WEBVTT
Kind: captions
Language: en
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Welcome to this fourteen lecture of lecture
series in pervious lecture we have seen that
00:00:26.140 --> 00:00:36.370
if we have a potentially unstable transistor
device then how to choose my gamma L and Gamma
00:00:36.370 --> 00:00:43.890
S to make it conditionally stable. Now that
time we raised the question can we judge from
00:00:43.890 --> 00:00:49.120
the S parameter can you determine whether
need to draw the stability circles are not
00:00:49.120 --> 00:00:57.879
the answer is fortunately yes people have
deviced some analytic test based on S parameters
00:00:57.879 --> 00:01:04.890
to test whether the device is unconditionally
stable or not if the device is unconditionally
00:01:04.890 --> 00:01:10.439
stable you go ahead with your actual amplifier
design you need not worry about the stability.
00:01:10.439 --> 00:01:19.640
If the test fails then you need to locate
or draw the stability circle on the smith
00:01:19.640 --> 00:01:25.399
chart and then determine the ranges from where
you will have to choose you gamma S and gamma
00:01:25.399 --> 00:01:30.359
L value and then you proceed to amplifier
design so we will see that test, . that test
00:01:30.359 --> 00:01:41.329
is called There are two test anyone you can
do called Rollets test or Rollest stability
00:01:41.329 --> 00:01:53.920
criteria. So Rollet you found out a parameter
called K which is given by 1 minus S11 square
00:01:53.920 --> 00:02:10.899
minus S22 square + delta square, delta is
the determinant of the S matrix 2 S12 S21.
00:02:10.899 --> 00:02:17.900
Such a simple parameter he proved, it can
be easily proved but that is you can see in
00:02:17.900 --> 00:02:24.569
books it is given but that is not so important
for the design, so you know that if this K
00:02:24.569 --> 00:02:30.900
value, you see everything here depends on
S parameters values S parameters. So you can
00:02:30.900 --> 00:02:44.710
find out K and if this K is greater than 1
as well as if delta which is nothing but if
00:02:44.710 --> 00:02:54.650
you see the S matrix it is S11 S22 minus S12
S21this magnitude that means these are real
00:02:54.650 --> 00:03:04.959
number. If these is less than 1then we say
that this Rollet test then passes you and
00:03:04.959 --> 00:03:19.280
device transistor is unconditionally stable
that means you can choose any passive load
00:03:19.280 --> 00:03:26.129
impedance source impedance and still you need
not worry your device wont go into oscillation.
00:03:26.129 --> 00:03:33.920
This is called Rollet test another test here
are two conditions that should be simultaneously
00:03:33.920 --> 00:03:46.579
satisfied another test this is also sometimes
called K delta test this is called Rollest
00:03:46.579 --> 00:03:57.909
constant or Rollets constant K another test
is called new test it depends on here you
00:03:57.909 --> 00:04:04.189
have K and delta both you have to test but
new test is an improvement of that it is recently
00:04:04.189 --> 00:04:14.219
people have found out and this test is that
you test this parameter Mu is given by 1 minus
00:04:14.219 --> 00:04:36.720
S11 square by S22 minus delta S1 star plus
S12 S1. So this mu is given by this here also
00:04:36.720 --> 00:04:51.280
you see this is a real number these is greater
than 1 then we can again say that device is
00:04:51.280 --> 00:05:01.500
unconditionally stable so any one of this
test you can do basically from this equations
00:05:01.500 --> 00:05:08.790
after manipulation can have this mu parameter.
So this is nothing new but people have made
00:05:08.790 --> 00:05:17.840
it more convenient if you test this you can
if not fails this test any of this test if
00:05:17.840 --> 00:05:31.490
it fails then you need to draw the stability
circle ok let us see the example of this . C
00:05:31.490 --> 00:05:39.750
in the rest part of this will see already
we have seen stability criteria, unconditional
00:05:39.750 --> 00:05:48.180
stable devices then we will see an example
if we get a conditionally stable devices how
00:05:48.180 --> 00:05:57.710
that can be if we have an potentially unstable
device how we can make conditionally stable
00:05:57.710 --> 00:06:05.160
amplifier design from that. . so this is from
Gonzalez book I already referred this book
00:06:05.160 --> 00:06:14.740
so in page hundred of my edition this problem
is given that suppose S parameter of transistor
00:06:14.740 --> 00:06:22.360
at 800 Megahertz is given like this are the
value given first determine the stability
00:06:22.360 --> 00:06:29.640
then if potentially unstable then draw stability
circles and then show how resistive loading
00:06:29.640 --> 00:06:34.880
can stabilize the transistor we see this is
the new thing we will discuss that with the
00:06:34.880 --> 00:06:44.040
example . so we will test Rolletâ€™s test
so K we can calculate and delta we can calculate
00:06:44.040 --> 00:06:53.020
so it turn out that k is equal to 0.547 and
delta is 0.504 and with the phase of this
00:06:53.020 --> 00:07:00.490
now since k is equal to 1 so we say that it
fells Rollets test so transistor is potentially
00:07:00.490 --> 00:07:06.650
unstable at eight hundred megahertz because
we know that unconditional stability K is
00:07:06.650 --> 00:07:14.430
greater than 1 and magnitude of delta is lies
than 1 so it as felt there so though magnitude
00:07:14.430 --> 00:07:20.930
of delta is less than 1 but K is less than
1 so it is potentially unstable also in the
00:07:20.930 --> 00:07:27.390
Mu test if you do that you see tha value turns
out to be 0.85 which is less than 1 but we
00:07:27.390 --> 00:07:35.370
know that mu needs greater than 1 for unconditionally
stable so it say that stability checking analytic
00:07:35.370 --> 00:07:41.720
stability indicate that this is a pontential
unstable device so if we want to use this
00:07:41.720 --> 00:07:49.090
device we need to find out or plot stability
circles and then determine what are the ranges
00:07:49.090 --> 00:07:56.660
of load impedance and source impedance, so
that we can make it stable amplifier. . So
00:07:56.660 --> 00:08:05.930
you see that first we have done input source
stability side cir iput circle CS expression
00:08:05.930 --> 00:08:13.520
I have already given so those are all I can
calculate that center will be 1.79 you know
00:08:13.520 --> 00:08:20.840
smith chart as a radius of 1 so the center
falls outside of smith chart that you see
00:08:20.840 --> 00:08:27.510
the center here this is outside smith chart
and 1.79. so these distance is 1.79 and this
00:08:27.510 --> 00:08:35.779
angle that it makes is hundred and twenty
degree so it is here and RS is 1.04 so partially
00:08:35.779 --> 00:08:45.959
you see this is 1.79 the radius is 1.04 so
it will partially interact intersect the
00:08:45.959 --> 00:08:51.350
smith chart intersect then we need to determine
which one is stable zone so values of delta
00:08:51.350 --> 00:08:57.150
S for which gamma out is equal to 1 this is
the input stability circle it is the border
00:08:57.150 --> 00:09:04.700
of stable and unstable region then as I said
assume that ZS is equal to 50 ohm so gamma
00:09:04.700 --> 00:09:13.710
S is equal to 0, so gamma out magnitude turns
out to be a S22 magnitude but what was S22
00:09:13.710 --> 00:09:34.200
magnitude you see S22 magnitude is 0.8 so
it is less than 1 so S22 magnitude is 0.8
00:09:34.200 --> 00:09:43.080
so I know this in input this cen cen the portion
of smith chart that is containing center that
00:09:43.080 --> 00:09:50.840
is the stable region we have marked as stable
region this black shading and this is the
00:09:50.840 --> 00:09:59.960
unstable region so I will have to choose the
gamma S from here . similarly we draw output
00:09:59.960 --> 00:10:06.250
stability circle this is the center again
outside the center is outside the smith chart
00:10:06.250 --> 00:10:12.400
and this time forty eight degree so this angle
is forty eight degree and you see the radius
00:10:12.400 --> 00:10:21.750
is 0.45 so 1.3 minus .45 there will be intersection
this redline is that locus of the circle a
00:10:21.750 --> 00:10:30.570
part of the circle arch and so again here
we can check that when will simply match the
00:10:30.570 --> 00:10:39.770
load side gamma in become S11 magnitude and
S11 magnitude in this case .65 you can again
00:10:39.770 --> 00:10:45.580
say that this input side is the stable one
so from this stable region we will have to
00:10:45.580 --> 00:10:56.020
choose out gamma L . So how we can do that
and also you see that if we want to give resistive
00:10:56.020 --> 00:11:03.560
loading suppose this is my input stability
circle this is my output stability circle
00:11:03.560 --> 00:11:14.880
now I see all this constant resistance things
so the one that is tangent to this input stability
00:11:14.880 --> 00:11:21.990
circle so there I can choose the corresponding
this is the constant resistant circle of resistance
00:11:21.990 --> 00:11:31.379
value normalize resistant value 0.18 so I
can out since this is input if I put this
00:11:31.379 --> 00:11:42.650
value you see I am further of here so I can
add a 0.18 0.18 normalize resistance so with
00:11:42.650 --> 00:11:51.850
the 50 ohm characteristic impedance the value
is here you see 9 ohm here I can add in the
00:11:51.850 --> 00:11:59.200
source side that means with the base of the
transistor I can add a 9 ohm resistance similarly
00:11:59.200 --> 00:12:05.140
from the output stability circle I can find
out tangent circle and find out that this
00:12:05.140 --> 00:12:14.770
is contact resistance circle this blue one
of 0.58. So upon unnormalizing I see that
00:12:14.770 --> 00:12:22.339
29 ohm so at the load side I can out 29 ohm
series resistance and that will make the whole
00:12:22.339 --> 00:12:30.640
thing stable . so it will stabilize the transistor
now sometimes at the output you want to have
00:12:30.640 --> 00:12:41.940
parallel loading so that we can see that this
0.58 if is just change it to the or this I
00:12:41.940 --> 00:12:53.330
can find out what is this point impedance
and I change it that becomes .11 so from that
00:12:53.330 --> 00:13:02.930
point 11 you know for admittance conductances
I need to divide it so five hundred ohm so
00:13:02.930 --> 00:13:12.180
instead of this previous case where I had
you see that previous I have given a 29 ohm
00:13:12.180 --> 00:13:18.750
series resistance instead I can also give
a five hundred ohm parallel resistance in
00:13:18.750 --> 00:13:24.970
the load side and that will also stabilize
the transistor so by this resistive loading
00:13:24.970 --> 00:13:33.300
you can add but obviously this will change
the overall S parameter or overall gain condition
00:13:33.300 --> 00:13:44.310
. so we need to check that whether this true
so we can do this that the series resistance
00:13:44.310 --> 00:13:49.750
that we have had or shunt resistance that
we have had we can find the ABCD parameter
00:13:49.750 --> 00:14:01.800
as you can see
that this one is the shunt element so it is
00:14:01.800 --> 00:14:07.410
in so if I know that ABCD of this because
I know that S parameter of this transistor
00:14:07.410 --> 00:14:13.410
so I can easily find out what is the ABCD
parameter of this then I can find out this
00:14:13.410 --> 00:14:20.190
shunt element what is the ABCD parameter I
can multiply this ABCD parameter to find the
00:14:20.190 --> 00:14:27.660
resistive loaded transistors ABCD parameter
from there I can go to again S parameter back
00:14:27.660 --> 00:14:33.120
composite S parameter and from that i can
again see whether this is resistive loading
00:14:33.120 --> 00:14:41.029
stabilize that this composite thing now needs
to pass the rollest stability criteria or
00:14:41.029 --> 00:14:48.300
MU test so that will be doing this are the
formulas . you can refer to so you can find
00:14:48.300 --> 00:14:58.620
out that when I have a series Z I know this
is the ABCD parameter so overall ABCD parameter
00:14:58.620 --> 00:15:04.710
in that case will be this similarly for the
shunt one i will have these because these
00:15:04.710 --> 00:15:12.350
then and in shunt with that I have this . so
overall ABCD parameters comes out to be this
00:15:12.350 --> 00:15:20.050
this are the transistor ABCD values and then
I can reconvert after doing this multiplication
00:15:20.050 --> 00:15:26.779
I can reconvert it to overall S parameter
overall parameter is this you see stability
00:15:26.779 --> 00:15:34.180
factor so I can calculate the rollest factor
that turns out to be 1 so you see and beta
00:15:34.180 --> 00:15:41.430
magnitude that is .409 so I know that I have
passed the test so above network is unconditionally
00:15:41.430 --> 00:15:47.810
stable at eight hundred megahertz so this
is the way by which I have an potentially
00:15:47.810 --> 00:15:56.529
unstable device still with loading proper
loading finding from Smith chart I can make
00:15:56.529 --> 00:16:05.000
it conditionally stable ok so after that we
will try to do that the actuals so this is
00:16:05.000 --> 00:16:24.250
stability checking after that we will see
the conjugate matching . so design of amplifier
00:16:24.250 --> 00:16:40.920
with conjugate matching because that is out
actual M so we will do that design with conjugate
00:16:40.920 --> 00:16:49.780
matching and you see that I have now made
that stability analysis I have chosen the
00:16:49.780 --> 00:16:58.370
gamma S and gamma L ranges etc if we refer
to our original diagram in previous class
00:16:58.370 --> 00:17:06.549
we have said that finally under conjugate
matching I will have a source then I will
00:17:06.549 --> 00:17:23.559
have a conjugate . then an input matching
network
00:17:23.559 --> 00:17:39.080
this is conjugate impedance matching this
will be connected to the transistor and then
00:17:39.080 --> 00:17:59.890
I will have output matching network and then
it will be finally looking at Z0 so I know
00:17:59.890 --> 00:18:06.139
that actual ZL is different but with output
matching it will appear as Z0 similarly actual
00:18:06.139 --> 00:18:13.169
source impedance is ZS but with input matching
network I can make it and that day we have
00:18:13.169 --> 00:18:23.240
also seen that transistor the transducer power
gain that comprises of these that you can
00:18:23.240 --> 00:18:31.530
say this generally we call ZS then this the
transistor side that we call Z0 and this is
00:18:31.530 --> 00:18:41.350
ZL and this Z0 is typically not in our hand
because transistor manufacturer he has made
00:18:41.350 --> 00:18:48.549
the transistor that transistor is giving me
some S21 values so we know that typically
00:18:48.549 --> 00:18:58.539
this Z0 is given by S21 square now and micro
wave designer and amplifier designer is job
00:18:58.539 --> 00:19:05.139
is to now design input matching network and
output matching network output matching network
00:19:05.139 --> 00:19:16.080
so that I will choose my proper gamma L gamma
Z and gamma S so that ZS that will appear
00:19:16.080 --> 00:19:24.309
as Z0 here to the source similarly the ZL
here was ZL but with this output matching
00:19:24.309 --> 00:19:31.269
network it will as Z0 then it will be conjugately
matched now what is conjugate matching that
00:19:31.269 --> 00:19:39.679
time we have seen the gamma in is nothing
but gamma S star and simultaneously because
00:19:39.679 --> 00:19:47.220
in general for any transistor unless and until
we do unilateral assumption so for bilateral
00:19:47.220 --> 00:19:57.029
transistor S12 and S21 are different values
S21 so in that case when I have a bilateral
00:19:57.029 --> 00:20:05.960
transistor this gamma in that is a function
of gamma L and gamma out is a function of
00:20:05.960 --> 00:20:12.580
gamma S so together simultaneously the input
and output should be matched for conjugate
00:20:12.580 --> 00:20:18.540
matching and conjugate matching says that
you choose gamma in is equal to gamma S star
00:20:18.540 --> 00:20:28.529
and gamma out is equal to gamma L star so
that you can become a conjugate match and
00:20:28.529 --> 00:20:35.919
you get maximum power transfer. so this is
already given by the device manufacturer so
00:20:35.919 --> 00:20:43.950
our job is to now design for this ZS and this
input matching network and output matching
00:20:43.950 --> 00:20:51.679
network so I can get maximum of this ZS and
this ZL now last time we found have out the
00:20:51.679 --> 00:21:00.830
expression for this ZS . so that expression
I am writing again that or before that this
00:21:00.830 --> 00:21:07.070
GT we have also found out the expression for
GT in terms of S parameters of gamma S and
00:21:07.070 --> 00:21:13.149
gamma L now under conjugate match under GT
will be maximized transducer power gain is
00:21:13.149 --> 00:21:19.909
maximized and that equation obviously we are
assuming that this input matching network
00:21:19.909 --> 00:21:26.789
and output matching network are lossless we
have seen in the basic building blocks of
00:21:26.789 --> 00:21:32.609
microwave engineering NPTEL lectures that
how to design lossless input matching network
00:21:32.609 --> 00:21:37.880
and output matching network those are not
lump elements L section lumped elements they
00:21:37.880 --> 00:21:44.039
are used for lower frequencies is but at high
microwave frequencies you need to have transmission
00:21:44.039 --> 00:21:53.460
lines and transmission lines stubs to design
lossless input matching network. So that GT
00:21:53.460 --> 00:22:04.410
max will become 1minus this expression we
have already found last time S21 square and
00:22:04.410 --> 00:22:19.179
1 minus gamma L square by 1minus S22 gamma
L square so with our this nomenclature GS
00:22:19.179 --> 00:22:42.419
Z0 GL basically this I can say is my GS maximum
this is my GL maximum so now we have enforced
00:22:42.419 --> 00:22:57.510
this condition so again start from this basic
conditions that I have gamma in is equal to
00:22:57.510 --> 00:23:14.139
S11 plus S12 S21 gamma L by 1 minus S22 gamma
L gamma out is equal to S22 plus S12 S21 gamma
00:23:14.139 --> 00:23:27.450
S by 1 minus S11 gamma S and you see as i
am always saying that gamma in is a function
00:23:27.450 --> 00:23:34.799
of Gamma L and also S parameters of the device
gamma out is a function of gamma S and S parameters
00:23:34.799 --> 00:23:41.269
of the device. Now for conjugate matching
I know that this gamma N so this is general
00:23:41.269 --> 00:23:50.679
now conjugate matching that time will make
this choice that this gamma in I will choose
00:23:50.679 --> 00:24:03.710
as gamma S star so gamma S star is S11 plus
S12 S21 gamma L by 1 minus S22 gamma L and
00:24:03.710 --> 00:24:21.009
gamma L star that will be S22 plus S12 S21
gamma S by 1 minus S11 gamma S so you see
00:24:21.009 --> 00:24:28.190
that by conjugate matching I have replaced
gamma in and gamma out with their appropriate
00:24:28.190 --> 00:24:35.000
gamma S and gamma L values so you see this
is two equations here all S parameters are
00:24:35.000 --> 00:24:45.359
known I have two unknown two equation so I
can write two equations two unknowns what
00:24:45.359 --> 00:24:55.179
are the unknown gamma S and gamma L so I can
solve for that let us solve first for gamma
00:24:55.179 --> 00:25:04.679
S from this equation if I solve for gamma
S then gamma S become something like this
00:25:04.679 --> 00:25:14.859
you see . just simple manipulation but obviously
complex number manipulation so that part you
00:25:14.859 --> 00:25:42.179
need to take care but otherwise it is trivial
and this becomes
00:25:42.179 --> 00:25:51.690
everyone do it write take it pen and paper
and do it so this is a you see solving so
00:25:51.690 --> 00:26:01.549
I think I should write it solving for gamma
S from this two equations two unknown so I
00:26:01.549 --> 00:26:08.220
can solve for gamma S that means basically
you replace all gamma L values and you get
00:26:08.220 --> 00:26:19.139
quantitive equations in gamma S so gamma S
you say this is some complex number this is
00:26:19.139 --> 00:26:28.259
a real number this is a coefficient here this
is a complex number this is real number this
00:26:28.259 --> 00:26:35.929
is again a complex number o we can know the
solution of quantitic equation we can find
00:26:35.929 --> 00:26:49.210
the routes that will be given by if I call
that if this one this whole thing i call minus
00:26:49.210 --> 00:26:59.820
b1 this one I call A1 and this one I call
C1 so I know gamma S I can write remember
00:26:59.820 --> 00:27:09.639
instead of B1 I have to get minus B1 so gamma
S can retained as B1 plus minus root over
00:27:09.639 --> 00:27:18.129
square minus 4 you see just simple manipulation
you write it you know this is . . formula
00:27:18.129 --> 00:27:26.840
indian contribution so 2C1 only thing here
is this A1 C1 they are complex numbers . . formula
00:27:26.840 --> 00:27:35.000
still holds now this what is then I should
write what is B1 values. B1 is this is minus
00:27:35.000 --> 00:27:49.159
B1 so B1 will be 1 minus 1 plus S11 square
minus S22 square minus gamma magnitude square
00:27:49.159 --> 00:27:59.070
and C1 we will see A1 and C1 they will just
complex conjugate of each other so I need
00:27:59.070 --> 00:28:08.849
not write it as so A1 is present in this because
actually A1 is complex conjugate or C1 so
00:28:08.849 --> 00:28:16.669
that is why you are getting a 4C1 magnitude
square and C1 will this one S11 minus delta
00:28:16.669 --> 00:28:35.019
S22 star ok so with this we will have solution
provided this part this B1 square minus 4C1
00:28:35.019 --> 00:28:41.700
square you see B1 minus B1 is a real number
so B1 is also a real number so this and this
00:28:41.700 --> 00:28:49.269
is also a real number if this is greater than
0 then only this will have a solution so people
00:28:49.269 --> 00:28:59.629
have found out that ok this it can be shown
so first this will have a solution and once
00:28:59.629 --> 00:29:04.309
that solutions is there we can easily because
all this are known quantities you see there
00:29:04.309 --> 00:29:09.809
are no unknowns no gamma L is involved in
this solution in this only I need to know
00:29:09.809 --> 00:29:15.619
the S parameter values delta is also dependent
on net parameter so I can immediately solve
00:29:15.619 --> 00:29:22.690
for this the route will exist route will have
a physical existence if this is satisfied
00:29:22.690 --> 00:29:34.690
similarly if I solve for gamma L solving for
gamma L . we get another equation quantitive
00:29:34.690 --> 00:29:43.249
equation and that solution is we call it 2
or B2 plus minus root over B 2 square minus
00:29:43.249 --> 00:29:59.259
4C2 square by 2C2 where B2 is given 1 plus
S22 square minus S11 Square minus delta square
00:29:59.259 --> 00:30:17.580
and C2 is S22 minus delta S11 star now it
will have solution if B2 square minus 4C2
00:30:17.580 --> 00:30:24.700
square that is greater than 0 so I got two
such conditions B1 square minus 4C1 square
00:30:24.700 --> 00:30:30.849
greater than 0 and B2 square minus 4C2 square
were is 0 where beyond B1 B2 C1 C2 are this
00:30:30.849 --> 00:30:40.580
people have shown that this conditions enforcing
this two simultaneously that B1 square minus
00:30:40.580 --> 00:30:46.240
4C1 square greater than 0 and B2 square minus
4 magnitude C2 square greater than 0 this
00:30:46.240 --> 00:30:54.409
is equivalent to that criteria that K which
was rollest stability factor k greater than
00:30:54.409 --> 00:31:07.070
1 and delta less than 1 so that mean what
is this mean that if I have made a conditionally
00:31:07.070 --> 00:31:15.759
stable amplifier then this things are automatically
satisfied and so gamma S and gamma L are values
00:31:15.759 --> 00:31:23.610
so I can always find the gamma S and gamma
L value and when the roots for chosen like
00:31:23.610 --> 00:31:39.419
this then we can maximize GT max and that
is given by S21 by S12 K minus root over K
00:31:39.419 --> 00:31:48.289
square minus 1 so this gain that maximum gain
for conjugate matching that sometimes is also
00:31:48.289 --> 00:31:56.700
called matched gain actually more scientifically
it should be call conjugatly . . matched gain
00:31:56.700 --> 00:32:10.630
there is also a simple match case now if we
have a devices for whom if we have devices
00:32:10.630 --> 00:32:19.739
. for whom k is less than 1 then conjugate
matching is not possible then we have but
00:32:19.739 --> 00:32:27.529
what we can do we can enforce K is equal to
1 that is the maximum possible gain maximum
00:32:27.529 --> 00:32:40.429
possible transducer gain so that is called
G maximum stable gain this name is maximum
00:32:40.429 --> 00:32:53.830
stable gain and that is given by simply S21
by S12 so if you have if you have failed you
00:32:53.830 --> 00:33:00.359
devices have failed the rollets test then
also you can make proper choice from these
00:33:00.359 --> 00:33:08.989
and that time you will get a gain of S21 by
S12 usually in any practical transistor device
00:33:08.989 --> 00:33:18.909
S21 is a quite high value S12 is low value
so you get good about of gain even from a
00:33:18.909 --> 00:33:29.629
device without this now this becomes further
simplified if we do it for unilateral case
00:33:29.629 --> 00:33:36.940
that time the input and output their coupling
that means gamma in depends on gamma L that
00:33:36.940 --> 00:33:44.169
is not there for unilateral case you know
unilateral means S12 is 0 or approximately
00:33:44.169 --> 00:33:55.489
0 under that case we know that gamma S under
becomes that means S11 star and gamma L becomes
00:33:55.489 --> 00:34:05.349
S22 star so you see with this formula we can
write the GT unilateral maximum gain possible
00:34:05.349 --> 00:34:13.990
that will be given by simply 1minus S1 square
instead of 1 minus those other complication
00:34:13.990 --> 00:34:26.270
S21 square and 1 my minus S22 square so this
for unilateral case which many times we will
00:34:26.270 --> 00:34:32.169
see that we can assume this and that time
we can achieve a transducer gain maximum like
00:34:32.169 --> 00:34:38.649
this for unilateral case design is much simpler
so I have talked out all the possibilities
00:34:38.649 --> 00:34:45.579
we will see since this lecture does not have
any more time in the next lecture we will
00:34:45.579 --> 00:34:49.280
see a conjugate match design as an example
thank you