WEBVTT
Kind: captions
Language: en
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Welcome to this third lecture, now the same
2 port reciprocal network we enforce symmetry
00:00:27.980 --> 00:00:37.100
in the network generally filters most of the
times we design symmetrical filters the 2
00:00:37.100 --> 00:00:45.199
port network is symmetrical . what do you
mean by symmetrical when I said this sorry
00:00:45.199 --> 00:00:55.770
this 2 port that I may say this is Z1 Z2 Z3
So in a T section as I already said that any
00:00:55.770 --> 00:01:01.700
2 port network whatever may be the interconnection
can be always represented like this now this
00:01:01.700 --> 00:01:11.909
Z1 Z2 different that is a symmetric network
if I enforce symmetry then Z1 is equal to
00:01:11.909 --> 00:01:26.660
Z2 so what happens to ZI1 my image impedance
ZI1 and you see the expressions of ZI1 and
00:01:26.660 --> 00:01:34.020
Z12 already derived the moment Z1 and Z2 are
same they becomes equal. So for a symmetrical
00:01:34.020 --> 00:01:47.310
network Z1 Zi2 are equal that we generally
call Z0 what is Z0 a very very important concept
00:01:47.310 --> 00:02:00.330
called characteristic impedance so for a symmetrical
network I have correct only one characteristic
00:02:00.330 --> 00:02:07.630
impedance for a non minus symmetrical reciprocal
network I have 2 image impedance but here
00:02:07.630 --> 00:02:13.860
it is one so what is the concept of characteristic
impedance that I terminate with characteristic
00:02:13.860 --> 00:02:22.290
impedance then the source with characteristic
impedance Z0 will see that input impedance
00:02:22.290 --> 00:02:30.890
is also Z0 so it can deliver maximum power
okay but now so we see that we have characteristic
00:02:30.890 --> 00:02:38.170
impedance let us see what is his value in
terms of ABCD parameters we already seen that
00:02:38.170 --> 00:02:47.450
when we enforce Z1 is equal to Z2 that mean
symmetry Z1 becomes ZI2 so that implies that
00:02:47.450 --> 00:02:58.490
we have already earlier derived ZI1 is equal
to AB by CD and this is BD by AC so that becomes
00:02:58.490 --> 00:03:06.510
A is equal to D. You all know into port parameters
transmission parameters if symma symmetrical
00:03:06.510 --> 00:03:14.000
network is equal to D it comes from here so
what happens to value of Z0 it is simply B
00:03:14.000 --> 00:03:27.350
by C and so now you see and that for a symmetrical
reciprocal network I have I can characterize
00:03:27.350 --> 00:03:35.880
the network by only 2 parameters One is that
this Z0 and another we have already seen it
00:03:35.880 --> 00:03:44.540
is gamma so Z0 and gamma are sufficient to
characterize the network so by Z0 we will
00:03:44.540 --> 00:03:50.330
understand what will be our impedance level
if there is an impedance mismatch etc and
00:03:50.330 --> 00:03:58.160
with gamma will enforce whether the frequencies
passed or not whether attenuated or not so
00:03:58.160 --> 00:04:06.810
let now express the ABCD parameters we are
so familiar with in terms of Z0 gamma that
00:04:06.810 --> 00:04:16.539
means from now on will specify only Z0 and
gamma for filter etc specification so we should
00:04:16.539 --> 00:04:29.050
have a mapping between ABCD etc so you can
easily do that express that ABCD in terms
00:04:29.050 --> 00:04:38.150
of Z0 and gamma because now our description
will be in terms of that so you can start
00:04:38.150 --> 00:04:47.720
from that AD minus BC is equal to 1 and also
A is equal to D so with this you can now find
00:04:47.720 --> 00:04:56.430
out that let us decide at what will happen
to them if we put this A is equal to D we
00:04:56.430 --> 00:05:09.320
get BC is equal to root over square minus
1 and we know that B by C is Z0 square you
00:05:09.320 --> 00:05:17.840
see this one we have found as B by C so from
this you can find out the value of C is root
00:05:17.840 --> 00:05:30.300
over square minus 1 by Z0 and B is Z0 root
over square minus 1 but and what is A to the
00:05:30.300 --> 00:05:38.800
power gamma A to the power gamma earlier we
have seen it is AD plus BC now in terms of
00:05:38.800 --> 00:05:47.300
this I can write it as A plus root over square
minus 1 so this equation I can solve so A
00:05:47.300 --> 00:05:54.900
will be expressed in terms of A to the power
gamma or gamma so by doing that I can see
00:05:54.900 --> 00:06:04.430
that a becomes A to the power 2 gamma plus1
by 2 A to the power gamma that is nothing
00:06:04.430 --> 00:06:11.820
but A to the power gamma plus A to the power
minus gamma by 2 which is nothing but COS
00:06:11.820 --> 00:06:21.870
hyperbolic gamma so A is nothing but COS hyperbolic
gamma now put this B is already expressed
00:06:21.870 --> 00:06:32.919
in terms of Z0 and this Z0 value you know
let me write root over B by C so you can call
00:06:32.919 --> 00:06:40.430
for and see that B will be equal to Z0 sin
gamma please all of you do this exercises
00:06:40.430 --> 00:06:46.210
I am just writing you cannot memorize all
these things but I am telling you the way
00:06:46.210 --> 00:06:53.240
so please do this manipulation it requires
a bit familiarity with hyperbolic functions
00:06:53.240 --> 00:07:08.139
and C is equal to 1 by Z0 SIN hyperbolic gamma
and D is equal to COS hyperbolic gamma so
00:07:08.139 --> 00:07:20.900
you see that this expression that means now
I can clearly write that A is equal to COS
00:07:20.900 --> 00:07:35.169
hyperbolic gamma B is equal to Z0 SIN hyperbolic
gamma C is equal to 1 by ZO SIN hyperbolic
00:07:35.169 --> 00:07:45.139
gamma and D is equal to COS hyperbolic gamma
so you see that actually here from also you
00:07:45.139 --> 00:07:51.570
can see that ABCD parameter was not needed
because A is equal to D for a symmetrical
00:07:51.570 --> 00:07:58.310
network and this is the description that any
ABCD parameters I can express in terms of
00:07:58.310 --> 00:08:06.120
2 quantities is gamma and Z0 so that is the
beauty you know that we are in the right path
00:08:06.120 --> 00:08:19.140
what we will do now will put the filter condition
so for a filter as I said what is the filter
00:08:19.140 --> 00:08:30.210
that it will allow some bands to pass it will
stop all other bands so we demand that now
00:08:30.210 --> 00:08:38.079
network 2 port network is in terms of gamma
and Zo what is our if it is to be a filter
00:08:38.079 --> 00:08:58.709
so gamma in pass band should be lossless
we want that gamma the propagation constant
00:08:58.709 --> 00:09:05.560
should be lossless then only the circuit wont
attenuate power also not store power it will
00:09:05.560 --> 00:09:19.260
be able to deliver it to the port 2 so but
you know that gamma is a complex quantity
00:09:19.260 --> 00:09:23.529
as I said that how we define that input volte
ampere and output volte ampere their ratio
00:09:23.529 --> 00:09:29.899
that we call it A to the power 2 gamma. So
gamma is that now you are also familiar that
00:09:29.899 --> 00:09:36.790
generally this complex quantity we write as
2 part alpha plus J beta, alpha is attenuation
00:09:36.790 --> 00:09:44.640
constant beta is the phase constant for any
wave for electromagnetic wave this is true
00:09:44.640 --> 00:09:51.170
so when we say alpha in the pass band should
be lossless what we basically say that alpha
00:09:51.170 --> 00:09:59.960
should be ) over the whole pass band that
there should not be any attenuation and what
00:09:59.960 --> 00:10:10.601
happens in stop band we want that it should
be lossy so atleast alpha should not be is
00:10:10.601 --> 00:10:20.459
equal to to zero ok and we have already decided
that to have this paaasss pass band alpha
00:10:20.459 --> 00:10:30.410
is equal to 0 we will use only reactive elements
now there is a theorem which will now see
00:10:30.410 --> 00:10:50.830
the theorem says that for a symmetric 2 port
reciprocal network of pure reactance’s.
00:10:50.830 --> 00:11:02.630
If the characteristic impedance is a pure
resistance that innovation constant is zero
00:11:02.630 --> 00:11:14.281
if the characteristic impedance is pure reactance
their attenuation constant is non zero so
00:11:14.281 --> 00:11:23.660
for this symmetric 2 port reciprocal this
is a theorem theorem of Everett symmetric
00:11:23.660 --> 00:11:45.090
2 port reciprocal network made of
pure reactances the theorem says that if Z0
00:11:45.090 --> 00:12:13.450
is pure cha pure reactance alpha is equal
to 0 if Z0 is pure reactants alpha is non
00:12:13.450 --> 00:12:29.420
minus zero let us prove this theorem so will
now prove this theorem since the network is
00:12:29.420 --> 00:12:45.329
made of pure reactances can I say that network
these are all instead of Z they are JX1 it
00:12:45.329 --> 00:12:57.529
is a symmetric network jx1and j let us say
X3 instead of Z1, Z2, Z3 now this may be plus
00:12:57.529 --> 00:13:10.950
minus whatever if this is the case then can
I say that Z0c and Zsc which we already saw
00:13:10.950 --> 00:13:27.570
they are also pure reactances ok we have seen
that Z0 is equal to root over Z0c into Zsc
00:13:27.570 --> 00:13:33.640
open circuit and short circuit cases so they
will be because we have seen their various
00:13:33.640 --> 00:13:43.750
combination of these they will be also pure
reactances so let us consider that Zoc let
00:13:43.750 --> 00:13:57.950
us call it either plus or minus the pure reactance
this is let us call plus or minus jxb that
00:13:57.950 --> 00:14:15.920
means XA and XB pure real positive numbers
you agree that it should be because since
00:14:15.920 --> 00:14:28.100
I have incorporated Jxa and xb are positive
real numbers now let us see there can be several
00:14:28.100 --> 00:14:41.620
cases. Case 1, 2 am proving the theorem, is
important for filter design now let me say
00:14:41.620 --> 00:15:01.240
what is Z0c what is Zsc then let me we know
Z not square is equal to Z0c, Zsc and we now
00:15:01.240 --> 00:15:17.800
10 the square is Zsc by Z0c so we will make
table. So case 1 is let us say that this Z0c
00:15:17.800 --> 00:15:34.910
and Zsc they are of same sign so it is plus
jxa it is minus jxb then what will be Z0 square
00:15:34.910 --> 00:15:44.840
the it will be simply XaXb and what will be
these Zsc by Z0c it will be minus XB by XA
00:15:44.840 --> 00:15:57.370
case 2 this is minus jxa this plus jxb this
will be XaXB this will be minus Xb minus Xa
00:15:57.370 --> 00:16:21.280
this is plus jxa plus jxb this is minus Xa
Xb this is Xb by Xa this is minus jxa minus
00:16:21.280 --> 00:16:32.720
jxb minus xaxb xb by xa so you see that time
I said that I have some proveb possibilities
00:16:32.720 --> 00:16:48.190
so I have taken all permutations of these
from here you can see that case 1 and 2 Z0
00:16:48.190 --> 00:16:56.050
is real that means Z0 it is a characteristic
impedance it is real means this is a resis
00:16:56.050 --> 00:17:06.470
pure resistance and TAN hyperbolic is imaginary
because TAN this square is minus so TAN hyperbolic
00:17:06.470 --> 00:17:17.490
will be an imaginary number and in case 3
and 4 I have marked that by rate you see Zo
00:17:17.490 --> 00:17:32.269
is imaginary and TAN hyperbolic is real so
let me write this thing that in case 1 and
00:17:32.269 --> 00:18:12.029
2 Z0 is real and TAN hyperbolic is imaginary
in our power cases cases 3 and 4 Z0 is imaginary
00:18:12.029 --> 00:18:22.710
and TAN hyperbolic gamma is real still I have
not proved the theorem because theorem said
00:18:22.710 --> 00:18:34.639
that if Z1 is real alpha will be zero if Z1
is imaginary there will be non zero alpha
00:18:34.639 --> 00:18:44.179
I have not reached there so I need to do that
to prove the theorem so part 1 is this then
00:18:44.179 --> 00:18:52.359
let me the write what is TAN hyperbolic gamma
? Tan hyperbolic gamma is nothing but TAN
00:18:52.359 --> 00:19:10.789
hyperbolic alpha plus J beta and that is equal
to SIN hyperbolic alpha COS beta plus G COS
00:19:10.789 --> 00:19:29.460
hyperbolic alpha SIN beta by COS hyperbolic
alpha COS beta plus J SIN hyperbolic alpha
00:19:29.460 --> 00:19:44.809
SIN beta now numerator denominator both are
complex number A plus JB so I generalized
00:19:44.809 --> 00:19:53.519
the determinator so that becomes after some
simple manipulation COS hyperbolic alpha COS
00:19:53.519 --> 00:20:06.920
hyperbolic alpha SIN hyperbolic plus J SIN
beta COS beta by COS hyperbolic alpha square
00:20:06.920 --> 00:20:18.980
COS square beta plus SIN hyperbolic square
alpha and SIN square beta so denominator is
00:20:18.980 --> 00:20:38.259
pure real now can talk of that When TAN hyperbolic
gamma is we have seen in this case TAN hyperbolic
00:20:38.259 --> 00:21:00.609
is imaginary so TAN hyperbolic is pure imaginary
COS hyperbolic alpha SIN hyperbolic alpha
00:21:00.609 --> 00:21:12.470
so TAN hyperbolic alpha to be in a imaginary
I require this part should go to 0 now here
00:21:12.470 --> 00:21:21.799
you see COS hyperbolic alpha always is greater
than zero so this is e to the power alpha
00:21:21.799 --> 00:21:28.639
plus e to the power minus alpha by e to the
power by 2 so it cannot be Zero COS hyperbolic
00:21:28.639 --> 00:21:38.450
alpha cannot be zero so the only possibilities
this implies SIN hyperbolic alpha is equal
00:21:38.450 --> 00:21:44.081
to zero what is SIN hyperbolic alpha that
means e to the power alpha is equal to e to
00:21:44.081 --> 00:21:53.429
the power minus alpha that alpha is equal
to zero so theorem theoooorem ha where is
00:21:53.429 --> 00:22:09.899
the theorem, said that when Z0 is pure resistance
alpha is equal to zero. You see Z) is pure
00:22:09.899 --> 00:22:20.989
resistance from this I got alpha is equal
to zero so, okay let us see the other option.
00:22:20.989 --> 00:22:38.729
When TAN hyperbolic gamma is pure real what
I get you see this thing is pure real that
00:22:38.729 --> 00:22:47.989
means this part should go to zero I demand
SIN beta COS beta should be zero and also
00:22:47.989 --> 00:22:59.349
COS hyperbolic alpha SIN hyperbolic alpha
nonzero because this this also may go to zero
00:22:59.349 --> 00:23:12.960
then whole thing will be zero but I require
that this is nonzero but COS as I said COS
00:23:12.960 --> 00:23:21.009
is always greater than one so no problem it
demands as COS this is always greater that
00:23:21.009 --> 00:23:38.029
equal to one so what we demand is SIN hyperbolic
alpha is nonzero so that means alpha nonzero
00:23:38.029 --> 00:23:43.999
means the attenuation constant is nonzero
that mean there is a attenuation so that proves
00:23:43.999 --> 00:23:55.479
the theorem that when so what it shows let
me go to my cases that it this case when Zo
00:23:55.479 --> 00:24:05.880
is real there will be no attenuation and when
Z0 is these there will be definitely attenuation
00:24:05.880 --> 00:24:13.970
so can I not say that if I enforce this condition
this is my pass band condition and this is
00:24:13.970 --> 00:24:29.289
my stop band condition of a filter so I will
do that so but you also observe here that
00:24:29.289 --> 00:24:38.950
so this is my filter pass band I will go from
here but here I have one demand that Z0c and
00:24:38.950 --> 00:24:50.419
ZSC they should be of opposites sign now ZOC
ZSC they are that means that or I will see
00:24:50.419 --> 00:25:05.109
that so let me now see my T section I can
also see it with PIE section but the so it’s
00:25:05.109 --> 00:25:14.999
a symmetrical and instead of earlier I was
calling Z1 now people call it Z1 by 2 because
00:25:14.999 --> 00:25:28.109
there are 2 in series so okay now remember
obviously Z1 and Z2 they are pure reactances
00:25:28.109 --> 00:25:35.389
we want to derive the propagation constant
of this gamma of this network already we have
00:25:35.389 --> 00:25:46.080
done TAN gamma so from there we will try to
see enforce the pass band condition so port
00:25:46.080 --> 00:25:58.649
2 let us make that this is open circuit condition
so that time we know we call it Z0c and what
00:25:58.649 --> 00:26:14.490
will be that Z0c is equal to Z1 by 2 that
is equal to Z1 plus 2 Z2 by 2 now I put that
00:26:14.490 --> 00:26:44.369
port 2 is shorted this is Z1 by 2, Z1 by 2Z2
and this is Zsc so what is Zsc it is you can
00:26:44.369 --> 00:26:57.570
easily see Z1 by 2 plus parallel of Z2 and
Z1 by 2 so that if you do it will become Z1
00:26:57.570 --> 00:27:10.440
by 2 plus and Z2 parallel Z1 by 2 and that
is equal to Z1 by square plus 4 Z1, Z2 by
00:27:10.440 --> 00:27:24.940
2 Z1 plus 2Z2 so I got Z0c and Zsc so I can
easily now find what is the characteristic
00:27:24.940 --> 00:27:34.679
impedance Z0 I know always this is true I
have expression for them say so if you put
00:27:34.679 --> 00:27:52.830
it becomes Z1 square plus 4 Z1 Z2 by 2 already
we have in that table that this alfff this
00:27:52.830 --> 00:28:01.679
gamma becomes pure imaginary so this is the
stop band there Z0c and Zsc there should be
00:28:01.679 --> 00:28:14.580
opposite side so, because that only makes,
you see Zoc and Zsc their opposite side makes
00:28:14.580 --> 00:28:26.820
the characteristics impedance real otherwise
they are of same type this is imaginary and
00:28:26.820 --> 00:28:31.600
what happens to the other, so characteristic
impedance we have find what is the other parameter
00:28:31.600 --> 00:28:39.889
of the 2 port symmetrical reciprocal network
gamma. So what is TAN gamma we know that is
00:28:39.889 --> 00:28:55.570
Zoc by Zsc and if you do this that will become
root over Z1 square plus 4 z1 z2 by Z1 plus
00:28:55.570 --> 00:29:16.629
2 Z2 square now this we want this TAN gamma
should be imaginary in the pass band so our
00:29:16.629 --> 00:29:34.450
demand that in the pass band Z1 square plus
4 Z1 Z2 by Z1 Z2 plus 2 Z2 square should be
00:29:34.450 --> 00:29:45.070
less than zero consider this case 1 in the
table will now confine ourselves to this case
00:29:45.070 --> 00:30:00.799
1 and 2. In case 1 Zoc is plus Jxa this is
minus Jxb so let us put it in case 1 case
00:30:00.799 --> 00:30:20.789
1 Zoc plus Jxa Zsc is minus Jxb so what happens
to 2Z1 plus 2Z2 by 2 this one Z1 plus 2 Z2
00:30:20.789 --> 00:30:38.489
by 2 becomes Jxa and Z1 plus so this 2Z2 whole
square that becomes minus 4XA squared we know
00:30:38.489 --> 00:30:54.820
Xa is positive real number so Z1 plus 2Z square
that is less than zero. So Tan beta that means
00:30:54.820 --> 00:31:05.419
in pass band we know this TAN beta value is
this now this is less than zero so this whole
00:31:05.419 --> 00:31:21.710
thing less than zero these need to be positive
to satisfy this say that TAN gamma imaginary
00:31:21.710 --> 00:31:39.820
to be imaginary Z1 square plus 4 Z1 Z2 that
needs to be greater than zero so this is from
00:31:39.820 --> 00:31:53.509
case one that means first case from pass band
we know this let us see case 2 there we have
00:31:53.509 --> 00:32:09.109
seen Zoc is minus Jxa this is case two Zoc
is minus Jxa Zsc plus Jxb so with this what
00:32:09.109 --> 00:32:18.590
happens to Z1 plus 2 Z 2 by 2 that becomes
minus jxa so what happens to the square of
00:32:18.590 --> 00:32:27.700
this 2 Z 2 square that is again minus 4 X
square that is again less than zero so same
00:32:27.700 --> 00:32:43.999
condition that here also TAN gamma to be imaginary
we require Z1 square plus 4 z1 z2 greater
00:32:43.999 --> 00:32:55.460
than zero. So the condition in the pass band
if we want to have pass band without any attenuation
00:32:55.460 --> 00:33:05.360
then condition is Z1 square plus four Z one
Z two greater than zero now Z1 and Z2 they
00:33:05.360 --> 00:33:16.330
are reactive elements so again there will
be various sub cases so I can say sub case
00:33:16.330 --> 00:33:28.099
one a sub case A because there will be various
cases of Z2 Z2 we should remember that Z1
00:33:28.099 --> 00:33:43.979
is equal to jx one let us chose this Z2 is
equal to jx two then we can write what happens
00:33:43.979 --> 00:33:52.460
to Z one square plus four Z one Z two that
is minus X square minus four x one X two greater
00:33:52.460 --> 00:34:04.659
than zero or X one square plus four X one
X two is less than zero so since X and X X
00:34:04.659 --> 00:34:12.050
one and X two are positive real numbers this
is not possible so I say that sub case is
00:34:12.050 --> 00:34:23.710
not possible similarly subcase B there will
be Z1 is equal to minus jx one is equal to
00:34:23.710 --> 00:34:34.659
minus jx two again we will get the same condition
X1 square plus four X one X two is less than
00:34:34.659 --> 00:34:49.909
zero so again not possible sub case 3 sub
case C here let me take here as ZX one Z two
00:34:49.909 --> 00:35:00.630
as opposite SIN jx two if you do this what
is this value Z one square always we are evaluating
00:35:00.630 --> 00:35:07.690
this Z one square plus four Z one Z two that
should be greater than zero. This is the condition
00:35:07.690 --> 00:35:17.369
to achieve the pass band. So we are doing
this and getting this four X one X two is
00:35:17.369 --> 00:35:30.549
greater than j X one square now as X one is
greater than zero So four X2 will be greater
00:35:30.549 --> 00:35:43.329
than X one or X one by four x two greater
than one, also x one is greater than zero
00:35:43.329 --> 00:35:58.500
so combining these the condition is that zero
less than X one by four X two less than one
00:35:58.500 --> 00:36:13.339
or I can say or zero less than JX one by four
X two is less than one or greater than j x
00:36:13.339 --> 00:36:25.260
one by four minus j X two is greater than
minus one or zero greater than Z one four
00:36:25.260 --> 00:36:34.970
Z two is greater than minus one so this is
possible that means here if we and enforce
00:36:34.970 --> 00:36:41.819
this condition it boils down to that Z one
by four Z two to this ratio should be between
00:36:41.819 --> 00:36:56.140
zero and minus one and let us see the sub
case D that will be Z one is equal to minus
00:36:56.140 --> 00:37:05.109
jx one Z two is equal to plus ZX two again
if you do you do you get the condition that
00:37:05.109 --> 00:37:23.990
zero Z one by four Z two n minus one so this
shows that you can have the pass band pass
00:37:23.990 --> 00:37:39.640
band of filter should be in the range this
Z one by four Z two this number should within
00:37:39.640 --> 00:37:55.509
this range and where is the stop band of the
filter stop band of the filter there are two
00:37:55.509 --> 00:38:02.490
possibilities for that we have already seen
that Z one by four Z 2 that is greater than
00:38:02.490 --> 00:38:16.880
zero or Z1 by four Z two that is less than
one this is mathematics we will see in the
00:38:16.880 --> 00:38:22.200
next lecture what is the implication of that
but what we have done here, we have found
00:38:22.200 --> 00:38:31.200
out that if you want to have pass band the
two reactances their impedances is Z one and
00:38:31.200 --> 00:38:39.269
Z two this Z one by four Z two should be within
this range then only you can have pass band
00:38:39.269 --> 00:38:48.950
and if you choose Z one by four Z two either
this or this you get stop band of filter Thank
00:38:48.950 --> 00:38:49.420
You