Transmission Lines
Transmission lines and waveguides may be defined as devices usedto guide energy from one point to another (from a source to a load). Transmission lines can consist of a set of conductors, dielectrics orcombination thereof. As we have shown using Maxwell’s equations, wecan transmit energy in the form of an unguided wave (plane wave) thoughspace. In a similar manner, Maxwell’s equations show that we can transmitenergy in the form of a guided wave on a transmission line.
Plane wave propagating in air Y unguided wave propagation
Transmission lines / waveguides Y guided wave propagation
Transmission line examples
Transmission Line Definitions
Almost all transmission lines have a cross-sectional geometry whichis constant in the direction of wave propagation along the line. This typeof transmission line is called a uniform transmission line.
Uniform transmission line - conductors and dielectrics maintain the same cross-sectional geometry along the transmission line in thedirection of wave propagation.
Given a particular conductor geometry for a transmission line, onlycertain patterns of electric and magnetic fields (modes) can exist forpropagating waves. These modes must be solutions to the governingdifferential equation (wave equation) while satisfying the appropriateboundary conditions for the fields.
Transmission line mode - a distinct pattern of electric and magneticfield induced on a transmission line under source excitation.
The propagating modes along the transmission line or waveguide may beclassified according to which field components are present or not presentin the wave. The field components in the direction of wave propagation aredefined as longitudinal components while those perpendicular to thedirection of propagation are defined as transverse components.
Transmission Line Mode Classifications
Assuming the transmission line is oriented with its axis along the z-axis (direction of wave propagation), the modes may be classified as
1. Transverse electromagnetic (TEM) modes - the electric andmagnetic fields are transverse to the direction of wave
zzpropagation with no longitudinal components [E = H = 0]. TEM modes cannot exist on single conductor guidingstructures. TEM modes are sometimes called transmission linemodes since they are the dominant modes on transmission lines. Plane waves can also be classified as TEM modes.
2. Quasi-TEM modes - modes which approximate true TEM modes for sufficiently low frequencies.
zz3. Waveguide modes - either E , H or both are non-zero. Waveguide modes propagate only above certain cutofffrequencies. Waveguide modes are generally undesirable ontransmission lines such that we normally operate transmissionlines at frequencies below the cutoff frequency of the lowestwaveguide mode. In contrast to waveguide modes, TEM modeshave a cutoff frequency of zero.
Transmission Line Equations
Consider the electric and magnetic fields associated with the TEMmode on an arbitrary two-conductor transmission line (assume perfectconductors). According to the definition of the TEM mode, there are nolongitudinal fields associated with the guided wave traveling down the
z ztransmission line in the z direction (E = H = 0).
From the integral form of Maxwell’s equations, integrating the line integral
1of the electric and magnetic fields around any transverse contour C gives
zwhere ds = dsa . The surface integrals of E and H are zero-valued since
z zthere is no E or H inside or outside the conductors (PEC’s, TEM mode).
0
0
The line integrals of E and H for the TEM mode on the transmission linereduce to
These equations show that the transverse field distributions of the TEMmode on a transmission line are identical to the corresponding staticdistributions of fields. That is, the electric field of a TEM mode at anyfrequency has the same distribution as the electrostatic field of the capacitorformed by the two conductors charged to a DC voltage V and the TEMmagnetic field at any frequency has the same distribution as themagnetostatic field of the two conductors carrying a DC current I. If we
1 2change the contour C in the electric field line integral to a new path Cfrom the “!” conductor to the “+” conductor, then we find
These equations show that we may define a unique voltage and current atany point on a transmission line operating in the TEM mode. If a uniquevoltage and current can be defined at any point on the transmission line,then we may use circuit equations to describe its operation (as opposed towriting field equations).
Transmission lines are typically electrically long (severalwavelengths) such that we cannot accurately describe the voltages andcurrents along the transmission line using a simple lumped-elementequivalent circuit. We must use a distributed-element equivalent circuitwhich describes each short segment of the transmission line by a lumped-element equivalent circuit.
Consider a simple uniform two-wire transmission line with itsconductors parallel to the z-axis as shown below.
Uniform transmission line - conductors and insulating medium maintain the same cross-sectional geometry along the entiretransmission line.
The equivalent circuit of a short segment Äz of the two-wire transmissionline may be represented by simple lumped-element equivalent circuit.
RN = series resistance per unit length (Ù/m) of the transmission lineconductors.
LN = series inductance per unit length (H/m) of the transmission lineconductors (internal plus external inductance).
GN = shunt conductance per unit length (S/m) of the media betweenthe transmission line conductors (insulator leakage current).
CN = shunt capacitance per unit length (F/m) of the transmission lineconductors.
We may relate the values of voltage and current at z and z+Äz bywriting KVL and KCL equations for the equivalent circuit.
KVL
KCL
Grouping the voltage and current terms and dividing by Äz gives
Taking the limit as Äz 6 0, the terms on the right hand side of the equationsabove become partial derivatives with respect to z which gives
Time-domaintransmission lineequations(coupled PDE’s)
For time-harmonic signals, the instantaneous voltage and current may bedefined in terms of phasors such that
The derivatives of the voltage and current with respect to time yield jùtimes the respective phasor which gives
Frequency-domain(phasor) transmissionline equations(coupled DE’s)
Note the similarity in the functional form of the time-domain and thefrequency-domain transmission line equations to the respective source-freeMaxwell’s equations (curl equations). Even though these equations werederived without any consideration of the electromagnetic fields associatedwith the transmission line, remember that circuit theory is based onMaxwell’s equations.
Just as we manipulated the two Maxwell curl equations to derive thewave equations describing the electric and magnetic fields of an unguidedwave (plane wave), we can do the same for a guided (transmission lineTEM) wave. Beginning with the phasor transmission line equations, wetake derivatives of both sides with respect to z.
We then insert the first derivatives of the voltage and current found in theoriginal phasor transmission line equations.
The phasor voltage and current wave equations may be written as
Voltage and current wave equations
where ã is the complex propagation constant of the wave on thetransmission line given by
Just as with unguided waves, the real part of the propagation constant (á)is the attenuation constant while the imaginary part (â) is the phaseconstant. The general equations for á and â in terms of the per-unit-lengthtransmission line parameters are
The general solutions to the voltage and current wave equations are
~~~~~ ~~~~~ +z-directed waves !z-directed waves
_ a
The coefficients in the solutions for the transmission line voltage andcurrent are complex constants (phasors) which can be defined as
The instantaneous voltage and current as a function of position along thetransmission line are
Given the transmission line propagation constant, the wavelength andvelocity of propagation are found using the same equations as forunbounded waves.
The region through which a plane wave (unguided wave) travels ischaracterized by the intrinsic impedance (ç) of the medium defined by theratio of the electric field to the magnetic field. The guiding structure overwhich the transmission line wave (guided wave) travels is characterized the
ocharacteristic impedance (Z ) of the transmission line defined by the ratioof voltage to current.
If the voltage and current wave equations defined by
are inserted into the phasor transmission line equations given by
the following equations are obtained.
Equating the coefficients on e and e gives! ã z ã z
The ratio of voltage to current for the forward and reverse traveling wavesdefines the characteristic impedance of the transmission line.
The transmission line characteristic impedance is, in general, complex andcan be defined by
The voltage and current wave equations can be written in terms of thevoltage coefficients and the characteristic impedance (rather than thevoltage and current coefficients) using the relationships
The voltage and current equations become
These equations have unknown coefficients for the forward and reversevoltage waves. These coefficients must be determined based on theknowledge of voltages and currents at the transmission line connections.
Special Case #1 Lossless Transmission Line
A lossless transmission line is defined by perfect conductors and aperfect insulator between the conductors. Thus, the ideal transmission lineconductors have zero resistance (ó=4, R=0) while the ideal transmissionline insulating medium has infinite resistance (ó=0, G=0). The equivalentcircuit for a segment of lossless transmission line reduces to
The propagation constant on the lossless transmission line reduces to
Given the purely imaginary propagation constant, the transmission lineequations for the lossless line are
The characteristic impedance of the lossless transmission line is purely realand given by
The velocity of propagation and wavelength on the lossless line are
Transmission lines are designed with conductors of high conductivity andinsulators of low conductivity in order to minimize losses. The losslesstransmission line model is an accurate representation of an actualtransmission line under most conditions.
Special Case #2 Distortionless Transmission Line
On a lossless transmission line, the propagation constant is purelyimaginary and given by
The phase velocity on the lossless line is
Note that the phase velocity is a constant (independent of frequency) so thatall frequencies propagate along the lossless transmission line at the samevelocity. Many applications involving transmission lines require that aband of frequencies be transmitted (modulation, digital signals, etc.) asopposed to a single frequency. From Fourier theory, we know that anytime-domain signal may be represented as a weighted sum of sinusoids. Asingle rectangular pulse contains energy over a band of frequencies. Forthe pulse to be transmitted down the transmission line without distortion,all of the frequency components must propagate at the same velocity. Thisis the case on a lossless transmission line since the velocity of propagationis a constant. The velocity of propagation on the typical non-idealtransmission line is a function of frequency so that signals are distorted asdifferent components of the signal arrive at the load at different times. Thiseffect is called dispersion. Dispersion is also encountered when anunguided wave propagates in a non-ideal medium. A plane wave pulsepropagating in a dispersive medium will suffer distortion. A dispersivemedium is characterized by a phase velocity which is a function offrequency.
For a low-loss transmission line, on which the velocity of propagationis near constant, dispersion may or may not be a problem, depending on thelength of the line. The small variations in the velocity of propagation ona low-loss line may produce significant distortion if the line is very long. There is a special case of lossy line with the linear phase constant thatproduces a distortionless line.
A transmission line can be made distortionless (linear phase constant) bydesigning the line such that the per-unit-length parameters satisfy
Inserting the per-unit-length parameter relationship into the generalequation for the propagation constant on a lossy line gives
Although the shape of the signal is not distorted, the signal will sufferattenuation as the wave propagates along the line since the distortionlessline is a lossy transmission line. Note that the attenuation constant for adistortionless transmission line is independent of frequency. If this werenot true, the signal would suffer distortion due to different frequenciesbeing attenuated by different amounts.
In the previous derivation, we have assumed that the per-unit-lengthparameters of the transmission line are independent of frequency. This isalso an approximation that depends on the spectral content of thepropagating signal. For very wideband signals, the attenuation and phaseconstants will, in general, both be functions of frequency.
For most practical transmission lines, we find that RNCN > GNLN. Inorder to satisfy the distortionless line requirement, series loading coils aretypically placed periodically along the line to increase LN.
Transmission Line Circuit(Generator/Transmission line/Load)
The most commonly encountered transmission line configuration is
Lthe simple connection of a source (or generator) to a load (Z ) through the
gtransmission line. The generator is characterized by a voltage V and
gimpedance Z while the transmission line is characterized by a propagation
oconstant ã and characteristic impedance Z .
The general transmission line equations for the voltage and current as afunction of position along the line are
The voltage and current at the load (z = l ) is
We may solve the two equations above for the voltage coefficients in termsof the current and voltage at the load.
Thus, the voltage coefficients in the transmission line equations can bedetermined given the voltage and current at the load. Once thesecoefficients are obtained, the voltage and current at any location on the linecan be determined using the transmission line equations.
Just as a plane wave is partially reflected at a media interface whenthe intrinsic impedances on either side of the interface are dissimilar
1 2(ç �ç ), the guided wave on a transmission line is partially reflected at theload when the load impedance is different from the characteristic
oLimpedance of the transmission line (Z �Z ). The transmission linereflection coefficient as a function of position along the line [Ã(z)] isdefined as the ratio of the reflected wave voltage to the transmitted wavevoltage.
Inserting the expressions for the voltage coefficients in terms of the loadvoltage and current gives
The reflection coefficient at the load (z=l) is
and the reflection coefficient as a function of position can be written as
LNote that the ideal case is to have à = 0 (no reflections from the loadmeans that all of the energy associated with the forward traveling wave isdelivered to the load). The reflection coefficient at the load is zero when
L o L oZ = Z . If Z = Z , the transmission line is said to be matched to the load
L oand no reflected waves are present. If Z � Z , a mismatch exists andreflected waves are present on the transmission line. Just as plane wavesreflected from a dielectric interface produce standing waves in the regioncontaining the incident and reflected waves, guided waves on atransmission line reflected from the load produce standing waves on thetransmission line (the sum of forward and reverse traveling waves).
The transmission line equations can be expressed in terms of thereflection coefficients as
The last term on the right hand side of the above equations is the reflectioncoefficient Ã(z). The transmission line equations become
Thus, the transmission line equations can be written in terms of voltage
o ocoefficients (V ,V ) or in terms of one voltage coefficient and the reflection+ !
ocoefficient (V ,Ã).+
The input impedance at any point on the transmission line is given bythe ratio of voltage to current at that point. Taking a ratio of the phasor voltage to the phasor current using the original form of the transmissionline equations gives
The voltage coefficients have been previously determined in terms of theload voltage and current as
Inserting these equations for the voltage coefficients into the impedanceequation gives
We may use the following hyperbolic function identities to simplify thisequation:
Dividing the numerator and denominator by cosh [ã(l!z)] gives
Input impedance at any point along a general
(lossy) transmission line
oFor a lossless line, ã=jâ and Z is purely real. The hyperbolic tangentfunction reduces to
which yields
Input impedance at any point along a lossless transmission line
L LSpecial Case #1 Open-circuited lossless line (*Z *64 , Ã = 1)
L LSpecial Case #2 Short-circuited lossless line (*Z *60, Ã = !1)
The impedance characteristics of a short-circuited or open-circuitedtransmission line are related to the positions of the voltage and current nullsalong the transmission line. On a lossless transmission line, the magnitudeof the voltage and current are given by
The equations for the voltage and current magnitude follow the crankdiagram form that was encountered for the plane wave reflection example. The minimum and maximum values for the voltage and current are
For a lossless line, the magnitude of the reflection coefficient is constantalong the entire line and thus equal to the magnitude of à at the load.
The standing wave ratio (s) on the lossless line is defined as the ratio ofmaximum to minimum voltage magnitudes (or maximum to minimumcurrent magnitudes).
The standing wave ratio on a lossless transmission line ranges between 1and 4.
We may apply the previous equations to the special cases of open-circuited and short-circuited lossless transmission lines to determine thepositions of the voltage and current nulls.
LOpen-Circuited Transmission Line ( Ã =1)
inCurrent null (Z = 4) at the load and every ë/2 from that point.
inVoltage null (Z = 0) at ë/4 from the load and every ë/2 from that point.
LShort-Circuited Transmission Line ( Ã =!1)
inVoltage null (Z = 0) at the load and every ë/2 from that point.
inCurrent null (Z = 4) at ë/4 from the load and every ë/2 from that point.
From the equations for the maximum and minimum transmission linevoltage and current, we also find
It will be shown that the voltage maximum occurs at the same location asthe current minimum on a lossless transmission line and vice versa. Usingthe definition of the standing wave ratio, the maximum and minimumimpedance values along the lossless transmission line may be written as
Thus, the impedance along the lossless transmission line must lie within the
o orange of Z /s to sZ .
Example
g g gA source [V =100p0 V, Z = R = 50 Ù, f = 100 MHz] is connectedo
to a lossless transmission line [L = 0.25ìH/m, C=100pF/m, l=10m]. For
L Lloads of Z = R = 0, 25, 50, 100 and 4 Ù, determine (a.) the reflectioncoefficient at the load (b.) the standing wave ratio (c.) the input impedanceat the transmission line input terminals (d.) voltage and current plots alongthe transmission line.
LR (Ù)(a.)
LÃ(b.)
s(c.)
inZ (Ù)
0 !1 4 0
25 !1/3 2 25
50 0 1 50
100 1/3 2 100
4 1 4 4
in LZ (0) = R because the transmission linein this example is an integer multiple ofhalf-wavelengths long (l = 5ë).
~
Writing the general case transmission line equations in terms of alossless line (ã = jâ) gives
oThe voltage coefficient V must be determined in order to plot the+
ovoltage and current along the transmission line. Since V is the+
voltage coefficient of the forward wave, it is independent of the value
oof the load impedance. For simplicity, we may determine V for the+
L o Lmatched case (R = Z = 50 Ù, Ã = 0). The input impedance seenlooking into the input terminals of the transmission line, undermatched conditions, is 50 Ù. The equivalent circuit at the input to thetransmission line under matched conditions is shown below.
~ ~ ~L max minR (Ù) *V(z)* (V) *V(z)* (V) *V( l)* (V)
0 100 0 0
25 66.7 33.3 33.3
50 50 50 50
100 66.7 33.3 66.7
4 100 0 100
~ ~ ~L s s smax minR (Ù) *I (z)* (A) *I (z)* (A) *I ( l)* (A)
0 2 0 2
25 1.33 0.67 1.33
50 1 1 1
100 1.33 0.67 0.67
4 2 0 0
Smith Chart
The Smith chart is a useful graphical tool used to calculate thereflection coefficient and impedance at various points on a (lossless)transmission line system. The Smith chart is actually a polar plot of thecomplex reflection coefficient Ã(z) [ratio of the reflected wave voltage tothe forward wave voltage] overlaid with the corresponding impedance Z(z)[ratio of overall voltage to overall current].
The phasor voltage at any point on the lossless transmission line is
The reflection coefficient at any point on the transmission line is definedas
Lwhere à is the reflection coefficient at the load (z = l) given by
Lwhere z defines is the normalized load impedance. The magnitude of thecomplex-valued reflection coefficient ranges from 0 to 1 for any value ofload impedance. Thus, the reflection coefficient (and correspondingimpedance) can always be mapped on the unit circle in the complex plane.
(1)
The corresponding standing wave ratio s is
The magnitude of the reflectioncoefficient is constant on anycircle in the complex plane sothat the standing wave ratio(VSWR) is also constant on thesame circle.
LSmith chart center Y *Ã * = 0 (no reflection - matched, s = 1)
LOuter circle Y *Ã * = 1
(total reflection, s = 4)
LOnce the position of à is located on the Smith chart, the location of thereflection coefficient as a function of position [Ã(z)] is determined usingthe reflection coefficient formula.
LThis equation shows that to locate Ã(z), we start at à and rotate through an
zangle of è =2â(z!l) on the constant VSWR circle. With the load locatedat z= l, moving from the load toward the generator (z < l) defines a growing
znegative angle è (clockwise rotation on the constant VSWR circle). Note
zthat if è =!2ð, we rotate back to the same point. The distance traveledalong the transmission line is then
Thus, one complete rotation around the Smith chart (360 ) is equal to oneo
half wavelength.
On the Smith chart ...
• CW rotation Y toward the generator
• CCW rotation Y toward the load
• ë /2=360 ë=720o o
• *Ã* and s are constant on a lossless transmission line. Movingfrom point to point on a lossless transmission line is equivalentto rotation along the constant VSWR circle.
• All impedances on the Smith chart are normalized to thecharacteristic impedance of the transmission line (when usinga normalized Smith chart).
The points along the constant VSWR circle represent the complexreflection coefficient at points along the transmission line. The reflectioncoefficient at any given point on the transmission line corresponds directly
Lto the impedance at that point. To determine this relationship between Ã
LLand Z , we first solve (1) for z .
L Lwhere r and x are the normalized load resistance and reactance,respectively. Solving (2) for the resistance and reactance gives equationsfor the “resistance” and “reactance” circles.
(2)
In a similar fashion, the reflection coefficient as a function of positionÃ(z) along the transmission line can be related to the impedance as afunction of position Z(z). The general impedance at any point along thelength of the transmission line is defined by the ratio of the phasor voltageto the phasor current.
nThe normalized value of the impedance z (z) is
Note that Equation (2) is simply Equation (3) evaluated at z = l. Thus, as wemove from point to point along the transmission line plotting the complexreflection coefficient (rotating around the constant VSWR circle), we arealso plotting the corresponding impedance.
(3)
Example (Smith chart)
A 50Ù lossless transmission line of length 5.25ë is terminated by aload impedance of (100 + j75) Ù. The line is driven by a source of 24V
L(rms) with an impedance of 50Ù. Using the Smith chart, determine (a.) Ã(b.) s (c.) the transmission line input impedance and (b.) the distance in
minwavelengths from the load to the first voltage minimum (d ).
L(1.) Locate the normalized load impedance z and draw constantVSWR circle through this point.
Draw a vertical line intersecting the leftmost point on theVSWR circle downward through the scales located below the
LSmith chart. The values of |Ã | and s are found using these
Lscales (s - left side, top scale, |Ã | - left side third scale from
Ltop). The phase angle for à is determined using the angle scale(degrees) surrounding the Smith chart.
(2.) Move 5.25ë (10.5 revolutions) toward generator (CW) to find
in innormalized input impedance z . Denormalize to find Z .
(3.) Rotate from the load toward the generator (CW) on the VSWRcircle to the location of the voltage minimum (the leftmost pointon the VSWR circle) using either the wavelength or degreescales. If degree scale is used, convert degrees to wavelengths.
Example (Smith chart)
inA 60Ù lossless line has a maximum impedance Z = (180 + j0) Ù at
La distance of ë/24 from the load. If the line is 0.3ë, determine (a.) s (b.) Zand (c.) the transmission line input impedance.
(a.)
in max(b.) [z ] occurs at the rightmost point on the s=3 VSWR circle. From
Lthis point, move ë/24 toward the load (CCW) to find z .
L in(c.) From z , move 0.3ë toward the generator (CW) to find z .
Quarter Wave Transformer
When mismatches between the transmission line and load cannot beavoided, there are matching techniques that may be employed to eliminatereflections on the feeder transmission line. One such technique is thequarter wave transformer.
o L If Z � Z , *Ã*> 0 (mismatch)
Insert a ë/4 length section of different transmission line (characteristicNoimpedance = Z ) between the original transmission line and the load.
The input impedance seen looking into the quarter wave transformer is
Solving for the required characteristic impedance of the quarter wavetransformer yields
Example (Quarter wave transformer)
Given a 300Ù transmission line and a 75Ù load, determine thecharacteristic impedance of the quarter wave transformer necessary tomatch the transmission line. Verify the input impedance using the Smithchart.
Stub Tuner
A quarter-wave transformer is effective at matching a resistive load
L o L oR to a transmission line of characteristic impedance Z when R �Z . However, a complex load impedance cannot be matched by a quarter wavetransformer. The stub tuner is a transmission line matching technique thatcan used to match a complex load. If a point can be located on thetransmission line where the real part of the input admittance is equal to the
o o in ocharacteristic admittance (Y =Z ) of the line (Y =Y ± jB ), the!1
susceptance B can be eliminated by adding the proper reactive componentin parallel at this point. Theoretically, we could add inductors or capacitors(lumped elements) in parallel with the transmission line. However, theselumped elements usually are too lossy at the frequencies of interest.
Rather than using lumped elements, we can use a short-circuited oropen-circuited segment of transmission line to achieve any requiredreactance. Because we are using parallel components, the use ofadmittances (as opposed to impedances) simplifies the mathematics.
l ! length of the shunt stub
d !distance from the load to the stub connection
s Y ! input admittance of the stub
tl Y ! input admittance of the terminated transmission line segment of length d
in Y ! input admittance of the stub in parallel with the transmission line segment
tl oY = Y + jB [Admittance of the terminated t-line section]
sY = !jB [Admittance of the stub (short or open circuit)]
in tl s oY = Y + Y = Y [Overall input admittance]
[Transmission line characteristic admittance]
oIn terms of normalized admittances (divide by Y ), we have
tl s iny = 1 + jb y = !jb y = 1
Note that the normalized conductance of the transmission line segment
tladmittance (y = g + jb) is unity (g = 1).
Single Stub Tuner Design Using the Smith Chart
L1. Locate the normalized load impedance z (rotate 180 to findo
Ly ). Draw the constant VSWR circle [Note that all points on theSmith chart now represent admittances].
L2. From y , rotate toward the generator (CW) on the constantVSWR circle until it intersects the g = 1. The rotation distanceis the distance d while the admittance at this intersection point
tlis y = 1 + jb.3. Beginning at the stub end (short circuit admittance is the
rightmost point on the Smith chart, open circuit admittance isthe leftmost point on the Smith chart), rotate toward the
sgenerator (CW) until the point at y = !jb is reached. Thisrotation distance is the stub length l.
Short circuited stub tuners are most commonly used because a shortedsegment of transmission line radiates less than an open-circuited section. The stub tuner matching technique also works for tuners in series with thetransmission line. However, series tuners are more difficult to connectsince the transmission line conductors must be physically separated in orderto make the series connection.
Example
Design a short-circuited shunt stub tuner to match a load of
LZ = (60 ! j40) Ù to a 50 Ù transmission line.
LFrom y , move toward the generator (CW) to the first intersection
tlwith the g = 1 circle (y = 1 + j0.75 at 69.8 ) to find the distance d.o
sShunt stub reactance must be y = !j0.75 (!106.3 )o
scTo find the stub length l, rotate from the short circuit admittance (y ,rightmost point on Smith chart, 0 ) toward the generator (CW) to theo
sstub reactance (y ,!106.3 )o
Power Flow on a Lossless Transmission Line
The phasor voltage and current on a lossless transmission line can bewritten as
The time-average power at any location on the transmission line can beexpressed in terms of the voltage and current as
The first two terms in the equation above are real-valued while the last twoterms combine to yield a purely imaginary result. This yields a time-average power which is independent of position on the line.
The first term in the average power equation is the power associated withthe forward wave on the transmission line while the second term is thepower associated with the reflected wave.
Transients on Transmission Lines
The frequency domain solutions for the voltage and current along atransmission line considered thus far are valid only for single frequencytime-harmonic signals under steady-state conditions. In many applicationscontaining digital or wideband signals, the transient response of thetransmission line is important.
As previously shown using the transmission line equivalent circuit,the instantaneous voltage and current on a lossless transmission line(RN=GN=0) must satisfy the following coupled time domain PDE’s.
These equations can be decoupled by differentiating one equation withrespect to time (t), differentiating the opposite equation with respect toposition (z), and combining the results. The resulting decoupled timedomain equations for the voltage and current are time domain waveequations.
The general solutions to the time domain wave equations are
where the individual terms in the general solutions represent forward andreverse traveling waves on the transmission line and u is the velocity ofpropagation on the lossless transmission line given by
Inserting the general solutions for voltage and current into the original timedomain PDE’s gives the following relationships for the forward and reversewaves.
Restating the voltage and current equations in terms of only voltagecoefficients gives
Consider the standard transmission line connection with a resistivetermination on the end of the lossless transmission line as shown below.
oLet the instantaneous generator voltage be a step of amplitude V at t = 0.
A forward wave is launched down the transmission line at time t = 0,traveling at a velocity u. Given only a forward wave on the transmissionline, the ratio of voltage to current at the transmission line input is equal to
oZ . Thus, the equivalent circuit seen by the
ogenerator is a resistance of Z . The voltage atthe transmission line input, by voltagedivision, is
The voltage and current as a function of time and position, prior to thearrival of the wavefront at the load, is then
L oAt the load (z = l), if R � Z , a reflected wave will be launched in thereverse direction. The voltage and current at the load are
The ratio of the voltage to current at the load connection must equal the
Lload resistance R , so that
Solving this equation for the reflection coefficient at the load gives
The reflected wave travels back toward the transmission line input. If the
g ogenerator impedance R is not equal to Z , another wave is reflected in theforward direction. The reflection coefficient at the generator is
This reflection process continues at both ends of the transmission line[unless one or both ends of the transmission line is (are) matched]. Eachsubsequent reflection is smaller than the previous such that the reflectionseventually become negligible.
If we denote the n forward and reverse wave amplitudes asth
the wave amplitudes in terms of the reflection coefficients are:
A convenient way of representing the transient voltage and current alonga transmission line as a function of time and position is the voltage bouncediagram and current bounce diagram.
Example (transmission line transients)
goConsider the lossless transmission line below with Z = 50 Ù, R =
L100 Ù, R = 200 Ù, u = 10 m/s and l = 100 m. Assume the generator8
waveform is a 12 V step voltage at t = 0. Draw (a.) the voltage bouncediagram, (b.) the voltage waveform at z = 0 for (0 # t # 4 ìs), (c.) thevoltage waveform at z = l for (0 # t # 4 ìs).
(a.)
(b.)
