Four Terminal Networks
1.1 Introduction In the complex communication systems, the d~signer may need to use different networks such as filters, attenuators, equalizers etc. All these networks may be realized in the different forms· of a four terminal network. It is essential to study different electrical properties of different types of the network. This chapter explains different electrical properties of symmetrical and asymmetrical networks and simple methods for deriving the expressions for the same. A four terminal network is a network having only one input port (or pair of terminals) and one output port. When the electrical properties of the network are unaffected even after interchanging input and output terminals, the network is called syl1,lmetrical network. When the electrical properties of the network get affected after interchanging input and output terminals, the network is called asymmetrical network or dissymmetrical network. .Since there are two pairs of terminals, a four terminal network is also called two port. network. Let us study the properties of the symmetrical and asymmetrical networks in detail.
1.2 Symmetrical Networks Any symmetrical network has two important electrical properties such as,
1) Characteristic impedance (Zo)
2) Propagation constant (y)
The two networks having the same electrical properties i.e. characteristic impedance
(Zo) and propagation constant (y) are called equivalent networks or identical networks. . in detail. Let us study the properties of the symmetrical networksesrmnotes.in|Class notes made easy.
(l)
Four Terminal Networks
2
Transmission Lines and Networks
1.2.1 Characteristic Impedance (Zo) Consider that infinite number of identical symmetrical networks are connected in cascade or tandem as shown in the Fig. 1.1 (a). The input impedance measured at the input terminals of the first network in the chain of infinite networks will have some finite value which depends on the network composition. This impedance i~ the important property of a symmetrical network. Thus the characteristic impedance of a symmetrical network is the impedance measured at the input terminals of the first network in the chain of infinite networks in cascade and it is represented by ZOo
I.
zo-1 ~ i b U ~ b J= ~ ";ty I
I
(a) Input impedance: at input terminals of first network of chain of infinite network I
~~~ J ~ d b J= ~~";~ I
(b) Input impedance at input terminals of second network of chain of infinite networks I
~-~1 i t
zo
I
I
(c) Input impedance of first network terminated in characteristic impedance
Fig. 1.1
If first network is disconnected from the chain as shown in the Fig. 1.1 (b), then also the input impedance measured at the input terminals of second network will be Zo again as number of networks in the chain are still infinite. That means we can replace this chain by impedance Zo at the output port of the first network as shown. in the Fig. 1.1 (c). Then the impedance at input terminals of the first network will be still ZOo Zo Vg
$
..
1
2
;
Symmetrical 4 terminal network
l'
~ 20
<
2'
Fig. 1.2 Correctly terminated symmetrical 4 terminal network
--
.....
Thus in general when any symmetrical network is terminated in its characteristic impedance Z01 the input impedance will also be ZOo This property is true for output impedance if the symmetrical
esrmnotes.in|Class notes made easy.
Transmission Lines and Networks
Min !Itt :!:he ~e
the ofa
mst
3
Four Terminal Networks
network terminated in Zo is driven by a generator with internal impedance equal to ZOo In such network, the output impedance will be Zo only. The network terminated in characteristic impedance at input as well as output terminals is said to be correctly terminated or properly terminated symm~trical network as shown in the Fig. 1.2.
1.2.2 Propagation Constant (y) Consider a chain of identical symmetrical networks connected in cascade as shown in the Fig. 1.3.
Fig. 1.3
The current leaving any section will be definite proportion of' that entering section and in general will be out of phase with it. Thus the relationship between the currents entering and leaving the section is a vector quantity with both modulus and angle. This quantity is represented in the form eY for convinence where y is a complex number given by 'Y = a. + jl3 . Let the ratio of input to output current be given by •
-I11s =
eY
... (1)
Since all the sections are identical, we can write
Is II 12 13 Y
= -==... =e II 12 13 14 . Thus,
I 2. Iz
=
I 2.. 13
= .
I I 2... ---l = eY • e Y = e ZY II I z I I I --E... ---l . 2 = eY • eY • eY = e3y 11 12 13
Hence for n identical sections connected is cascade the ratio of input to output current is given by
~ =. eny
IR
...
(2)
Note that input current is represented by sending end current, Is; while the output current is represented by receiving end current, IR . Above equation can be written as Is = e(a + jJl) = e na . ejnll
I R
esrmnotes.in|Class notes made easy.
,
....
Transmission Lines and Networks
'I ~ IR
I
Four Terminal Networks
4
= e nu
~COS2 n~ + sin 2 n~ L tan -1 sin n~
= e no
Ln~
Tran
I
cosn~
... (3)
where e nu gives ratio of absolute magnitudes of sending end current to receiving end current, and n~ gives the phase angle between these two currents.
~
If the network is correctly terminated, then we can write,
Is _ Is' 2 0 _ E s
1; - I R · 2 0 - E R
Also
E
_5 ER
=
= eno
e ny
L n~
... (4) ;nUl
The real part et of the propagation constant y is called attenuation constant and it is measured in nepers. eU
=
I~II
,I R
... for one section i.e. n = 1 (5)
et = In I :: I neper Similarly for n-sections, nu
e
=
I:: I
net = In
... for n-sections
I:: I neper
The imaginary part ~ of propagation constant y is called phase constant and is equal to the angle in radians by which output current leaving section lags that input current entering section. For n-sections, the phase constant will be n~ radians.
1.3 Asymmetrical Networks An asymmetrical network has following electrical properties, 1) Iterative impedance 2) Image impedance 3) Image transfer constant Let us study the properties of asymmetrical network in detail one by one.
1.3.1 Iterative impedance (Zo) Consider that infinite asymmetrical networks having identical electrical properties are connected in cascade as shown in the Fig. 1.4 (a) and (b). The iterative impedance is the impedance measured at one pair of terminals of the network in the chain of infinite networks as shown in the Fig. 1.4 (a) and (b). This is the impedance measured at any pair of terminals of the network when other pair of terminals is terminated in the impedance of the same value as shown in the Fig. 1.4 (c) and (d). esrmnotes.in|Class notes made easy.
,~
Four Terminal Networks
5
Transmission Lines and Networks
(a)
(b)
1((4))1
II Z'01
<'>
Asymmetrical network
"> Z'01 <:
<
Asymmetrical network
Z'02 <' ;>
(c)
-Z'02.
(d)
Fig. 1.4 Iterative impedances of asymmetrical network
The iterative impedances for any asymmetrical network are of different values when measured at different ports of the network. The iterative impedances are represented by 2 'th and 2 02 respectively at port land port 2.
1.3.2 Image impedances (Z;)
,
Similar to the iterative impedances, the image impedances are also of different values at different ports. Let the image. impedances be denoted by 2iJ and Ziz' Consider that the asymmetrical network is terminated with image impedance of port 2 Zi2 at its output pair of terminals then the impedance measured at its' input pair of is terminated terminals will be image impedance of port 1 i.e. Zi 1 . Similarly if port one • in the image impedance of port 1 i.e. Zil then the impedance measured at port two will be the image impedance of port 2 i.e. Ziz . These conditions are illustrated by the Fig.1.5 (a) and (b). 1
~
Asymmetrical network
1'
<
2'
(a)
2
1
2
Asymmetrical network
~
<;:
2'
1' (b)
Fig. 1.5 Image impedances of asymmetrical network
When an asymmetrical network is terminated in image impedances at both the ports the network is called correctly terminated asymmetrical network as shown in 'the Fig. 1.6.
esrmnotes.in|Class notes made easy.
1
Four Terminal Networks
6
Transmission Lines and Networks
'2
'1
+
+
Zj
,
Asymmetrical --Z'2 E2 I network
--E Zi2 1
E
~
"v
2
-
..
::2 Zi2 <
2
I~II
Fig. 1.6 Properly or correctly terminated asymmetrical network
1.3.3 Image transfer constant (e 8 )
\1
When an asymmetrical network is terminated in its image impedances at both the . ports as shown in the Fig. 1.6, then the ratio of currents ratio of voltages
:1.
:1
will be different from the
2
2
Hence image transfer constant eO
)E 1 1
=
e is defined
as
1
E 2 12
The real part of image transfer constant is called image attenuation constant; while the imaginary part is called image phase constant. After detailed discussion of various properties of symmetrical and asymmetrical networks let us study the properties of some of the important symmetrical and asymmetrical networks including symmetrical T, 11, lattice networks and asymmetrical networks such as half sections, L sections. Let us start with analysis of the symmetrical networks first.
.l
1.4 Symmetrical T Network
Z1
Z1
2
2
'=C' Z2
1'~2'
In line transmission theory, the symmetrical T network is the most frequently used network. The condition in the symmetrical T network is that the total series arm impedance and shunt arm impedance must be ZI and Z2 respectively. To have a total series arm impedance of ZI' the two
series arm impedances must be selected as
~1
each as shown in the Fig. 1.7. Fig. 1.7 Symmetrical T network
Let us derive the expressions for the characteristic impedance (Zo) and propagation constant (y) in terms of the network elements. esrmnotes.in|Class notes made easy.
"0-.."....
-
I:
-~ i
t \
l, '\
·~. l!.· ,
.
.
!It
Transmission Lines and Networks .
Four Terminal Networks
7
1.4.1 Characteristic Impedance (Zo) A) In terms of series and shunt arm impedances
Consider a symmetrical T network terminated at its output terminal with its characteristic impedance as shown in the Fig.1.8. By the property of the symmetrical network, the input impedance of such network terminated in Zo at other port is equal to ZOo Fig. 1.8 Symmetrical T network terminated in Zo
ZlN = Zo =
The input impedance of a T network is given by,
~l + [Z2 II (~l
+ Zo )]
Z/lZl +Zo) Zl 2 + ---'----=- 2 Zl Z2 +2+Zo
-
Zo( Z2 +
~l
+ Zo
)= ~1
(
Z2 +
~1
+ Zo )+
Zl~2 + Z2 Z 0.
Z Z 2 1 2 0 + Z2 - 2 1 Z 2 Z~ Zl Z 0 Z1 2 2 Z Z . 2 0 + 2 0 - -2- + 4 + -2-- + - 2 - + 2 0
Z6
=
Z2 -t + Zl 2
2
'" (1)
Zo
=
~~L + ZlZ2
... (2)
B) In terms of open and short circuit impedances
For symmetrical networks, impedances measured at any pair of terminals with other pair of terminals either open circuit or short circuit are of same value. Consider symmetrical T network with terminals 2-2' either open circuit or short circuit as shown in the Fig. 1.9. Z1
'
Z1
.z,c~n
l'
2' (b)
Fig. 1.9 Open and short circuit impedances of symmetrical T networknotes made easy. esrmnotes.in|Class
Four Terminal Networks
8
Transmission Lines and NetworkS
Consider Fig. 1.9 (a), ZlOe
J
=
.
Z
Zzoe
= Zoe = _1 . 2 + Z2
Zzsc
= Zsc =
Consider Fig. 1.9 (b), ZISC
~
Zse
= =
... (3)
i +[Zd
IIZ2]
I
Tra.,.
I
Fro
I
f.n
Z1 2 2 -2
Zl
-+--
Zl Z -+ 2 2
2
Z~ ZlZ2 Zl Z 2 -+--+- Zsc -- 4 2 2 + Z2
(~l
... (4)
J
Multiplying equations (3) and (4), we can write,
.
z
Zoe' Zsc =
( -1- +Z2
~
ZocZsc
=
)
1-+ Z2
Z + Zl 2
. (Zl -+Z2 2
l
)j =
Z2 Z -++Zl 2
Z6
Zo = ~Zoc' Zsc ... (5) Hence in any symmetrical network, the characteristic impedance 2 0 is the geometric mean of open and short circuit impedances measured at any pafr of terminals. Ex. 1.1: Find the characteristic impedance of a T section shown in the Fig. 1.10. VerifiJ the value of impedance with the help of open and short circuit impedances.
100.\1
100.\1 2 ~ 400.\1
ii 1'
'I t
I:
2'
Fig. 1.10
Sol. :
The characteristic impedance of the symmetrical T network is given by Zo =
22
~T +Zl Z 2 esrmnotes.in|Class notes made easy.
,'-
-..works Transmission Lines and Networks
Prom the network,
... (3)
Zo
Zl- = 100 n =
~ (20Z)
= 200 nand Z2 = 400 n
i.e. Zl
2
=
= 300 n
+ (200) (400)
From Fig. 1.10,
Zoe = (100 + 400) = 500 Zse
Four Terminal Networks
9
II
(100 + [100
n
400])
= 100 + 80 = 180 n.
By property of the symmetrical network,
Zo
=
= ~ (500) (180)
/Zoe' Zse
= 300
n
Ex. 1.2: A symmetrical T network consisting pure resistances has following open and short ... (4)
circuit impedances. Zoe = 800 L 0° n.
Zse == 600 L 0° n.
Design symmetrical T network
Sol. :
Consider Pig. 1.8 (a)
Zoe
=
Consider Fig. 1.8 (b)
Zse ... (5)
.....11JlIr:ir
=
Z
_1
2
i
.
+ Z2 = 800
+ ( Z2 II
... (i)
~1 = 600
... (ii)
)
Squaring equation (i), we can write,
mean
z~ 2 4+ Z2 + Zl Z 2
= 640000
. Vm,fu the
t
Z; = 640000
Z;
= 640000 -
Z2 -t+2 2 1
] 2
Z5
Z~ = (640000) - (Zoe' Zsd
Z;
= [(640000) - (800) (600)]
Z~ = 160000
2 2 = ± 400 Q.
..
But as the network is of pure resistance, neglecting negative value of Z2' Z2 = 400 n
..
... (iii)
Putting value of Z2 in equation (i), we can write,
i
+ 400 = 800
esrmnotes.in|Class notes made easy.
Transmission Lines and Networks
Four Terminal Networks
10
Z1 == 400.0
.. , (iv)
2
Thus a symmetrical T network can be drawn as shown in the Fig. 1.11.
J
Fr
~= 400Q ~= 400Q 2
~
2
'~~2 :::OOQ
I
II I
TranSl
l'
-02'
co"'"
Fig. 1.11
1.4.2 Propagation Constant (y) 21
Is 1
"2
21
+"2 _ 2
r
'R
~
ys-I R )
Zo
Z2
l'
entering •and
The currents leaving network are assumed to be Is and IR respectively.
2'
Fig. 1.12 Correctly·terminated symmetrical T network
- II{ -(
Consider properly terminated symmetrical T network as shown in the Fig. 1.12.
i )-
Applying KVL to the outer loop,
II{ . (Zo) + (Is - I R ) . Z2 == 0 (Z2) Is == ( 2 2 + ~1 + 2 0 ) IR
Hence
[
Z1 2 2 +--+2 0
IR
2 2
eY == 2- ==
eY == 1 +
2
Z
_1_
22 2
2 +---.9_
.. , (1)
22
The propagation constant of T section is therefore given by
z; J
Z1 - + 2 0 y == in [ 1 + 22 2
... (2)
Putting value of Zo in equation (1), we can write,
I~-+-2-]-2-2
1 + _1_ + --,~_---=,.---_ _
Z
e Y ==
22 2
2 2
esrmnotes.in|Class notes made easy.
I
.I
Transmission Lines and Networks
Four Terminal Networks
11
... (3)
From above equation we can write, 1
... (4)
,., (5)
. , (1)
tanh y =
",'12)
tanh y =
20
( 2; + 2
_ 2
) -
~-Z~~ Zoe
= Gsc_
~ Zoe
Above equation is very useful as it enables y to be calculated from open and short circuit impedances.
esrmnotes.in|Class notes made easy.
~
--'..-..-_~---.,,=~-
Four Terminal Networks
Transl
Thus both important electrical properties, the characteristic impedance (Zo) and the propagation constant (y) of a network can be determined from open and short circuit impedances. From this, we can conclude that the two networks will behave similarly, if they have the same Zoc and Zsc.
SiJ
12
Transmission Lines and Networks
:t
pror a ;
1.4.3 Series and Shunt Arm Impedances in Tenns of Characteristic Impedance (Zo) and Propagation Constant (y)
H and p
Consider equation (5) from section 1.4.2,
Z
cosh y = 1 + 1_ 2Z 2
t
Z1
cosh y - 1
2Z 2
!
2 sinh2 r = Z1 2 2Z 2 sinh2
'h Y
2:
sm
Z1 42 2
r2
"'IJ
rz;
~'
= {4~
'" (1)
Consider equation (6) from section 8.4.2,
.
sinh Y .
,
2smh
y y cosh ~ 2 2
cosh
cosh
r2
i=
=
2
_0 22
s.L
= --22 0 2
20
22 2 sinh ~
12.)~~12 = J~~2
2 2
'" (2)
Dividing equation (1) by (2), tanh
r2
sinh~ --
cos h
i
) 21
42 2
.
-/2;Z; = ~ 20
22 0
Thus each series arm impedance, in terms of, the characteristic impedance and propagation constant is given by 21 = Zo' tanh '2Y ... easy. (3) esrmnotes.in|Class notes made
-:c
Transmission Lines and Networks
Four Terminal Networks
13
Similarly, shunt arm impedance in terms of the characteristic impedance and propagation constant is given from equation (6) in section 1.4.2. Zo Z2 == - - ... (4) sinhy
lII'lilCtleristic
Hence T network with components expressed in terms of characteristic impedance and propagation constant is as shown in the Fig. 1.13.
Fig. 1.13 Symmetrical T network impedances in terms of
and YT
Ex. 1.3: If the measurements made on a box enclosing a two port network are ZlOC = 40 Loon and 2 1SC = 20.3 L 29. 8°n Find values of characteristic impedance and propagation constant alongwith attenuation constant and phase constant if the network is symmetrical. Sol. : As network is symmetrical,
... (1)
2 10e
'" (2)
ZOT
I I
=
Z20e
= Zoe and
2 1se == Z2SC == Zsc Hence we can write,
2 0 == ~ Zoe . Zse
20 ~ [40LOO] [20.3L 29.8°]
2 0 ==
J 812 L 29.8°
Zo == 28.495 LI4.9°n
The propagation constant can be expressed in terms of open and short circuit impedance as, tanh y
I Zse
'Iv Zoe
tanh y =
..
tpedance and
20.3L29.8° 40LO°
tanh y 0.7123 L14.9°
tanh y == 0.6884 + j 0.1831
But
... (3)
tanh y ==
0.6884 +jO.1831 _ N
e 2y -1 == 2y -D 1 e +1 esrmnotes.in|Class notes made easy.
-
-------
-
J
,
Four Terminal Networks
14
Transmission Lines and Networks
Ex. 1.5'
Using componendo and dividendo, N+D _ 2e 2r _ 1.6884+jO.1831 N - 0 - ~ - -0.3116 + j 0.1831 e
1.6884+jO.1831_ 1.6981L6.1289° - 0.3116-jO.1831 - 0.3605L-30.fn
e2:r
= 4.71 L36.32°
2y
Transm
Sol. :
_
0
The
Taking natural logarithm on both the sides of above equation,
2y = In [4.71] + j 36.32° ... In [a L bj
:=
In [aj + jb
1lJJlIe
y = 0.7748 + j 18.16
Thus, attenuation constant a=0.7748 N
Phase constant 13 = 18.16° = 0.3169 rad
Ex. 1.4: Design symmetrical T section to have Zo
= 600 nand y = 0 + j i
JIll
Sol. :
I
1,.5
I
= j 248.528 n
The shunt arm impedance of symmetrical T network is given by
Zo- = 600 600' . h('e) Z2 = =-... sm J sinhy . sm J'4 J5m '4
h(.TI) .. (TIl)
. e
= J.sm 1!'
- j " ~~~" = - j 848.528 n Hence, the symmetrical T network to have 2 0
= 600 nand y = 0 + j ~ is as shown
in the Fig. 1.14.
1.5.' Z1
Z1
2"
'2
1~~2 j248.528Q j2~~.~~~~_
-l
~ - j848.528Q
Fig. 1.14
esrmnotes.in|Class notes made easy.
~
I
Transmission Lines and Networks
Four Terminal Networks
15
Ex. 1.5: A line is composed of T section of pure resistances. Calculate characteristic impedance and propagation constant if each series arm is 50 0 and shunt arm is 5000 O.
~
Sol. :
= 50 0 Le. 2 1 = 100 0 and 2 2 = 5000 0
The characteristic impedance is given by .
.
rzr-,-V4+Z,Z2 =V--4-+(100)(5000) =708.870
r(iO-O)-='?:-
20 =
The propagation constant is given by, y = In
[1 +~_ +~] = In [1 +_1~ +!~8.87l 2Z Z 2 (5000) 5000 J 2
= 0.1413
Note:
2
+ jO = 0.1413
From above solved examples Ex. 1.4 and Ex. 1.5 it is clear that when network is composed of pure reactive elements, attenuation constant becomes zero and such network provides only the phase shift (Le. y = 0 + j P). Similarly when the network is composed of pure resistive elements, then such network provides attenuation only, with no phase shift. (y = a + jO)
1.5 Symmetrical
11:
Network
cr--.,...-...J\f'\IV\v~---,---o2
1'o----'---------.J...---o 2 '
Fig. 1.15 Symmetrical
1l
The symmetrical n network is another important network in line transmission fulfilling the conditions of total series and shunt arm impedances as 2 1 and 2 2 respectively. Thus the series arm impedance of a n network is selected as 2 1 and to have a total shunt arm impedance. of 2 2, each shunt arm impedance is selected as 22 2 as shown in the Fig. 1.15.
network
Similar to the symmetrical T network, let us derive the expressions for the characteristic impedance (20) and propagation constant (y) of the symmetricaln network.
··1.5.1 Characteristic Impedance (Zo) A) In terms of series and shunt arm impedances
Consider a symmetrical n network terminated at its output terminals with its characteristic impedance 2 0 . as shown in the Fig. 1.16.
FiQ,. 1.16 A symmetrical
1t
network terminated with Zo
esrmnotes.in|Class notes made easy.
n Trans .;..;.-
Us
Four Terminal Networks
16
Transmission Lines and Networks
By the property of the symmetrical network, the input impedance of such network terminated with Zo at other port is equal to ZOo The input impedance of a symmetrical n network is given by ZIN = Zo = 2Z2 II [Z1 + (2Z 2 II Zo)J Z
, I
Z
\
o o
1
= =
Zo
=
20
=
2Z
2
2Z2
II
[z
1
+ 2Z 2 Z 0 ] 2Z 2 +Z 0
II [Z~ (22 2 + 2 0 ) + 2Z 2 Z 0 ] 22+2 2
0
422 2 6 + 2 126
So
=
l
2Z [2212 2 + 2 12 0 + 2 2 22 0 2 2 Z2 + 2 0 J Z 22 + 2Z1Z 2 + Z1 0 + 2 2 22 0 2 2 Z2 + 2 0 2Z 2 (2Z 1Z 2 + Z1 Z 0 + 2Z 2 Z 0 ) 2Z 2 (2Z 2 +Zo)+2Z1Z 2 +Z1 Z 0 +2Z 2Z 0
42~ 2 0 + 22 226 + 2212 22 0 + 2 126 + 22 226
Ex.
.
= 4Z12~
+ 221Z22 0 + 4Z~20
4Z1Z~
26 (ZI + 4Z 2 ) = 4Z 1 Z~
,
26
2 1 +42 2
r~
~H
4
26
=
r ~
Z 12 Z22 2
~.+Z1Z2 4
Taking square roots on both the sides,
2 1 Z2
Zo
'" (1)
~Zl +Z1 Z2
h~u
t\
/7
Multiplying numerator and denominator by the factor Z1,
\'~1: ~,
= 4212~
Writting 2 0T for the characteristic impedance of a T section and Zo" for the characteristic impedance of a n-section having the same series 'and shunt arm impedances,
Zo"
or
=
2 1 2}
2
... (2)
0T
... (3)
2 0T 2 0" = 2 1 2 2
B) In terms of open and short circuit impedances
Consider symmetricaln network shown in the Fig. 1.17 (a) and Fig. 1.17 (b).
esrmnotes.in|Class notes made easy.
Transmission Lines and Networks
Four Terminal Networks
17
Consider Fig. 1.17 (a), Z1 o-----.--'VV'V\r---.----()2 <
-
2Z z
2Z z
..
<
1 'o---~-----'-----()2· 1'
2'
(a)
(b)
Fig. 1.17 Open and short circuit impedances of symmetrical
II (Z +
Zoe = 2Z 2
1
1t
network
2Z) = 2Z 2 (Zl +2Z 2 ) 2 2 1 +42 2
... (3)
Consider Fig. 1.17 (b),
Zse
=
2Z 2
II Z = 1
2Z 1 2 2 (2 1 + 22 2 )
'" (4)
i\tultiplying equations (3) and (4), we can write, 2Z 2 (2i + 2Z;; 22 1 2 2 Zoe' Zse = 2Z
(z-':t
(2 1 +42 2 )
Zoe' Zse
__
Zoe' Zse =
r)
4,ZlZ~
. Zl + 4Z 2
(~1
) Z21 Z22 4Z1Z~ '--= = Z~ Z2 Zl + 4Z 2 Z1 Z 2
(~1 )
T+
Thus the characteristic impedance of symmetrical n network is given by
Zo
=
ZOn
=) Zoe' Zse
... (5)
Ex. 1.6: Find characteristic impedance of a symmetrical
section shown in the Fig. 1.18.
(800+jO)Q
·.. (1)
O------.-~'VVVlA.r---r---o2
(1000+jO)Q
z'h
>
<
for the shunt arm
1.11 (b).
1t
(1000+jO)Q
1,o---~----~----02'
Fig. 1.18
... (2) ... (3)
Sol. :
2Z 2 = 1000 n i.e. Z2 = 500 .Q (800) (500) = 534.52 .Q (800) 2 + (800) (500) 4
esrmnotes.in|Class notes made easy.
-----
- ~
18
Transmission Lines and Networks
Tn
Four Terminal Networks
1.5.2 Propagation Constant (y) Consider correctly terminated symmetricaJ
1t
tot
network as shown in the Fig. 1.19.
:::;,
i
EttEE1
l
I
~~
s~,
1~
2'
1'
Fig. 1.19 Correctly terminated symmetrical
It
network
. As the network is symmetrical, by definition,
Is E v
- E s - e' I R
R
By potential divider rule;
E
-
R
E
-
S
I
(2 Z2 II Zo )
]
l Zl + (2 Z211 Zo)
~2 Z2 Z 0
ER =
.
r 2 Z2 + Zo Es 1-----
lZl
.,.?,
+ 2 Z2 Z 0 2 Z2 + Zo
E I 2 Z2 Z 0 I s Zl (2 Z2 + Zo) + 2 Z2 Zo
ER
l
ES
J
e Y == [2 Z2 Z 0 +2Z 1 Z 2 +Zl Z 0 l 2 Z2 Z 0
J
ER
eY =
Zl Zo
Zl 2 Z2
1+-+~-
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eYrt
:=:
Z + __ Z 1
1 +_1_ 2Z 2
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i.e.
. I
.....
Zo"
Y"
:=:
In[l +_3L +~] 2 Z2 ZOT
e Y"
:=:
1+~+
e Yrt
=
Y"
=:
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l
2Z 2
YT
Zl (Zl Z2) ZOT
+ ZOT 22
=:
... (1)
... as Zo"
Zl Z 2 ZOT
=-
e YT . .. (2) esrmnotes.in|Class notes made easy.
Transmission Lines and Networks
Four Terminal Networks
19
Thus propagation constatnts of symmetrical T and 11: networks are same because the total series arm and shunt arm impedances are same in the assumed symmetrical T and 11: sections. ' . Hence all other expressions for Y derived for symmetrical T network apply to the symmetrical 11: network also.
1.5.3 Series and Shunt Arm Impedances of Symmetrical Terms of ZOn and Yn :
TC
Network in
For symmetrical T network, each series arm is given by, Zl YT 2 = ZOT tanh 2 Putting
2 0T
=
Z Z
and YT
- z 2l
= Yn'
On
= -ZOn --
... (1)
tanh h
2
Similarly shunt arm of symmetrical T network is given by 2 0T Z2 = --- sinh YT Putting
ZOT
Zl Z2 -Z-- and YT = Yn' On
2 1 Z2 ZOn
sinh Yn Zl = ZOn sinh Yn '" (2) Hence n: network with series and· shunt arms expressed in terms of 2 0n and Yn is as shown in the Fig. 1.20.
(11 v
~2)
Fig. 1.20 Symmetrical
1t
network impedances in terms of Z Orr and y 7tnotes made easy. esrmnotes.in|Class
20
Transmission Lines and Networks
t
Four Terminal Networks
jl-c calculate its characteristic
Ex. 1.7
For a symmetrical n network, Zl == j wL,
Sol. ;
impedance at 500 Hz and 1000 Hz if L == 0.1 Hand C == 2j.l F. The symmetrical n network is as shown in the Fig. 1.21.
Z2 =
1
~
Z1
lO' L=O.1H
=:= C=2~F
2Z 2
--
,
C=2fl F=*2Z2
Fig. 1.21
Given
Z1 == j wL 1 Z2 jwC
2 2 Z2 =JUlC
The characteristic impedance of the symmetricaln network is given by, Z ZOrr ==' l Z2 Z Z1 Z2
_f{
-t+
( 1 ) (jwL)ljwC
ZOrr
lJW~) .
2
(
1 )
+(jw L) jwC
L C
ZOrr
~_w
2
e
4
1) At £ == 500 Hz, w == (2 x n x 500) == 1000 n
0.1
2x 10-
- - -6
ZOlt ==
r
1
0.1 _ (2x nx 500)2 x (0.1)2l 2 2xl0-6 J . 4
l
= 314.18 n
" --.
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84
2.6.4.3 Variation of Z or and Z On I Attenuation Constant ( a ), Phase Constant~ Frequency The variations of ZOT and ZOn' attenuation constant (a) and phase shown in the Fig. 2.23 (a), (b) and (c) respectively. Consider that f1 cut-off frequencies and Ro is the design impedance of the band eliminati Zo
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Filters
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Transmission Lines and Networks
Attenuation (0.)
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Ex. 2.5
r'f) ( Y
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(Z0)'
2.8.3.1 Variations of Characteristic Impedance Shift (~) with Frequency
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The variations of characteristic impedance (2 0 ), attenuation constant (a) and phase shift (~) vvith frequency are as shown in the Fig. 2.34 (a), (b) and (c) respectively.
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C
"
I
~)
f
I
. (a) Variation of Zo with frequency (f)
Pha,,,h;ftIP)
co
I
..
Frequency (t)
(b) Variation of a with frequency
1
: m HPF CHARACTER
-T- - -i - ---
.
1!C \ - - - - - -
I
01
I
I
I
I I I
I I I
I - - - - + - - - - I I I I
C
I
l:t
.w
f
f
-1!
'"
Frequency .. (f)
C
(c) Variation of
p with frequency esrmnotes.in|Class notes made easy.
Fig. 2.34
As the characteristic impedance for both the sections i.e. prototype section and m-serived section, the variation of 2 0 in m-derived section is similar to that in the prototype filter section. The attenuation constant a in the attenuation band is given by,
a " 2 cosh-'
r fJ ~ m(
I
l"~
p
11
-( f", f
Y
J
J
lor fro < 1< fe
c~
__
LJ..
1 I
--------
a
esrmnotes.in|Class notes made easy.
.~
L
zf .
().:y:)
LI Cl - ,
!!~
UJcP (/
.L~ WcP
/_yA2
J
[~o.J
[ f; -I]
1
L~::. ~I<- U1.--~lJ
Ct
.11
[Jc;"~I] to·
7\
1cP
~ !" - ~
LJ.CI
1
Ir 7f F,d I I L.
Ir. 7f -{ 1 -{1-
(r-yY\~
~4
L
L
L...::o
A 7\L.i":r
W((,A) 1r7i.1, 1....
f1.--J,_ hr~-t,J~
esrmnotes.in|Class notes made easy.
II J
F ')L
1;:.
JcP
~L
~ ~
'l.-
')L
I
".
-f ~ ( f v
k -:::
)l-! cP-'- _ l cf' U,-·i 0
- "'i- .f
I
.f 1J
!L L))t L
+- b:>( -e L
~ 0
~:± {h1--hu--t "-- ~O\~ 0...
b
~
~
:>L
-e'2- -4,J
L ':. -:)L
~Jt.
rJt
-
-
~
(j"Y""~
/1.--1 J +
--
'2-..:>'L
- -
--
AI
/1-_
-
:± JCJ.a- ~ J '- i
--.
IvJr .
J.tJC'- f,-/L
M2
t
')l~ ~L ')t 1--..:
(j>-_1,),.L -+~0t I t ) t 2-
L /- V\l\1,..,J
~
--
NrJ/~ ~
;).,~ J-~'fm (,)"d~"
~U1{ ')"Our (.$ q.,t~h1f
'/kI..,. }:J.--41 . sa Joe ~ {I-;;-'" A(.I.s c:> 1'1 e l-\: 1/ ~) 'IcJu..~.J Onte (-vt) I/Ju-t!. • IcJ eA. VffI'J tN·s 1. CNJ.,..~-1 ifJ U,7l. esrmnotes.in|Class notes made easy.
J 'DoOl, :::-
f({l--IJj ~
V J; ( f _ Wl0
-t
J, 'J
.{~,
,
- ~;L-JIJ
co{
, I
~ {J-~>l
b
dcP.1:O fCl>---fJj ~-t'j!l ,. Civ~,J V IrLI-~0 ~~)-)'V\t:
-
~ 1--~ ~~
~ ~ll-1
C-y\.
1
-!cJJ
both. h"d e .
I
JrY'J
C- Vt) Cr: ve)
»
~ 1~ Y'" -~
1
I
.,
f
.
I
,
)!
I
f
f
I
I
.,
,
I
1"\1
/~
1
I" :
1
I
I
1
\
. \
1~
.1 1
BPP
j-
esrmnotes.in|Class notes made easy.
Zo
Filters
100
Transmission Lines and Networks
00
rf)
'I
1 1 1
t~1
1
I 1
1
ZOn
-1-----+------ I I \.l. : ZOT
RO
Frequency (f)
0
.
'
f 100
f1
f 200
f2
fo
(a) Variation of Zowith frequency
Alte""ation (a)
1
'*'
00
00
/i\
1 1 I I
1 1 1
1 1
,I
,I
m derived BPF CHARACTER
1\\
I
1 I I I
~ ~,"
i
I
,~
1
I
1 1
1 1 1
1 1 1 1
I
1 I
I I
f 100
0
Frequency (f)
I
1
f2
fo
f 200
(b) Variation of attenuation (a) with frequency
, Phase shift
W) I
c
.. ~~:'
I
1 1
I 1
I
1
:
:
1 1
:
+-I
I I
I I
I I
1 1
I
I
I
1
I
I
1
C
f
-rr
1------
1
1 1
~
tN
1
----~-----:-----t----
+rr
o
1 1 I 1 1
100
f
1
f
o
f
2
(c) Variation of phase shift
Fig. 2.37
1
f ~
_
00
200
Frequency (f)
with frequency
esrmnotes.in|Class notes made easy.
If the frequency of resonance is fo then it is the geometric mean of two cut-off frequencies as well as of two frequencies of infinite attenuation. ,"'
fa =
~,
~
~(£1) (£2)
Simplifying equation (I),
~1 _ m 2 = (£200 - f1(0 ) 'Jf
(£2 - f1)
=
~(£hJ) (£200)
...
(2)
--~-~,
(J;Y(I ~
I
~) ~I
_o----L----~..)A. ---t
!-= yv'l L
21
hW' (J
!c;.:>
_
I
=--
5e
0
~ "- -~ I ~r,=: /-)"A ---
~. 1-
tA 0J('fY'
vi rJ1 ')'Y\ 1.. J 1.
--r
ShU. V\rf
A
-!.0 i'l'"
CJ..' 'f'{\ (eJ"'S"!
r
S
trl Se (l'tA arP LCIWl h iYW n'oN"- d'f LIJ. A 'fV\ C~. ,A ~ f cA~,) (a.rlIJ, vQ4h"" .
L yv1
(jrf\~jS r.s. fcJ.YvJJJ. (p.M""Wlh~'
11'
:::
I
r1
l' :;;j
J- {
.J
L~;;JL{
esrmnotes.in|Class notes made easy. \
c:P \
cP
Filters
m derived BSF character
ol two cut-off ... (2)
esrmnotes.in|Class notes made easy.
ntages of m derived filters nenuation rises near cut-off frequency
fc
and its slope
IS
adjustable by
e characteristic impedance will be uniform in the pass band when m derived half . n having m = 0.6 is connected at the ends. omposite filters can be constructed for any desired frequency and attenuation cteristics using m derived filters.
posite Filters composite filter is a combination of constant k filters, m derived filters and m ed half sections. Fig.l.23. Terminating
HalfiL Section m=0.6
Constant-k Prototype Filter fc == fC1
Constant-k Prototype Filter f c == f C2
m Derived Filter m=m,
~
~
Terminating m Derived I-Half/L Filter Section m ==m2 I-m == 0.6
--0
~
Fig. 1.23. Block diagram ofcomposite jilter
composite filter, constant k filters provide high attenuation away from cut-off DCy, in stop band, m derived filters provide high attenuation close to f,YJ and ved half sections with m == 0.6 provide uniform characteristic impedance
111
band.
Performance
~ i I ,
Composite ~
I
I
I
t
I
'\.---- Prototype :
I
i I
c
I 0 '----'--f---::'---f
_
:e",eJ
et
Fig. 1.24. Attenuation Characteristics olfilters
1.24 Shows the attenuation characteristic of constant k, m derived and ite filters. Constant k filter has low attenuation near the cut-off frequency fc
esrmnotes.in|Class notes made easy.
11.40 I
Transmission Lines and Waveguides
and attenuation increases with increase of frequency. In derived filter has high attenuation near to f 00' But composite filter provides uniform attenuation in the pass band. esrmnotes.in|Class notes made easy.
Crystal Filter
Crystal filter is made up ofPiezo electric crystal with very high Q. It acts as a very narrow band filter. The equivalent circuit and reactance curve of crystal filter is shown in Fig.1.25. I
Cs
i
+
Rs
! I !
r'l
I
xoll Cp
Iii ::
.
I
(b) Reactance Curve
(a) EqUivalent Circuit ~
~~~
~
__~
!
~_~~J
Fig. 1.25. Crystal Filter
:
~
•• ;; ~
c
The crystal has R s L s C s series in parallel with capacitance C p. The resonant frequency of series circuit is
II
8: :~ 1;: ~
1
fa
21t1.! L s C s
i!t ,II!
ant
iIIlIl S
The resonant frequency of parallel circuit (anti resonance) is
I
_ r:-7S'\
'.
•
-
fA
=
fa\}
l+lc~j
_'iIt
~ :~;
Filter gives maximum output at resonance and minimum at anti resonance.
ail
.•
.11
iii, Iai:
"'~
6
1m
~IIII
lB
-•
III'
Filters
OVERVIEW OF SESSION 37 TO 45
SUMMARY
Decibel dB
=
20 log
i
311Cts as a very
stal. Elter is
VI
I -V
2
I = 20 log 1I I[ I I = 10 log : Pi I! I·
12 !
I
-
P2 I
Characteristic impedance of symmetrical network For T network
For 1t network 20e";.. ~ +'2.2,.
,..
Zsc:. 21 ..... 2, Z'-I'-. :t.
Propagation constant . The resonant
~ = 'Z-o hA. '" ~ ""1.2 ~
z.a
Z~::: ~f
I
... hY
teJ."
Pass band region,
a+j~
Y
.... J~
Y
Z (2
\2
~2 j +
=
l z~ ~ Z
Zoe
- 1
Stop band region, Constant k prototype low pass filter
----
~nance.
k
I
Cut-off frequency, 201-:: 2c)?f -:..
R~j 1 4-.J 1. RJ!t ~ \- ~J~
Ie
=
R
2 k
L
=
C
1
1t
-fIE
=
1t
1 R~ C
esrmnotes.in|Class notes made easy.
l'
,iti
11.421
,~
Transmission Lines and Waveguides
"
I
Z1 /2
Z1 /2
Zl
!~
',."
"
I~
'.
I
--,-
,ill
Ii
I
!0 I I
~
~rZ2
C
0
2Z
2I:'
n-filter
T-filter
l
"at "' ,~
Constant k prototype high pass filter
..
Cut-off frequency,
I 7- o r=
Ie
Rk.] I-~fJ
; \! . iii!
1'1
1:1 I
Ii :1
-
~K::
XI<.
-
{ J -(J;)~
,
4n-{LC
L=~ 4nIc
: :!
1 4n R k C
1 =
and
C
1 4 nIc R k
II ltantk
~----~----------
Z1 /2
Z1
0----1'
B.dpm
I
01-= :lloS~-1 (~)
p ~ J."',, '" (Sc.) 1 -I
2 Z2 0
I
0
T-filter
I
n-filter
---.-JI
_~
m derived low pass filter
mU2
(1 - m 2 ) C
mU2
4m
mC...L
~T
m
~
mL
...LmC
r--: ;
-0 -(j~)' 1
esrmnotes.in|Class notes made easy.
c
'~1!lII
Lines and Waveguides
Filters
-~\ II
..L" C/2
Z'l"""-
m derived high pass filter
I I I
I
I
2 C/m
2 C/m
II---.,.----l I
o
0
-
T~
2L m
) -4m -C \1-m 2
I
i o----I
0
m
=
----!
~ l-(j;)' !t) ~ 1, !:L ~{)
Constant k prototype
Band pass filter L[
Ll C, Rj( 7t (j~ - J;)
C2 =
(1 - m 2) C ----~
4m
I~
(P1)
L.:t C:t
~
C:z.
...
!::!= .CJ
41t~h
RJi/; - J;) 47tJ;h 1
7t ~(f2 - J;) 1
I . .. derived band pass filter I
,-----
I
'lI1IIII,l
=.
h-J;
Cj L2
10 -= .[.1, d;
I
:~1~/m2C~,
1 - m2
--L1
4m
~------~
4m
,_m~C~
~
mC2
m
,
0
--'--
---<0
esrmnotes.in|Class notes made easy.
-
lIIIO . .
..If
!1.441
2....- -.. Q
'
Transmission Lines and Waveguides
Constant k prototype band elimination filter
.!- I., is illl"
..
'
.iIi 15
L
!;J$ ,II;
Cj
c'
-1,' -~'
LJ -= R\< ;, -;K
C~ ~
L
1\ ,t.t<1d"
=
&At2 -I.) 1[1.12 1 4n~-J;)
=
2
tJ,.-I,)
i" -J I
j
2C,
1 .' 4 n 21'1. C2 ../
2C,
l-C2
~---~-- L
0
(12 -I.)
C ..(/1 Y 1t RI<
L2
,
01
---.J
1.1;
./
m derived band elimination filter m L1/2
I_ m~
[/JeO -i,,,oJ.1.
m L,/2
i
I:. --I' L2/m 2 C 1/m
2 C,/m
(1 - m2/4ml L1
(4m/1 - m2 ) C,
I
o
i
I
0 --'
esrmnotes.in|Class notes made easy.
Crystal filter ·ls
fa
j~
'~
21t-'V/ L s C s
= .' f.
\rc; f 1 +~C p
I
c~
11----'
Cp
--------------
-~-----
,ii,lIes and Waveguides
Filters
SOLVED PROBLEMS
I Example 1.1 I Find characteristic impedance of T section shown in figure. Verify impedance with the help ofopen and short circuit impedance. Zl = 100 2
'-J
,------~-------.
(200)2
4
+ 200 x 400
I
I Zo
=
Zoe
= 2 + Z2 = 100 + 400 = 500 n
300n ZJ
Zl Zse
= 2+
(~I) Z2 ZI
2+ ZZ 00 + 100 (400) 1 . 500 Zo
1800
-0 Zoe Zse
= ~ (500) (180)
IZo
300n
I
I Example 1.2 ~ A symmetrical T-network ofpllre resistance has following open lind short circuit impedance, Z oc = 800 LO°
n,
Z sc
=
600 LO°
a
Design
symmetricaIT-network. Given:
800 n, Zse = 600 n
Zoe Z\
2+Z2
=
800n
... (1)
esrmnotes.in|Class notes made easy.
11.461
Transmission Lines and Waveguides
Zl Zz 2 Zl
_ Zl Zse -2+
II 11'.
ZZ+2
lIS
t·
Zl Zz Z ._ . 2 -+ Z
,
600
....
·1' S'
2
Z
Squaring equation (l)
.
(~' + Zz
J
... (2)
Z +_1
2
=
640000
=
640000
Z
Zl
4
2
+ Zz + Z I Zz z
~+ Z\ Z2 J
Zz
640000 - (
Zz
~ 640000-Z~
Zz
~ 640000 - Zoe Zse
Zz = ~ 640000 - 800 x 600 ±4000
Zz
Z
From equation (1),
-t +
.
800
400
Zz::: 400
Zj -
I Example 1.3 , A
1----------------, i Z1/2::: 400 f.! Z1/2 ::: 400 n j
2
=
n
4000
line is composed of T-section of pure resistances. Calculate
characteristic impedance and propagation constant if each series arm is 50 nand shunt arm is 5000 Q. Given:
Zl
-2 = 50 0 ' Z z = 5000 0 Z,
=
1000 ZZ
Zo = -\ I
-t +
Z I Zz
esrmnotes.in|Class notes made easy.
$$Ion
Lines and Waveguides
(iooi 4 + (l00) (5000) 708.87
n ZiJ l
. Z\
in [
y
1 + 2 Z2
+ Z2
J
... (2) 708.87l
100
in [ 1 + 2 (5000) + 5000 Y
=
J
0.1413
I Example 1.4 I Find characteristic impedance ofthis symmetrical ~section: Given:
ZI 2 Z2
111
Z2
)
800n
-----, 800 Q.
n soon 1000
2 1000 Q.
ZI Z2 " Z~ 1 4+ Z1 Z2
Zo
800 x 500 (800)2 4 + (800) (500)
~ Z2 =400 Q. I
1000 Q.
l'
=
2'
534.52
n
! Example 1.5 I For a symmetrical ~network, Zj = jOJL, Z2 = j~C. characteristic impedance at 500 Hz and 1000 Hz if L
, resistances. Calculate
0.1 Hand C = 2 pF.
=
Z1 L=O.1 H
Calculate
!
I~I
snies arm is 50 nand
12~22"FJ
L Given:
2 jmC
esrmnotes.in|Class notes made easy.
11.481
Ii
Transmission Lines and Waveguides
ZI Z2 Z2 \ 4+ Z \ Z2
ZOn
..,"'"'
B_ ill ri_ '
11;1 (jroL)
lET Ii
r. =a._
c~c)
~ ~ (jOlL)'
~
;mE
4
+ jroC
L C
~ ~_ ro:L
1
(i)
f
=
500Hz
ro
=
27tf
= 2 7t X 500 = 1000 7t 0.1 2 x 10-6
ZOn
~
I Ql----= (27t
X
\j2x1O-6 314.18.Q
ZOn
(ii)
f
500)2 (0.1)2 4
=
1000 Hz
ro = 2 7t X 1000 = 2000 7t
r
ZOn
L2 =
0.1 2 x 10-6 0.1 X
_ (2000 7t)2 (0.1)2 ]112
10---6
4
-j226.58 Q esrmnotes.in|Class notes made easy.
" . . Lines and Waveguides
Filters
I Example 1.6 , If Zoe
= 40
.Lf)0
D and Zse = 20.3 L29.8° D of a 2-port
network, find characteristic impedance and propagation constant for symmetrical network. Given:
40LO°
Zoe Zo
n;
Zse = 20.3 L29.8° n
'.} Zoe] ZSCI './ 40 LO° [20.3 L29.8°]
Zo = 28.495 LI4.9° n tanh y
Y
=
VE ZOCl
= tanh- l (
20.3 L29.8 ) 40LO°
= 0.6884 + j 0.1831 y
= 0.7123 LI4.9°
I Example 1. 7 I Design symmetrical T-section, Z
0=
600 D,
.0+ 4 [i!!:.]
Z I = 2 x 600 tanh
Zl
2 = j248.528
2
+
t;-.
:.. . . fz:'~:'M8:528 2 1/2
.
r= 0
2 1/2-= j 248.528
n Q
(a) T-network
Zo
sinh y
600
- j 848.528
n esrmnotes.in|Class notes made easy.
Transmission Lines and Waveguides
11.50 I
Constant-k filter
I Example 1.8 I Design a low pass filter both T and ~section having cut-off frequency of 2 kHz to operate with a terminated load resistance of500 n. Given:
. Ie = 2 kHz
R k = 500 Q;
/<~ ...
@
=~C
Rk
/o\(
\ ~\
Rk
L=- 1t
1
C
Ie
1t
500 L =- 1t X 2000
1t X
L = 79.6mH L/2
=39.8 mH
L/2
Rk
C
=39.8 mH
=
Ie
1 500 x 2000
0.318 JlF
L =79.6 mH
1~2~
:, i
1
I
c
=0.3'8"F,
0
0 I2 =0.159 "F
I
;0 =0.159.F
02
T-section
It-section
I
I Example 1.9 I Determine cut-offfrequencyfor low pass filter shown.
~v
I
L/2
=40 mH
L/2
=40 mH
L =10 mH
!~.~
I
I
C
I 0 I
~T2=0.5"F
=0.5",
. I,=0.5.F
0
0
(b)
(a)
I
(aj T-network Given:
L
2"
= 40 mH, C = 0.5 JlF
L = 80mH
Rk =
~
C = \
3
80x.l0 0.5 x 10-6 = 400 Q esrmnotes.in|Class notes made easy.
-
.'.... -_I
•I· •
[1.50 I
Transmission Lines and Waveguides. ilfiirers
i ii Iii
Constant-k filter
-~='"'" c·
I Example 1.8 I Design a low pass filter both
Ii""
(5' _
N
-
,""
IiII!
frequency of2 kHz to operate with a terminated load resistance of 500 Q
.g
ii
'l!!l Ii,
C
Given:
Rk
L~'
~
':;)'11'
ratll
1:0-:
'l!!l
1t
L =
1t X
I'.J
_
0".,.;.,
_""',
is '" ......
.....
~ 39.8 mH
81 "'Ill '1/1' iii
III
;g :i:
:Ii:
c=
1
I
,
II
Ie
I 500 x 2000
0.318j.tF
~' o
1
2 =0.159"F
I
2 =0.159"F
02
0
L
.'~ ,11&,
Rk
L = 79.6 m-H---~---I
.
Ll2
TC=0.318~F,
, •
1t
1t X
=39.8 mH
1~2
,
Gil',ea:
1
c
L = 79.6mH
::.:.;m =0
'0
tIJ'J' :IF-netw/l
500 2000
0;
e,
; fc = 2 kHz
Ie
L =
...
~I'
i
500 n
Rk
-,""'- -..., IiII!
=
Rk=~
110>..:'
om
T and ~section having cut-ofJ
0
n-section
T-section
I Example 1.9 I Determine cut-offfrequency for low pass filter shown.
:t
I '·'·
I
(V
Ll2 = 40 mH
L = 10 mH
L/2 = 40 mH
!~~
T
-.l ~ ~r" 1T c2 =_ 0.5 ~F T
: . 1 C = 0.5 ~F I I
L 0
0
(a)
C
2
=0.5 ~F
~ I (b) ______J \1
(aJ T-network Given:
L
"2
= 40 mH, C = 0.5 j.tF
L = 80mH
Rk =
~
C = \
3
80x 10 0.5 X 10-6 = 400
n
esrmnotes.in|Class notes made easy.
· 'j"
Lines and Waveguides
Filters
Rk
Ie
nL
::::::>
Ie
400 n x 80 x 10- 3
::::::>
Ie =
1591.54 Hz
'on having cut-off ' 0/500 n.
(b) n:-network: Given:
L
10 mH,
C
1 j.lF
Rk
{f
::::::>
Z = 0.159 ~F
0.5 j.lF
10 x 10-3 1 x 10- 6
Rk , C
C
"2 =
100 n
Rk
Ie =
Rk nL 101)
nx10x10- 3
= 3183.09Hz
IExample 1.10 I Design a constant-k low pass filter having cut-offfrequency 2.5 kHz and design resistance R o = 700 D. Alsofindfrequency at which this filter produces attenuation of 19.1 dB. Find characteristics impedance and phase constant at passband.
G
2= 0.5 ~F ,
I
I
...-
1
®
Ie a
2.5 kHz, R o = 700 n 19.1 dB 19.1 8.686
--
= 2.199 Nepers =
.+00
n
L
R 700 -onle n x 2500
89.12 mH esrmnotes.in|Class notes made easy.
11.521
Transmission Lines and Waveguides
c;=
1;=
1t
cosh-I
Rolc
1t x
=
2 cosh-I
(L) Ie
=
a 2
=
cosh
I
=
Ie
I
=
2500 x oosh
I
=
4.170 kHz
Zar
III
(t)
a
i
I !Dut2'e 1.1/ I.
1 700 x 2500 ;= 0.182 Jl-F
lICJft1f lIH:..
~:
(i)
cosh (
~ )
.
(2.~99)
~ Ro~ l-UJ ~ 700~ 1-( ~I;)' ~ j932.75 =
ZOll
Phase constant
p
=
II;AIII&1t
R0
~ 1-(£)' - ~ 1-( iIi)' ~ j530.30 2 sin- J . -I
2 sm
(i)
L"Q=
(4.17)· 2.5
=
. -I
2 sm (2.668)
Pis imaginary. I
U2
=44.56 mH
U2
=44.56 mH
L
~
=89.12 mH
~ .~ T I T 2 • 0.091 .F
o
2.=0.091.F
0
T-section
freQoe:Id
700
_
ll-section
I ~
esrmnotes.in|Class notes made easy.
ian Lines and Waveguides
2500
=
0.182 /-IF
I Example 1.11 I Design prototype T-section band pass filter having cut-off quency of 1 kHz and 4 kHz and impedance of 600 D. Find resonantfrequency.
Ji = 4 kHz,
J; = 1 kHz,
Given:
.•:v
Rk
600
1t (4 - 1) x 10 3
1t (fi - J;)
C\
Rk = 600 Q
= 63.66 mH
(4-1) X 10 3
Ji -Ji = 41t R J;Ji = 4 1t X 600 x 4 x 10 6
k 0.099/-lF
Rk (fi - Ji) 41tJ;Ji
L2
=
600 (4 - 1) x 10 6 41tx4x10
35.8 mH 1
1
C z = 1t R k (fi - Ji)
= /932.75
Resonant frequency,
100
= 10 =
tm======:::;::-2 = j530.30
-
1t X 600 x (4 - 1) x 10 3
0.1768 /-IF
W
= ~ 4 x 10 6
m_(~l?)
= 2kHz
" .... :l
I
I
:" -h
3
L1/2 = 31.83 mH
2 C1
=0.198 !J.F
2 C1
=0.198!J.F
L1/2
=31.83 mH
I
!
--o.
I
~1------r----1 ~
(~.668)
i
L2 ---I.._ __=_3_5_.8_rn_H
O-
I Example 1.12 I Design a T-section and ~section constant-k I,igh-pass filter c = 0.091 "2
~F
'ng cut-off frequency of 12 kHz and nominal impedance R o = 500 D. Also : (i) Z 0 ' (ii) Phase constant at 24 kHz, (iii) Attenuation at 4 kHz. /"':~ Given: Shunt arm inductance,
Ie = L
12 kHz, R o = 500 Q
~ 41t/c
Q~~~~/
esrmnotes.in|Class notes made easy.
11.541
Transmission Lines and Waveguides
=
4 x 1t
500
3
x 12 X 10
'ers
~ i:'\
Attenuation at 4 kii
L = 3.316 mH
Series arm capacitance,
C =
1
4rc R k
Ie
~ Example 1.13
0.0132 ~F 12 C
=0.026 IlF
I 0------11
2C I
toad of 600 f1. C = 0.0132 IlF
=0.026 !iF I~
C
2 L =6.632 mH
L =3.316 mH
i
2L"6.6~mH
I
! I
i
I
ZOT
~ R o~ 1 - ( ; )'
3)2
1 ~(12 x 10 24 x 10 3
500
2 C = 0.26
0-----11
433 Q Ro Z01t
~l-(Y)' 500 1_(12XI0 24 x 10
3J2
=
577.3Q
3
Phase constant in pass band at 1= 24 kHz (i)
Il'h
~
. -1 = 2 SIn
(Ie) I 3
. _1(12 X l0 ) 2 SIll 24 x 10 3
= 60 esrmnotes.in|Class notes made easy.
~F
2
J
!iicm
Lines and Waveguides
Filters
1.55 '
(ii) Attenuation at 4 kHz,
U
-I
2 cos
(Ie) 'I
12) 2 cosh- 1 ( 4 3~525
neper
I Example 1.13 I Design a high pass filter having cut-offfrequency of 1 kHz _h load of 600 n.
Given:
6000
2 l'" 6.632 mH
L
c ...
11, = 3 41tIc R k 41t x 10 x 600
=
0.133 lJ-F '
I,
',,"
!, 2 C
I
I
= 0.26!lF
517.30
= 0.26 ! l F C = 0. 133 !lF
r-----o
!
L'" 47.74 mH
2 L =95.48 mH i
0------11----,------l
I
ILl
2C
T_-s_e_c_tio_n
lt_-s_e_ct_i_o_n
I
~,-~ j
~ Example 1.14 Design a high pass filter having fe=: 1 kHz with load of
n. Given:
~
'
Ie
1 kHz,
L
k --
L
47mH
R
41tIe
R k = 6000 , 600 3 • 41txlxl0'
C
1 R k 41tIc 1 600 x 4 1t x 10 3
C
0.132 lJ-F esrmnotes.in|Class notes made easy.
11.561
Transmission Lines and Waveguides
m-derived Filter
I Example 1.15 I Design an m-derived LPF having cut-offfrequency of 1 kHz,
'II ~
design impedance of400 n and resonant frequency 11 00 Hz.
iiS
,-,
"", Ii .... ""....
Rk
Given:
...""
('\
j..5~/
.,.;, N
'\..?""."
~C-rk
m
=
fC = 1000 Hz, foo = 1100 Hz
400 Q,
~ ~ 1 J~ foo
v;"
",.
e:c
Ii
1 - (1000)2 1100
=
=!
""
~,.
=:J
.;>.
'"
",.
~
5
...
". ~
~
'-'
:='~
L = Rk = 1t f e
C
1t
1
Rkf
e
=
1t
127.32 mH
1 x 400 x 1000 = 0.795 IlF
m lf2
=26.48 mH
'w'
rO
. ... nl
,6,
:.
~1; :2)) L= 63.27 mH
~
"".
~,
"'"'"""" "'" o
-
iCIll'
0' -.jjl
T Network
is ..
;l
2:
.111, II!:'
am..
••
•
Ifl,
0.416
400 x 1000
m U2 = 26.48 mH
:; ~
\.,)) :ml
=
1t
=
1 '•
((1;
mC
-2- = 0.16 JlF
1
:2))
C:: 0.395
~F
m L:: 52.96 mH
mC =0.16JlF 2
1t Network
esrmnotes.in|Class notes made easy.
r-----,
i 1.57:
Fillers
m L _ 0.416 x 127.32 2 2
fn!.quency of 1 kHz,
(
mC
0.416
mC 2
0.165 IlF
mL
52.96 mH
X
0.795
X
l)
14-mm2) L
(1-(0.416 4 X 0.416
l)
(1-(0.416 1 m2) ( - ;m C == 4 X 0.146
=
26.48 mH 6
10- = 0.33 flF
X
127.32 - 63.27 mH
X
0.795
=
0. 395 1lF
I Example 1.16 I For LPF of figure find m-derived section to have resonant ;..frequency oJ}!OO Hz and 3300 Hz for T and ~respectively. 80/2 mH
: ·
Of' .(\ e y \ ".~
i
10 mH
80/2 mH
(l~
I
L-f l-;
~~:
.ln~ ~
L"F
0
F 0:.Lr - -~
0
l-~_
i
I__________________J ~section:
T-section: Given:
{" .Ie
Given:
1571 Hz
Ie
3183 Hz
L
80mH
L
10mH
C
0.51lF
C
1 IlF
100 m
100
1700 Hz
~ l-(;~)' 1
m
_( llli Y 1700
0.3523
.'11& IIF
mL 2
0.3523 2
3300 Hz
~ l-(j~)' 1 _(illlY 3300 0.2639
X
80mH
14.09 mH ./.
mC 2
0.2639 x 10- 6 2
= O.1 32 1lF esrmnotes.in|Class notes made easy.
11.581
Transmission Lines and Waveguides
0.3523 x 0.5
mC =
1- m2 )
( 4m
0.176
~
0.2639 x 10 mH
mL
F
2.63 mH
~F
2
(1...=.0.3523)21 1 - m2 ) . 4m C ( L 4 x 0.3523 ) 80 mB
1 - (0.2639) 0- 6 4 x 0.2936 x 1
_
0.88
49.72 mB m L/2
=14.09 mH
m L/2
4. I\J-.~)J
'11
~F
=14.09 mH
.J
~ 2
(
1 m )
4m
l- = 49.72 mH .
m2)
1( ~ C=0.88MF
m2c 0013:"'
I Example 1.171 Design
I
m L" 2.63 mH
I~O 0132"F
m-derived T-section HPF with frequency cut-off
10 kHz. Design impedance 200 D;
m
=
0.4. Determine frequency of infln
attenuation. Givell: ,
f e = 10 kHz,
R k =200Q,
m = 0.4
,.G'
""
'C ,'--
. ~/'
; /\'i, :'
\
Rk 200 L = 4nfc = 4nx lOx 103 = 1.59mB
~/ . / '
C = 2C m
1 1 = 3 = 0.039 F 4nR k fe 4nxlOxlO x200 ~ 2 x 0.039 = 0.195 0.4
~F
esrmnotes.in|Class notes made easy.
L m
0.2639 x 10 mH
1.59 0.4 = 3.975 mH
2_63 mH 1- jO.~639l x 10- 6 4 A 0 ..... 936
4 x 0.4 4m ) 2 x 0.039 ~lF = 0.0735 ~F ( 1 _ m2 C. = 1- (0.4)
m
~ l-(j;)'
100
2 [(1- m 2)] x/ e
100
[1 - (0.4)2] x 10 x 10 3
lex
9.16 kHz
Ie
5000 Hz,
0.88 !IF
_ L -
Rk
1tle
Rk
600 5000
_
-
1t
x
1
C
1t
=
38.2 mH
1 , x 600 x 5000 = 0.10 6 ~f 1 (
m
-
- frequency cut-off a frequency of injinit mL
2
mC I-m 2 4m L
100= 1.25 Ie
= 600 n,
=
~
=
11.46mH
1-0.8
0.063
2
=
~F
10.18 mH ,
Ie )2
1.25 Ie
0.6
mL
22.92 mH
mC - 2 -- 0.03111F r I-m 2 4m C
0.028
~F
esrmnotes.in|Class notes made easy.
11.60 I
Transmission Lines and Wavegui, m U2 = 11.46 mH
m L/2 = 11.46 mH
mC
l~/I~
=0.0636 J.lF
m2)
1( ~ L=10.18mH
m L= 22.92 mH
I
I-¥ L-
2- " 0031
L"
4-=fR2) C" 0.028 MF
(1
-L
J
-LmC
0.031 MF
I Example 1.19 I Design a band pass filter with characteristic impedance 100 and pass bandfrom 4. 8 kHz to 5.2 kHz. Given:
@
Rk Lj
100 0,
J;
=
4.8 kHz,
Rk n (h - J;)
12 =
5.2 kHz
100
n (5.2 - 4.8)
x
10 3
79.58 mH
Cj
12 - J; n RJ;12
3
(5.2 - 4.8) x 10 3 n x 100 x 4.8 x 10 x 5.2 x 10 3
12.7 nF
L2
R ((2 - J;)
4nJ;12
3
100(5.2 - 4.8) x 10 4n x 4.8 x 10 3 x 5.2 x 10 3
0.127 mH
C2
1 1t
R (h -,f;)
1 1t x
100 x (5.2 - 4.8) x 10 3
7.96 j.lF esrmnotes.in|Class notes made easy.
i.;~
........ tllL
il_._1_. ~
~~U _ _ ~'~.I~~~il=~=~~ 'I L 1 /2 = 39.79 mH
\11
c
I
0
i
2 C 1 = 25.4 nF
II ~
'OOW
1
L /2 = 39,79,m__ H_2 C = 25.4 n: 1
'OOW-
'I'
I I
I o
I1
C_2_=_7_,9_6_Il_F_Y_L-_L2_=_O_.1_2_7_m_H
o I
TWO MARK QUESTIONS AND ANSWERS ----------------
-----~
Define Neper. me IIF .2 = 0031 . ,.
The natural logarithmic of current ratio or voltage ratio can be expressed in neper
~
is expressed in neper.
tDistic impedance 100
52kHz
dB
=
10 log
(~~ J
1010g(~~)
20 log
: -.., x , ;0(
)
__
10 3
C~)
network is said to be symmetrical if the two series arms of a T network or shunt arms of a 'IT network are equal. Jf7zat is the characteristic impedance of (i) T and (ii)
1C symmetrical
network?
~ Z, Z, ( I + 4Z~J 4.81xlO
3
Zl Z2
Zj
1+ 4z 2
esrmnotes.in|Class notes made easy.
11.621 5.
Transmission Lines and Waveguide.
If the short circuit impedance is 100
n and open circuit impedance 400
what is the characteristic impedance ofsymmetrical network. Zse
=
~ Zse Zoe
= './
100 x 400
2000
6.
Mention the condition for stop band and pass band ofafilter. Stop band
Zl
<- 1
4Z 2
Pass band
7.
Zl
-1 <--<0 4 Z2
Define cut-offfrequency ofa filter. The frequency at which the network changes from a pass band to a stop band vice-versa is called cut-off frequency.
8.
What is the condition for occurrence ofcut-offfrequency ofafilter? (i)
Zl
4Z 2
(ii)
= 0
Zl
- 1 4 Z2 - -
:. Zl
=
0
:. ZI
=
-4Z 2
Where Z 1 and Z2 are opposite types of reactance.
9.
What is low passfilter? Low pass filter is a network which allows only low frequencies at below cut-o frequency and attenuates high frequencies.
t
I
~
[
Gain
L
lJit:6
JIifigh pass fillar
ilall!ueDIC)'. '
= 1000
Zoe = 4000 Zo
:1flIuJt is
fc frequency ~
esrmnotes.in|Class notes made easy.
What is high passfilter? High pass filter is a network which allows only high frequencies above cut-off frequency and attenuates low frequencies. ~--------------I
G!in
f--~~~~-----band
Pass band
I I
I I
I
i
I
jiIIu.
I
fc
~
frequency -
I 1
• What is band pass filter? A band pass filter is a network which allows frequencies between two cut-off frequencies (band offrequencies)Ji andJ2 and attenuates all other frequencies.
If
Gain I
I !
of a filter?
i
I
f2 frequency --
~
What is band elimination filter? A band elimination filter is a network which attenuates frequencies between two cut-off frequencies.li and J2 and passes all other frequencies.
-i t
----, ~~~~ ,-
Gain
II
r i e S at below cut-off
:: I
I
I
,
I
I
: :,
i i
.
L
I i
I !
I
f~_ _f2~~enCy -
I
What are the advantages ofconstant k prototype filter? The advantages of constant k prototype filter are as follows: (i) Attenuation does not increase rapidly beyond cut-off frequencies. (ii) Characteristic impedance varies widely in the pass band from its desired
value.
esrmnotes.in|Class notes made easy.
11.64)
Transmission Lines and Waveguides
14. Write down the expressions for frequency of infinite attenuation for m-derive wwpa~andh~hpa~fifter~
For low pass filter,
t
Ie
100
=
~ I _ m2
where
I
Ie
I
=
o<
:at
For high pass filter.
I
,,,
fro
=
~ I-m 2 fe
where
Ie
1t
-V LC
m<1 1
=
4 1t -{LC
15. Write down the expression for m ofm-derived low pass and high pass fifters.
,II:
II
For low pass filter,
m
For high pass filter,
m
.. , '"
; ~ if
~
S
,
~
I6i
~ l-(JJ ~ l(j;)'
where fe
where
Ie
1 -
1t-[LC
1
41t-{LC
16. Mention the advantages ofm-derivedfifters.
:: :i
The advantages of m-derived filters are
~l
~
(i) Attenuation rises near cut-off frequency fe and its slope is adjustable by ~
!ill ";;
I,
varying
100'
(ii) The characteristic impedance will be uniform in the pass
m-derived half section having m = 0.6 is connected at the ends.
esrmnotes.in|Class notes made easy.
~ ~
:i :i
S
i!
:;::1
1»1 II·
17. What are the salientfeatures ofcrystalfifter? Crystal filter is made up of Piezo electric crystal with high Q factor. (i) It acts as a narrow band filter. (ii) It gives maximum output at resonance and minimum at anti resonance.
