WEBVTT
Kind: captions
Language: en
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Welcome to this second lecture on concept
of image impedance Now I hope you agree with
00:00:31.580 --> 00:00:39.870
me that any 2 port network . can always be
represented by either a T section or a PIE
00:00:39.870 --> 00:00:47.690
section without losing a generality I take
that my 2 port network that means that ab
00:00:47.690 --> 00:00:58.960
issh transmission matrix characterization
is ABCD that is a T section the same analysis
00:00:58.960 --> 00:01:16.170
hold for PIE section so now I have a T section
this is Z1 this is Z2 this is Z3 all this
00:01:16.170 --> 00:01:25.110
are impedances complex impedances so this
is many port 2 this is my port 1 this is the
00:01:25.110 --> 00:01:41.970
internal description of the network now I
want when I will excide the port 1 with that
00:01:41.970 --> 00:01:51.940
voltage BS and I have some internal impedance
let us call that internal impedance Zi1last
00:01:51.940 --> 00:02:01.009
time I called it ZS this time I am calling
it ZI1 is simply change of numb…ain this
00:02:01.009 --> 00:02:08.640
is my port 1 this is my port 2 now you see
all of you are familiar with maximum power
00:02:08.640 --> 00:02:16.590
transfer theorem maximum power transfer theorem
says that if the load impedance complex is
00:02:16.590 --> 00:02:24.400
a complex conjugate of the source impedance
then maximum power gets transferred I think
00:02:24.400 --> 00:02:30.880
that you have noticed that in low frequency
particularly this VLSI people etc when they
00:02:30.880 --> 00:02:37.930
work upto gigahertz range they do not give
any consideration to these maximum power transfer
00:02:37.930 --> 00:02:46.290
theorem because in baseband unless you go
upto radio frequency and transmit it you have
00:02:46.290 --> 00:02:52.870
plenty of power so you are more concerned
with your voltage maximization thats why you
00:02:52.870 --> 00:03:03.680
design a good C amplifier with very high voltage
gain but voltage gain does not necessarily
00:03:03.680 --> 00:03:13.000
mean a maximum power gain but we when we go
to radio frequency we know power microwave
00:03:13.000 --> 00:03:19.140
power is very precious to produce microwave
power lots of complete complicated circuits
00:03:19.140 --> 00:03:26.790
are required So also when power is received
by a receiver in radio frequency it is very
00:03:26.790 --> 00:03:35.700
small amount so its life and death for RF
engineer or a RF circuits to maximize power
00:03:35.700 --> 00:03:43.900
So maximum power transfer theorem always is
the design of RF circuits always aa ba we
00:03:43.900 --> 00:03:52.570
try to pay regard to the maximum power transfer
theorem So can I have this whole thing suppose
00:03:52.570 --> 00:04:00.820
this I will terminate by some load impedance
let that load impedance is called ZI2 this
00:04:00.820 --> 00:04:08.630
one I called ZI1 source impedance is ZI2 now
the idea of image impedance is at this point
00:04:08.630 --> 00:04:18.321
obviously if I look at I will get some impedance
Now if this impedance is equal to ZI1 then
00:04:18.321 --> 00:04:29.330
I can from the source I can have maximum power
transfer ok now according to maximum power
00:04:29.330 --> 00:04:39.400
transfer I suppose these I am looking at some
Zin or something Zin now if ZI is complex
00:04:39.400 --> 00:04:48.020
conjugated ZI1 star then I know that maximum
power transfer will takes place But as I said
00:04:48.020 --> 00:04:58.240
that our consideration now is filter which
is lossless network so this Z1 Z2 Z3 there
00:04:58.240 --> 00:05:10.410
is no R involved ideally So they are complex
but they are generally they will be either
00:05:10.410 --> 00:05:18.350
real or not real either all of them will be
pure imaginary term there will be pure reactance
00:05:18.350 --> 00:05:28.199
So in that case I say that now I say that
I will look into here ZI1 then ZI1 is equal
00:05:28.199 --> 00:05:36.590
to ZI1 not comp not ness suppose any complex
conjugate for pure imaginary things it is
00:05:36.590 --> 00:05:46.680
equal as you know that suppose I have two
complex number A is equal to B star suppose
00:05:46.680 --> 00:05:52.310
A is complex number B is a complex number
if A is pure real and B is pure real then
00:05:52.310 --> 00:05:59.130
I can say A is equal to B Similarly if A is
pure imaginary and B is pure imaginary then
00:05:59.130 --> 00:06:07.250
only I can say A is equal to B Since we know
that this will be my all these are pure reactances
00:06:07.250 --> 00:06:17.030
this may be a pure resistance so that is why
I can call that my demand is ZI1 here should
00:06:17.030 --> 00:06:24.900
be equal to ZI1 here That means source impedance
and these input impedance looking at this
00:06:24.900 --> 00:06:32.300
port should be equal but you see this ZI1
is a function of this load resistance ZI2
00:06:32.300 --> 00:06:39.490
But now so why it is called this ZI1 is called
image impedance if I can find an impedance
00:06:39.490 --> 00:06:45.100
load impedance Zi2 so that if I terminate
this network this is already given network
00:06:45.100 --> 00:06:52.060
and I look at here and see that the input
impedance is ZI1 then I know that I can transfer
00:06:52.060 --> 00:07:01.300
maximum power from the source to this network
this to this port one of this network now
00:07:01.300 --> 00:07:08.639
why ZY ZI1 is called an image impedance because
if at this plane I look so I looking at this
00:07:08.639 --> 00:07:14.720
side I am getting to the right side and getting
and ZI1 looking at left side I am seeing the
00:07:14.720 --> 00:07:22.280
impedances ZI1 So the this side is image of
these that is why this is an image impedance
00:07:22.280 --> 00:07:36.400
it so ZI1 is an image impedance Same thing
here and here I want that if I look at here
00:07:36.400 --> 00:07:48.240
I should look at some output impedance that
should be equal to Zi2 So at port 2 if I look
00:07:48.240 --> 00:07:56.330
to this side I am getting ZI2 if I look this
side I should get ZI2 So if I can find PR
00:07:56.330 --> 00:08:06.199
to ZI1 ZI2 so that if I terminate by ZI2 then
I get here ZI1 impedance similarly here if
00:08:06.199 --> 00:08:14.720
I terminate by ZI1 and excite here I should
see here is ZI2 these two pairs are called
00:08:14.720 --> 00:08:21.050
image impedance Since this is an a symmetrical
network because ZI1 is not equal to ZI2 I
00:08:21.050 --> 00:08:39.889
will have two impedance ZI1 and ZI2 Let us
see that whether this image impedance can
00:08:39.889 --> 00:08:51.070
be represented in terms of this impedance
of this network . so the same diagram I can
00:08:51.070 --> 00:09:10.009
write that ZI1 or do like this so this ZI1
I can write as what is ZI1 obviously it is
00:09:10.009 --> 00:09:38.629
Z1 plus you see Z3 parallel to Z2 plus ZI2
Likewise what is ZI2 it is Z2 plus Z3 parallel
00:09:38.629 --> 00:09:53.249
to Z1 plus ZI1 Now these two equations you
see ZI1 here I have ZI2 I have ZI2 here I
00:09:53.249 --> 00:10:04.570
have ZI1 I have two equations so I can solve
for ZI1 and ZI2 in terms of Z1 Z2 Z3 If I
00:10:04.570 --> 00:10:23.369
do that upon solving this I get ZI1 is equal
to root over Z1 plus Z3 into Z1 Z2 plus Z2
00:10:23.369 --> 00:10:56.089
Z3 plus Z3 Z1by Z2 plus Z3 and ZI2 is Z2 plus
Z3 into Z1 Z2 plus Z2 Z3 plus Z3 Z1 So you
00:10:56.089 --> 00:11:05.319
see that this image impedance can be represented
in terms of the component impedances of the
00:11:05.319 --> 00:11:14.939
T section Now always we won’t be knowing
Z1 Z2 Z3 as I said that let us consider two
00:11:14.939 --> 00:11:23.339
port network as a black box But we can do
measurements and always find these image impedances
00:11:23.339 --> 00:11:33.329
How you know that any measurement requires
either an open circuit or short circuit of
00:11:33.329 --> 00:11:41.230
the one of the port So for any impedance measurement
you need to do this also you have seen that
00:11:41.230 --> 00:11:46.879
if you want to find any 2 port parameter you
need some port condition either short or open
00:11:46.879 --> 00:12:02.050
etc etc . So if we measure measurement of
image impedance
00:12:02.050 --> 00:12:16.959
let us say that one port one
we measure impedance
00:12:16.959 --> 00:12:30.910
when port 2 open We call that measurement
as Z1 since we are doing at port 1 open circuit
00:12:30.910 --> 00:12:42.639
Z1OC means I am measuring the impedance input
impedance at port 1 with port 2 open So if
00:12:42.639 --> 00:12:52.860
we look at the circuit if I open circuit this
what will be Z1OC it will be simply Z1 plus
00:12:52.860 --> 00:13:06.860
Z3 Similarly if we measure impedance at port
2 with port 2 with sorry imp the imp again
00:13:06.860 --> 00:13:21.480
port 1 measure impedance when port 2 is short
it now let me short this port So I call Z1SC
00:13:21.480 --> 00:13:31.019
second port is shortened you look at the circuit
If I short it it will be Z1 plus Z2 Z3 parallel
00:13:31.019 --> 00:13:52.689
So Z1 plus Z2 parallel Z3 Ok now what is this
this is Z1 Z2 plus Z2 Z3 plus Z3 Z1 by Z2
00:13:52.689 --> 00:14:04.629
plus Z3 Now you observe the image impedance
terms already I have solved ZI1 can I just
00:14:04.629 --> 00:14:20.299
compare can I say that Zi1 is equal to Z1OC
into Z1SC So by measurement I can always find
00:14:20.299 --> 00:14:29.609
Z1 OC I can find Z1 SC I know what image impedance
is immediately I can calculate from these
00:14:29.609 --> 00:14:37.529
Similarly instead of port 1 if i measure in
port 2 by once open circuiting port 1 find
00:14:37.529 --> 00:14:45.690
the input impedance at port 2 and then again
you short the port 1 and measure the input
00:14:45.690 --> 00:14:54.930
impedance and port 2 . So port 2 things if
we do you will see the same thing that ZI2
00:14:54.930 --> 00:15:10.329
can be expressed as Z2 OC into Z2 SC So this
shows that in image impedances can be always
00:15:10.329 --> 00:15:22.079
obtained from short or open circuit measurements
on any network So we can easily do this suppose
00:15:22.079 --> 00:15:31.499
I am given a network I can always find this
Z1 OC Z1 SC Z2 OC Z2 SC and find out tis Zi1
00:15:31.499 --> 00:15:40.850
Zi 2 and then I choose a source with that
internal impedance that I want and terminate
00:15:40.850 --> 00:15:51.919
or choose the load as Zi 2 I know I can achieve
maximum power transfer
00:15:51.919 --> 00:16:02.389
Already I said so that means I can have maximum
power transfer is guaranteed if I use image
00:16:02.389 --> 00:16:13.309
impedance as the terminating impedance at
both the ports Now
00:16:13.309 --> 00:16:21.549
now we know we have said that we will be using
the two port network as only lossless components
00:16:21.549 --> 00:16:34.249
that means you do not use any hard so their
own many internal loss there So by terminating
00:16:34.249 --> 00:16:45.990
with image impedances I assume maximum power
transfer no loss in the circuit lossless So
00:16:45.990 --> 00:16:54.959
image impedance is an important thing performance
measure of the power transport power transmission
00:16:54.959 --> 00:17:09.310
that is taking place to a network so you see
that we can specify something on it later
00:17:09.310 --> 00:17:17.380
when we will design a filter So instead of
ABCD we can specify image impedances and that
00:17:17.380 --> 00:17:28.020
will solve one many of our problems but think
one point that I have image impedance here
00:17:28.020 --> 00:17:36.810
I have image impedance here also you see with
this I require to know that ok by this Zi
00:17:36.810 --> 00:17:45.190
1 and Z Zi 2 terminations Zi 1 here and Zi
2 here I have to ensure that I am giving maximum
00:17:45.190 --> 00:17:53.390
power am delivering to this load But I am
assuming that here there is no loss but the
00:17:53.390 --> 00:17:58.790
power is flowing in this direction it may
so happen since I am using reactive elements
00:17:58.790 --> 00:18:06.950
power may be locally confined that is not
flowing there So I need to also see how propagation
00:18:06.950 --> 00:18:14.990
is taking place inside this 2 port network
So we need to have the transmission of power
00:18:14.990 --> 00:18:26.450
to the network also that we will next see
that this is called propagation of power . So
00:18:26.450 --> 00:18:45.030
what we define that again two port network
I have this V1 I have V2 I have I1 I have
00:18:45.030 --> 00:18:57.480
I 2 as before now let me define V1 I1 by V2
I2 VI is volt ampere the concept you have
00:18:57.480 --> 00:19:07.950
learnt in you electrical circuit class So
input volte ampere these are complex quantities
00:19:07.950 --> 00:19:20.580
input volte ampere and output volt ampere
What is the what is this ratio that will be
00:19:20.580 --> 00:19:30.130
something now I want to ensure that that is
fully making the power transmission possible
00:19:30.130 --> 00:19:35.090
So we call this I can name it any number this
will be some number you see input by output
00:19:35.090 --> 00:19:43.550
volt ampere but we have certain advantage
if we instead of defining any number here
00:19:43.550 --> 00:19:52.390
we write it as some exponential factor e to
the power two gamma why because you see when
00:19:52.390 --> 00:19:59.500
I will cascade many such networks this one
will have some this ratio e to the power 2
00:19:59.500 --> 00:20:07.340
gamma 1 this will have E to the power 2 gamma
2 another will have E to the power 2 gamma
00:20:07.340 --> 00:20:16.610
3 etc Now from this input to this input if
I want to find what is this transmission ratio
00:20:16.610 --> 00:20:24.060
of all the volte ampire if I express it exponential
factor the final thing will E to the power
00:20:24.060 --> 00:20:31.710
2 gamma 1 plus 2 gamma 2 plus 2 gamma 3 but
if I do not use this exponential factor If
00:20:31.710 --> 00:20:39.100
I just write it as gamma suppose then I will
have to work out and I will have to work out
00:20:39.100 --> 00:20:44.630
and I will have to find out what is the magnitude
and phase all these things here But exponential
00:20:44.630 --> 00:20:49.530
factor makes simply be an addition in if it
is an absolute value it would have been some
00:20:49.530 --> 00:21:00.550
multiplication We always prefer addition to
multiplication that is why it initially people
00:21:00.550 --> 00:21:07.220
did like that they put this as propagation
constant But now with after some learning
00:21:07.220 --> 00:21:14.290
people understood that if we represent this
ratio is it exponential factors and also you
00:21:14.290 --> 00:21:19.680
see I have taken a factor two here why because
many times will be interested to see what
00:21:19.680 --> 00:21:26.130
is the voltage ratio what is the current ratio
But this is actually a volt ampere ratio which
00:21:26.130 --> 00:21:33.860
is for that it is actually product of voltage
and current So I have taken two gamma this
00:21:33.860 --> 00:21:44.160
gamma is called propagation constant So what
is the definition of propagation constant
00:21:44.160 --> 00:21:56.800
you see gamma is equal to half LN V1 I1 by
V2I2 a very important definition propagation
00:21:56.800 --> 00:22:03.870
constant you see it shows that how input power
is propagating through the network inside
00:22:03.870 --> 00:22:14.480
the 2 port network . so my job is now to find
out what is this e to the power 2 gamma ratio
00:22:14.480 --> 00:22:26.610
is equal to V1 I1 by V2 I2 is equal to in
terms of ABCD parameters AV2 plus BI2 into
00:22:26.610 --> 00:22:41.530
CV2 plus DI2 by V2 I2 also I know V2 is equal
to ZI 2I image impedance it is terminated
00:22:41.530 --> 00:22:48.020
with image impedance so if you do that finally
you can solve that this ratio will turn out
00:22:48.020 --> 00:22:57.540
to be this simple manipulation put thus and
you know the value of Zi1ZI2 So you will get
00:22:57.540 --> 00:23:08.850
this will be simply this or e to the power
gamma is equal to root AD plus root BC Now
00:23:08.850 --> 00:23:17.890
here you see this propagation constant I have
expressed in terms of ABCD parameters One
00:23:17.890 --> 00:23:24.310
more thing is remaining I have already said
about characteristic impedances is characteristic
00:23:24.310 --> 00:23:31.530
impedances also expressible in terms of ABCD
parameters . let us see I have the same 2
00:23:31.530 --> 00:23:45.610
port network I have I1 here I have V1 here
and I want this should be ZI1 and here this
00:23:45.610 --> 00:23:58.710
should be terminated by ZI2 and this is V2
this is my I2 So I can write I know this is
00:23:58.710 --> 00:24:16.310
ABCD so V1I1 is equal to ABCD the definition
of transmission parameters also I have V2
00:24:16.310 --> 00:24:30.640
is equal to I2 ZI2 and V1 is equal to I1 ZI1
So put these equations and find out what the
00:24:30.640 --> 00:24:43.930
Zi1 you will see you will get AZI2 plus B
by CZi2 plus D let me call this for timing
00:24:43.930 --> 00:24:59.080
equation 1 . Now reverse the picture that
same transmission line this time I am putting
00:24:59.080 --> 00:25:11.280
the excitation here So I am looking at it
here I will get ZI2 and this is my V2 dashed
00:25:11.280 --> 00:25:23.360
as before this is my I2 dashed as before and
from here I am taking terminating you to Zi
00:25:23.360 --> 00:25:36.690
1 this is my V1dashed this is my I1 dashed
so here again I can write that V1 dashed I1
00:25:36.690 --> 00:25:51.700
dashed is equal to ABCD V2 dashed minus I2
dashed and what about the ports V1dashed is
00:25:51.700 --> 00:26:06.300
equal to ZI1 I1 dashed V2 dashed is equal
to ZI2 I2 dashed Then find out that what is
00:26:06.300 --> 00:26:18.360
your ZI2 So or you find what is your ZI1 which
is nothing but V1 dash by I1 dash that will
00:26:18.360 --> 00:26:35.270
turn out to be AZi2 minus B by minus C ZI2
plus D So this let me call equation 2 you
00:26:35.270 --> 00:26:40.290
have equation 1 you have equation 2 to solve
for Zi1.
00:26:40.290 --> 00:27:01.460
If you solve even so from 1 & 2 you can solve
for ZI1 and that will be equal to
00:27:01.460 --> 00:27:27.190
or ZI1 will be equal to AB by CD and ZI2 will
be equal to root over BD by AC Now I am happy
00:27:27.190 --> 00:27:38.520
because I know that ZI1 one of the image impedance
can be expressed in terms of four ABCD parameters
00:27:38.520 --> 00:27:46.840
ZI2 also I can express in terms of ABCD parameters
and I have already seen that propagation constant
00:27:46.840 --> 00:28:00.750
gamma you see this propagation constant that
also I can express in terms of ABCD parameters
00:28:00.750 --> 00:28:08.400
ABCD parameters completely characterization
2 port network I say equivalently I can say
00:28:08.400 --> 00:28:21.580
two image impedance ZI1 ZI2 and e to the and
gamma these three also characterizes a network
00:28:21.580 --> 00:28:32.110
But what is the beauty if I have ZI ZI 2 I
know what is i impedance level of the excitations
00:28:32.110 --> 00:28:36.590
of the network that means what is the source
impedance what is the load impedance they
00:28:36.590 --> 00:28:42.350
are according to the power matching So that
no maximum power sorry they are according
00:28:42.350 --> 00:28:49.490
to the maximum power will flow and by putting
conditions and gamma I will be able to say
00:28:49.490 --> 00:28:55.510
whether these frequency will pass or not So
instead of ABCD parameters this is a better
00:28:55.510 --> 00:29:04.140
description of a 2 port network if I want
to design a filter And already I have seen
00:29:04.140 --> 00:29:21.800
that I can that I can do the yes I can do
the measurement of image impedances that time
00:29:21.800 --> 00:29:30.559
I said in terms of the by opening and shorting
the port I will also have to prove that I
00:29:30.559 --> 00:29:37.630
can do this for propagation constant also
because this is a new thing that time I didn’t
00:29:37.630 --> 00:29:43.920
say these So that I will do now that measurement
of image impedance and propagation constant
00:29:43.920 --> 00:29:56.300
. So what we will do the same network this
is port 2 this is open circuit and I am looking
00:29:56.300 --> 00:30:13.070
here at let me call this Z01 So I know V1equal
to AV2 plus BI2 I1 is equal to CV2 plus Di2
00:30:13.070 --> 00:30:22.380
etc and open circuit means I2 is equal to
0 So if you enforce that is ZO1 that will
00:30:22.380 --> 00:30:34.470
be A by C then you short circuit so or open
circuit this port purpose you open circuit
00:30:34.470 --> 00:30:48.620
and measure here ZO2 so ZO2 that will be turn
out to be D by C then you do that short circuit
00:30:48.620 --> 00:31:00.120
port one and measure the from this port you
measure ZS2 you will see ZS2 will turn out
00:31:00.120 --> 00:31:16.809
to be B by A And
which one I missed ZS2 ZS1 this so you short
00:31:16.809 --> 00:31:31.290
circuit this port and measure here ZS1 ZS1
will be B by D . Once you have that you can
00:31:31.290 --> 00:31:41.460
immediately write because already we have
seen ZI1 is equal to the ZO1 into ZO2 etc
00:31:41.460 --> 00:31:53.430
So you will get that is equal to AB by CD
and that is nothing but ZO1 ZSL similary Zi2
00:31:53.430 --> 00:32:05.550
is equal to root over BD by AC and that is
D by C into B by A that is nothing but ZO2
00:32:05.550 --> 00:32:16.310
ZS2 And you see what is TAN gamma TAN hyperbolic
gamma all of you are familiar with hyperbolic
00:32:16.310 --> 00:32:25.680
functions So this is E to the power gamma
minus E to the power minus gamma by E to the
00:32:25.680 --> 00:32:41.080
power gamma plus and that is nothing but BC
by AD that is ZS1 by ZO1 and or ZS2 by ZO2
00:32:41.080 --> 00:32:51.840
So you see that gamma can be expressed completely
in terms of short circuit and open circuit
00:32:51.840 --> 00:33:02.170
measurement So I can measure image impedance
by open circuit short circuit measurements
00:33:02.170 --> 00:33:08.990
I can also measure gamma by open circuit short
circuit measurements Previously I showed that
00:33:08.990 --> 00:33:15.860
Zi1 Zi2 and gamma the completely characterizes
the network reciprocal 2 port network Now
00:33:15.860 --> 00:33:20.410
I have now I have shown that they also can
be measured so you do not have a difficulty
00:33:20.410 --> 00:33:29.860
any 2 port reciprocal network lossless network
you can represent like this So an alternate
00:33:29.860 --> 00:33:38.232
description for characterization of 2 port
network is in terms of ZI1 and ZI2 and gamma
00:33:38.232 --> 00:33:49.340
I think in the next class will introduce another
criteria all symmetrical network and we will
00:33:49.340 --> 00:33:51.669
simplify this procedure Thank you