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okay so we come to this final lecture on microwave
filter so here will see some tutorial problems
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we have seen a long journey we have started
from the image impedance concept and how based
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on image impedance the low frequency filters
were designed then we have seen that if insertion
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loss based filter design in image impedances
already you are we have introduced the concept
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to high light the importance but those problems
generally you can solve yourself also in the
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tutorial sheets will give some problems but
let us see some insertion loss based design
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and suppose if we want to have let us say
maximally flat . a maximally flat
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low pass filter is to be designed for cutoff
frequency 8 gigahertz minimum attenuation
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of twenty db at eleven gigahertz so how many
elements are required elements means how many
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inductance capacitance etc whichever elements
how many required so what you said i require
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that minimum attenuation should be 20 db and
at what frequency 11 gigahertz what is your
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your cutoff frequency is given as 8 gigahertz
it is a maximally flat filter a low ass filter
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so what i am saying is this is my cutoff this
is my plr this is my omega c but i am specifying
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at certain other omega let us say 1 am specifying
that it should be at least this. so you should
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achieve this is omega. so let us see that
what is said that what is omega c is 2 pie
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into 8 into 10 to the power 10 hertz sorry
8 gigahertz so 10 to the power 9 gigahertz
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and what we want at omega = omega 1 which
is equal to 2 pie into 11 into 10 to the power
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of 9 hertz, not hertz omega cannot be hertz.
omega is radiant per second, this is also
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radiant per second i required that plr should
be 20 db basically this omega 1 by omega c
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that ratio is nothing but 11 by r we know
that for butterworth type of polynomial this
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is a maximally flat means it is actually butterworth
function we can use to synthesis so for omega
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greater than omega c we know that plr sorry
plr can be approximated as omega by omega
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c whole to the power 2n that means one time
i can neglect. so, if i put it in db plr db
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that will be 20 in log omega by omega 1 by
omega c so 11 by 8 so you solve for this it
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is given as 20 so 20 is equal put it as 20
so that gives you any 7.23 now n needs to
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be integer now obviously it says atleast 20
db so if i take n is equal to 7 it wont be
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atleast 20 db. so i know that since it is
increasing here so since atleast 20 db required
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i should take n should be equal to 8 so there
should be 8 order here, this is a simple one
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but you see this approximation of the function
at higher value that will help us to find
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out what is the order of the filter . next
one let us say that again another problem
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that specification as maximally flat filter
low pass filter omega c 2 gigahertz load impedance
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50 ohm and at at least 15 db insertion loss
at 3 gigahertz you see cut off frequency to
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gigahertz far away i want may be at 3 gigahertz
there is some interfering signa or into another
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signal source is there so there i want to
put it at least you should have a high insertion
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loss of 15 db there now again the first thing
is like before also this is first we can find
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the order n by previous method that plr should
be 1 + omega by omega c whole to the power
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2, so that is 10 to the power 15 db so solving
that n we are getting 4.22 so we will take
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n to be greater than equal to 5 now let us
take 5 and from 5 you find the prototype from
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butterworth filter see the table for n is
equal to 5 prototype values will be g = 0.618,
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g2 = 1.618, g3 is from the table directly
you need not when that this will be given
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in a problem got or as an engineer you can
always refer a tables and find out what are
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the values g6 = 1.0 now you scale up you know
that from these values the first value so
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i can draw that this will be the circuit i
have the source then the source impedance
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that has been said to be 50 ohm this is a
butterworth filter so in our prototype assumed
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as 1 that means g0 is always 1 but here it
is 50 so you need to scale with 50 at appropriate
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time that means our unknown value multiplier
will be 50 and then we know the first element
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we can choose as l1so let me call l1 since
it will be scaled both in impedance in frequency
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type is same so no type change so i am calling
it l1 dash then this will be c2dhased this
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will be ll3 dashed this will be c4 dashed
then this will be l5 dashed then i need to
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stop and the load impedance will be again
rl = 50 ohm. so what will be my value for
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l1 dashed if you see your notes this will
l1 r0 omega c and what is l1? l1 is nothing
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but your g1 this is 0.618. so 0.618 into 50
by omega c which is 2 pie into 2 into 10 to
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the power 10. so this will give you a value
of 2.46 nano henry then c dashed will be c2
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by r0 omega c now c2 you can see that it will
be 1.618 from the table 1.618 by 50 into 2
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pie into 2 into 10 to the power 9 = 2.575
picofarad similarly i think now you can go
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on you own l3 dashed. you can get the value
7.958 nano henry c4 dashed will be 2.575 picofarad
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and l5 dashed is equal to 2.459 nano henry
and so this is the example of a prototype
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filter design but we also scaled up prototype
was first from the table we could have found
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the prototype but we have scaled up in impedance
and frequency . now we will see a type change,
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type of thing so our third problem is this
that design a band pass filter having a .5
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db = repel response it is given that you take
n = 3 the center frequency that means our
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w0 is 1 gigahertz bandwidth percentage bandwidth
is 10 percent and impedance level is 50 ohm
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so we can you see n = 3 said so basically
we will have how many the 4 things from table
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g0 always this one. so what value this okay
so i can take the circuit is like this first
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element if i take l1 dash second because this
is band pass so i know any l1 is basically
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l1 c1 dashed then l2 is l2 dashed c2 dashed
then l3 test c3 dashed and that 3 = 3 finally
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it will be terminated by 50 ohm now where
from my get all this l1c1 values from the
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it is equal repel so chebyshav so i see chebyshav
is equal to 3 i get g1 = 1.5963 that is l1
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l1 of the prototype g2 that is 1.0967 = c2,
g3 = 1.5963 again from the table l3 and g4
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remember it is chebyshav with n0 so no problem
g4 will be 1 and that is nothing but your
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rl. now you can easily find out l1 dash c1
dash from l1 what is that accha what is this
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that delta value, delta they say 10 percent
that means 0.1. so what is my l1 dashed l1
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dashed is l1 z0 by omega 0 delta so it is
l1 i have seen 1.5963 1.5963 into fifty divided
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by 2 pie 10 the power 9 into .1 that comes
127 nano henry. lc1 dashed again it will depend
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on l1 the formula is delta by omega 0 l1 z0
so that you can put and it will be 0.199picofarad
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l2 dashed again is delta z0 by omega zc2 and
that will be 0.726 nano henry and c2 dashed
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will be c2 by delta omega 0 z0 and that will
be 34.91 picofarad similarly l3 dashed it
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will be same as this 127 nanohenry and c3
dash will be 01.99 picofarad. so if you can
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plot his you can with this you can plot this
attenuation obviously that require some computer
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programming this attenuation curve goes like
this, this is 0.75 this is in gigahertz frequency
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0.75, 1.0, 1.25 roughly the curve is some
ripples when it comes to 1.25 and generally
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this ripple is around 0 db approximately.
so this type of graph so this is a band pass
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to see now let us see a another problem that
suppose that was i think four let us say ah
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3 I think band pass filter design we have
seen now implementation wise you see that
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design let us say a low pass filter cut off
frequency 4 gigahertz third order equal ripple
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that means chebyshav with n = 3, then impedance
is 50 om and pass band ripple is 3 db so easily
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third order so we can find out low pass prototype
so here from table we can see from chebyshav
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third order again you see the final one rl
should be 1 so g1 is 3.3487 and that we can
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call l1. first one will take l1 series element
then then g2 is 0.7117 let us call that c2,
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g3 from the table is 3.3487 that we call l3
and g4 again i said chebyshav with n card
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third order so this is 1 and 0 this is rl
let me draw the circuit but this is one prototype
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this is my l1 this is my c2 and this is my
l3 then i have this as 1 good this is prototype
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lpf prototype now from this these are all
lumped elements i know the values but i need
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to do richard transformation so i need to
put those richard transformation value. if
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you do that then richard transformation . will
look like this if you are known that this
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first one is l1 so that is a series stub i
will have a series stub, this is a series
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stub and that is l1 so i have to short it
so what will be the value you see that time
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we said that this l will be the characteristic
impedance the length of the stub will be unit
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length lambda by 8 so that we are calling
that this characteristic impedance let me
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call like this is that this is z0 is equal
to 3.3487 and this length is l then we will
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go the c2 will have to put c2 is shunt so
will have to put a shunt stub with open circuit
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and this l again will be that same lambda
by 8 at the cut of frequency that means lambda
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by 8 at 4 gigahertz so what is the value of
l, is 4 gigahertz lambda = 30 by 4 centimeter
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that is 7.5 centimeter so l = lambda by 8
is 0.9275 probably if am correct centimeter
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or you can say 9.275 millimeter so all these
lengths are this and what is the characteristic
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impedance of this c2 that this is g2c2 value
is 0.7117 will have to take inverse of that
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so that will become z0 is 1.045 after c2 i
have a series stub so will have a series stub
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and then will have a short head it is a impedance
inductance so and then from there this is
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a resistance 1 ohm so what is this value again
let me say that this is l and z0 is 3.348
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so now this is this is after richard transformation
now if it is implemented okay but if we think
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that okay why do not have uniform type of
shunt things because making shunt stubs are
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easier than making series stubs so if we want
to do that you see that series stub was where
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in koroda’s identity . yes so see kuroda’s
identity so series stub i want to make shunt
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stub so kuroda’s second identity i will
use so using kuroda’s second identity so
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let me write applying kuroda’s second identity
this new circuit becomes very systematic procedure
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this thing so instead of this i will make
a shunt stub then again already there was
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this one so then so what we have done that
here we have put because after the stub if
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you can see the second identity that you put
a unit element and then a but your stub so
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that means sorry here actually . let me redraw
because the element should be drawn first
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so first will have to make one unit element
and then that element will having anything
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then we will have our original l1 then again
we will put another unit element and then
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will push that so you see this is 50 om these
are now i will say l length all are l lengths
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l is lambda by 8 so there are three search
and this is also length l unit length this
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is also unit length l so all lengths are l
only they differ by their characteristic impedance
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so that first one z0 you see that what happens
to n square. n square is 1 + z2 by z1 and
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you see our z1 our z1 is 3.3487 so 1 + z2
is 1 and z3.3487 that gives the ratio as 1.299
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so with this the first one will have a characteristic
impedance of 1.299 then this one if you that
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the z0 for is unit element that will be with
that ratio 4.35 then this one z0 = 1.405 and
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this one is z0 =4.35 and this one is z0 = 1.299
this is thus this is the original capacitor
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so that you see that is why this is difference
but these towards same now you can have the
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this instead of this one ohm i have impedance
level as 50 ohm so 50 ha this was 1 ohm, now
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i need to scale upto 50 ohm so all these values
i will scale up and i will write this is 50
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ohm then we will have this stub length again
l and z0 is this 64.9 ohm just scale up by
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50 into this then i will have
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again this is of length l a length l and this
is z0 = 70.3 ohm a.4 into 50 ohm then you
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go another accha here there was one l and
this l has a z0 of 217.435 so 217 by 5 ohm
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then here again oh this is l and again the
z0 z0 is 7.3 acha acha l will be from here
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but here to here there will be another line
and there the length is l z0 = 217.5 om another
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unit cell then we have that the from here
another stub so that again is length l oh
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sorry this length l and this stub will have
z0 = 64.9om and then we will have this 50
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om line okay so you see i have now see these
stub and also per unit cell 2 unit cells stub
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but they are not stub they are transmission
lines these towards these three are three
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open circuited stub this is for the first
inductance this is for the capacitance this
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is for the inductance l3 and this are unit
cells so now with length sizes are manageable
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i have separated them now these unit cells
are called redundant sections they are matched
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so that is why they are because they are impedances
are both sides they are matched to one that
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is why they do not disturb the operation of
the filter but what they do they become separates
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the two so that two stub elements there is
enough separation between them so that any
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in the discontinuity any high even as a modes
that gives generated that can be tackle there
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particularly at higher level frequencies haaah
if any in discontinuity those points there
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is any other modes generated that comes because
this we are implementing by transmission lines
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but in other microwave filters we need to
do it by web guides etc there that type of
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problem may come so this completed the microwave
filter design hope from the very basic point
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of filters just from rc filter we have started
then we have seen need to have lc filters
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for radio frequencies because rc filters are
lossy so went into lossless lc filters there
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we have seen how to design from image impedance
comcept then we came to the concept of characteristic
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impedance we said that out network will be
symmetric will have instead of two impedance
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many impedances one characteristic impedance
that characteristic impedance will try to
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control but we saw that there are lot of problems
of impedance matching etc that can be tackled
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but another viewpoint is if we tackle the
insertion loss if we specify the insertion
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loss of the whole filter then we can synthesis
any desired haaa insertion loss characteristics
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and by that we have seen first we need to
do a low pass prototype filter design and
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once we have low pass filter design we can
scale it up either in impedance or in frequency
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or in type we can change the type and we can
have any desired baah type of filters with
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any desire it then we have seen that if we
want to implement it as ha high frequency
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of microwave type of filter then we need to
implement not to lumped the circuit elements
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but distributed circuit elements so how to
do that we have seen thru richards transformation
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you can do that distributed hmmm transmission
line implementation and then to have some
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realizable gaps or to convert one stub to
another how to do that so for that we have
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introduced 4 kurodo’s identity any identity
you need to choose for you particular application
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and you apply the method is straightforward
it can also be implemented on a computer and
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you can do that whole things in you think
that type of simulators etc are available
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you can use that but once you know the basic
principle design then you use that then it
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becomes meaningful otherwise you won’t be
understanding what is happening what is the
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principle microwave principles formula this
is there so we hope that with this you have
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got a thorough knowledge of microwave filter
both it is design and implementation also
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you will be able to analyze any filter design
from this view point after this we will switch
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over to another very important microwave circuit
thing that is amplifier it will be an active
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based device design this is filter was passive
but microwave amplifiers there will active
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device base we will see that in the rest of
this lecture series. thank you