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Welcome to this lecture on filter design micro
filter design by insertion loss method Now
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as I said that we can specify there are some
restrictions on specifying the insertion loss
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now once that is done and we chosen some way
to represent that then we need to know how
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to design and implement the filter . So I
am now giving you the whole process flow that
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first from the filter specification you need
to understand that what will be the PLR or
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insertion loss requirement what type of filter
I will do etc etc that we have already seen
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once we have that we do it this is an important
step so this will understand that whatever
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may be the filter filter may be band pass
filter may high pass filter may be band stop
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filter may be low pass we always first do
a low pass prototype design So the first step
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is important one low pass prototype
this is unlike the image parameter method
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Here whatever may be the thing we have decided
about what will be the insertion loss which
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with polynomial will do that means what type
of filter I will use is and the pass band
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stop sand which will characterize many times
it is not low pass but we always start with
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low pass prototype design that is why we are
calling it prototype Once we have done that
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then we will do ome operations called scaling
and conversion to make to the specification
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that the specifier user as given So this is
another important step in this microwave filter
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design and finally we will see how to implement
it implementation of the that means on a actual
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microwave circuitry how we implement So filter
design by insertion loss method mainly have
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these four important blocks in any design
you have this at from specification you start
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to decide something then here you do first
a loss pass prototype then you have some scaling
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and conversion operations and then finally
you think how to implement it if required
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will for implementation and we need to some
minimization is required about the length
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etc that will see So we this low pass prototype
design here in this part already we have decided
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about type of filter that means whether it
is butterworth or a chebyshav or elliptic
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or the linear phase that we have decided so
we are going here Now first we start with
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butterworth low pass prototype now please
see that here we have a source then we have
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a source assistance that we assume to be always
one And then we have
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and LC circuit and then we terminate it with
a resistance source Source resistance is one
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this is one ohm so we can say that source
we have assumed in the prototype always source
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impedance as one later we will see that we
can go to any actual source impedance but
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in prototype we have this is L this is a C
in the series . . in the shunt term I put
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C and this Now this is a single section if
required you have multiple sections etc So
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source impedance is one Ohm and also in prototype
design the cut off frequency in low pass is
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taken as one hertz so omega C is one hertz
So our and you see we have design L and C
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so that we get a Low pass circuit with this
cut off frequency and you see the number of
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elements N is equal to 2 So that will determine
the PLR what was the PLR expression for the
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maximally flat you see for maximally flat
if that it was 1 plus K square omega by omega
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C to the power N omega C to the power 2N So
here this PLR will take as since we are taking
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N is equal to 2 and omega is already one so
one plus omega to the power 4 So PLR we have
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specified and it is realizable because it
is an even function of Omega as ah hmm not
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only that function of omega square which is
also required that even powers of omega So
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it is omega 4 so it is satisfies I can be
realized now let us look at what is the input
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impedance I am looking from the source So
I can write Zin will J omega L well then parallel
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combination of C and R once I have Zin what
is the characteristic impedance immediately
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ii What is the reflection Coefficient I can
immediately write that reflection coefficient
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will be Zin minus1 by Zin plus 1 and once
I have this I can find out PLR it is 1 by
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1 minus 4 square that is 1 by 1 minus Zin
minus 1 if you simplify you get Zin plus 1
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whole square by 2 Zin plus Z in star now any
complex Z is a complex Zin is a complex quantity
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Zin plus Zin star will be twice real so we
know that
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we already have the expression for Zin so
if we put Zin plus Zin plus you see only the
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real term is this one because this is imaginary
this is also imaginary So we can immediately
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write this will be 2R by 1 plus plus omega
square R square C square and another term
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is this Zin plus magnitude square that immediately
we can write that
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so now once I have this I have this PLR expression
. I can immediately find out what is PLR PLR
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turns out to be 1 plus omega Square R square
C square by 4R R by 1 plus omega square R
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square C square plus 1 whole square plus omega
L – minus omega CR square by 1 plus omega
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square C plus plus whole square to the power
4 So this is the PLR expression from the circuit
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from my this circuit I have found the PLR
Now this we can little bit simplify and see
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that this I can write as because ultimately
my actual PLR the butterworth PLR that is
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1 plus omega 4 So let me write it in terms
of omega square omega 4 and omega independent
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term etc If we do that you will get like this
so you compare this with our PLR with our
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specified plr which is 1 plus omega 4 this
is butterworth this is the thing I got so
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now by comparing I can say that what is immediately
is observable is what is the value of R because
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this is omega square this is omega 4 so 1
is equal to this that means we should be this
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term 1 by 4 4R 1minusR whole square that should
go to zero if we compare this two so r becomes
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1 ok Then if I omega square terms I see that
his also will go to zero because in our butterworth
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thing we do not have this omega square here
we have that this should go to zero so I should
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get r square c square plus L square – 2
lcr square is equal to 0 and by comparing
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omega 4 I can see that l square c square r
square should be 1 so these are the three
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equations are ah r or this is r already solved
so I can put this values so I have two equations
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I need to solve LC that I can do and if you
do that you get C is L is root 2 so I have
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solved because see in my assumed circuit there
are unknowns LCNR and I have found that R
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is equal to now 1 C is equal to root 2 and
L is equal to root 2 So this way you can do
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you can see that always this part is true
I have taken this this is N is equal to 2
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Suppose I have N is equal to 4 If I specify
that N is equal to 4 so that time will have
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like this is 1 ohm then I have I have another
LC N is equal to 4 then this side again I
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will have this R this is L this is C I will
find out what is Zin from Zin I will find
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out what is gamma in from gamma in what is
PLR So I wil have PLR as a function of omega
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square here and I know that if I have N is
equal to 4 what will be the pLR specification
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from the butter worth so that I know it will
be 1 plus omega by omega C whole to the power
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r 2 N So depending on N here is equal to 4
means it will 1 plus omega ssh 8 omega C is
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1 So it is 1 plus omega 8 so you will have
this expressions here you compare so you will
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see that the non omega term the DC term that
should be 1 and omega square omega 4 omega
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to the power 6 will be 0 and the components
and the coefficient of omega 8 term that should
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be 1 so by that you have solve But obviously
you see this is becomes a bit cumbersome the
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process is valid but it becomes very cumbersome
if I go for higher and higher order So for
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the large N the method is nor very practical
but what people have done people have tabulated
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these values that people have solve it because
polynomial is known that is known so prototype
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you see that we know all the source impedance
1 ohm omega C 1 hertz so I can solve for this
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So that table it is tabulated now that tabulation
need a bit understanding that is why let (refer
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tim. us see that people have tabulated it
in micro books you will see that in . . book
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it is given for butterworth filters I have
and they have given the valure suppose if
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I have only one element if I have 2 element
if that tables are available now you see that
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when I view it was LCR but here are G1 G2
G3 etc So how to interpret that so that concept
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is this that this is called ladder prototype
So you have seen in any synthesis problem
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you have this type of prototype because this
is your reference with reference to this you
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design So you know first element is like this
then
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in this previous case how I designed the chart
teekay yes so you see that what I designed
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that is called as ladder prototype now there
are 2 varieties You see this in the first
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one I can have a series inductance or I have
a shunt capacitance now if I have series induction
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this is called as ladder prototype beginning
with a series element and in this case you
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see the nomenclature is this I know that this
is actually the resistance but they call it
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to have this table general this is your first
element G1ah sorry this is not first one because
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in prototype always this value is known R0
is equal to G0 is equal to 1 so this they
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call it G0 then this one they call it G2 So
basically I can say G2 will be your if that
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G0 so this G1 this is L1 then this one they
should call G2 this id G3 this G4 and the
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last one will be how much because we are starting
here there N capital N order so this is an
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extra one so GN plus 1 so this is ladder prototype
beginning with the series element so you see
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that this is your L1 now this will be worth
this you can call as C2 this is your L3 this
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is your C4 like that There can be another
variety (refer time . that ladder prototype
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beginning with a shunt element so in that
case the circuit will be like this
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so here we see that R0 is equal to G0 is equal
to 1 then this will be your G1 this will be
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G2 this will be G3 this will be G4 this will
be G5 this will be GN plus 1 So we can now
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summarize that so in both these cases what
we get is Gn or G0 . . for what it is the
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generator resistance for first type so let
me call that this first type and this one
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let us call that means second type So please
not that when we have series element type
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we call it second type when we have shunt
element we call it first type So with reference
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to that you can write that so in the first
type when I have shunt here what is this circuit
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this ah so I need the change this circuit
because here generally they in the second
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type they do like this that instead of a voltage
source they represent a current source here
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so that means here we will have G0 is equal
to G0 is equal to 1 So in the second type
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this is a conductance so we say that generator
conductance in the second type then what is
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GN plus 1 this is in the first type it is
load resistance if GN is a shunt capacitance
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here load conductance if GN is a series capacitance
please understand that here it is ladder network
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so this GN plus 1 what it is it depends on
depending on the number if I have finished
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with a previous one GN if it was a shunt capacitance
then this will become a load resistance if
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it is a series capacitance shunt capacitance
and series sorry this is not series capacitance
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series inductance If this is a series inductance
last one then this will be conductance ok
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And in general these are the two N values
what is in between let us call that GK where
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is K is varying from 1 to N so this is inductance
for series inductors and capacitance for shunt
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capacitors So the moment that has been done
you see this generalization now looks meaning
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they are specified various values as will
take now it is up to us which design we are
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following accordingly we will choose whether
it is an inductance if it is inductance this
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is capacitance this is induct if this is capacitance
this is inductance capacitance like that it
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alternate So will see suppose haaaaah you
can have a design that a maximally flat low
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pass filter is to be designed with a cut off
frequency of 8 gigahertz and a minimum attenuation
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of 20 db at eleven gigahertz you know that
means specifying that what is your pass band
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how much and in stop band how much is the
thing how many elements are required So you
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can go on do that we will have a tutorial
there will see this problem Similarly you
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can have chebyshav filter design prototype
filter . so let me just briefly say this at
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chebyshav low pass
prototype so again omega C is 1 only thing
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is the PLR is now 1 plus K square TN swuare
omega now TN this chebyshav polynomial they
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have a property that TN0 is equal to 0 IF
N odd and plus minus N even so accordingly
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your PLR will be one at omega is equal to
zero for N odd you see because N odd TN is
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0 so PLR becomes 1 and this will N even it
will be 1 plus k square at omega is equal
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to 0 for N even So it is up to you that if
you choose the N value because N values will
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come from all those specifications that in
stop band I want this much attenuation in
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pass band I want these from that N will come
the moment N comes you can choose what is
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the PLR values from there So for the N is
equal to 2 prototypes you can now find out
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that suppose if we choose N is equal to 2
let us say so for N is equal to 2 let us say
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that T2X the chebyshav polynomial second order
that is 2xs square – 1 so PLR will be immediately
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1 plus you see it is N is equal to2 so N even
so TN is plus minus 1 and it is 1 plus K square
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You see N even 1 plus K square 2 omega square
– 1 whole square So that is 1 plus k square
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4 omega 4 – 4 omega square plus 1 So it
already we have seen that own change these
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our basic this if we have assume this design
this ah ah ah this one this the same design
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so from there we have chosen that Zin etc
so that expression finally we have derived
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what is the PLR ha this expression PLR this
is for that design So this is one side and
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then for the Chebyshav we have this so you
can equate that means I am writing that one
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plus this one this equal to I can write that
one plus already that time we have derived
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this 1 – R whole square plus R square C
square plus L square – 2 LC R square omega
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square plus L square C square R square omega
4 this already we have done so now we can
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compare and from that you can find out various
values .So at you know at omega is equal to
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0 k square is equal to 1 minus R whole square
by 4R so from here you can solve for R as
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a function of K square for K So K is what
this is an N even K is what K is a ripple
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level so that will be specified that I want
ripple not to cross this once I know that
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I have K value so I can find out R similarly
can do So these things can be done it hand
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so may be in exams etc will tell you to do
for small values but if larger once we do
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N is equal to 3 N is equal to 4 N is equal
to N will give you the tabulated values the
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element G1 G2 G3 values will get ok But you
understand so that you can find out from the
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table which G1 to apply what is your ladder
prototype etc So now you see that if you specify
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some K value here then R is not 1 as was the
case in butterworth in the chebyshav R is
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not 1 so R as some value that means you see
from the basic design there is a problem that
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ah ah where is that yes so I have taken these
as 1 ohm source impedance is 1 ohm But your
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this r that is not coming as 1 ohm it as some
value so there will be some reflections here
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also depending on that R value Zin may not
be 1 so there may be reflections here So filters
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generally not matched you have faced the hmmmm
music in image parameter base method here
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also there will be mismatch But I again refer
you to the ninth lecture of basic tools of
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microwave engineering where we have learnt
how to tackle impedance matching problem So
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you can always design with this filter quarter
wave transformer or you can also add an extra
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additional filter element Suppose you are
having any even you see this problem comes
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in N even but it N is odd then this is not
there R is equal to 1 that we have already
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seen So you can also do that because for odd
N R is 1 for chebyshav so you try to make
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N odd there Sooooo so you can design for various
ripple values these tabulations are available
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you should know you should not memorize or
think this But if you understand this simple
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concept by which you can design this way for
other filter types also the tables are available
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or you can make your own table once by programming
etc you can first make and then you can go
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on designing whenever the need arises So this
makes the whole thing step forward that whatever
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may be the filter you first design the prototype
Now we have not covered that actual filter
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wont be like that it will be either high pass
or it will be band pass or stop band it also
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its cut off frequency is not always 1 hertz
it will have some other values So source impedance
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also need not be 1 ohm as we have assumed
in the prototype it can be any 50 ohm 100
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ohm those are typical values so now from this
design once we have got this prototype low
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pass filter We will learn in next lecture
how to scale up that and how to get the actual
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filter thing that is called filter transformation
that will take up in the next class