This is the forty-fifth lecture and we are to going to discuss the rest of RC synthesis and introduce
also RL 1 port synthesis 2 element kind. Let us recall the properties of RC impedance and
admittance because they are not similar they are not identical.
The properties are different. Let us make a make a short review of the properties of RC impedances
and admittances. You recall that starting from the most general faster 1 form of RC
impendence that is impedances of this type. And the most general form for RC admittance
which is R in parallel with C, and then a number of RC in parallel. Starting from this
we derived the general forms for RC impedance and RC admittance.
The general form was K0 by s due to this capacitor plus K infinity due to this resistor plus
summation of terms like Ki divided by s plus sigma i. Where Ki’s and sigma i’s are
expressible in terms of the individual resistors and capacitors. And therefore, all these coefficients
K infinity Ki and sigma I they are real end positive. On the other when you too K an RC
admittance we saw that it is of the form K0, because of this resistance plus K infinity
s, because of this capacitor plus summation of terms like K is divided by s plus sigma i.
This is the difference there occurs an s here
because R and C in series if you take the admittance where admittance is of this form.
In other words these Ki’s are not the residues of YRC and you also notice that ZRC and YRC
only differ by a multiplication by s. That is, if you multiply ZRC by s the form is the
same as that of YRC. Another way of expressing this is that YRC by s is of the same form
same form a ZRC all right. And from these 2 expressions from these 2 expressions, and
this similarity we derived several properties of RC impedances and RC admittances and these
are as follows.
I keep the expression here, so as to facilitate, facilitate appreciation of the properties
that I am going to mention K0 by s plus K infinity plus summation Ki by s plus sigma i.
Then K0 plus K infinity as just multiply this by s and you get the general form for
YRC. It is Ki s divided by s plus sigma i. And the first thing you notice is that poles
of ZRC that is this sigma I since they are real in positive and the other pole is possible
at s equal to 0. The origin, so poles must be on negative real
axis and the same applies here poles on negative real axis and what are poles of YRC at the
zero’s of ZRC. Therefore, I can say poles and zero’s of RC admittances must lie on
the negative real axis all right. Then we look at the; we look at the situation at infinity.
There is no way that ZRC can have a pole at infinity there is no way.
Yes Fine that is also negative real axis. Infinity
is a single point whether you go from this side or that or this side or that side infinity
is a single point and when convenient we consider it on the negative real axis when convenient
in the Lc case for example, we consider it at at plus j infinity. Since we do not know
it we can put it anywhere it is exactly like a concept of God or Heaven. Infinity is a
very similar to that concept in various religions and so on.
What you said was if what are poles of YRC at the zero’s of ZRC. So, we conclude that
poles and zero’s of RC admittances all lie on the negative real axis. Then the situation
at infinity there is no way that ZRC can have a pole at infinity. No pole at infinity. Pole
at infinity is not permitted all right. At infinity what can happens is that be value
can be Either K infinity either a constant or 0 at
infinity. On the other hand YRC can have a pole at infinity pole or if it as no poles
then it is a constant that infinity. And similarly, at the origin it can ZRC can have a pole because
of this term or a constant at origin at s equal to 0. On the other hand for YRC at s
equal to 0 it is either a constant which is equal to K0 or 0 at s equal to 0.
While if you look at look at the properties of ZRC this property at YRC is obvious if
ZRC has a pole at infinity at s equal to zero; obviously, YRC has to have a 0 right. So,
it is the reciprocal property, but never the less we list them out separately.
Yes. This point we did mention earlier, K0 can
is not a residue because no pole. There cannot be a pole at the origin pole at the origin
is not permitted K infinity is a residue. This K infinity is the residue of the pole at infinity.
However these Ki’s are not residue’s they are residue’s of YRC by
s not of YRC all right.
We continue this discussion ZRC YRC K0 by s plus K infinity plus summation Ki. This
is the kind of a Bible for RC impedance or admittance synthesis. If you remember this
everything else follows. That is why I am repeating this in every slide K0 plus K infinity
s plus K is divided by s plus sigma i. Now, since the question of residue came residue’s
here K0 K infinity K they are all residue’s are real.
K is not a residue. Is not a residue, so K0 and ki’s they are
all real and positive. Whereas if you find the residues of YRC not YRC by s you can show
that they are real, but negative. We shall demonstrate this with the help of examples,
but we will not prove it, it will come out to be negative there is no other way. We also
showed by taking the value ZRC on the real axis that is sigma and by differentiating
this with respect to sigma we showed that this has to be strictly negative.
If you do the same thing with respect to this that is you put s equal to sigma and differentiate
you can show that YRC of sigma with respect to sigma, would be strictly positive. It would
be strictly positive and as a corollary to this property that is the slope is strictly
monotonic. It is either strictly negative or strictly positive it follows that poles
and zero’s in either case have to alternate there is no other way.
Poles and zero’s alternate this is a consequence of this slope property to have to alternate.
In the addition, in addition for ZRC the first critical frequency first critical frequency
has to be for ZRC has to be. A pole…
A pole first critical frequency has to be a pole and it follows that for YRC the first
critical frequency must be a 0. Last critical frequency must be a 0 and last critical frequency
here last critical frequency Lcf, this must be a pole all right. If you plot the sketches
of YRC verses sigma it is very easy to show that the first critical frequency must be
a 0 for YRC and it must be a pole for YRC. And it also follows that the infinite frequency
value of ZRC ZRC infinity is less that ZRC 0. And, because admittance is reciprocal YRC
of 0 must be less that YRC of infinity. If the 2 values are constants if ZRC at ZRC
has a pole at the origin; obviously, it is infinity; the value is infinity. If YRC has
a pole at infinity; obviously, it is infinitely large. Now not only that
Yes. So, the value is infinity which you do not known what the value is. Nevertheless
never the if it is a constant, then this constant much bigger than this constant. That is what
it means if it is a pole then we do not bother. Now, not only that, for ZRC we showed that
ZRC of infinity is the minimum value of the real part of Z of j omega, we showed this.
And in an exactly similar manner we can show that YRC of 0 is the minimum value of the
real part of YRC of j omega. Which means that, in the synthesis problem,
you can always remove ZRC infinity without converting the rest of the function or the
remainder function non positive real? Similarly, here from YRC you can always remove the dc
value the rest of the remainder function shall still remain positive real.
And then as for a synthesis concern ZRC YRC the faster 1 form can be synthesized from
the partial fraction expansion of ZRC. Where is faster 2 requires the partial fraction
expansion of YRC by s this has to be remembered. Cover 1 did we discuss cover 1 in the last
class. We concentrate our attention on the pole at infinity or the point at infinity.
Let’s look at cover 1 and cover 2 then we shall illustrate by means of an example.
Cover 1 synthesis I am in repetition even if I even if I did this. ZR, ZRC cannot have
a pole at infinity right RC impedance cannot have a pole of infinity. Recall the form ZRC
is of the form K0 by s plus K infinity plus summation k is divided by s plus sigma i.
This is the form. If you remember this all properties all synthesis are should be obvious.
How can I make that mistake no s all right? Now, cover 1 concentrates on the point at
infinity is at infinity what ZRC can have is either a constant or a pole. Now, this
constant can be removed without affecting the rest of the function. So, I remove K infinity
if I remove this then what will be the behavior of ZRC at infinity it will be a 0. And therefore,
YRC shall have a pole at infinity which I can remove as a capacitance all right. If
I remove that as a capacitance then I take the reciprocal of that function.
That will again be a constant and which I can remove and I can proceed this way till
the function is exhausted that is simple cover one. Cover 1 simply focuses on the point at
infinity. Now, the starting point is important do you start from RC impedance or RC admittance.
You see if you remove the infinite frequency value of RC admittance then you have made
a great mistake. Because you cannot remove them from RC admittance what you can remove
is YRC of 0 not infinite frequency values. So, 1 must remember that if you want to develop
in cover 1 you can remove 2 things are permitted either the constant value from ZRC the; the
infinite frequency constant value from ZRC or a pole from YRC. These are the 2 things
that have to be done alternately. Which means that, what you actually do is to expand ZRC?
In continued fraction Starting with the highest powers all right starting with the highest
powers. The caution is that you can do this for Lc functions either in impedance or admittance
it does not matter. But in the case of RC you must start from the impedance function
all right. This is cover 1, it would be more obvious when we take an example.
As far as yes… The last part, question is suppose we had
a suppose you invert this YRC? YRC is D by N can I do a continued fraction expansion
starting with the highest powers with YRC no. What would that mean? If I do this what
would that mean? It would mean removing the constant value at infinity from YRC. Which
you cannot do right what we can remove from YRC is it Dc value not D infinite frequency.
Cover 2 focuses the attention on the point at the origin s equal to 0 this is the point
which you look at. Now, at s equal to 0 what is it that we can
remove from an RC remittance function; obviously, it is constant from ZRC or YRC.
YRC YRC therefore, cover 2 must start from YRC.
There is no other way what you do is you look at YRC which is K0 plus K infinity s plus
summation k is divided by s plus sigma i. You can remove from YRC the constant value
K0. And , then if you remove this if you remove K0 then YRC has a 0 at the origin. So, ZRC
shall have a pole at the origin which you can remove as a series capacitor and you proceed like this.
So, what you get is shunt resistors and series capacitor. This is cover 2.
Once again cover 1 has to be started with impedance cover 2 with admittance. And this
can be formalized by making a continued fraction expansion starting with the lowest powers
starting with the lowest power. That you divide P on to Q, but write them in reverse order.
That is the constant term comes first here also the constant terms comes first. What
is the logic? Logic is that we are removing the Dc value. That is the constant divided
by constant value.
Let us take an example to illustrate this and this example we shall continue throughout
today. Our function is s plus 1 s plus 3 we carry this out all through s plus 2 s plus
6. The question is realize this function as a driving point driving point function it
does not see impedance or admittance all right. The question is realizing this function as
a driving point function of a 2 element kind network; obviously, since the first critical,
first is realize ability is it realizable by an RC network.
First critical frequency is is 0 1 2 actually minus 1 minus 2 minus 3 minus 6. So, they do alternate.
The first critical frequency is 0 last is a pole and therefore, if at all
realize it should be an YRC; not ZRC because for ZRC the first critical frequency must
be a pole last critical frequency must be a 0. And therefore, for faster 1 you have
to take the impedance function Z of s which is equal to s plus 2 s plus 6 s plus 1 s plus
3 and expand in partial fraction. 1 has to be careful; does the function have a pole
at the origin? No, it does not have a pole at the origin.
So, have to take the infinite frequency value K infinities. So, general form has to be K0
by s, if it is impedance plus K infinity plus summation Ki by s plus sigma i. What is K infinity?
No, I it is 1. S square and s square that is 1 plus terms of this form we have
2 of them s plus 1 s plus 3 this is K1 and K2. And I told you the trick to find K1 is
multiply this by s plus 1 and put s equal to minus 1.
So, we get s plus 2 s plus 6 divided by s plus 3 under the condition of s equal to minus
1 and you can do this very quickly 1 5 2; therefore, 5 by 2 2.5. We will not use the
decimal fractions till we can because decimal might require a truncation which you do not
want to do. We will continue fractions, rational numbers as long as possible; if it becomes
very large then you go ahead. Did; what is the danger? Suppose, you truncate were what
the harm is, the harm is that in continued fraction things will not cancel.
If you truncate somewhere in continued fraction things will not cancel. Because in faster
1 is an end by itself you do not have to do anything further. But in cover 1 once you
cover 1 or cover 2 once you do truncate your interest. So, do not truncate.
Similarly, you can find K2 is equal to if I am not mistaken 3 by 2. Therefore, my Z
of s is 1 plus 5 by 2 divided by s plus 1 plus 3 by 2 divide by s plus 3. Now, do not
try to place mart here because 5 by 2 may become 2 by 5 in the element value and things
like that. So, have the patience to write this as 1 over 2 s by 5 plus 2 by 5 then element
value shall be obvious. Similarly, here you write 2 s by 3 plus s
here 2 then you draw the network K1 ohm and then you have s a parallel RC. Another parallel
RC corresponding to this term and that goes back this is my faster 1. The capacitor here
would be 2 by 5 since this is impedance the reciprocal would be admittance and the resistor
here would be 5 by 2 not 2 by five. This is cs plus gs sc plus g similarly this will be
2 third and this would be Half…
Half that is the correct value all right this is faster 1.
Let us look at faster 2. The function is YRC is equal to s plus 1 s plus 3 s plus 2 s plus
6 and in faster 2 we have to divide by s. So, we divide by s and expand this into partial
fraction. The first quantity would be K0 by s K0 here is 3 divided by 12. So, 1 quarter
s 1 quarter divided by s which is 1 by four s is that. K0 should be obvious put s equal
to 0 except for this term then 1 into 3 3 2 into 6 12 3 by 12 in 1 quarter. So, I have
written 1 by 4 s. I could have written 1 quarter divided by
s this is Kz0ero, but I have saved 1 step plus you shall have some let us say M 1 I
used K earlier, but it does not matter. M 1 by s plus 2 plus M 2 by s plus six and you
can find out I will skip this steps M 1 in my case it came to 1 8 and M 2 was 5 8. So,
I can write down YRC is equal to 1 quarter plus now, I multiplied by s 1 by 8 s divided
by s plus 2 plus 5 by 8 divided by s plus 6. And as I said do not try to play smart
write this in terms of realizable quantity that is here it would be 8 agree.
5 by 8. Yes 5 by 8 s multiplies by s thank you. So,
I get 8 as plus 16 divided. So, 8 plus 16 divided by s plus what do I get here 8 by
5 plus 48divided by 5 s agreed.
Accordingly, now I can draw the faster 2 network as we always indicate resistance values. So,
this resistance should be four ohms and then we shall have 2 RC series networks in parallel.
The resistance value would be 8 agreed reciprocal of admittance is impedance here also the resistance
value would be 8 by 5 and then R plus 1 over sc. So, this capacitance will be 1 by 16 and
this would be 5 by 48 and your faster 2 is complete faster 2.
Now, suppose we forgot that we have to divide by s let us see what happens it is worth doing
at least once s plus 1 s plus 3 divided by s plus 2 s plus 6. Suppose I expand this in
partial fraction , then you see if I write this as what are the things that I should
write. I should write the infinite frequency value; that means, 3 divided by 12 which will
be 1, this is 1 plus some K1 by s plus 2 plus K2 by s plus 6.
Let us see what K1 comes out to be it would be s plus 1 s plus 3 divided by s plus 6 under
the condition s equal to minus 2 which becomes negative. Similarly, you can show that K2
is also negative it can be strictly through we do not need to expansion of Y.
Sir what about K0? What about K0 K0 is 1 by 4, but shall we take
K0 out yes we can, you should not necessarily; actually this expansion should not be done.
So, let us not discuss it further, it is like learning spelling 1 does not learn spelling
by giving you a wrong spelling because that gets in printed on your mind. So, let us not
consider this further. You must remember that YRC as sacred time sacred not is to be tied
around by division by s. Otherwise the expansion is not valid.
Let us look at cover 1 now, for cover 1 we have to take the impedance function because
it is RC we have to take the impedance function and write it as a rational function not in
terms of the factors. But in terms of a polynomial and I get s square plus 8 s plus 12 s squared
plus four s plus 3. Is that all right? Now, I start continued fraction expansion s square
plus 8 s plus 12. The logic is that we are removing the infinite frequency constant value
from 0 phases. So, I get 1 s squared plus 4 s plus 3 let
me bring the remainder here 4 s plus 9 that divides s squared plus four s plus 3 do not
forget to indicate the dimensions. This is a Z. Then I get s by 4 s squared plus 9 by
4 s and this an y. So, I bring the remainder here 16 minus nine is 7 by 4 as plus th3ree
divides 4 s plus 9 the number start getting 16
16 by… 7
7 I might have made a mistake. This is impedance so 4 s plus 48 by 7 48 by 7, so 63 minus 15.
15 by 7 I did make a mistake 15 by 7 that divides
7 by 4 s plus 3 So, I get how much?
49 by 60 49 by 60 correct
Yes, is it not? That needed s is needed. This is a y. So, I get 7 by 4 as the remainder is 3.
remainder is three
3 divide 15 by 7 So, 5 by 7 this is a Z impedance.
In other words my network shall be while let me write this continued fraction this is 1
plus 1 over s by 4 plus 1 over 16 by 7 plus 1 over 49 by 60 s plus 1 over 5 by 7. Is it correct?
And the network should now be obvious my network would be 1 then a capacitor of
value 1 quarter then a resistance of value 16 by 7 then a capacitor of value 49 by 60.
Then a capacitor a resistor of value 5 by 7 and that is it. This is Z of s and this
is my cover 1. Have you started this with y of s you would get negative values and that
would show that you are you have to go back, but I hope you would not make a mistake.
If it is a ZRC, you start continued fraction would be highest covers all right. If it is
to be cover 1 no it is other way down. If it is to be cover 1 then you starts continue
fraction of ZRC starting with highest power.
Let us look at cover 2. Cover 2 you have to take Y of s it is RC you have to take Y of
s and you write the rational function in the reverse order. That is you write 3 plus 4
s plus s squared divided by 12 plus 8 s plus s squared. And , then you start continue fraction
because you can remove Y of 0 YRC of 0. Then 3 plus 4 s plus s squared if you make a mistake
if you try the other way round you will immediately be correct it, but you lose time in the process.
The first quotient would be 1 quarter is it correct 3 plus 2 s this should be 2 s plus
s squared 12 plus 8 s plus s squared. So, you get 6 divided by s this is y this is Z
it has to be because this will be s series capacitor 6 by.
What is this? How can we ignore that s square by 4, so there
would be another term here? No, the coefficient of s squared would be.
3 by 4 s square 3 by 4 s square
So, I is this 6 by s where would be 12 plus how much would be this 18 by 4 which is 9
by 9 by 2 s. No s square. So, I get 16 minus so 7 by 2 s plus s squared divides 2 s plus
three quarter s squared. What we are getting here 4 by 7 and this is admittance. So, I
get 2 s plus 4 by 7 s squared the numbers are not too bad the remainder would be.
How much 5 by 28 s square this divides 7 by 2 s plus s square and the remainder is the
quotient is 90 That is good 98 by 5 s
Yes, 7 by 2 s and this is an impedance remainder s squared the last step is very nice 5 by
28 and this is an admittance 5 by 28 s square the remainder is 0.
The network can then be immediately drawn if I write the continued fraction expansion
like this Y equal to 1 quarter plus 1 over 6 by s plus 1 over 4 by seven plus 1 over
98 by 5 s plus 1 over 5 by 28. At least in the first few problems that you workout you
should write this is instructive. So, that a its full prove after a little bit
of experience you may not write the continued factor you may write the element values write
from the continued fraction expansion. So, what we get is our network becomes our network
becomes a resistance of value 4 then a series capacitor of value 1 6 right
Yes sir Then a resistance of value 7 by 4 then a capacitor
of value 5 by 98
f by 98 and finally, a resistance. 28
28 by it is done, we have done cover 1 cover 2; all the 4 canonic forms have now been done.
Now, once we have we know how to how to design and synthesize an RC network it turns out
that its RL networks are very simple to synthesize. Once we know how to do it for RC.
For RL network let us look at the lets go back and add and issue the most general form
the most general form would be this a number of R’s RL networks in parallel. This should
be the most general form for faster 1 configuration. And you see that if you call this R 0 if you
call this L infinity and you call this Li and Ri. Then the expression would be expression
for impedance would be R 0 plus s times L infinity plus summation, the equivalent of
this is Ri sRi Li divided by sLi plus Ri. Which I can write is R 0 plus sL infinity
plus summation if I take Li common from here then I get sRi divided by s plus Ri divided
by Li. So far so good the general form therefore, obviously, is K0. You understand why we write
this is R 0 at DC at DC all the inductors are short for this is the value and DC and
we wrote L infinity because at infinity it is this inductance which dominates this is
an open, but it is shunted by Ri. So, at infinity it would be sL infinity plus
a resistance which is finite and therefore, the value would be infinity. So, the form
is K0 plus K infinity s plus summation k is divided s plus sigma i. Where all K’s are
real and positive and so are sigma i’s. And if you a pause a for a minute you notice
this is exactly the same form as YRC. And therefore, whatever we have said about YRC
is applicable to a Z RL impedance. For example if you want to expand an RL impedance
in faster 1 you have to first divide by s all right. All that we have said about YRC
now applies to ZRL and by the same token YRL therefore, shall be of the same form as ZRC.
Just take the reciprocal of both sides all right. Therefore whatever we have said about
ZRC that is apply to YRL also. For example the example that we took.
Faster 2 no for faster 1 we have to divide there because faster 1 is an expansion of
impedance. And since the impedance is of the form of YRC that is Ki they are not be residue
you have to first divide by s then to it. If a faster 1 is a series connection faster
2 is a parallel connection that is the difference between them. Cover 1 is the point at infinity
cover 2 is the point at the origin. So, for example, if you want to expand an RL impedance
into force into cover 1. Then how would you start the continued fraction
next 1. If you want to expand a ZRL. Can you can you start with ZRL?
No cover 1 I am talking You have to start from YRL.
Why? RL has pole at infinity.
Why RL has pole at infinity or ZRL had the pole at infinity?
So, can we start from with the starting with highest powers?
Why let’s look at this let us look at the same example may be things would be clear
Our example was s. First thus these qualify to be a ZRL can it be can this be a ZRL.
Yes because a first critical frequency is a 0 all right. So, now I want to I want to
get a cover 1 for this. Can I start expansion with the highest power that is the question
I am asking? s square plus 4 s plus 3 s squared plus 6 no 8 s plus 12. Can I go ahead and
start continues fraction expansion? No sir.
No because if I do that I shall be removing the infinite frequency value of ZRL which
is not permitted. Therefore I must take the reciprocal and then start continue fraction
this is the point that. If you do it blindly you will come back let say do not do it again.
Because you will get negative from yes you will get 8 s here so, minus 4 s and so on.
Then you know here made an mistake, you have to come back is say is the point clear. Similarly,
if you want to cover 2, cover 2 can you start it ZRL? Yes you can.
That’s you can start it ZRL. So, in cover 1 and cover 2 you have to be very careful.
Similarly, faster 1 and faster 2 you will also have to be careful which 1 is to be divided
by s. Z no…
ZRL ZRL or Ys, without any further time spending
any further time on RC and RL and we take further problems in the in the problem session.
Can we go to an introduction to RLC? That is if all the three elements are there then
what you do? In general, RLC 1 pole synthesis. In general, it has been found very unfortunately
that the partial fraction expansion and continued fraction expansion which works. So, beautifully
for RC RL and LC network was does not in general work for RLC. And if you are lucky it may
be a combination of partially fractions and continued fractions.
Or may be continued fraction starting with the higher power then switching over to lowest
powers then switching over to highest powers again. But it requires a lot of engine with
you be you cannot you cannot say right by looking at the function that it will work.
In RLC 1 pole synthesis in general partial fractions and continued fractions or any combination
of them may or may not work. And therefore, this remained a problem this remained a problem
for every long time till in 1939 a gentleman; I think have mentioned his name Otto Brune.
a gentleman settled in the United States Otto Brune in his PhD thesis at MIT Massachusetts
Institute of Technology proved that the most general synthesis of RLC 1 port; most general
and canonic synthesis. You see all the synthesis that you have been talking about. So, far
2 element kinds is canonic the number of elements is exactly equal to the number of specs Nothing
more nothing less. You cannot do it less you can do it more of
course, Otto Brune proved that given a let us say a bi quadratic let us say a bi quadratic
like this s square plus a 1 s plus a 0. Take the simplest form, s square plus b 1 s plus
b 0 where a 1 a 0 b 1 b 0 are constants restricted only by the positive real condition. That
is they are all positive and real and they satisfy the real part condition what is the
real part condition. That a 1 b 1 should…
Greater than or equal to square root a 0 minus square root b 0 whole square. This is all
that it is satisfy nothing else otherwise is completely general. Brune proved that is
you require 5 elements, Brune proved that if you require a canonic synthesis you cannot
do it without a transformer. You cannot do it without a transformer that is a transformer
less canonic synthesis is not possible. Let me state this is beside this is beyond
the scope of this course. But I think you should know this you should know the beautiful
elegant history that is behind 1 pole synthesis. Transformer less canonic synthesis of of RLC
1 port is impossibility. Now, in between the in between the advent of telephone where filters
were required end 1939 there have been lot of attempts at doing this. And there have
been a PhD thesis written showing in possibility of synthesis at all.
That such things under possible you cannot synthesis in RLC 1 pole unless you are lucky
unless a very special case occur. Therefore since 1939 since 1939 lot of attention gone
to prove to prove or disprove Brune a more to disprove Brune, but none succeeded. And
therefore, the next step for to us when do not insist on canonic. It requires 5 may be
we are we are willing to give see a transformer is not a very favored element.
A transformer has lot of problem shielding coat it becomes heavy it becomes bulky it
picks ups things magnetic induction and thinks likes an transformer. Where, you cannot do
it out a transformer you have power lines for example: transmission distribution Then
in matching for audio frequencies and so, but in general if you can avoid a transformer
nothing like it. And this is why in power amplifiers is complimentary transistor that
was invented and it was a boom because you could produce a power amplifier without using
any transformer. And it is very small size and without any
of though shielding and other problems. So, people…
Biquadrate is the simplest form you see bilinear we know bilinear s plus alpha by s plus beta
it would be either in RC or in RL. So, biquadrate is the first complication that arises and
people concentrated on bi quadratic, because if you know how to design a bi quadratic you
can possibly find out how to design others. So, since 1939 people had been saying all
right let us let us leave of this first or insistence on canonicity. I am willing to
you 6 can you give a transformer like synthesis.
While unfortunately this question remains I am answered till 1948 when a 2 mathematicians
the names are: Fialco no I beg your pardon Bott and Duffin; Duffins excuse me A Bott
and R J Duffin 2 mathematicians. I do not remember where they are working, but Duffin
is a more than hundred is still alive and he has a special office at the University
no at Carnegie Mellon at Pittsburgh yes Carnegie Mellon University. He has an office and Professor
for life there till he lives. The beauty of this they were both of them
are mathematicians they were told by some of their electrical engineering friends that
such a problems exists. And they looked at it and in 1948 they published a half a column paper.
Half a column means, if you take a quarter size paper a well of half of this
of this size that much print with 1 figure in it. They prove that if you keep the canonicity
then transformer less synthesis is possible. Unfortunately, the number of elements that
have to be used was 7 not 6, 7 and till to date there hasn’t been any improvement on this.
So, transformer less canonic synthesis still
remains a mysterious problem whereas, practical networks are possible. You can built network
you can analyze you can built network without transform, but given s is given a a function
it is very difficult to do it. The Bott and Duffin procedure then after 1948
once again was subject on intensive investigation, because did the demand for filters started
going up with an expansion of communication. The demand for filters grow up and there are
many attempts and under very special circumstance 1 could reduce the number to 6, but not in general.
It still remains a mysterious problem. We will continue this discussion in the next
class at 12.