Transmission line theory and parameters

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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=Özy for symmetrical T network

Z=R+jwL and Y=G+jwC

3.Define propagation constant.

Propagation constant (g) is defined as the natural logarithm of the ratio of sending end current to the receiving end current.

g=ln(I_S/I_R)

g=a+jb

where,

a®attenuation constant in neper

b®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=E_RcoshÖzys+I_RZ₀sinhÖzys

Where,Z₀®characteristic impedance

S®distance from the receiving end

E_R&I_Rare 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=w/b

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 a 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

coshNg¢ =ZL/(2Z₀)sinh(Ng)+cosh(Ng)

where,

N®number of miles between the loading coils

¡¢®Propagation constant for the loaded line

Z_L®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=(Z_R-Z₀)/(Z_R+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=ln½(Z₁+Z₂)/(2ÖZ₁Z₂)½

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¢ =½2ÖZ₁Z₂/(Z₁+Z₂)½

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/K_S) + ln(1/K_R) –ln(1/K_SR) +al

Where, K_S,K_R & K_SR 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/_½k_½)

k®reflection coefficient

20.Give the equation for input impedance interms of Z₀ & k

Z_S=Z₀ [(e^{g l}+ke^-^gl)/(e^gl-ke^-^gl)]

21.What are the different types of transmission lines used in practice?

open wire line
cables
co-axial cables
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 P section equivalent parameters of a ine of length l terminated in Z₀

T section

Z₁/2=Z₀ tanh(gl/2)

Z₂=Z₀/sinh(gl)

Psection

Z₁=Z₀ sinh(gl)

Z₂=Z₀ coth(gl/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=dw/db

25.State the disadvantages of reflection?

reflected wave appears as echo at the sending end
the efficiency is reduced
the output reduces as load rejects part of the energy
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

Derive the relations to find the voltage and current at any point on the transmission line at a distance S from the receiving end.
Derive an equation for a & b 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 a&b 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?

At very high frequency, skin effect is considered. Hence the internal inductance become zero.
Due to skin effect, resistance R increases with f^(1/2). But the line reactance wL increases directly with frequency f ie, wL>>R
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+jwL

Consider R is slightly small with respect to wL or R is completely negligible as compared with wL . 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 wL,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)[ m/m1+4ln(d/a)] H/m

L= 10^(-7)[ m1+9.21log(d/a)] H/m

Where

aàradius of round conductors

dàdistance 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) mF/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.

1-½K½

S= --------

1+½K½

1+S

½K½ = -------

1-S

12.Differentiate line at radio and power frequencies

the power transmission lines are electricallyshort in length not exceeding l/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 = VICosj Expressed in Watts,Kwatts.

14.Define reactive power

Q=VIsinj 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,g

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?.

d = ----- meter

Ö ( Pfms)

m--conductor permeability

s--conductivity of the conductor.

19.What is the expression for input impedance of the liner.

( 1+| K| ëj -2bS)

Zin = Zs=Ro----------------------

( 1-| K| ëj -2bS)

20.Distinquish clearly between open and short circuited line

A short circuited line means having zero receiving end voltage and infinite impedance,i.e. Z_R=¥ and E_R= 0

A open circuited line means having infinite receiving end voltage and zero impedance,i.e. Z_R=0 and E_R= ¥

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 E_max or I_max 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,

Z_S=Z_R

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 r_iaxis 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 l/4

7.What are the uses of quarter wave line?

The expression for input impedance of a quarter wave line is given by, Z_S=R₀²/Z_R.Hence the quarter wave line is considered as a transformer to match impedances Z_R & Z_S.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 bl circles?

Constant bl 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:

Single stub matching is useful for a fixed frequency. So as frequency changes, the location of single stub will have to be changed
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(a)and phase constant(b) as

BW=Df=2a/b*fr

Then the selectivity or the quality factor of resonant line given by,

Q=fr/BW=b/(2a)

Generally , the value of a 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 Z_R is given by,

Zin=R₀²/Z_R

This expression is similar to the expression for impedance matching using transformer. If the line has characteristics impedance R₀then the impedance match between the load ie, terminating impedance and input impedance can be achieved by using relationship,

R₀=½ÖZ_RZ₀½

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:

measurement of input impedance and load impedance
measurement of standing wave ratio(SWR)
measurement of reflection coefficient k in polar form
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

circle the relationship between standing wave ratio S and the reflection coefficient magnitude?

½k½=(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,

½Zin½=R₀

For quarter wave line,

Zin=R₀²/I_R

18. Why half wave line is considered as an one to one transformer?

The input impedance of half wave line is given by,

Zin=Z_R

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?

Low pass filters
High pass filters
Band pass filters
Band elimination filters

7.Define LPF, HPF, BPF and BEF

LPF: it is a filter passes all frequencies up to cut-off frequency(fc) and attenuation all frequencies after fc
HPF: it is a filter passes all frequencies after fc and attenuates all frequencies below fc.
BPF: it is a filter passes all the frequencies between two frequencies and attenuates all frequency.
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=jwl, shunt arm impedance z2=-j/wc

Ro=z1*z2=(l/c)^ (1/2)

To get the cut-off frequency x1/4+x2=0,z=jx,so x1=wl, x2= -1/wc

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/wc, z2=jwl

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, f_LPF.>f_HPF

BEF is constructed connecting HPF and LPF in cascade Where, f_LPF<f_HPF

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(P_in/P_out).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 equalizeràthe 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 equalizeràthe 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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