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welcome to the forth lecture . we have already
seen that the z1 by port z2 an important ratio
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z1 is the impedance of the series elements
or a z1 by 2 the series elements and z2 is
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shunt elements that impedance that ratio in
the pass band it should be in this range and
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in stop band it should be in this range you
know impedance is always a reactive element
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is a function of frequency so in pass band
we need to enforce this stop band we need
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to enforce this . so this boils down to this
if i plot z1 by 4z2 i have to point 0 and
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1 so any z1 by 4 z2 greater than 0 is a stop
band so this entire thing up to infinity stop
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band so let me write this is stop also z1
by 4 z2 greater than minus1 is stop band so
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this is also stop band so i have only this
zone this is my pass band okay so now we got
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a filter we know that what i how i will have
to choose the z1 and z2 so i get pass band
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but another thing remains that we have found
out that in pass band alpha is 0 and in stop
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band alpha is not 0 but what is this value
that we have not found so we will have to
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find the value of alpha in the stop band also
we will find what is the value of beta in
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the stop band because beta is phase constant
so as the wave moves per unit length how much
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phase changes that also is a curiosity and
in pass band we have seen that alpha is equal
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to 0 but what happens to beta here how beta
changes in pass band so that we have already
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seen because everything is embodied here ah
sorry tan hyperbolic square gamma or let me
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start in a new . tan hyperbolic gamma square
is z1 square plus 4z1 z2 by z1 plus 2z2 square
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we have already seen this number i think this
so this already . we have derived this that
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tan gamma is here now we will try to see that
under those stop band pass band conditions
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that means under that z1 by 4 z2 that range
what happens to alpha what happens to beta
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both in stop band and pass band so this is
the number now let us do a bit manipulation
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and from this we can find out that cos gamma
these are all manipulation so that the thing
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become a bit simplified cos gamma from 10
square i can always find cos gamma that will
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be 1 plus z1 by 2 z2 but we want to express
in terms of zi 1 by 4 z2 because our condition
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is this condition is given by 4 z2 say bit
more manipulations and we get that sin hyperbolic
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gamma by 2 is this number z1 by 4 z2 so from
here i came here from here again here so now
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i neither wholly gamma . so here you put this
sin hyperbolic gamma means alpha plus j beta
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by 2 is equal to root over z1 by 4 z2 let
me explain now sin hyperbolic alpha by 2 cos
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beta by 2 plus j cos hyperbolic alpha by 2
sin beta by 2 is good now i now designate
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the two bands and because i have 2 stop bands
every time instead of saying that let me call
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this is my stop band 1 and this is my stop
band 2 i have only one pass band so no problem
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so i have got this let me see one by one stop
band one in stop band 1 what is the definition
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of stop band one what is stop band 1 z1 by
4z2 is greater than 0 so z1 by 4z2 is greater
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than o and real so this is greater than 0
and real so can i say that if this is real
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this is equal to this and this is 0 so my
conditions are that this implies that sin
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hyperbolic alpha by 2 cos beta by 2 is equal
to root over z1 by 4 z2 this is one and cos
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hyperbolic alpha by 2 sin beta by 2 is equal
to 0 this is 2 so these two conditions now
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00:07:29,840 --> 00:07:40,979
from two if this is there we know that cos
hyperbolic alpha cannot be 0 so sin beta by
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2 should be 0 and what is sin i beta by 2
0 i know that beta then becomes beta by 2
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i know that beta then becomes 2n pi where
n is an integer and from so beta value i know
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so what is cos beta by 2 now cos beta by 2
is plus minus 1 so let me write what is cos
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beta by 2 that will be minus 1 whole to the
power n so putting this into putting this
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value to 1 . that will say that alpha is 2
into minus 1 whole to the power n sin hyperbolic
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z1 by 4 z2 so i got the non zero alpha value
in the stop band 1 now let us see the stop
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band 2 stop band 2 what is stop band 2 this
is my stop band 2 that means z1 by 4z2 is
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less than minus 1 so stop band to z1 y 4z2
is less than minus 1 so this is imaginary
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this root over z1 by 4z2 is pure imaginary
so let me see this quantity so this is pure
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imaginary so this should go to 0 and this
should be equal to this so i can write sin
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hyperbolic alpha by 2 cos beta by 2 is equal
to 0 let me call that earlier i want to set
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so three and j cos hyperbolic alpha by 2 sin
beta by 2 is equal to root over z1 by 4z2
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this let me call 4 okay so these three implies
what that we know that if sin hyperbolic alpha
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by 2 that cannot be 0 here so cos beta by
2 is 0 that means beta is 2n minus 1 this
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is 1 pie n belongs to integer so if that is
case what is the value of sin beta by 2 you
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know that it will be one so j of this is this
so can i say that j so now putting it here
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j putting this value here j cos hyperbolic
alpha by 2 is plus minus root over z1 by 4z2
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so that becomes alpha is equal to 2 cos hyperbolic
root over z1 by 4z2 so this is the value of
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alpha in pass band 2 ok . now we see what
happens to pass band we know alpha is 0 that
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we already seen but what happens to beta so
you see this again alpha is 0 so cos hyperbolic
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alpha that this term become 1so we have this
term only so j sin beta by 2 is equal to root
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over z1 by z2 so sin beta by 2 is z1 by 4z2
magnitude so i know what is beta in pass band
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beta is 2 sin inverse z1 by 4z2 so in pass
band alpha is this beta is this in various
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stop bands i have so once i have this now
let us design a prototype filter so let us
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take again i am taking t network you can also
take a pie network . so in paa for pass band
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we know that the z1 and z2 should be opposite
sign let let us take z1 is inductive that
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means the picture will that i have inductance
here capacitance here again inductance here
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let us call this l by 2 this is l by 2 this
is c so z1 is what z omega l z2 is 1 by j
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omega c and what is the that important number
z1 by 4 z2 this will be minus omega square
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if you just put this omega square lc by 4
in pass band we know this number is so pass
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band is where this number is between -1 and
0 okay so you write that one is greater than
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these greater than 0 so you can write 1 from
here you can follow that 1 is greater than
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4 pie square f square lc by 4 greater than
0 omega is 2 pie f so we have written that
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or 1 by pie square lc is greater f square
greater than 0 now this whole thing i can
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name as what fc so fc is what fc is 1 by pie
root over fc this is lc square so this fc
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square so fc is this all of you know this
from prototype so what type of filter is you
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see that in passband my pass band extend from
fc 0 to fc so the frequency 0 to fc are unattenuated
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and over fc it is suppressed or starts attenuating
so this is a low pass filter if instead of
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this i had interchanged this the some capacitance
with some inductance i could have a high pass
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filter all those you know that then we can
have band pass low pass but what i did from
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image impedance concept i have brought you
here let me finish now that
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what happens when we have seen the alpha is
zero in pass band .so between 0 to f to fc
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the alpha is 0 and what is beta we already
derived you see your expressions i can write
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it as 2 sin inverse root over z1 by 4z2 i
know now z1 4z2 values so this becomes 2 sin
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inverse f by fc very good so i can draw that
versus f i can plot alpha so before plotting
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let me see the stop band that what happens
when f is greater than fc f and f greater
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than fc z1 by 4z2 is minus omega square lc
by 4 so you put in our case this so this is
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f is greater than fc we at which stop band
stop band 2 do you see that you will here
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because of this minus omega square lc so actually
we are z1 – 4z2 minus omega square lc so
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we are in this part and in this part we know
that what is alpha alpha will be 2 cos hyperbolic
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inverse z1 by 4z2 and this is 2 cos hyperbolic
inverse f by fc so now you can plot alpha
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tha there will be fc so upto this fc alpha
is 0 this is your pass band this is 0 but
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then after that it will have this type of
variation cos hyperbolic f by fc so that is
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like this so alpha is rising here with a cos
hyperbolic inverse variation now what happens
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to beta we have seen in pass band where is
beta .we have already seen derived that in
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paaaaah pass band beta is 2 sin inverse root
over z1 by 4z2 so and in stop band 2 beta
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will be 2n – 1 pie so you can take that
beta is simply pie so if i plot fc so these
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2 sin inverse variation so this will be something
like this and then it will stay at pie this
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is beta this is your pie so this will be the
beta variation what does that mean beta variation
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means you see that if i have if my base band
signal or if my rf signal that as a frequency
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spread that means instead of that means instead
of a single tone if i have a multi tone things
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a band of frequencies present in my voice
and suppose it is getting transmitted if you
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put this filter there will be dispersion because
this is not a linear function of frequency
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so various frequency components will be reaching
haha va the transmission thing but since the
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length of filters are small it is not a problem
but if you go to higher in frequency suppose
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you cross v gauge range then even this is
significant and you will start having dispersion
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etc so you need to design it from a micro
filter view point this is the problem of these
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low low frequency filter design and what is
z0 also another important point these we have
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seen alpha beta but now we’ll have to see
what happens to z0 but characteristic impedance
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. because our whole impedance matching is
depending on that we said that the filter
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is terminated by z0 filter is both in the
source and load side what is the z0 value
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so we have already seen that z0 is z1 square
+ 4z1 z2 by 2 what is this if you put our
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those values that z1 is equal to j omega l
z2 is equal to 1 by j omega c you will see
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that it is l by c into 1 – f by fc square
z0 is real in pass band
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that we have already seen from that theorem
so 0 to f to fc now at f is equal to 0 what
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is z0 you see from that expression oh 1 minus
it is root over l by c and that generally
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is this filter design people rf filter design
they call it k a constant you see lyc is constant
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and z0 here is independent of frequency but
that is only at dc but at any other frequency
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even in pass band z0 is changing as we are
increasing f as f is increased z0 you see
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getting decreased f you are increasing this
part is increasing so that z0 is decreasing
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and at f is equal to fc what happens to z0
f is equal to z0 is 0 so i know that variation
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i can plot it and also let me complete that
in pass band in stop band what will be z that
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means when f is greater than fc z0 will be
j k is constant into f by fc whole square
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-1 just put that in this expression for mine
when f is greater than fc you will have bring
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this j out and from here you get f greater
than fc z0 is this so if we see the plot this
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plot will be suppose i am plotting z0 in the
x axis f here there will obviously a fc here
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so we will see the graph as so this you see
that this is your stop a pass band so it is
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starting from the value of k gradually decreasing
to 0 and then increasing so it is a dangerous
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thing you see at fc it is very good resonator
it is a good filter it has good output etc
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but in other places it as a characteristic
impedance which is not a constant value k
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so you will have the load impedance and source
impedance chosen according this characteristic
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impedance of the filter so at that only frequency
it is matching all other there is mismatch
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so there will be power loss due to reflection
in the circuit this is a problem of this constant
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k filters low pass filter so tricky what they
do they put try to improve this impedance
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matching that means instead this graph so
they try to flatten it that is why m derived
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section they can flatten it to very good amount
but you need some extra hardware for that
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so this is the problem so what we will do
we will see that if i try replicate the design
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to microwave filters i will have problem due
to these i can have m derived etc that may
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solve the problem but then it is bulky what
will see that you can design the whole thing
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completely from a different stand point that
is called insertion loss view point insertion
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loss filter design and all modern micro wave
filter design are based on these but i introduced
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this concept because in btech sometime this
though this is part of the text syllabus in
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00:27:45,600 --> 00:27:53,530
our present day this is not taught with so
emphasis but these are the impedance image
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impedance concept characteristic impedance
concept all comes from here which later we
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use for transmission line and other things
so unless and until this impedance part is
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clear propagation constant is clear later
understanding the microwave filter design
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00:28:10,039 --> 00:28:17,049
microwave amplifier that will become problem
that is why i took so much classes extra to
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revise but before completing i say that i
have always taken t section you can always
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take an pie section so if we take a pie section
. just what happens let me see this constant
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k filter pie so my filter will be like this
again i decide this is l so this is l and
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these two i take c by 2 this is a symmetrical
pie filter so it is z0 expression if you now
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00:29:00,909 --> 00:29:26,570
derived it will be who taught it f arrow top
it f z0 pie actually that time i have said
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z0 you can call i z0 t section also z0 pie
that will be fc so this part here also note
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that upto this part this is possible this
is real and this is the reactants part because
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it is pure imaginary but in the same graph
since we are showing we should that this portion
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is reactive this portion is real to resistive
and here this portion is reactive this portion
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is real this is in pass band so you see here
in the loap t section it was coming like this
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in pie section increasing like this and you
can find that when f is greater than fc this
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z0 that is even by from here just when f is
greater than fc so l by c into 1 by root over
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00:30:55,130 --> 00:31:05,490
f by fc whole square minus 1 you just take
j out that become this so other baah alpha
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00:31:05,490 --> 00:31:13,769
beta’s are same but these characteristic
impedance of a pie section is different from
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characteristic impedance of a t section ok
so you know that if we change this l and c
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when we get high pass filter then we put cascade
of low pass filter with high filter you get
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00:31:32,029 --> 00:31:43,320
band pass filter or in one arm instead of
single l we can take a series of l and c here
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00:31:43,320 --> 00:31:50,570
also we can take a series of l and c parallel
combination of l and c and that gives different
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filters notch filters etc of stop band filters
as it is sometime called notch filter that
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thing that design you can easily do but all
this at low frequency is ok rf frequency is
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ok upto 1 gigahertz i say 12 gigahertz is
ok but above that if you go see these will
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pose a series problem you need to have impedance
matching m section for that etc etc which
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created lot of problem we wont go into that
we will see that instead of you can simply
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do it if you consider insertion loss because
whatever we say ideal filters in pass band
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it should not be should not have loss but
always it will have loss if we calculate that
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loss and enforce that in passband i need insertion
loss to be within these and in stop band i
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needed to be enormously high compared to pass
band that is filter so how in power terms
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00:32:54,009 --> 00:33:01,369
we are saying that here we are saying that
in terms of that propagation constant should
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00:33:01,369 --> 00:33:08,570
be lossless that means attenuation constant
should be zero so instead of that i can that
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00:33:08,570 --> 00:33:15,629
in terms of power that insertion power so
we will have to define what insertion loss
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00:33:15,629 --> 00:33:21,799
that means in power transfer what is the loss
happening if i in introduce this filter and
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that if we can keep insertion loss to minimum
over a band that is that will be pass band
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and very high at other band so we can do microwave
filter design by that even all low frequency
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00:33:36,460 --> 00:33:43,279
filter can design by that so that is the more
modern technique and then you can have various
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00:33:43,279 --> 00:33:50,141
functions to specify your pass band and stop
band so any arbitrary specification you can
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make and you can synthesize the filter of
that that is the huge story of research still
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00:33:57,249 --> 00:34:03,080
it is evolving that various specifications
because when you go to space when you go for
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00:34:03,080 --> 00:34:11,850
various modern electronic equipment you have
abaaah haaa . . of specifications of filters
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i want these to be cut to this much i want
nearby it should have this you given this
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so various shapes are available and you know
if you know how to do by insertion loss you
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can incorporate all these things we will see
some typical cases some chevyshav butter worth
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etc shape which or if various elliptic shapes
that how people so design filters from that
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insertion point from here the next class so
this more or less classes completes the low
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frequency part or rf part of the filter design
microwave which starts from 12 gigahertz and
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mandatory for above 3 gigahertz you cannot
have this design because of this problem also
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the your tm mode apart from that other mode
starts propagating there te tm modes etc and
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also you cannot consider that your distributed
line is considering as i said that tm om other
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waves will start coming so you need to consider
real wave picture voltage and current does
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not have any meaning there in equivalent since
you have voltage current but microwave thing
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are based on electric field and magnetic field
etc so filter design there will be very complicated
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if we transform this network theory concept
voltage current concept but as i said in terms
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of insertion loss it will be very easy also
we will see that baaah that makes the whole
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thing very computer program friendly and by
that now a days all modern design are insertion
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loss thank you