Circuit Theory
Prof. S. C. Dutta Roy
Department of Electrical Engineering
Indian Institute of Technology, Delhi
Lecture - 46
Elementary RLC One - Port Synthesis and Introduction to Two - Port Synthesis
All right, forty-sixth lecture elementary RLC 1 port synthesis and introduction to 2 port
synthesis, this is the topic. As I told you in the last class RLC 1 port in general possess a
difficult problem and if at all solvable it is solvable by using a transformer .
(Refer Slide Time: 00:45)
If the use of a transformer is a allowed then there is no problem or you use a non canonic
synthesis of Bott & Daffin or some of its modifications. But as I said in some special
cases if you are lucky it is possible to synthesize a an RLC RLC impendence function.
And we shall take a few examples to check the the tricks of the tray.
If these tricks work perfectly all right if they do not work then they do not work that is it.
One of the examples that I take first is s square plus 2s plus 2 divided by s squared plus s
plus 1 you see that the poles if this is an impendence functions. You see that the poles are
neither on the imaginary axis nor or the real x axis they are complex, and therefore if at
all realizable it should be realizable as RLC. But, before you check the RLC you must
check the realizability condition.
There is only 1 in this case, because there are no poles on the j omega axis that is a1 b1
which is 2 in to 1 and square root a 0 a square root 2 minus square root b 0 square root 1
square this is 2 and this is 2. So, with greater than equal to it is equal to therefore, the real
part condition is satisfied. You understand the shortcut procedure because by inspection
we see all all coefficients are real and positive there are no poles in the j omega axis. So,
all that you have to do is to check whether a1 b1 is greater than equal to square root of 0
minus square root b0.
Sir, greater than strictly, greater than or equal to.
Greater than or equal to it can equal to.
(Refer Time: 02:40)
How come what did
0.1 sir 2 into 1 and square.
Sir, root 2 minus root 1 is 0.4 0.4 square.
Yes i great root 2 minus root 1 is point.
41
414 square, so it is greater than I make that obvious mistake, but you see if you if you
make a continued fraction expands let us see if it works. S squared plus s plus 1 blindly I
mean I do not know how else I could to it. So, we will try continue faction s square plus
2s plus 2 1 this is an impedance s square plus s plus 1. So, I am left with s plus 1 divides
s squared plus s plus 1 s s squared plus s 1 s plus 1. Let us see this is admittance then s s
1 1 1 1 0, this is impedance and this is a admittance. So, it does work.
(Refer Slide Time: 04:11)
You have got all questions which are which are real and positive coefficients and
therefore, I can draw the network as 1 ohm. Then an admittance of so 1 farad then an
impedance as; that means, a an inductor of value 1 Henry and then an admittance 1 that
is this. So; obviously, it is an RLC network and that is why it was neither it did not
satisfy the condition of RC RL or LC. But suppose, we do it the other way down that is
starting with the lowest powers all right. 1 plus s plus s squared 2 plus 2s plus s squared
2.
So, 2 plus 2s plus 2s squared and minus s squared we cannot go further. Suppose, we do
it the other way down that is admittance we take the admittance let us see what happens.
Sir lowest power
Again lowest powers
(Refer Slide Time: 05:10)
So, 2 plus 2s plus s squared divides 1 plus s plus s squared half. So, 1 plus s plus s square
by 2 we are left with s squared by 2 this is an admittance divides 2 plus 2s plus s square
well four by s squared. We do not know how to realize this isn’t that right. So, we do not
proceed further.
Similarly, you can try other procedures and see if any of them works you will have to try
this there is no established procedure to do this. We take another tricky problem, but you
assure that if 1 synthesis exists there rather indefinite number. So, may be something else
will succeed all right.
(Refer Slide Time: 06:20)
Now, this problem is slightly tricky s plus 2 s plus 3 s plus 1 s plus 4; suppose this is
given as an admittance y of s. Now, the poles in zeros do not alternate first critical
frequency is a pole, but then you have 0 another 0 adjacent 0 then a pole first is a pole
last is also also a pole. So, it cannot be RC or RL it is not LC at all. But let us see
whether its positive real or not s squared plus 5s plus 6 s squared plus 5s plus 4 a1 b1 is
25 a 0 square root a 0 minus square root b 0 minus 2 whole squared. How much is this? 2
point approximately 2.4 so; obviously, greater than sign is satisfied.
So, it is peer there is nothing; obviously, in this which can make it non peer. Therefore
we proceed by let us say start with partial fraction y of s. Well in partial fraction you take
the infinite frequency value is 1 plus. Let us say K1 divided by s plus 1 plus K2 divided
by s plus 4. Let us try does this work K1 multiply by s plus 1 and put s equal to minus 1
you get 2 by 3 that is positive good 2 by 3
Then K2 minus 2 by 3, so this does not work if it is negative; obviously, we do not know
how to do it. So, we give up.
(Refer Slide Time: 08:10)
We try Y of s by s whereas, that might work let us see s plus 2 s plus 3 divided by s plus
1 s plus 4 s. So, it is K0 by s 6 by 4 3 by 2 s plus all right plus K1 by s plus 1 plus K2 by
s plus 4 what about K1 s plus 2 s plus.
Minus there does not work
It is minus
(Refer Time: 09:00)
Does not work, similarly, you can try the reciprocal of this Z of s.
In either case you see if Z of s works then we know we know how to how to find, but Z
of s also does not work 1 of the residues comes negative. If Z of s by s work because if it
is Z RL if it is RL then you know Z of s by s shall work it does not work there either. I
give up I would not would not repeat this procedure, but then we ask if it does not work
what is it that we are missing? and the clue these are tricks of the tray there are no set
patterns.
The clue is the following let me show you this 1.
(Refer Time: 09:47)
That also we will try we will try, but let us exhaust my partial fraction. The clue is here
you see K1 is positive, but K2 is negative. Now, is it possible to combine a part of this is
it possible to combine a part of this constant with this term which is a negative negative
residual to make it positive that is the clue. But if this was not there then of course,
nothing could be done. But since, there is a constant terms i could take a part of this and
add here so that negative coefficients disappear.
Sir that means, here taking out the Z value which greater than
No it is not RC it is not RL either and therefore, there those considerations do not apply
here. What we can remove that consideration does not apply here because this is neither
pure RC nor pure RL all right.
(Refer Slide Time: 10:56)
So, we argue like this that perhaps what we are missing is because it cannot be RC
because it cannot be RL. It is not pure LC either it is RLC therefore, perhaps what we
have to do is we have let us say some constant k 0 let us say or no let us let us put it
something else. Something constant K1 plus K2 divided by we found 1 of the 1 of the
residues positive. So, let us make it as plus 1 plus let us say some K3 by s plus 4 K3
cannot be negative why do not we try K3 s by s plus 4 this is what I am trying here.
Yes.
A resistance a parallel combination of resistance and capacitance and this will be a
parallel combination of resistance and inductors. Now, nobody told me nobody told me
before hand whether K3 will be positive or not. And there is nothing sacred about
assigning the pole at s equal to minus 4 to K3 s by s plus 4. I could have done it the other
way round. For example, I could have written K1 plus K2 s by s plus 1 let us do this first
plus K3 by s plus 4. Why do not I do this?
K3.
Because to find K3 the value of K3 is found by the same procedure and K3 was found to
be negative. So, you do not try this, do you follow this. These are various steps 1 is to
apply they said there is no set pattern you have to exercise your. You see if I write it like
this then how do you find K3 you multiply the function by s plus 4 and put s equal to
minus four that you had already found to be negative. So, we do not try this.
Now, if I go by this then what is K3 s plus 4 y of s divided by s all right, at s equal to
minus 4. And this comes out to be one-sixth is that correct which is very hopeful which
is very encouraging that it is positive therefore, all my 3 constants: K1, K2, K3 are
positive.
Instead of doing this procedure so we will replace 1 by say 3 s by 4 by 3 s plus four and
then we have added the 2.
Besides that is what I am doing you see taking a part of the constant and adding to that
the negative coefficient negative resident term. That is precisely what I am doing, but I
am doing it in the more systemic manner I argue that 1 of the parallel combinations
perhaps is RL. Which can it be; obviously, the 1 with a pole at s equal to minus 4. Why;
obviously, because otherwise we found K3 to be negative.
Sir how can we share about K1
How can how can we be share about K1? Why should K1 be there?
(Refer Time: 14:06)
There is no guarantee you have to find out K1.
What is K1?
1 minus 1 plus.
(Refer Slide Time: 14:17)
So, K1 is now now wait wait a second there must be systemic procedure s plus 2 s plus 3
divided by s plus 1 s plus 4 this i had said K1. K2 is the same as in the previous case how
much was it?
2 by 3
2 by 3 divided by s plus 1 then I find out 1 sixth s divided by s plus 4. So, what is K1 and
how do you find?
(Refer Time: 14:47)
Put s equal to I could put s equal to 0 also.
(Refer Time: 14:54)
What dear no it is not it is not that difficult you see K1 if I put s equal to infinity
becoming prefer that all right. We will do that if I put s equal to infinity this goes to
infinity. So, I get 1 sixth K1 plus 1 sixth must be the infinite frequency value; that
means, equal to 1 and therefore, K1 is 5 by 6. I could put s equal to 0 also if I put s equal
to 0 then I get K1 plus 2 third equal to 3 in to 2 6 divided by 4 it would give the same
result.
So, my final network is 5 by 6 resistance then an RC do not you see the 5 by 6 that 1
sixth is now added to that negative residue term.
Sir 6 by 5 resistance because it is the admittance
This is admittance yes 6 by 5. So, it is not this fracture at all it is a faster 2 type structure.
So, I have 6 by 5 resistance then can you give me the values 1 RC can you give me the
values.
There by 2 is L.
3 by 2 is L, so I have R and L why is 3 by 2 an L this is admittance you are right 3 by 2
is the inductance and the resistance is 3 by 2. Now, what about the capacitance
1 by 24
Capacitance is 1 by 24 and the resistance is 6 that is a correct structure. Is there an
alternative can you find an alternative another structure foster type? Why not take the
impedance?
(Refer Slide Time: 17:05)
Take the impedance as Z of s equal to s plus 1 s plus 4 divided by s plus 2s plus 3. And
expand like this means sum K1 plus K2 by s plus 2 plus K3 by s plus 3. Will this work?
And why not what is.
K2 be negative.
Why should K2 be negative? The reason is obvious you have to see
(Refer Time: 17:39)
You see if you assume this form; obviously, you are assuming an RC network this is not
an RC function and therefore, 1 of the residues must be negative.
And which 1 is negative any 1 say.
K2 is negative.
K 2 is negative if K2 is negative then; you assume this to be of the form k prime plus K2
prime s divided by s plus 2 plus K3 by s plus 3 all right. You do not find K1 to start with
you find K3 in the usual manner you find K2 prime by multiplying by s plus 2 and
dividing by s and putting s equal to minus 2. Then you put either s equal to 0 or s equal
to infinity to find K1 prime is this point clear.
So, there is the structure that you will get is the faster 1 in which 1 parallel circuit is RC
the other parallel circuit is RL you agree. Now, let us see is there any question as I said
this has to be trial and error. But a bit of commonsense has to be applied then you can
avoid some unnecessary wastage of time and efforts.
(Refer Slide Time: 19:14)
Now, let us try continued fraction. Our function is s squared plus 5 plus 4 divided by s
squared plus 5 s plus 6 all right this is by impedance function. And blindly, let us start
continued fraction expansion with highest powers
(Refer Time: 19:34)
You will get negative so this term you are. What about starting with the lowest powers
let us see y of s 6 plus 5 s plus 6 four plus 5 s.
(Refer Time: 19:49)
S square s squared 4 plus 5 s plus s squared let us see if this works 2 third oh no 3 by 2 or
2 by 3.
2 by 3.
4 plus 10 by 3 s plus 2 by 3 s squared, I am left with 5 by 3 s plus 1 by 3 s squared 6 plus
5 s plus s squared the quotient is …
(Refer Time: 20:30)
18 by 5 s. Good enough this is y this is z. So, I get 6 plus yeah 18 by 15 is that right.
(Refer Time: 20:49)
Which 1 is 6 by 5 s this is 6 by 5 all right all right. So, i get 25 minus 6 19 by 5 s plus s
squared 5 by 3 s plus 1 third s squared. Now what?
Sir how is the first 1 1.
Where is the first 1 because we started then no we did not invert it this is Z, this is y.
How much is this?
Sir 25 by 57
25 by 57, so 5 by 3 s plus I need not go further because the remainder is negative. So,
this does not work and you can you can try taking 1 it.
(Refer Time: 22:00)
It depends on where did you start with you see in here here my Z of s was this and I
started with lowest power 6 pus 5 s plus s squared in the denominator therefore, the
coefficient is impendent. Now, there comes now a a trick or exercising some ingenuity or
trial and error.
Excuse me sir.
Yeah
(Refer Time: 22:33)
This 1 yes sir we did try this isn’t it did we try?
Yes sir.
We did it did not work, what happened to my well it did not work.
(Refer Time: 22:48)
Same page.
On the top sir.
On the top we did try you see the; obviously, the the remainder will be negative. Now,
what we ask is the following: it is at this stage that the continued fraction expansion falls
to ground. Suppose, at this stage we reverse the order in other words we come up to this
we come up to this. Then I do continued fraction expansion starting with highest powers.
(Refer Slide Time: 23:29)
Let us do that, s squared plus 19 by 5 s divide 1 third s squared plus 5 by 3 s as I said
there is no set procedure. The first quotient is 1 third 1 third s squared plus 19 by 15 s.
(Refer Time: 23:55)
19 by 15 s is it, so I get 15.
6 by 15
6 by 15 that is right 30 now how did you get 6 3 times 5 15 6 by 15 s then s squared plus
19 2 by 5 s all right Then I get 5 by 2s 5 by 2s s squared 5 by 2 is all right then I get 19
by 5 s 2 by 5 s; it does work all right. I do not have to proceed further I know it works all
right. So, it is not necessary that you continue continued fraction expansion now why did
you decide that we will do it at this stage at this stage.
Because, the next stage was coming negative so we do something to the previous stage to
correct. If it did not work if it did not work then what would you have done you have
said about it we cannot do it.
1 by 3
Because the other way round it did not work starting with the highest part of …
What about the values of impedance and admittance? So, the dimensions
Dimensions are starting with highest power or lowest power that does not change the
dimension, dimensions are ok. There may be other variations inside if it does not work
for example, if we take the reciprocal and then start with highest power or lowest power
then the dimension would have changed.
(Refer Time: 25:58)
We can
Sir then we change the dimension.
Then we change the dimension the next term would not be impedance then it would be
admittance.
(Refer Time: 26:07)
By this 1 this is also possible that is at some certain stage you can either go from lowest
to highest or highest to lowest or you can take the reciprocal and go. But as I said purely
trial and error there is nothing, nothing that ensures that there exists a continued fraction.
(Refer Time: 26:30)
That is right we will try that again.
(Refer Time: 26:37)
There are many possible cases many possible cases.
Sir even if the coefficient is positive then we can take the reciprocal or change that.
Still we can change the reciprocal if it works you see, why where we encouraged to try
reversing the order. There must be a reason why were we encourage encouraged to try
this procedure out that instead of.
S square cannot to be negative, so we can reverse it.
No, that is a mechanical thing we try reverse, but why did not we give up at this stage we
had a hope that because 1 circuit has been found out; 1 possible circuit has been found
out. We know that there exists an indefinite number of circuit and therefore, there is no
reason why partial fraction works with a slight twist. Why continued fraction would not
work with another slightly twist?
Excuse me sir.
Yes.
(Refer Time: 27:43)
Correct, what is the other part what do you mean other part?
Sir, initially we are started with removing the lowest 1 sir.
Correct.
Then we get negative term from there means if too remove the highest power sir.
We can start with either highest power or you can take the reciprocal
Sir if again we get a negative term can we shift to the lowest start moving the lowest 1.
Yes why not anything is permissible so long as you do not change the dimensions. So,
long as you do not change the numbers.
(Refer Time: 28:19)
You can change you can keep changing after every step it will be a mixture of cover 1
cover 2 and yes.
(Refer Time: 28:27)
Right, so instead of the quotient being an impedance it will again been admittance. So, 2
admittances will come in parallel or 2 impedances will come in come in series.
(Refer Time: 28:44)
It will not be a how do you make the circuit? At that stage the 2 quotients if they have
the same dimension then they will come either in parallel or in series. They will not
come as 1 in parallel and the next 1 is series that will not happen.
Yes, there is a question.
(Refer Time: 29:02)
Circuit realization let me complete it then or tell me what this, what this 2 by 19, so 2 by
5 s this is 0. Now, if we come back (Refer Slide Time: 19:14) this is a z, so this is y.
No sir 1 by z.
What is a z?
(Refer Time: 29:30)
This is not there this is a z no we took a z we started a z 6 plus 5 s plus s squared divides
4 plus 5 s plus s squared. So, it is a z z y this is z this is z.
(Refer Slide Time: 30:02)
So, my circuit would be this two third then what would this be inductors 5 by 18 the next
1.
25 by 27
No, before that there is resistance of 1 third next 1 is next 1 is capacitor 5 by 2 that was
the effect of reversing the order. The next 1 is a resistance 2 by 19 and then the
remainder is 0, impedance 0 means a short circuit. So, this is the total circuit. Since, a
serious consideration of 1 port synthesis RLC is beyond the scope of this particular
course. We now turn attention to a glimpse into a transfer function synthesis and we
recall a few facts.
As you shall see by definition a transfer function is a network function that relates the
output at 1 port to the input at some other port. That is definition of transfer function
right. Otherwise if the port is the same because an effect at the same at the same port
then you call it a rhyming point function. A transfer function therefore, relates the effect
or the response at 1 port to the cause or the excitation at some other port and you have
already seen example.
(Refer Slide Time: 31:48)
A current generator I 1 and you measure the voltage here V 2. Then what is V 2 by E 1 I
1 it is a transfer impedance and how is it related to the parameters of the circuit and
which parameter.
(Refer Time: 32:05)
z12 or z21.
z21.
Z21 this is described by the z matrix this is z21 all right. Of course I know that if this is a
reciprocal network which we have been discussing then this will also be equal to z12.
Now, suppose we take we take this network with a source impedance with a voltage
source and you keep this open and measure the voltage. Then V2 by V1 is the open
circuit voltage transfer function and this is expressed in terms of z21 and z11 z21 by z11.
It is very simple write V 1 equal to I1 z11 plus I2 z12 and V2 equal to I1 z21 plus I2 z22,
but I2 is 0 therefore, the ratio simply becomes z21 by z11.
I can also express this in terms of y parameters. And if you do not recall this is y21 by
y22 with a negative sign and it comes because I2 equal to V1 y21 plus V2 y22 and I2 is
0 that is how it come. These derivations I mean how you get this you should be able to
remember simple things like this. Now, you complicate matters a little bit.
(Refer Slide Time: 33:50)
.
Suppose, we have a network N passive reciprocal network N driven by a voltage source
V1 and terminated in a resistance R the current is I2. And what we are interested in is V2
by let us say I1 this may be a current source. Now, let us put is as a source it may be a
current source it may be a voltage source.
Suppose, we want V2 by I1 you recall we had used a symbol for this this is a transfer
function. But not, the z parameter of the network N because N is being augmented by
capital R, we call this a transfer impedance and we used a capital Z for this. z21 that is
the response point to the excitation point and this is simply equal to z21 R divided by z22
plus R. Do you recall how this is done? How this is derived how?
(Refer Time: 35:04)
Z21 no this is z21 no capital; z21 is the transfer impedance of the terminated network
there is a small z21 here, which relates to N.
(Refer Time: 35:25)
No, this is the actual condition terminated under terminated condition. Now, how did we
obtain this give me a simple method?
V2 equal to minus I2R
V 2 equal to minus I2R and
(Refer Time: 35:48)
I1 z21 plus I2 z22, that is wonderful absolutely wonderful. If you go back to the basics
there is no you cannot make a mistake, well in a there is a simpler method if you do not
want to write the equation you look at the Thevenin equivalent and derived by a current
source. What will be the open circuit voltage? I1 z21 and the output impedance would be
z 2 2 because, current source. And there will be a division between z22 and R that is it. Is
the point clear not clear?
This would be I1 z21 this will be the open circuit voltage then opens the Thevenin
impedance would be z22 and you have an R this is V 2. So, there are simple ways I have
done this in detail I am simply reviewing them, but it it is necessary that you recall at
least a simplest of that.
(Refer Slide Time: 37:02)
We also treated the most general case that is we have a VG generator in series with a
resistance R and network N and a termination of R2 V2.
We have found out that V2 by V1 is equal to z21 R2 divided by z11 plus R1 this is the
effective z11 of this network multiplied by z22 plus R2 effective z22 minus z12 z21. Of
course if you remember this then you can derive every other case as a special case of this
this is the most general case terminated in both the ports. And at 1 port is the excitation
at the other port is this point. The point to notice from all the simple examples is that the
transfer functions. Yes.
V2 by VG.
V2 by VG V1 is here thank you for the correction that is wondering way people have not
notice this is there any other mistake all right.
Silent says that you have may be able to detect which means in all probability there is no
other mistake. The thing that I want you to notice is that in in in computing this transfer
function simple transfer functions we have always struck to z and Y parameters and this
is 1 of the philosophies of synthesis. Z and Y parameter came fast in to the picture
scattering came much later. And the H parameters came the H parameters gained there
prominence when transistor was brought in into the field of electrical engineering.
So, let us say around 1949 or 1950 the H parameters struck gained their prominence. The
transmission parameters gained prominence when people wanted to build systems as a
cascade. But, basic transfer functions we work only in terms of z and Y parameters due
to historical reasons. The other reason is that it is more convenient to use because
impedance and admittance we know the properties.
You see all these transfer functions involve z12 which is equal to z21 and then these
driving point functions z11 z22. We already know the properties of this. At least 2
element kind we know completely 1 element kind we know that they have to positive
there. Either this or you can express in terms of Y parameters that is minus Y12 and
minus Y21 then Y11 and Y22.
We also know, the properties of driving point admittances they must also be positive
real. So, if we now know, if we know investigate the properties of this transfer
impedance and this transfer admittance that is Y21 and Z21. If we know the properties of
this then we should be able to know the properties of a transfer function properties of a 2
port all right. So, we shall now spend some time on the properties of this transfer
impedance and this transfer admittance. But, before that, some general observations you
see whatever the transfer function is some general observations.
(Refer Slide Time: 40:44)
Let us call, the transfer function as T of s the transfer function could be could have be
dimension of an impedance could have be dimension of admittance or could be a
dimensionless transfer function. If it is dimensionless then it could be a voltage ratio or a
current ratio. But whatever it is, if s is real what can you say about T of s it must also be
real. Why because, T of s contains z12 z21 z11 and z22. Z11 and z22 we know they are
real for s real. What about z12 z21 or Y21 Y12? They must also be real they are rational
functions and there is no reason why they should not be real for s real.
So, if s is real T of s is real this should be if you get a transfer function having a complex
coefficient you say, we cannot realize this. Second property irrespective of what the
transfer function is can you have a pole in the right half plane? No, no poles in right half
plane because we are talking of passive networks. Passive networks are always stable all
right and therefore, there cannot be any poles in right half plane.
If there are poles on the j omega axis can we have multiple poles on the j omega axis?
No, no multiple poles on the j omega axis and what is the logic now because this leads to
instability. That is, even if the input is bounded the output shall not be bounded. Give an
example. Suppose, we have 2 pairs of poles at plus minus j omega naught then the time
domain response shall be of the form t times sign omega naught t or cosine omega
naught t. And obviously, as t goes to infinity these oscillation amplitudes grow
indefinitely.
So, no multiple poles on the j omega axis this must be shared by all transfer functions if
there is a multiple pole then you give up, you do not you do not proceed further. Next 1
is a tricky question. Suppose, T of s is written as P of s by Q of s what can be the
maximum difference in degree between the 2? Well I said it is a tricky question. So,
maximum difference maximum difference; obviously, can be only 1. Isn’t that right? If
the difference is 2, we shall have a multiple pole at infinity, which is a point on the j
omega axis.
If the if Q of s well I think I have said what I wanted to say there is the maximum
difference cannot exceed 1. But can they be equal?
Yes sir.
Yes it can be equal can know I made a wrong statement I take that. Maximum difference
in degree can be more than 1, let me tell you.
(Refer Slide Time: 44:55)
I should what I should have said is that degree of P the numerator degree cannot exceed
denominator degree by more than 1. This is the correct statement because, Q of s degree
can be three and P of s degree can be 0. For example, you take this network you know
that is can you write down its transfer function by inspection by just looking at it. That is
I have a voltage source and this is the voltage that I that is my concern V2 by V1. What
would be the form of this?
(Refer Time: 45:25)
1 by s square plus let us say B1 s plus yes what should be this coefficient. Constant b 0 is
there a restriction on b 0 or it must be positive. But do not we know its value. I appeal to
your commonsense. What is the DC transfer function if you apply DC here?
(Refer Time: 46:02)
Of course, it is a constant. But, what is the value of the constant?
Constant.
1 it has to be 1 because a DC this is open this is open and therefore, whatever voltage
you apply here will come no current is flowing. Therefore, b 0 has to be 1 and you see
that the numerator degree is 0 the denominator degree is 2. They are differing by 2 they
can differ by any number in fact, but the restriction is that the numerator degree cannot
exceed the denominator degree by more than 1. And the reason is that if it exceeds by
more than 1 we shall have a multiple pole at infinity.
Here, what is what is wrong what is right here and what is wrong here? There is a
multiple 0 at infinity that is permitted for a transfer function; it is not permitted for a
driving point function. Why not why is such a thing not permitted in a driving point
function? Because, the reciprocal of a driving point function is also a driving point
function and if there is a multiple 0 at infinity there shall be a multiple pole at infinity of
another driving point function. Which, therefore ceases to qualify as a driving point
function, the reciprocal of a transfer function is not necessarily a transfer function.
For example, the reciprocal of this cannot be realized why not because it has a multiple
pole at infinity is the point clear. So, this is the first difference between a driving point
function and a transfer function. That if a transfer function is written as P by Q the
degree of P can be at the most 1 greater than the degree of Q. But the degree of Q can be
any number greater than the degree of P any number depends on the complexity of the
transfer function.
Just 1 more 1 more observation; obvious property then we will see we will call it a class.
(Refer Slide Time: 48:17)
Suppose, T of s is written as m1 plus n1 divided by m2 by n2 we have always been doing
this with driving point function. Now, I appeal to your commonsense and ask you what is
T j omega squared in terms of m1 of j omega and n1 of j omega? M1 no m1 of j omega
square agreed then plus n1 of j omega square divided by n2 I set you a trap and you have
fallen into it.
(Refer Time: 49:05)
No, I am trying finding a magnitude squared magnitude square should be real part
squared plus imaginary.
(Refer Time: 49:15)
Any other brilliant suggestion
(Refer Time: 49:26)
Say it again.
(Refer Time: 49:31)
It won’t be into j omega.
This omega sir it is not a function of j omega.
All right we put s squared equal to minus omega square that is not the inaccuracy there is
something else something else grossly wrong. And I have cautioned against it several
times. You see n1 is the odd parts suppose n 1 is equal to so then what is n1 of j omega it
is j omega. Would you bring here n1 squared j omega?
(Refer Time: 50:06)
That is it that correction is there, you see the imaginary part is n1 j omega divided by j.
(Refer Time: 50:19)
(Refer Slide Time: 50:37)
We take a function let us say s squared plus s plus 1. Then 1 minus omega squared is the
real part and plus j omega. So, the magnitude squared would be 1 minus omega squared
whole square plus omega square. If I insist on taking this I must add a negative sign
minus j omega whole square. Isn’t that right? Please be careful about it, I have cautioned
you again and again that when you take the odd part you and put s equal to j omega.
Then the imaginary part is that quantity divided by j because j comes into the picture.
(Refer Time: 51:18)
You can do that you can take the mod. Now, the question that I was asking is
(Refer Time: 51:29) M 1 squarer minus M 1 square is the (Refer Time: 51:32)
That is correct well it should have it should have been clear also from the consideration
very simple consideration. I was hood winking I was trying to take to detour and confuse
you and you have been confused, please come out of it.
(Refer Slide Time: 51:57)
You see what you have to do is to find out the magnitude square take m1 plus n1 and
then you take m1 minus n1 j omega. This is this is the numerator P j omega and this is P
minus j omega, which will make the numerator magnitude squared. So, T j omega
squared magnitude squared is equal to the numerator squared divided by the denominator
squared.
(Refer Time: 52:30) negative sine (Refer Time: 51:34) negative sign
Now, but if multiply the 2 is it not m1 squared minus n1 square.
Sir (Refer Time: 52:39)
Let it do that also, you let it do that also. The question that I am asking now is I have
discussed this earlier what can you say about this magnitude squared function? It must be
purely even. Is it not right?
Yes sir.
Magnitude squared function cannot be negative; it only contains even powers of omega
in the numerator as well as the denominator. What about the magnitude? That is the
square root of this magnitude that must that also be an even function. And what about the
argument angle of T j omega? Must be an odd function; this we have said again and
again and you will see the deflection of this evenness and oddness in this synthesis
procedure they play a dominant part. We will continue this in next class.