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Unformatted text preview: Peterson: Transmission Lines for Electrical and Computer Engineers 12/00 Lecture 15 Lines with Arbitrary Terminations Objectives: Extend the development of frequencydomain analysis to
transmission lines with arbitrary impedance loads. Introduce the reﬂection
coefficient, the standing wave ratio, and various ways of visualizing the
standing waves. Also discuss the timeaverage power relationships arising
on transmission lines. Figure 15.1 shows a lossless transmission line terminated with an impedance ZL. In
previous lectures we employed a coordinate system whose origin is at the generator end of
the line. As depicted in Figure 15.1, we now introduce a new system in terms of a variable
u, whose origin is located at the load end (Figure 15.1). This is nothing more than a
change of variable, with u = z — L. In this new coordinate system, the voltage and current
phasors can be expressed as V(u) = v; eJ‘F’“ + v; eﬂ‘B“ (15.1)
I(u) = (V1720) eJ‘B“ — (Vf/Zo) eﬂ‘ﬁ" (15.2) where the complexvalued coefﬁcients are denoted {V1+, Vl‘} to distinguish them from
those used previously with the origin at the generator end. The change of coordinates from the zsystem to the usystem is actually a widelyused
procedure known as a shift of reference plane. In any frequency domain analysis, the
phase of all the signals can be shifted arbitrarily by an equal amount since a phase shift is
equivalent to a change in the time origin, which is usually arbitrary. In a similar manner, in
transmission line problems it is convenient to choose a location somewhere on the line, and
deﬁne the relative phases of the various coefﬁcients with respect to an origin at that
location. This reference plane can subsequently be adjusted by the addition of an
appropriate amount of phase to the relevant quantities. In changing from the usystem back
to the z—system, the coefﬁcients are modified according to V(u) = V1+ ejB(z—L) + Vl e+jl3(z—L)
= (Vl+ e+JI3L) 6sz + (Vl eJ'BL) e+JBZ
= v; ejﬂz + V," em” (15.3) Therefore the representation in either coordinate system is equivalent, with the coefficients
related by V0+ = V1+ e+jBL (15.4) v; = V; eJ‘BL (15.5) Peterson: Transmission Lines for Electrical and Computer Engineers 12/00 Locating the origin at the load end of the line simpliﬁes some of the algebra to follow.
At the load end of the line, we evaluate (15.1) and (15.2) to obtain
V(u=0) = VL = V,+ + Vl’ (15.6)
I(u=0) = IL = (V1720) — (Vl’IZo) = VL/ ZL (15.7)
Eliminating VL from these two equations yields
V1'= 1"LV1+ (15.8) where 11 = (ZL—Zo)/(ZL+ 20) (15.9) is the reﬂection coefﬁcient at the load. This deﬁnition is consistent with the reﬂection
coefﬁcient studied in previous lectures, in conjunction with time domain analysis. For example, at a short circuit load ZL = 0 and FL = —1. At an open circuit load, ZL —) co and FL = +1. At a matched load (ZL = Z 0), FL = 0 and there is no reﬂected wave. However,
when working in the frequency domain, we often encounter a complex—valued load impedance. If ZL is complex valued, FL will also be complex valued. By rewriting equations (15.1) and (15.2) in terms of the reﬂection coefficient, we obtain
the alternative representations V(u) = V,+ {eiﬁ“ + 1“L 61‘3"} (15.10)
101) = (Vi/20) {e'j”“ — FL e*“’“} (15.11) These expressions are convenient to work with because they only involve one coefﬁcient,
which is determined by the applied generator voltage. By adding and subtracting a factor FL 6”“ to the expression in brackets in (15.10), we obtain
V(u) = V1+ {(1 + FL) 5113“ + FL (eﬂ‘Bu _ e—jpu)}
= V? {(1 + 11) e43“ + 2er sin(Bu)} (15.12) We observe that the voltage consists of two parts, one with the form of a traveling wave (615“) and the other with the form of a standing wave (sin Bu). This situation should be compared with the special cases considered in the previous lecture: for a shortcircuit or
opencircuit load, the voltage consists of a pure standing wave. For a matched termination,
the voltage consists of a pure traveling wave. However, in the more general case exhibited Peterson: Transmission Lines for Electrical and Computer Engineers 12/00 in (15.12), the waves on the transmission line are a mixture of traveling and standing
waves. Figure 15.2 depicts a typical standing wave pattern. In common with a pure
standing wave, we observe that voltage minima are separated by half wavelength distances,
as are voltage maxima and current minima and maxima. The voltage minima occur at the
same location as the current maxima, and vice versa. An alternative representation for the voltage on the line is obtained by an application. of
Euler’s identity to (15 . 10), yielding V(u) = V1+ {(1+ FL) cos(Bu) —j (1— FL) sin(Bu)} (15.13) The load voltage can be obtained by evaluating either (15.10) or (15 .13) at u=0, to obtain + VL=V(0)=V1+ (1+1“L)=vl (Id 1.) (15.14)
The load current, obtained from a similar evaluation of (15.11), is given by
a.
I =V+ 1—r /z =v+ ) 15.15
L1< 001(zuz. < > If the load is purely resistive, with RL > 20, FL will be positive (between 0 and 1) and a
voltage maximum and current minimum occur at the load. If the load is resistive, with RL < Zo, FL will be negative and a voltage minimum and current maximum occur at the load. In the general case of an arbitrary ZL, as illustrated in Figure 15.2, the load voltage is
neither a minimum nor a maximum. If the load is resistive, the voltage magnitude can be obtained from equation (15. 13) as
[V(u)] = [V11 (1+ I‘L)2 cos2([3u) + (1— FL)2 sin2([3u) (15.16)
The voltage phasor traces out an ellipse in the complex plane, as illustrated in Figure 15.3. In summary, an arbitrarily terminated transmission line may exhibit a pure traveling wave,
a pure standing wave, or a mixture of traveling and standing waves. To characterize the
wave behavior, it is convenient to deﬁne a parameter known as the standing wave ratio
(SWR), or more commonly as the voltage standing wave ratio (VSWR). This is the ratio
of the maximum voltage magnitude to the minimum voltage magnitude, as depicted in
Figure 15.2. Obviously, a pure traveling wave has VSWR = 1, while a pure standing
wave exhibits an infinite VSWR. For a general load, the standing wave ratio lies between
these extreme values. It can be shown that the standing wave ratio and the reﬂection
coefﬁcient are related by lvuul __ ‘4' IV.‘ \—\v¢.l «m (15.17) V5009. = \V' 51% 15.3 Ellyn. W 17 unifatal. [vbaw ‘M a...
W gﬁ flung?» u, m aL/ouless ﬁne.)
A Wu, can. y a, may“; #44! Wk R1280. Peterson: Transmission Lines for Electrical and Computer Engineers 12/00 The current has the same standing wave ratio as the voltage. The standing wave ratio is an
easily measured quantity, as it involves a ratio of voltages and is relatively insensitive to
measurement calibration. Example A resistive load RL is placed at the end of an airﬁlled line having Z0 = 50 (9). Measurements indicate that the resulting standing wave ratio is 3.0, and the voltage
minimum occurs at the load and at repeated distances of 3 (cm) from the load (Figure
15.4). Find the frequency of the generator and the unknown load RL. The frequency of the generator can be determined from the spacing of the voltage minima, which occur at halfwavelength intervals. In this case 71. = 6 (cm), corresponding to a
frequency of 5.0 (GHz) for an airfilled line. Given the VSRW, equation (15.17) can be inverted to yield the magnitude of FL: VSUQ I
V5w¢ +1 “1 '—‘ (15.18) Consequently, II”: 0.5. Because the load voltage is a minimum, it follows that RL < 20,
and FL will be negative. Thus, FL = — 0.5. Then, using 11 = (R — Zo)/(RL + 20) (15.19) we determine that RL = 16.67 (Q). Power Relations The instantaneous power associated with a voltage v(t) and current i(t) is W) = v(t) i(t) (15.20) If the signals are sinusoids, a more practical measure is the timeaverage power obtained by integrating the instantaneous power over one period of the sinusoid:
T Pa, = ; (v(t) 1(1) dt (15.21)
0 The timeaverage power is a constant, not a sinusoid in time. If the voltage and currents are described by phasors, the timeaverage power may also be obtained from (Wed! “(9
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Pay: (1/2) Re{V 1*} (15.22) where the asterisk denotes the complex conjugate. This result is easily veriﬁed by
substituting sinusoids into (15.21) and performing the indicated integration. It is important to note that Pav is not a phasor! [V and I represent sinusoids at radian frequency to, while
Pav is a constant in time] The factor of (1/2) in (15.22) stems from our deﬁnition of phasors. We deﬁne phasor
quantities in terms of their zerotopeak values, not their rootmeansquare (RMS) values.
Some texts use RMS values for phasor amplitudes, and therefore omit the factor of (1/2) in
their power definitions. For a transmission line wave described by (15.1) and (15.2), the timeaverage incident
power is obtained from P3, = (1/2) Re{V,+ e—J'B“ (V,*/Z0)* eﬂ'ﬁu} = [Wr/ (22°) (15.23) while the timeaverage reﬂected power is given by
P}, = (1/2) Re{Vl' €315“ (_.V1/Zo * e—jBu} = —‘V1‘‘/ (220) = _ in V3"? (22,) (15.24) Thus, the reﬂected power is proportional to the square of IFJ. The negative sign in (15.24) indicates that the reﬂected power is ﬂowing in the —u direction. On a lossless transmission
line, conservation of power dictates that PM + P'av = (1/2) Re{VL IL*}, or equivalently that
the incident power must be accounted for by the power reﬂected and the power absorbed
by the load. Example Figure 15.5 depicts a transmission line system, involving a 10 (V) sinusoidal generator, a
line with characteristic impedance Z0 = 50 (£2), and load impedance ZL = 50 + j 50 (Q).
The transmission line has electrical length [SL = (3/8) 7». We would like to determine Vin,
1m, VL, IL, and the time average power delivered to the load. The electrical length can be specified in several ways. In this case, the length in
wavelengths can be converted into degrees or radians of phase using the relation Peterson: Transmission Lines for Electrical and Computer Engineers V 12/00 7. = 360° = 2n (rad) (15.25)
Thus, (3/8) x = (3/8) 360° = 135°. Similarly, (3/8) A = (3/8) (21:) = 311/4 radians. To determine the voltages, we compute the input impedance using equation (14.1). This
yields 504550—550}
5 ~——————— = 2 ‘ \ :1. 15.26
in»: 0{ 5° ‘5 5. II. so 9+0 9 h ) ( ) It follows that
Vi. = VG Zn. /(Z.n + ZG)
=10(20+j 10)/(45+j 10)
= 4.851 19140“ (V) (15.27)
and
Iin = vin / zin = 216.9 gin53° (mA) (15.28) The reﬂection coefficient at the load is
FL =j 50/ (100 +j 50) = 0.4472 (96143" (15.29) By evaluating the voltage expression in (15.10) at the generator end of the line (u=—L), and
equating it with the input voltage, we obtain Vin = V1+ {6ij + FL e'jl’L}
= V1+ {e“”° + 0.4472 cm” e‘j135°}
= 4.851 ej‘4‘°4° (V) (15.30)
from which we determine the coefﬁcient
V1+ = 7.670 e'j‘39'4°° (V) (15.31)
Evaluating (15.10) at the load end (u=0) produces VL =Vl+ {1 +11} Peterson: Transmission Lines for Electrical and Computer Engineers 12/00 = (7.670 e‘j‘39'4°°) (1 + 0.4472 €614?) = 9.701 @1209” (V) (15.32)
and it immediately follows that IL = VL / ZL = 137.2 6116597" (mA) (15.33)
The timeaverage incident power is given by P; = [V1177 (220) = 0.588 (W) (15.34)
while the reﬂected power is P‘a‘,=—l"LV1"r/(ZZo =—O.118 (W) (15.35) The sum of these is the timeaverage power delivered to the load, which can also be
obtained using Ploadav = Re{VL IL*}
= (1/2) Re{9.701 e—jIZO.97° 0.1372 e44155970}
= 0.666 cos(45°) = 0.470 (W) (15.36) Since the line is lossless, the power absorbed by the load is the same as the power
delivered to the input end of the line. Thus, the result in (15.36) could also be obtained
using Vin and 1m. ~s.<+,.m.w«wv», t E xa MB is ‘\l A sinusoidal generator is attached to a length of lossless transmission line as shown: Find numerical values for: a) 11, the reﬂection coefﬁcient at the load
 s 0 n3 5 o , ' ro_ 2::30
L' 50550 2;." 29 v
— b) 1}”, the generalized reﬂection coefﬁcient looking into the line i
 '2. e , m1; ,‘1‘U3  sir/(p ' SO
r: 5 l7. 6 1 ﬂ = e j ‘3 3
6 ‘ 8 s e. s v0.866tj 0.? c) Zin, the input impedance ’ M!“  ‘ 13442.
Ztu' 29 ‘r‘”‘ ' 3, '
am. = v5 2““ =21: gm. = mam v
R3?Zu~ e) the timeaverage power ﬂow at a point halfway down the transmission line (i.e., lat/2) .\7 20.0 ( \ossl¢$s “we, Yuck“ led) ,— PM“;M 15.! The transmission line shown below has a 9.3 MHz sinusoidal generator connected to it. Find the
following quantities assuming that both cables are lossless and use polystyrene (s = 2.6 so where so = 8.854 x 10‘12 F/m and u=uo= 1.256 x 10'6 H/m) as the insulating material between the
conductors. I”: SOJL Pom b POA" t B (a.) The wavelength on these transmission lines
7» = meters (b.) The total impedance (wave impedance) at Point A and at Point B
Z(Point A) = ohms
Z(Point B) = ohms (c.) The Voltage Standing Wave Ratios (V SWR) on Lines 1 and 2 VSWR1 = VSWR2 = PML’LQ‘M 15, Z M A resistive load RL is placed at the end of a teflonﬁlled (8,,=2.2) coaxial transmission line having Z0=75 Q. Measurements of the voltage along the line have been performed and are
shown below. ‘ \ 1) What is the VSWR on the line? 2) What is the reﬂection coefﬁcient F? 3) What is Rm 4) What is the frequency of the generator? 5) Plot the current along the line. ...
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 Spring '08
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