FWLec15 (1) - Peterson: Transmission Lines for Electrical...

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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 frequency-domain 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 time-average 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; e-J‘F’“ + v; efl‘B“ (15.1) I(u) = (V1720) e-J‘B“ — (Vf/Zo) efl‘fi" (15.2) where the complex-valued 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 z-system to the u-system is actually a widely-used 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 u-system back to the z—system, the coefficients are modified according to V(u) = V1+ e-jB(z—L) + Vl- e+jl3(z—L) = (Vl+ e+JI3L) 6-sz + (Vl- e-J'BL) e+JBZ = v; e-jflz + 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; e-J‘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,+ {e-ifi“ + 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 short-circuit or open-circuit 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 uni-fatal. [vb-aw ‘M a... W g-fi 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 air-filled 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 half-wavelength intervals. In this case 71. = 6 (cm), corresponding to a frequency of 5.0 (GHz) for an air-filled 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 time-average power obtained by integrating the instantaneous power over one period of the sinusoid: T Pa, = ; (v(t) 1(1) dt (15.21) 0 The time-average power is a constant, not a sinusoid in time. If the voltage and currents are described by phasors, the time-average power may also be obtained from (Wed! “(9 c M -3<; M u :o owe, M 15 4‘1 S MA {5: 0L. “,3 A»: vs: \Njfl PM - +69 mgjtoel h,- we. 2' o"- S .n. \l (r: to a“ L—Wb—j + 2 BC” 3 $0 .3.“ k +- V -1. Z L: 501»; s 0 PM ‘5 I 3 \‘S O 1h Peterson: Transmission Lines for Electrical and Computer Engineers 12/00 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 zero-to-peak values, not their root-mean-square (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 time-average incident power is obtained from P3, = (1/2) Re{V,+ e—J'B“ (V,*/Z0)* efl'fiu} = [Wr/ (22°) (15.23) while the time-average 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 504-550—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 1914-0“ (V) (15.27) and Iin = vin / zin = 216.9 gin-53° (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 time-average 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 time-average 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' 50-550 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 v0.866-tj 0.? c) Zin, the input impedance ’ M!“ - ‘ 13442. Ztu' 29 ‘-r‘”‘ ' 3, ' am.- = v5 2““ =21: gm. = mam v R3?Zu~ e) the time-average 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 teflon-filled (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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FWLec15 (1) - Peterson: Transmission Lines for Electrical...

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