1
00:00:18,800 --> 00:00:27,550
Welcome to this fifth lecture of this lecture
series, now in the last lecture we have seen
2
00:00:27,550 --> 00:00:39,160
how to design constant K filter section using
image parameter method today. We will point
3
00:00:39,160 --> 00:00:49,790
out 2 major drawback of this design . we look
the graph yesterday we have drawn these graphs
4
00:00:49,790 --> 00:01:04,270
that the characteristic impedance of the section
for, if we take a T section the characteristic
5
00:01:04,270 --> 00:01:13,740
impedance falls from a constant value K at
zero frequency to zero at cut off frequency.
6
00:01:13,740 --> 00:01:22,520
So characteristic impedance is changing so
it is a problem for matching the filter section
7
00:01:22,520 --> 00:01:30,410
impedance matching similarly this is the graph
for the PIE section characteristic impedance
8
00:01:30,410 --> 00:01:37,410
and you see here also from constant K it goes
to very high value so it is very difficult
9
00:01:37,410 --> 00:01:43,630
to match this so if we cannot match there
will be reflection and some power will be
10
00:01:43,630 --> 00:01:51,140
lost so even if we are using a lossless filter
due to impedance mismatch between source and
11
00:01:51,140 --> 00:02:00,679
load side there will be some problem and that
needs to be rectified also if we look at the
12
00:02:00,679 --> 00:02:06,470
attenuation constant graph we see that ok
in the passband from that means frequency
13
00:02:06,470 --> 00:02:14,160
from 0 to FC the filter is ok this is low
pass filter the it is ok there is no attenuation
14
00:02:14,160 --> 00:02:20,940
but once it crosses the pass band the stop
band there is attenuation that is also desirable
15
00:02:20,940 --> 00:02:29,910
but you see that near this cut off frequency
near this zone the attenuation value is not
16
00:02:29,910 --> 00:02:37,200
much so if I have another channel very near
to this cut off frequency or another radio
17
00:02:37,200 --> 00:02:45,220
signal then the attenuation won’t be much
once it is separated by reasonable distance
18
00:02:45,220 --> 00:02:51,890
then it goes to high value and it is also
desirable that when omega is very high the
19
00:02:51,890 --> 00:02:58,980
attenuation is infinite that means this portion
that is ok but this portion in some application
20
00:02:58,980 --> 00:03:06,130
may not be acc app acceptable because we want
that stop band there should be a good amount
21
00:03:06,130 --> 00:03:12,840
of attenuation but near the stop bandage for
this constant K section the attenuation is
22
00:03:12,840 --> 00:03:19,550
not much that is why this is also another
problem. So these 2 problem motivated people
23
00:03:19,550 --> 00:03:41,530
to design a better filter section that design
is called m-derived filter section. . So what
24
00:03:41,530 --> 00:03:53,290
is m-derived filter almost same as our constant
k filter I I start with PIE I start with T
25
00:03:53,290 --> 00:04:00,319
section so this is same as T section it is
a semmetrical section reciprocal element,
26
00:04:00,319 --> 00:04:15,370
lossless element all these are reactances
so I call now the new series impedance as
27
00:04:15,370 --> 00:04:25,599
Z1 dash by 2 so I am using a dash or primed
quantities to represent this and this I call
28
00:04:25,599 --> 00:04:46,939
Z2 dashed so this is empty right now. What
was our constant K? that was Z1 by 2 , Z1
29
00:04:46,939 --> 00:05:02,630
by 2z2. This was constant K today we are saying
it is Z1 dashed now is M-derived so how this
30
00:05:02,630 --> 00:05:19,639
m terms come the actually we let Z1 dashed
is equal to M Z 1that means we just M is a
31
00:05:19,639 --> 00:05:27,539
parameter we multiply Z1 dashed with this
parameter and then we want to keep the characteristic
32
00:05:27,539 --> 00:05:35,759
impedance for same with contact K so what
was the characteristic impedance of constant
33
00:05:35,759 --> 00:05:43,770
K. If I am calling that Z0 this time I am
using a subscript k to remind you that this
34
00:05:43,770 --> 00:06:02,139
is constant that was root over Z1 square plus
4 Z1 Z2 by 2 and the new cons m-derived filter
35
00:06:02,139 --> 00:06:14,490
characteristic impedance will be Z1 dashed
square plus 4 Z1 dashed Z2 dashed by 2 now
36
00:06:14,490 --> 00:06:22,949
we do not want to disturb this characteristics
impedance so we equate this 2, so we say that
37
00:06:22,949 --> 00:06:31,309
this 2 are equal once we say that then we
can find out Z2 from here because Z1 dashed
38
00:06:31,309 --> 00:06:44,039
we have expressed in terms of M Z1 so by solving
I get Z2 dashed is equal to Z2 by M plus 1
39
00:06:44,039 --> 00:06:59,020
minus M square BY 4 M Z1 so if you remember
that Z1 and Z2 they are the impedances of
40
00:06:59,020 --> 00:07:06,190
the series of the shunt term of the constant
K section and if you remember that always
41
00:07:06,190 --> 00:07:13,069
should have interaction they should be of
opp opposite nature that means if one is chosen
42
00:07:13,069 --> 00:07:20,289
as L and another is C and depending on whether
I choose L in the series are . . see in the
43
00:07:20,289 --> 00:07:28,689
shunt term I get a low pass or high pass action
etc so now you see that this equation says
44
00:07:28,689 --> 00:07:39,589
that Z2 dashed it will be now 2 elements one
is the scaled version of Z2 and in series
45
00:07:39,589 --> 00:07:48,749
with is a scaled version of Z1. Ok so the
dim new m-derived version of section I can
46
00:07:48,749 --> 00:08:24,749
derive as you so this will be Z2 by m and
this will be 1 minus m square by 4m Z1. So
47
00:08:24,749 --> 00:08:36,950
it is a series circuit in the shun term. So
this is the new design so we have seen the
48
00:08:36,950 --> 00:08:43,310
propagation haah ha characteristic impedance
now we have already known that filter has
49
00:08:43,310 --> 00:08:51,709
2 fundamental characteristics. 1 is characteristic
impedance another is propagation constant
50
00:08:51,709 --> 00:08:59,040
let us see what happens the propagation constant
. in this case so this propagation constant
51
00:08:59,040 --> 00:09:14,690
if you remember that yesterday we have derived
this TAN hyperbolic gamma is equal to root
52
00:09:14,690 --> 00:09:30,970
over Z 1 square by 4 Z1 Z2 by Z1 plus 2Z2.
So from here we will have to find our e to
53
00:09:30,970 --> 00:09:37,250
the power gamma it is easy to do because this
TAN hyperbolic gamma is equal to SIN hyperbolic
54
00:09:37,250 --> 00:09:42,410
gamma by COS hyperbolic gamma and you know
SIN hyperbolic gamma is e to the power gamma
55
00:09:42,410 --> 00:09:47,440
minus e to the power gamma by 2 e to the power
minus gamma by 2 plus COS hyperbolic gamma
56
00:09:47,440 --> 00:09:54,500
is e to the power gamma plus e to the power
minus gamma divided by 2. So if you that and
57
00:09:54,500 --> 00:10:02,589
then some component or dividend as a break
manipulation you get the value e to the power
58
00:10:02,589 --> 00:10:16,329
gamma will turn out to be simple mathematics
1 plus Z1 by 2 Z2 plus root over Z1 by Z2
59
00:10:16,329 --> 00:10:27,250
1plus Z1 by 4 Z2 this is the expression we
always we get now we are looking for m-derived
60
00:10:27,250 --> 00:10:40,810
section put prime quantities here e to the
power Z1 dashed okay. So this is the expression
61
00:10:40,810 --> 00:10:51,230
for e to the power gamma from these since
we know the values of Z1 and Z2 dashed. We
62
00:10:51,230 --> 00:11:00,480
can find out for the particular type loss
pass or high pass that what happens to gamma
63
00:11:00,480 --> 00:11:06,920
and once we have this expression for gamma
we can enforce the stop band and pass band
64
00:11:06,920 --> 00:11:12,600
specification and from that we can find out
what is the stop band pass band etc., So let
65
00:11:12,600 --> 00:11:44,560
us do that for . m-derived low pass filter
so you know we need to choose Z1 in constant
66
00:11:44,560 --> 00:11:55,810
K we choose Z1 is equal to j omega L Z2 is
equal to J omega C this is constant K so Z1
67
00:11:55,810 --> 00:12:09,750
dashed for m-derived section will choose as
J omega LM and Z2dashed as 1by j omega CM
68
00:12:09,750 --> 00:12:21,920
plus I minus m square by 4m j omega L. Okay
so now I can draw the in the component level
69
00:12:21,920 --> 00:12:50,060
this will look like this that I have so this
will be ML by 2 and this will also be ML by
70
00:12:50,060 --> 00:13:02,740
2 this will be M into C and this will be 1
minus M square by 4M into L. Okay so this
71
00:13:02,740 --> 00:13:10,980
is a m-derived section so find out as I said
what is the propagation factor now if we look
72
00:13:10,980 --> 00:13:19,399
at the propagation factor expression there
is a this important factor Z1 dashed by 4Z2
73
00:13:19,399 --> 00:13:25,900
dashed. If I know that I can find out a lot
because these factor always you see at Z1
74
00:13:25,900 --> 00:13:35,980
dashed by Z2 dashed, Z1 dashed by 2Z2 dashed
so if I can find this then I can have various
75
00:13:35,980 --> 00:13:43,180
I can easily find out the propagation factor
so let me find out what is this value Z1 dashed
76
00:13:43,180 --> 00:13:47,879
by 4 Z2dashed you will be able to do this
because Z1 dashed expression is given Z2dashed
77
00:13:47,879 --> 00:13:57,920
expression is given Z2dashed expression is
given if you do this it turns out to be like
78
00:13:57,920 --> 00:14:14,980
this good. So now once we have this term we
can enforce pass band we know that always
79
00:14:14,980 --> 00:14:25,629
for this image parameter method this pass
band is like this pass band . is when minus
80
00:14:25,629 --> 00:14:37,100
1 less than Z1 dash by 4 Z2 dash less than
zero. So this if we manipulate we get the
81
00:14:37,100 --> 00:14:56,610
4 minus 1 minus M square omega square LC by
M square LC greater than omega square greater
82
00:14:56,610 --> 00:15:06,089
than zero. So as we do that you can put this
part that means the pass band is extending
83
00:15:06,089 --> 00:15:12,920
from zero omega, omega is equal to zero 2
omega is equal to this value we can say as
84
00:15:12,920 --> 00:15:21,230
omega C and then we can solve for that and
that will turn out to be omega C is 2 by root
85
00:15:21,230 --> 00:15:32,041
over LC or FC root cutoff frequencies 1 by
PIE root over LC so you see that for M derive
86
00:15:32,041 --> 00:15:42,779
section also the cutoff frequency same so
the pass band extends from zero to f FC that
87
00:15:42,779 --> 00:15:55,990
means zero so alpha will be something like
this so there is no change in the low pass
88
00:15:55,990 --> 00:16:07,649
filtered structure. So then, what is the stop
band is as before you know that again since
89
00:16:07,649 --> 00:16:16,450
I have this quantity so I can enforce that
if this quantity is either it is greater than
90
00:16:16,450 --> 00:16:25,630
zero or it is less than minus1 I have pass
band. So from that I can say that the stop
91
00:16:25,630 --> 00:16:33,249
band .16:25) according to our yesterdays nomenclature
stop band one will come out to be when F is
92
00:16:33,249 --> 00:16:40,790
less than zero but we cannot have less than
zero so it is not possible so we did not bother
93
00:16:40,790 --> 00:16:52,240
about this stop band and the stop band 2 that
will be greater than FC obvious that’s why
94
00:16:52,240 --> 00:17:00,149
I have done like this knowing that these FC
will be like this now then what we have gained
95
00:17:00,149 --> 00:17:11,209
in m-derived section the cutoff frequency
is same etc the we have enforced the aaah
96
00:17:11,209 --> 00:17:16,590
characteristic impedance to be same as constant
K section. So then what we have gained to
97
00:17:16,590 --> 00:17:28,590
see that look at Z1 dashed by 4 Z2dashed expression
carefully I derived that again here that get
98
00:17:28,590 --> 00:17:41,240
Z1 dashed by 4Z2dashed this is omega square
m square LC by 4 minus omega square, 1 minus
99
00:17:41,240 --> 00:17:50,870
m Square LC now look at the dominator there
is a chance that denominator may go to zero
100
00:17:50,870 --> 00:17:55,880
and if that goes to zero this whole thing
will become infinity so that is a crucial
101
00:17:55,880 --> 00:18:02,430
point it was not present in constant K let
us solve this let us say that when frequency
102
00:18:02,430 --> 00:18:08,661
what if happens let us call that omega infinity
because this will go to infinity there. So
103
00:18:08,661 --> 00:18:17,490
what is omega infinity if we solve we get
that omega infinity is equal to 2 by root
104
00:18:17,490 --> 00:18:35,420
over 1 minus m square LC. So what is my omega
C for constant K 2 by root over LC omega infinities
105
00:18:35,420 --> 00:18:53,570
2 by 1 minus M square LC okay. You see this
is omega C so if I enforce that M is a parameter
106
00:18:53,570 --> 00:19:01,540
that I will choose I will keep it within 0
to 1 it is in my hand designer fan if keep
107
00:19:01,540 --> 00:19:13,130
it like this then omega infinity is greater
than omega C or F infinity is greater than
108
00:19:13,130 --> 00:19:23,940
omega C that means if infinity is in the paa
stop band 2. So in this band there is a this
109
00:19:23,940 --> 00:19:33,080
thing but what happens to the alpha there
let us see the e to the power gamma expression
110
00:19:33,080 --> 00:19:41,080
that will indicate what happens there so you
see that e to the power gamma expression now
111
00:19:41,080 --> 00:19:50,710
here this term is going to infinity now obviously
at that point this term will also high so
112
00:19:50,710 --> 00:19:55,830
you see this is infinity then this is also
infinity so the whole thing is infinity so
113
00:19:55,830 --> 00:20:09,450
can I say that at that point gamma at omega
is equal to omega infinity my gamma goes to
114
00:20:09,450 --> 00:20:21,470
infinity so gamma is very high. Now what is
gamma gammais alpha plus J beta we have proved
115
00:20:21,470 --> 00:20:31,410
that we are at stop band 2, so at stop band
2 what is the we know that beta beta is a
116
00:20:31,410 --> 00:20:39,320
constant is equal to 2M minus PIE so let us
say PIE so beta is constant and but alpha
117
00:20:39,320 --> 00:20:50,610
plus J that is very high so that means we
have alpha also is going to infinity there.
118
00:20:50,610 --> 00:20:59,430
So that was the thing we were looking for
when I said that the problem of this constant
119
00:20:59,430 --> 00:21:11,150
K section is near FC near FC I do not sufficient
alpha but I see here that I can place a pole
120
00:21:11,150 --> 00:21:17,060
very near to FC because that I can control
by choosing this parameter M because ah where
121
00:21:17,060 --> 00:21:25,950
is FC this is where is F infinity with respect
to FC that’s is in my hand because I can
122
00:21:25,950 --> 00:21:34,900
play with this M value and I can put anywhere
and then I can modify that ok the attenuation
123
00:21:34,900 --> 00:21:46,920
constant graph alpha this is FC suppose I
put so obviously here oh sorry that it will
124
00:21:46,920 --> 00:21:56,290
come like this but when I have a choice that
if I put suppose F infinity here then I can
125
00:21:56,290 --> 00:22:07,450
force the curve like this assume . .. So I
can give a very high attenuation just when
126
00:22:07,450 --> 00:22:12,610
I enter the stop band which was desirable
because many application says that do not
127
00:22:12,610 --> 00:22:18,910
disturb your neighbor channel so there I can
put sufficiently high attenuation. So how
128
00:22:18,910 --> 00:22:26,880
we are able to do these F infinity. You look
at this structure of M derive section basically
129
00:22:26,880 --> 00:22:34,670
I have a series circuit this series circuit
if it goes to resonance this series circuit
130
00:22:34,670 --> 00:22:44,850
if it goes to resonance that gives that high
attenuation. So it becomes a resonance circuit
131
00:22:44,850 --> 00:22:51,170
let us check whether we are physically it
is correct or not so we know that if we have
132
00:22:51,170 --> 00:23:02,170
a series like this what will be the resonant
frequency . F resonant for that will be a
133
00:23:02,170 --> 00:23:15,770
F resonance for that will be FR is equal to
what that 1 by 2 PIE root over MC into 1 minus
134
00:23:15,770 --> 00:23:28,500
M square by 4 M into L this M goes and so
I get 1 by 2 PIE root over 1 minus M square
135
00:23:28,500 --> 00:23:50,180
LC which is nothing but my F omega. F omega
is this by PIE this so that shows me that
136
00:23:50,180 --> 00:23:59,920
that shows me that F physically that is the
case since I have this series resonance circuit.
137
00:23:59,920 --> 00:24:07,200
I can make it resonance resonated F infinity
and by that I can put very high attenuated
138
00:24:07,200 --> 00:24:21,790
in the stop band but the price I pay for that
is unfortunately these thing that after F
139
00:24:21,790 --> 00:24:32,590
infinity the curve is like this so alpha gain
falls so I also want that at very high values
140
00:24:32,590 --> 00:24:38,040
or all further values attenuation should be
high but you see attenuation is falling due
141
00:24:38,040 --> 00:24:43,910
to this M derive section. So I can solve 1
problem but then in constant K this was not
142
00:24:43,910 --> 00:24:53,240
a problem in constant K it going exponentially.
So in constant K where is the constant K ah
143
00:24:53,240 --> 00:24:58,391
yes, so in constant k it was rising so at
very high frequencies it was high but due
144
00:24:58,391 --> 00:25:06,480
to my paying I could solve the problem here
but I have created another problem here so
145
00:25:06,480 --> 00:25:20,460
this is a problem that so it intro once again
the attenuation constant of M derive section
146
00:25:20,460 --> 00:25:32,890
that there will be FC there will be infinity
now I can put it shows that I come here then
147
00:25:32,890 --> 00:25:41,410
it goes to very high value and then it falls
like this, this is a M derive section alpha
148
00:25:41,410 --> 00:25:51,820
attenuation constant so what is the remedy,
remedy is we need to put in cascade this these
149
00:25:51,820 --> 00:26:01,670
a constant K filter section, let us see that
is the constant K filter that we have seen
150
00:26:01,670 --> 00:26:09,380
that time that it is thing something like
this so here you see this is a very sharp
151
00:26:09,380 --> 00:26:15,450
increase I wont have any problem this 2 if
they in the cascade when I put alpha will
152
00:26:15,450 --> 00:26:21,100
be almost like this and here also it is a
very sharp decrease so it will be dominated
153
00:26:21,100 --> 00:26:28,250
here also like this but then here you see
that this constant k section this is constant
154
00:26:28,250 --> 00:26:39,860
K, Constant K this is M derived. So, if I
combine then I get something which is called
155
00:26:39,860 --> 00:26:52,500
composite filter . so I get a attenuation
constant frequency that let me mark FC, let
156
00:26:52,500 --> 00:27:15,420
me mark F infinity it is these then it and
it goes almost like this then it comes and
157
00:27:15,420 --> 00:27:27,070
then goes like this so this is called composite,
composite of M derived in cascade with a constant
158
00:27:27,070 --> 00:27:34,490
K. So there you can see that I will have to
be careful about these part, that these alpha
159
00:27:34,490 --> 00:27:40,200
should accept this part for the application
that means nearby neighbor or next to that
160
00:27:40,200 --> 00:27:47,610
channel there is a low attenuation but if
this is sufficiently high this level for that
161
00:27:47,610 --> 00:27:54,340
application I can go with it I am dash dash
that is henceforward it will be higher and
162
00:27:54,340 --> 00:28:05,250
higher attenuation so these you see that I
have started with 2 problems that means this
163
00:28:05,250 --> 00:28:11,130
alpha was not very high here so that at least
with its composite design I have solve this
164
00:28:11,130 --> 00:28:17,170
problem but these problem is still unfinished
because we have enforces M derived section
165
00:28:17,170 --> 00:28:26,320
to be have same Z0 variation with some Z0
variation with constant K. So, characteristic
166
00:28:26,320 --> 00:28:32,470
impedance problem I have not solve M derived
section has not solve the problem but what
167
00:28:32,470 --> 00:28:38,660
people do then to match the section there
are some extra matching section at the load
168
00:28:38,660 --> 00:28:46,780
end and the source end they put so that makes
this whole filter think a bit bulky and all
169
00:28:46,780 --> 00:28:54,910
so that is the problem with image parameter
method but something is done and throughout
170
00:28:54,910 --> 00:29:05,070
aaah bbb this development for last uaah haaa
sixty seventy years people have used that
171
00:29:05,070 --> 00:29:13,050
in telephony in radio, radio reception etc
this filters were designed from this image
172
00:29:13,050 --> 00:29:22,870
parameter method but as we sure that this
problem can be solved if instead of all these
173
00:29:22,870 --> 00:29:30,080
problems that means we try to find out. Okay
if we choose these how would happen if we
174
00:29:30,080 --> 00:29:37,350
can say that from a specification I will specify
what is my stop band attenuation I will specify
175
00:29:37,350 --> 00:29:45,060
what is my stop band baah phase sorry pass
band phase I will specify what is my pass
176
00:29:45,060 --> 00:29:53,410
band attenuation that should be zero. So all
these if I space always zero is not haaa required
177
00:29:53,410 --> 00:29:59,000
because in application I require that okay
to a sufficiently low attenuation so I will
178
00:29:59,000 --> 00:30:05,990
specify that and that If we can find out that
what type of filter components that means
179
00:30:05,990 --> 00:30:13,110
what LC values how they are interconnected
there if I can find that means are basically
180
00:30:13,110 --> 00:30:19,760
a synthesis type of thing that I will specify
some attenuation characteristic and from that
181
00:30:19,760 --> 00:30:27,100
you find out which section which components
can achieve that so that is called insertion
182
00:30:27,100 --> 00:30:35,570
loss based filter design. All high frequency
filters microwave filters are based on that
183
00:30:35,570 --> 00:30:45,430
so that will take up in the next lecture but
these upto this this image parameters based
184
00:30:45,430 --> 00:30:52,900
filter design that give us insight about what
happens in a filter what is the characteristic
185
00:30:52,900 --> 00:31:01,120
impedance thing how it is very important for
matching the filter and also how to specify
186
00:31:01,120 --> 00:31:08,930
pass band stop band of a filter from the propagation
constant point of you. So that is the basic
187
00:31:08,930 --> 00:31:17,501
foundation on which we will see how insertion
loss base filter design can be attempted come
188
00:31:17,501 --> 00:31:18,480
to the next class.