TRANSMISSION LINE AND NETWORKS (EC335)
UNIT I
TRANSMISSION LINE THEORY AND PARAMETERS
PART A
1.What do you mean by a transmission line?
Transmission lines are considered to be impedance-matching circuits
designed to deliver power from the transmitter to the antenna and maximum signal
from antenna to the receiver.
2.Define characteristic impedance of a transmission line
characteristic impedance is an important electrical property of a
symmetrical network and it is defined as the input impedance measured at the input
terminals when the output terminal is terminated in the same impedance.
Z0=zy for symmetrical T network
Z=R+jL and Y=G+jC
3.Define propagation constant.
Propagation constant () is defined as the natural logarithm of the ratio of
sending end current to the receiving end current.
=ln(IS/IR)
=+j
where,
attenuation constant in neper
phase constant in radian
4.Give the equations for voltage and current at any point on the transmission line at a
distance S from receiving end.
E=ER coshzys+IRZ0sinhzys
Where,Z0characteristic impedance
Sdistance from the receiving end
ERIR are the receiving end voltage and current respectively
5.Define phase velocity
Phase velocity or velocity of propagation(v) is defined as the velocity which the
wave is transmitted in the transmission line. It is given by,
V=
6.Define loading: What are the two types of loading?
The transmission line produces distortion in the signal, which is to be transmitted,
and it is undesirable. The condition for distortion less line is RC=LG, ie, R/G>L/C.In
order to achieve this condition, extra inductance is added with the line which is called
loading. There are two types of loading namely, 1.continuous loading 2.lumped loading
7.What is continuous loading?
Continuous loading is the process of adding inductance by distributing it
uniformly along the line. This is accomplished in submarine cables by winding it with a
high permeability steel tape such as permalloy.
8.What is lumped loading?
Lumped loading is the process of adding inductance in the line at uniform
intervals. This lumped loading is used in open wires / load lines or cables to improve the
transmission. The core of the coil is toroidal and made by perm alloy.
9.What are the advantages of continuous loading?
i)The attenuation is independent of frequency and it is same for all frequencies
ii) The value of  can be reduced by increasing L provided R is not increased greatly
iii) The increase in the inductance up to 100mH per unit length of the line is possible.
10.What are the disadvantages of continuous loading?
i) This method is very expensive. The existing lines cannot be modified by this
method
ii) Extreme precision care should be taken while manufacturing continuous loaded
cable. Otherwise it becomes irregular
iii)for an ac signal there will be large eddy current and hysterisis loss. Eddy
current loss varies direcly with square of frequency while the hysterisis loss varies
directly with frequency.
11.Write the Campbell’s equation
coshN =ZL/(2Z0)sinh(N)+cosh(N)
where,
Nnumber of miles between the loading coils
Propagation constant for the loaded line
ZLInductance of the loaded coil
Z0characteristic impedance
12.What are the advantages of lumped loading?
i)There is no practical limit to the value by which the inductance can be increased
ii)The cost involved is small
iii)With this method, the existing lines can be tackled and modified
iv)Hysteresis and eddy current losses are small
13.What are the disadvantages of lumped loading?
i)After particular frequency the line acts as a low pass filter and above the cutoff
frequency the attenuation increases
ii)The cutoff frequency must be at the top of voice frequency. Hence fractional
loading must be used. Care must be taken while installing the lumped inductors so as to
maintain the exact balancing of the circuit
14.What do you mean by smooth line?
Finite line terminated in Z0 without having any reflection is called smooth line.
The waves travel smoothly along the line and energy is absorbed in the load Z0 without
any reflection
15.Define reflection coefficient
The ratio of amplitudes of the reflected and incident voltage waves at the
receiving end of the line is frequently called the reflection coefficient
K=reflected voltage at load / incident voltage at load
K=(ZR-Z0)/(ZR+Z0)
16.Define reflection loss
The reflection loss is defined as the number of nepers or decibels by which the
current in the load under image matched conditions would exceed the current actually
flowing in the load
Reflection loss,nepers=ln(Z1+Z2)/(2Z1Z2)
17.Define reflection factor
It is the ratio of current actually flowing in the load to that which might flow
under image matched conditions
K=I2/I2 =2Z1Z2/(Z1+Z2)
18.Define insertion loss
The insertion loss of a line or network is defined as the number of nepers or
decibels by which the current in the load is changed by insertion.
Insertion loss in nepers=ln(1/KS) + ln(1/KR) –ln(1/KSR) +l
Where, KS,KR & KSR are reflection factors
19.Define return loss
Return loss indicates the ratio of the power at the receiving end due to incident
wave to the power reflected by the load
Return loss=10 log(P1/P3)=20 log(1/k)
kreflection coefficient
20.Give the equation for input impedance interms of Z0 & k
ZS=Z0 [(e l +ke-l)/(el-ke-l)]
21.What are the different types of transmission lines used in practice?
1. open wire line
2. cables
3. co-axial cables
4. wave guides
22.Give the relation between neper and dB and prove the relation
1 neper=8.686 dB
let m be represented in neper and dB,
neper=ln m------------(1)
dB=20 log m----------(2)
ln m=2.303* log m
log m=(1/2.303) ln m
20 log m=(20/2.3030) ln m
dB=8.686 neper
23.State the T and  section equivalent parameters of a ine of length l terminated in Z0
T section
Z1/2=Z0 tanh(l/2)
Z2=Z0/sinh(l)
section
Z1=Z0 sinh(l)
Z2=Z0 coth(l/2)
24.Define group velocity
The velocity which is produced by a group of frequency traveling along the
system is called group velocity. It is defined as,
Vg=d/d
25.State the disadvantages of reflection?
1. reflected wave appears as echo at the sending end
2. the efficiency is reduced
3. the output reduces as load rejects part of the energy
4. if generator impedance is not Z0, then reflected wave is again reflected at
sending end as new incident wave. This continuous back and forth till all
the energy is dissipated as the line losses
PART B
1. Derive the relations to find the voltage and current at any point on the
transmission line at a distance S from the receiving end.
2. Derive an equation for    for a continuous loaded cable.Derive the
Campbell’s equation for lumped loading
3. Derive an expression for reflection loss, reflection factor and insertion loss
4. Derive an expression for input impedance interms of z0 and k
5.Derive the expressions for  for a distortion less line
UNIT:II
THE
LINE AT RADIO
AND POWER
FREQUENCIES
Part A
1Define skin effect?
At very high frequency, skin effect is considerable. Skin effect is
defined as the effect in which the current may flow on the surface of conductor . Now the
internal inductance of conductor becomes zero.
2.What are the standard assumptions made for the analysis of the performance of
transmission line?
1) At very high frequency, skin effect is considered. Hence the
internal inductance become zero.
2) Due to skin effect, resistance R increases with f^(1/2). But the line
reactance wL increases directly with frequency f ie, L>>R
3) The line at radio frequency is constructed such that the leakage
conductance G may be considered zero.
3.What is zero dissipation line?
Z=R+jL
Consider R is slightly small with respect to L or R is completely negligible as
compared with L . If R is neglected completely, then such a line is termed as zero
dissipation line.
4.What are the parameters of the line at high frequency?
The four important electric parameters are
resistence®,inductance(L),capacitance(C ) and conductance(G).
5.What is small dissipation?
When R is small compared with L,then such a line is called as
small dissipation line. In the applications where line is considered as a circuit element or
properties of resonance are involved , this concept of small dissipation line is very much
useful.
6.What is the inductance L for open wire line and coaxial line?
For open wire line :
L=10^(-7)[ /1+4ln(d/a)] H/m
L= 10^(-7)[ 1+9.21log(d/a)] H/m
Where
aradius of round conductors
ddistance between the centers of 2 parallel conductors.
For coaxial line:
L=10^(-7)[2ln(b/a)+(2c^4ln(c/b))/(c^2-b^2)^2-c^2/(c^2-b^2)] H/m
7.What is the capacitance C for open wire line and coaxial line?
For open wire line :
C=3.14E/ln(d/a) F/m=12.07Er/log(d/a) F/m
For coaxial line:
C=2*3.14*e/ln(b/a) F/m=24.14Er/log(b/a) F/m
8.Define nodes and anti nodes.
The points along he line where magnitude of voltage or current is zero are called
nodes. while the points points along the line where magnitude of voltage or current is
maximum zero are called anti nodes.
9.Define standing waves.
If the line is terminated in a load other thanR0 the distri bution of voltage at a
point along the length of the line consists maximum and minimum values of voltage are
called as standing waves
10.Define SWR.
The ratio of the maximum to minimum magnitudes of voltages or currents on a
line having standing waves is called as SWR.
11. State the relation between SWR and reflection coeff.
1-K
S= -------1+K
1+S
K = ------1-S
12.Differentiate line at radio and power frequencies
the power transmission lines are electricallyshort in length not exceeding /10.
Power efficiency of power transmission lines is high as compared to other sources
The design considerations are simple for power transmission systems as frequency of
operation is fixed either 50Hz or 60Hz.
13.Define real power.
P = VICos Expressed in Watts,Kwatts.
14.Define reactive power
Q=VIsin Expressedin Vars,Kvars
15.What is network analyzer?
The network analyzer is a versatile device which can be employed in system
studies of power flow and generating plant loading of existing and future system. It gives
protective breaker settings under fault conditions. The stability limits also determined
using network analyzer.
16.What are the primary and secondary constants of zero dissipation line?.
Primary constants—L,C
Secondary constants—Zo,
17.what are the advantages of zero dissipation lines?.
The concept is useful when the line is used for transmission of power at high fr.
And the losses are neglected completely.
Resistance can be neglected completely.
18,What is a effective cross section of a conductor?.
1
----- meter
 ( f)
--conductor permeability
--conductivity of the conductor.
19.What is the expression for input impedance of the liner.
 1+ K  -2S
Zin = Zs=Ro--------------------- 1- K  -2S
 =
20.Distinquish clearly between open and short circuited line
A short circuited line means having zero receiving end voltage and infinite
impedance,i.e. ZR = and ER= 0
A open circuited line means having infinite receiving end voltage and zero
impedance,i.e. ZR =0 and ER= 
PART B
1.Derive the voltages and currents on dissipationless line.
2.Derive the relation between SWR and reflection coefficient.
3.Derive the input impedance of dissipation less line.
4.Derive the input impedance of open and short circuited lines
5.Describe power and impedance measurement on a line
6.Explain the operation of network analyzer with suitable diagrams.
UNIT-III
IMPEDENCE MATCHING AND IMPEDENCETRANSFORMATION
1.what are the applications of quarter wave line?
It provides impedance transformation in coupling atransmission lineto a resistive load
such as an antenna.
In the line with load which is not pure resistive the impedance of the line at the point
where the Emax or Imax the resistive impedance of the line is either Sro or Ro/S.
It may be used to provide the mechanical support to the open wire lineor the center
conductor of the coaxial cable.
2.State the applications of half line
The expression for the input impedance of the line is given by,
ZS=ZR
Thus the line repeats its terminating impedance. Hence it is operated as one to one
transformer.Its applicaion is to connect load to a source where they can not made
adjacent
3.What is the use of tapered line section?
The tapered line section are useful in making transitions between lines and
configuration in long lines without appreciable reflections
4.What are constant -S circles?
The constant S circles are the circles drawn with center of the chart as center of circle
and radius equal to the value of the standing wave ratio S. These constant-S circles
cut the real axis in two points. The point which is to the right hand side of center of
the chart gives the voltage maximum as well as the value of standing wave ratio S
while the point to the left hand side on the real axis gives the point of voltage minima
nearest to the load. The load point and generator pointer, both are located on S-Circle
5.Explain how smith chart can be used as an admittance chart?
If the smith chart is to be used for admittances, the ri axis becomes gi axis while
Xi axis becomes bi axis. Then above real axis the succeptance is capacitive which
is negative The extreme left point on the real axis represents zero conductance
while the extreme right point on the real axis represents infinite conductance.
6.What are the advantages of double stub matching?
Without actually using slotted line sections for measurement of SWR is possible as
in double stub, the stubs are located arbitrarily but the adjustment is done by the
length of the stubs . Also the distance between the two stubs is generally equal to 
7.What are the uses of quarter wave line?
The expression for input impedance of a quarter wave line is given
2
by, ZS=R0 /ZR. Hence the quarter wave line is considered as a transformer to match
impedances ZR & ZS .It is used as a impedance matching section.
The important application of quarter wave line is to couple a transmission
line to a resistive load such as antenna.
A short-circuited quarter wave line can be used, as an insulator to support
an open wire line is coaxial line conductor
8.What are the constant l circles?
Constant l circles are the circles indicating the distance interms of wavelengths
9.Double stub matching is preferred over single stub due to what reasons?
Due to following disadvantages of single stub:
1. Single stub matching is useful for a fixed frequency. So as frequency changes,
the location of single stub will have to be changed
2. The single stub matching system is based on the measurement of voltage
minimum . Hence for co-axial line it is very difficult to get such voltage
minimum without using slotted line section
10.Explain selectivity of resonant line
The bandwidth of the resonant line can be expressed in terms of attenuation
constant()and phase constant() as
BW=f=2fr
Then the selectivity or the quality factor of resonant line given by,
Q=fr/BW=
Generally , the value of  is very small, hence the selectivity of the resonant line is very
high
11. How is impedance matching achieve utilizing quarter wave transformer?
The input impedance of a quarter wave line terminated in ZR is given by,
Zin=R02/ZR
This expression is similar to the expression for impedance matching using
transformer. If the line has characteristics impedance R0 then the impedance
match between the load ie, terminating impedance and input impedance can be
achieved by using relationship,
R0=ZRZ0
A quarter wave transformer can transform low impedance into high impedance
and vice versa
12.What do you mean by reflection loss?
when there is a mismatch between line and load, the reflection takes place.
Because of this the energy delivered to the load by a properly terminated line. This loss in
power is called reflection loss.
13.Discuss the importance of smooth line:
A line terminated in its characteristic impedance R0 is called properly terminated line
which acts as a smooth line. Because of proper termination, there is no mismatch of
impedance hence no reflection takes place. Thus no standing waves are produced. Then
the maximum power transfer from generator to load is possible
14. Mention two applications of smith chart:
1. measurement of input impedance and load impedance
2. measurement of standing wave ratio(SWR)
3. measurement of reflection coefficient k in polar form
4. location of voltage minimum, maximum
15.What are advantages of smith chart?
 Without actually calculating the quantities using formulae, we can get
directly input impedance, load impedance, SWR ,k location of voltage
minimum etc, with the help of smith chart
 Analysis of single and double stub matching systems is very much
simpligied with the use of smith chart
16. circle the relationship between standing wave ratio S and the reflection coefficient
magnitude?
k=(S-1)/(S+1)
17. Circle the expression for input impedance of eighth-wave line, quarter-waveline
and half wave line
For Eighth wave line,
Zin=R0
For quarter wave line,
Zin=R02/IR
18. Why half wave line is considered as an one to one transformer?
The input impedance of half wave line is given by,
Zin=ZR
From this eqn, it is clear that a half line wave line repeats the terminating impedance
half wave line is considered as one to one transformer
19. How will you find out the location of voltage maximum and minimum?
The intersection of the S circle with the horizontal U axis, gives voltage
minimum points. The intersection point located on the left of the center represents a
voltage minima while that on the right of the circle represents a voltage maxima
PART-B
1.Derive the input impedance of
(i)The eighth wave line
(ii).The quarter waveline
(iii) The half waveline.
2.Explain how the exponential line is used for impedance matching.
3.Describe single stub matching
4.Describe double stub matching
5.problems.
UNIT IV
FILTERS
PART A
1.Define pass band, attenuation band and cut-off frequencies?

Pass band: The range of frequencies in which the attenuation is zero.
Attenuation band:The range of frequencies in which the attenuation is infinity by filter.
Cut-off frequencies: Frequencies which separates a pass and stop band.
2.What are the types of filter?

Active filters
 Passive filters
3.What are the advantages of active filters over passive filters?

Active filters eliminate inductors, which are bulky and are very much
expansive at lower frequencies.

Active filters offer gain which may be variable.

It is easy to tune active filters and can drive low impedance loads.
4.Characteristics of filters?

Pass band characteristics

Stop band characteristics

Cut-off frequency characteristics

Characteristic impedance
5.What are the uses of filters?
 In video frequency telegraphy
 In telephony and TV broadcasting
 In AM detection
 In radio and TV receivers
 In audio amplifiers
6.What are the types of passive filters?
1. Low pass filters
2. High pass filters
3. Band pass filters
4. Band elimination filters
7.Define LPF, HPF, BPF and BEF
1.
LPF: it is a filter passes all frequencies up to cut-off frequency(fc)
and attenuation all frequencies after fc
2.
HPF: it is a filter passes all frequencies after fc and attenuates all
frequencies below fc.
3.
BPF: it is a filter passes all the frequencies between two
frequencies and attenuates all frequency.
4.
BEF: it is a filter eliminates all the frequencies between two
frequencies and passes all the frequency.
8.What are constant-K filters, derive the expression for Ro and fc for constant-k LPF?
Constant-K filters are prototype filters.
Ro and fc and const-k LPF:
Series arm impedance z1=jl, shunt arm impedance z2=-j/c
Ro=z1*z2=(l/c)^ (1/2)
To get the cut-off frequency x1/4+x2=0,z=jx,so x1=l, x2= -1/c
Sub x1 and x2 we get fc=1/(3.14*squareroot(lc))
9.What are design equations of constant-K LPF?
L=Ro/(3.14fc), C=1/(3.14Rofc)
10.Derive the expressions for attenuation(a) and phaseshift(b) for constant-k HPF?
Sinh(p/2)=(z1/4z2)^(1/2),z1= -j/c, z2=jl
Sub z1, z2 we get
Attenuation=2cosh-1(fc/f)
Phase shift=2sin-1(fc/f)
11.Give variations of attenuation const and phase shift const of conatant-k HPF with
frequency?
Attenuation=2cosh-1(fc/f)
Phase shift=2sin-1(fc/f)
If frequency increases, attenuation const decreases and phase shift decreases
to 0
12. Give the expression for Ro and design equations of constant-k BPF?
Ro=(L2/C1)^(1/2) or Ro= ( L1/C2)^(1/2)
L1=Ro/(3.14(f2-f1))
C1=(f2-f1)/(4*3.14f1f2Ro)
L2=(Ro(f2-f1))/(4*3.14f1f2)
C2=1/(3.14Ro(f2-f1))
13.derive the design impedance and cut-off frequency of constant-k BEF?
Ro=(L2/C1)^(1/2) or Ro= ( L1/C2)^(1/2)
Fo=(f1f2)^(1/2)
14.write the design equations of constant-k BEF?
L2=Ro/(3.14*4(f2-f1))
C2=(f2-f1)/(3.14f1f2Ro)
L1=(Ro(f2-f1))/(3.14f1f2)
C1=1/(3.14*4Ro(f2-f1))
15,Disadvvantages of constant-k filters?
 Attenuation does not increase rapidly beyond fc
 Zo varies widely in the transmission or pass band from the desired value
16.why we go for m-derived filters?
Since const-k filters have following disadvantages
1.Attenuation does not increase rapidly beyond fc
2.Zo varies widely in the transmission or pass band from the desired value
3.There is no sharp cut-off frequency.so to overcome this and to improve attenuation const
we go for m-derived filters.
17.Derive m for m-derived LPF?
M=(1-(fc/finfinity)^2)^(1/2)
18.what is composite filters? what are the functions of each section used on it?
Composite filter is a series combination of const-k section,m-derived section and
terminating half sections.
(1)one or more of prototype const-k sections to produce cut-off or transmission b/n
transmission band, the stop band at a specified fc.
(2)one or more of m-derived sections to give infinite attenuation at a frequency in the
neighborhood of the cut-off frequency.
(3) Two terminating m-derived half sections,with m=0.6 to give reasonably const i/p and
o/p impedances
19.What is the important feature of terminating helfsections?.
Using Half sections the composite filters can be terminated properly.It is observed
that m=.6 the characteristics impedance Zo remainsalmost constant throughout the pass
band.
20. Explain how BPF andBEF can be constructed using LPF and HPF.
BPF is constructed connecting LPF and HPF in cascade Where, fLPF.>fHPF
BEF is constructed connecting HPF and LPF in cascade Where, fLPF<fHPF
PART B
1.Explain LPF in detail 2. Explain HPF in detail
3.Derive the expression for design equations of BPF?
4.Drive the exp of Ro and fo of BEF
5.Problems.
UNIT: V
ATTENUATORS AND EQUALIZERS
Part A
1).Why Attenuator is used?
Attenuators are used to reduce the signal level by an amount whereas amplifiers
are used to increase the signal level by an amount. It is also used in laboratory when it is
necessary to obtain small value of voltage or current for test purpose.
2).What is attenuator?
Attenuator is a four terminal network usually resistive network designed for the
purpose of introducing a given loss and is inserted between specified impedances.
3).What is the forms of attenuator?
An attenuator can be in symmetrical or Asymmetrical in form. It can be either a
fixed or balance type. A fixed attenuator is often referred as pad.
4).What is the types of attenuators?
1. Symmetrical an d asymmetrical attenuators.
2. Resistance and capacitive attenuators.
5).What is resistance attenuators?
It will have only resistive components and no phase shift and propagation constant
is equal to attenuation constant.
6).What is the use of resistance attenuator?
1. Matching impedances.
2. Volume control in broadcasting stations.
3. Attenuation distortion of a transformer is avoided.
7).what is the use of capacitance attenuator?
It is used for some application at high frequency.
8).Define symmetrical attenuator and its types?
It is used between two equal impedances mainly the characteristic impedance. The
attenuation is given in decibels and nippers.
The types are T-type, pi-type, lattice and bridged type.
9).What is power ratio, voltage and current ratio?
Power ratio’s=input power/output power
=|p1/p2|
Voltage ratio’s=20log10|v1/v2|
Current ratio’s=20log10|i1/i2|
10). Define attenuation?
It is the power loss in transmission line.
D=20log10|v1/v2| db
N=.5*ln(Pin/Pout).nepers
Attenuation should satisfy the condition: correct input and output impedance
And specified attenuation.
11).What are the design equations of symmetrical T and PI attenuators?
and their differences?
For T-type,
R1=2Ro(N-1)/(N+1).
R2=2NRo/(N^2-1).
For PI type,
R1=Ro(N^2-1)/2N.
R2=Ro(N+1)/2(N-1).
Where R1&R2 are design impedances and Ro is characteristic impedance and N is the
attenuation in nepers.
In the T attenuator the shunt arm impedance becomes small while the series
arm impedance in PI attenuator becomes large, when the attenuation is high.
12).Explain bridged T attenuators?
The design equations are
R1=Ro;R2=Ro/(N-1);R3=Ro/(N-1)
Where R1,R2,R3 are the design impedances ,Ro is the characteristic impedance and N is
the attenuation in nepers.
It is economical and convenient and to be used as the variable attenuator.
13).Explain lattice attenuators?
The design equations are
R1=Ro(N-1)/(N+1)
R2=Ro(N+1)/(N-1)
Where R1,R2,R3 are the design impedances ,Ro is the characteristic impedance and N is
the attenuation in nepers
It is comparatively important type because it can be converted to any other symmetrical
attenuator by Bartlett’s bisection theorem.
14).Explain asymmetrical attenuators and its types?
It is used to work between two unequal impedances. The values of the arms
can be calculated for the values of two image impedances and relation for attenuation
.N=sqrt(p1/p2)=Is/Ir*sqrt(r1/r2)
Where N-attenuation
P1, p2=input and output power
Is,Ir=sending end and receiving end current
R1, R2-image impedance
It can also be designed by using a matching network along symmetrical attenuator.
Total attenuation is the sum of both attenuations.
The types are L-type and T type and PI type.
15).Explain L, T, PI type attenuators?
The design equations are
For L-type,
R1=Ro’(N-1)/N=sqrt(Ri1(Ri1-Ri2))
R2=Ro’/(N-1)=sqrt(Ri1Ri262/(Ri1-Ri2)
Where R1,R2 are design impedances and Ro’ is characteristic impedance Ri1,Ri2 are the
image impedance ,N is the attenuation in nepers.
For T-type,
Ra=N1((N^2-1)/(N^2+1))-2sqrt(R1R2)n/(N^2-1)
Rb=2sqrt(R1R2)N/(N^2-1)
Rc=r2(N^2+1)/(N^2-1)-2sqrt(R1R2)N/(N^2-1)
For PI type,
Ra=R1(N^2-1)/(N^2-2Nsqrt(R1/R2)+1)
Rb=sqrt(R1R2/2)(N^2-1)/N
Rc=R2(N^2-1)/(N^2-2Nsqrt(R2/R1)+1)
Where Ra,Rb,Rc are the design impedances and N is the attenuation in nepers.
16).circuit diagrams of different attenuators.
EQUALIZERS:
17).define equalizers.
Equalizers are electrical networks designed to counteract the attenuation or
phase distortion occurring in any part of the circuit.
18).State the type of equalizers.
There are two types.
1. Attenuation or amplitude equalizerthe network that counteracts the attenuation
distortion. It is of 2 types i.constant resistance ii) nonconstant resistance.
It exhibits a prescribed amplitude change between input and output as a function of
Frequency.
2. Phase or distortion equalizerthe network that counteracts the phase distortion.
It exhibits a prescribed phase change between input and output as a function of
Frequency. it is made up if reactive elements.
19).what are the uses of amplitude equalizer?
1.carrier telephone systems.
2.transmission lines and transducers and amplifiers
3. Recording and reproduction of speech signals.
20).what is the uses of phase equalizer?
1. Preservation of the waveform of the input signal
2. in facsimile and television signal transmission over lines.
21).What is the applications of equalizers?
1. Communication circuits
2. Carrier telephone systems
3. Design of servomechanisms
4. as a lag or lead network to improve stability response and performance of a
system.
22).Define inverse impedance.
Two impedances are said to be inverse if the geometric mean is a real
number.ie,sqrt(z1z2)=R^2 .
23).what is an inverse network?
The structure constructed on the basis of inverse impedance. The resistance is
converted to another resistance that of an inductance a capacitance and viceversa.the
inverse of a series connection is a parallel connection and viceversa.
24).What is meant by series and shunt equalizers?
It is a two terminal network connected either in series or shunt with the sending are
receiving end of the network to be connected. Their impedances varies with frequency
and are used in audio frequency circuits.
25)Explain four terminal equalizer.
The resistance becomes constant. And avoid mismatch condition. The types
are:
1.full series equalizer
2. Full shunt “
3. Bridged T “
4. Lattice
“
26).What is the characteristics of equalizers?
1. Insertion loss
2. Attenuation varies with frequency
27).Define attenuation in equalizer?
It is the ratio of maximum power delivered to the load without equalizer by power
delivered to the load with equalizer. It is denoted by N.
28).circuit diagrams of different equalizers.
29).design equations of different equalizers.
Part B
1).Describe the following in detail about symmetrical I) T attenuator. I) PI type attenuator
iii) lattice attenuator IV) bridged T attenuator.
Solutions:
2).Describe asymmetrical attenuators and its types
3) Describe the configuration of equalizers?
4).describe attenuation equalizers in detail.
Solutions:
15).Problems in attenuators and equalizers.
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