current determines this voltage.
So, it is a controlled source and by a similar by a similar argument the equivalent circuit
at port number 2, is that we shall have a z22 here and another voltage source which
would be plus minus and the value would be I1 z21. This is the term that we have to use. c. Intrinsic. Yes, the network parameter no termination now is this is this clear? This is a truly
4 terminal truly 4 terminal and it does that assume either reciprocity or non reciprocity
we have used z21 and z12 we have not assumed them to be equal. And therefore, this is a
general equivalent circuit. What have we achieved through this equivalent circuit nothing much,
we have simply represented instead of equations we have represented this by means of a circuit.
The circuit contains 2 controlled sources the 2 circuits although shown physically isolated
from each other another e not really isolated. Why? Because the coupling comes to the control
sources you see this current control space. So, the 2 circuits are not decoupled from
itself. Although, physically there is no connection, but there is a connection through the control
of a voltage source by a current source similarly control of this source by the current source.
So, we have not achieved much we have simply represented the equations by means of an equivalent
circuit. However, sometimes this equivalent circuit is of great help great simplification
as we shall show in 1 or 2 examples. But by a similar token you can represent the y parameters. I1 equal to V1 y11 plus V2 y12 and I2 equal
to V1 y21 plus V2 y22 you can represent this by a dual circuit. That is what you do is
V1 I1 I1 is V1 times y11. So, introduce an admittance y11 here plus another current and
this current would be V2 y12, it is a current source controlled by the voltage at port 2.
It is a current source controlled by voltage at port 2 V2. In a similar manner for the
other port V2 and I2 I2 is V2 y22. So, we have an admittance y22 here and another current
source controlled current source whose value would be V1 y21 all right. This is an exact dual of the z parameter equivalent circuit in the case of z parameter there was
a series connection of a voltage source and impedance. Now, you have a parallel connection
of admittance and a current source. As I said you have not achieved much you have only represented
2 equations by means of a circuit. But, a circuit a picture is a word 1000 words as
we shall see in a few examples. Suppose we have a current source 1 example
of application of this suppose we have a current source and network and the termination ZL.
The output voltage is V2 and what we want is V2 by I1. What is it? A transfer impedance
and we should represent it by Z. 21. 21 that is correct Z21 this is what you want. Now, instead of going into any other any other
part if we simply represent if we simply take, the z parameter equivalent circuit. What was
the output equivalent circuit you have a V2. You have a series impedance z 2 22 this is
the current I2 and a voltage source. What is the value? I1 z21. I1 z21 agreed and what we have done here I
do not have to draw the other part because I do not need it what I need is a relation
between V2 and I1 I1 is already here. So, what I have here is a an impedance let us
say ZL and what I have to find out is the ratio V2 by I 1; obviously, V2 is equal to
I1 z21 times ZL divided by z22 plus ZL. Simply a potential division this is the voltage source
and there is a potential division between z22 and ZL all right. Therefore, in 1 stroke of the pen we find
out the transfer impedance z21 as V2 by I1 as equal to z21 ZL divided by z22. Distinguish
between small and I always cross the middle draw a horizontal line in the middle to indicate
z and do not do it for Z; this is how I differentiate. Suppose, in this example it is not V2 well
suppose the quantity of interest is I2 that is the current show this impedance I2.
Can I find out the transfer function I2 by I1? Very simply I2 is related to V 2.
No, I do not need that V2 is equal to well I2 is equal to V2 divided by ZL with a negative
sign agreed. So, all that I have to do is yeah what did you say? Sir z 1 z21 plus z22 by to ZL. There is a mistake. z21 upon There is a minus sign all I have to do is
to divide this transfer function by minus 1 by ZL all right. So, ZL ZL cancels the negative
sign this you must not forget all right. The last example we have a voltage source
V1 the network N and an admittance YL. Our quantity of interest is V2 and I what I want
to find out is V2 by V1. Now, in this situation also the y parameter equivalent circuit the
4 terminal comes into help. If you recall from the load end from port number 2 what
I have is an admittance y22 and a current source. How much y21 times? I1 V1. V1. So, I do not have to draw the other part
at all I do is connect a YL here. So, V2 is obviously, equal to V2 is the drop
due to a current y21 V1 flowing into a parallel combination of y22 and YL agreed. This is
a current source which flows to this parallel combination to produce a drop of V 2. So,
V2 is y21 V1 current multiplied by impedance. There is a minus sign here minus sign, because
V2 and y2 and V1 they do not agree. Therefore minus y2 and V1 divided by y22 plus YL agreed. Let me write it down again V2 by V1 is equal
to minus y21 divided by y22 plus YL. And what was the situation? N there is a V1 here and
there is a YL here this is V2. Now, I want to recall I want you to recall that, if YL
was equal to 0. If YL was equal to 0 which means that, the termination is open circuit.
Then the transfer function open circuit voltage transfer function was precisely this was derived
earlier minus y21 by y22; which checks all right it checks. The question that I leave you is that is there a preferred method of solution. Given a problem
is there a prepared method which parameters to use, which equivalent circuit to use and
so on. Is there a preferred method? My answer is no and yes. What does it mean? There is
a preferred method which comes there is no prescription. You cannot say if this, this,
this are to be found out and this are given you follow z parameter. No, there is no prescription
it comes by experience. That means, you solve more and more problem then given a problem
it will be obvious to you which parameter should we use all right. Next occasion we shall take more examples and then go about to interrelations.

Author: Kevin Mason

3 thoughts on “Lecture – 26 Two – port Network parameters

  1. why saying 'Zee" while it is "zed". You know Americans also say "soccer" for "football".
    It is ENGLISH and not American.

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