45787-UlabyISMCh02 - 25 Chapter 2: Transmission Lines...

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Unformatted text preview: 25 Chapter 2: Transmission Lines Lesson #4 Chapter -- Section: 2-1, 2-2 Topics: Lumped-element model Highlights: TEM lines General properties of transmission lines L, C, R, G 26 Lesson #5 Chapter -- Section: 2-3, 2-4 Topics: Transmission-line equations, wave propagation Highlights: Wave equation Characteristic impedance General solution Special Illustrations: Example 2-1 27 Lesson #6 Chapter -- Section: 2-5 Topics: Lossless line Highlights: General wave propagation properties Reflection coefficient Standing waves Maxima and minima Special Illustrations: Example 2-2 Example 2-5 28 Lesson #7 Chapter -- Section: 2-6 Topics: Input impedance Highlights: Thvenin equivalent Solution for V and I at any location Special Illustrations: Example 2-6 CD-ROM Modules 2.1-2.4, Configurations A-C CD-ROM Demos 2.1-2.4, Configurations A-C 29 Lessons #8 and 9 Chapter -- Section: 2-7, 2-8 Topics: Special cases, power flow Highlights: Sorted line Open line Matched line Quarter-wave transformer Power flow Special Illustrations: Example 2-8 CD-ROM Modules 2.1-2.4, Configurations D and E CD-ROM Demos 2.1-2.4, Configurations D and E 30 Lessons #10 and 11 Chapter -- Section: 2-9 Topics: Smith chart Highlights: Structure of Smith chart Calculating impedances, admittances, transformations Locations of maxima and minima Special Illustrations: Example 2-10 Example 2-11 31 Lesson #12 Chapter -- Section: 2-10 Topics: Matching Highlights: Matching network Double-stub tuning Special Illustrations: Example 2-12 Technology Brief on "Microwave Oven" (CD-ROM) Microwave Ovens Percy Spencer, while working for Raytheon in the 1940s on the design and construction of magnetrons for radar, observed that a chocolate bar that had unintentionally been exposed to microwaves had melted in his pocket. The process of cooking by microwave was patented in 1946, and by the 1970s microwave ovens had become standard household items. 32 Lesson #13 Chapter -- Section: 2-11 Topics: Transients Highlights: Step function Bounce diagram Special Illustrations: CD-ROM Modules 2.5-2.9 CD-ROM Demos 2.5-2.13 Demo 2.13 CHAPTER 2 33 Chapter 2 Sections 2-1 to 2-4: Transmission-Line Model Problem 2.1 A transmission line of length l connects a load to a sinusoidal voltage source with an oscillation frequency f . Assuming the velocity of wave propagation on the line is c, for which of the following situations is it reasonable to ignore the presence of the transmission line in the solution of the circuit: (a) l 20 cm, f 20 kHz, (b) l 50 km, f 60 Hz, (c) l 20 cm, f 600 MHz, (d) l 1 mm, f 100 GHz. Problem 2.2 Calculate the line parameters R , L , G , and C for a coaxial line with an inner conductor diameter of 0 5 cm and an outer conductor diameter of 1 cm, filled with an insulating material where 0 , r 4 5, and 10 3 S/m. The conductors are made of copper with c 0 and c 5 8 107 S/m. The operating frequency is 1 GHz. Solution: Given combining Eqs. (2.5) and (2.6) gives 0 25 2 m 0 50 2 m 1 109 Hz 4 10 7 H/m 2 5 8 107 S/m 0 788 /m R 1 2 f c c 1 a 1 b 1 10 b 1 0 2 cm 0 50 10 2 m a 0 5 2 cm 0 25 10 2 m 1 10 0 33 (nonnegligible) 0 40 (nonnegligible) 0 01 (borderline) 1 33 Solution: A transmission line is negligible when l lf 20 10 2 m 20 103 Hz l (a) up 3 108 m/s l lf 50 103 m 60 100 Hz (b) 8 m/s up 3 10 l lf 20 10 2 m 600 106 Hz (c) up 3 108 m/s 3 m lf 1 10 100 109 Hz l (d) up 3 108 m/s 0 01. 10 5 (negligible). 34 From Eq. (2.7), CHAPTER 2 From Eq. (2.8), From Eq. (2.9), Problem 2.3 A 1-GHz parallel-plate transmission line consists of 1.2-cm-wide copper strips separated by a 0.15-cm-thick layer of polystyrene. Appendix B gives c 0 4 10 7 (H/m) and c 5 8 107 (S/m) for copper, and r 2 6 for polystyrene. Use Table 2-1 to determine the line parameters of the transmission line. Assume 0 and 0 for polystyrene. Solution: 3 Problem 2.4 Show that the transmission line model shown in Fig. 2-37 (P2.4) yields the same telegrapher's equations given by Eqs. (2.14) and (2.16). Solution: The voltage at the central upper node is the same whether it is calculated from the left port or the right port: vz z t 1 2R z i z z t 1 2L z vz 1 2 z t vzt 1 2R z i z t 1 2L z izt t iz t z t C 0 r 26 w d w d 10 9 36 G 12 15 10 10 2 1 84 10 L 10 (F/m) R 2Rs 2 f c 2 109 4 10 7 w w c 1 2 10 2 5 8 107 7 3 d 4 10 1 5 10 1 57 10 7 (H/m) w 1 2 10 2 0 because 0 1 2 1 38 (/m) C 2 ln b a 2r 0 ln b a 2 45 8 854 ln 2 10 12 F/m G 2 ln b a 2 10 3 S/m ln 2 9 1 mS/m 362 pF/m L b ln 2 a 4 10 7 H/m ln 2 2 139 nH/m CHAPTER 2 R'z 2 L'z 2 R'z 2 L'z 2 i(z+z, t) 35 + i(z, t) + v(z, t) G'z C'z v(z+z, t) - z - Figure P2.4: Transmission line model. v z 1 z t 2 t From both of these equations, the proof is completed by following the steps outlined in the text, ie. rearranging terms, dividing by z, and taking the limit as z 0. Solution: From Eq. (2.22), 0 788 /m From Eq. (2.33), up 2 109 44 5 19 6 j0 030 1 41 108 m/s Z0 R G jL jC 0 788 /m j 2 109 s 1 139 10 9 H/m 9 1 10 3 S/m j 2 109 s 1 362 10 12 F/m Thus, from Eqs. (2.25a) and (2.25b), From Eq. (2.29), 109 10 3 j44 5 m 1 0 109 Np/m and 44 5 rad/m. 91 10 3 S/m j 2 109 s 1 362 10 j 2 109 s 1 R jL Problem 2.5 Find up , and Z0 for the coaxial line of Problem 2.2. G jC 139 10 9 H/m 12 F/m izt iz z t G z v z 1 2 z t C z Recognizing that the current through the G C branch is i z t Kirchhoff's current law), we can conclude that i z z t (from 36 CHAPTER 2 Section 2-5: The Lossless Line Problem 2.6 In addition to not dissipating power, a lossless line has two important features: (1) it is dispertionless (p is independent of frequency) and (2) its characteristic impedance Z0 is purely real. Sometimes, it is not possible to design a transmission line such that R L and G C , but it is possible to choose the dimensions of the line and its material properties so as to satisfy the condition Such a line is called a distortionless line because despite the fact that it is not lossless, it does nonetheless possess the previously mentioned features of the loss line. Show that for a distortionless line, Solution: Using the distortionless condition in Eq. (2.22) gives Hence, Similarly, using the distortionless condition in Eq. (2.29) gives Problem 2.7 For a distortionless line with Z 0 50 , up 2 5 108 (m/s), find the line parameters and at 100 MHz. Z0 R G jL jC L C R L G C j j L C 20 (mNp/m), R C L LC up LC j R j L C R L C L LC R L j R L j LC R L j G C j R jL G jC j R RG C L LC Z0 L C 1 LC RC LG (distortionless line) CHAPTER 2 37 Solution: The product of the expressions for and Z 0 given in Problem 2.6 gives With L known, we use the expression for Z 0 to find C : The distortionless condition given in Problem 2.6 is then used to find G . and the wavelength is obtained by applying the relation Solution: From the equations given in Problem 2.6, Solution: Given an arbitrary transmission line, f 125 MHz, Z 0 40 , 0 02 Np/m, and 0 75 rad/m. Since Z 0 is real and 0, the line is distortionless. From Problem 2.6, L C and Z0 L C , therefore, 2 L Z0 0 75 40 125 106 38 2 nH/m Problem 2.9 A transmission line operating at 125 MHz has Z 0 40 , (Np/m), and 0 75 rad/m. Find the line parameters R , L , G , and C . 2 4 Z0 L C R G 2 10 1 2 100 RG 2 2 10 4 1 2 2 10 2 (Np/m) Problem 2.8 Find and Z0 of a distortionless line whose R G 2 10 4 S/m. p f 2 5 108 100 106 25m 2 /m and G RC L 1 80 10 2 10 7 12 4 10 4 (S/m) 400 (S/m) C L 2 Z0 2 10 50 2 7 8 10 11 (F/m) 80 (pF/m) L Z0 up 50 2 5 108 2 10 7 (H/m) 200 (nH/m) and taking the ratio of the expression for Z 0 to that for up 1 L C gives 0 02 R Z0 20 10 3 50 1 (/m) 38 CHAPTER 2 L C, and Problem 2.10 Using a slotted line, the voltage on a lossless transmission line was found to have a maximum magnitude of 1.5 V and a minimum magnitude of 0.6 V. Find the magnitude of the load's reflection coefficient. Solution: From the definition of the Standing Wave Ratio given by Eq. (2.59), min Solving for the magnitude of the reflection coefficient in terms of S, as in Example 2-4, Problem 2.11 Polyethylene with r 2 25 is used as the insulating material in a lossless coaxial line with characteristic impedance of 50 . The radius of the inner conductor is 1.2 mm. (a) What is the radius of the outer conductor? (b) What is the phase velocity of the line? Solution: Given a lossless coaxial line, Z 0 50 , r 2 25, a 1 2 mm: (a) From Table 2-2, Z0 60 r ln b a which can be rearranged to give b aeZ0 1 2 mm e50 2 25 60 r 60 4 2 mm S S 1 1 25 25 V 1 1 S max V 15 06 25 0 43 G 2 R 0 02 Np/m 0 8 /m 2 0 5 mS/m R RG R G RG From R G and R C LG, L C Z0 0 02 Np/m 40 0 6 /m C Then, from Z0 L 2 Z0 38 2 nH/m 402 23 9 pF/m CHAPTER 2 (b) Also from Table 2-2, 39 Problem 2.12 A 50- lossless transmission line is terminated in a load with impedance ZL 30 j50 . The wavelength is 8 cm. Find: (a) the reflection coefficient at the load, (b) the standing-wave ratio on the line, (c) the position of the voltage maximum nearest the load, (d) the position of the current maximum nearest the load. Solution: (a) From Eq. (2.49a), (b) From Eq. (2.59), (c) From Eq. (2.56) (d) A current maximum occurs at a voltage minimum, and from Eq. (2.58), Problem 2.13 On a 150- lossless transmission line, the following observations were noted: distance of first voltage minimum from the load 3 cm; distance of first voltage maximum from the load 9 cm; S 3. Find Z L . 9 cm 3 cm 6 cm 4 Solution: Distance between a minimum and an adjacent maximum lmin lmax 4 3 11 cm 8 cm 4 1 11 cm lmax r 4 n 2 79 8 8 cm rad n 8 cm 4 180 2 0 89 cm 4 0 cm 3 11 cm S 1 1 1 1 0 57 0 57 3 65 0 57e ZL ZL Z0 Z0 30 30 j50 j50 50 50 j79 8 4. Hence, up c r 3 108 m/s 2 25 20 108 m/s 40 min CHAPTER 2 Problem 2.14 Using a slotted line, the following results were obtained: distance of first minimum from the load 4 cm; distance of second minimum from the load 14 cm, voltage standing-wave ratio 1 5. If the line is lossless and Z 0 50 , find the load impedance. or and From this we obtain Also, S S 1 1 15 15 1 1 02 0 2 rad 36 0 r 2lmin n 2n 1 rad 2 10 rad/m 0 04 m 2 2 rad/cycle 20 cm/cycle 2 lmin 1 lmin 0 lmin 1 lmin 0 2 20 cm 10 rad/m rad Solution: Following Example 2.5: Given a lossless line with Z 0 lmin 0 4 cm, lmin 1 14 cm. Then 50 , S 1 5, ZL Z0 150 1 1 Hence, 0 5 e Finally, j 2 j0 5. 1 1 j0 5 j0 5 90 which gives r 2. S S 1 1 3 3 1 1 2 4 r 2 2 8 05 j120 or 24 cm. Accordingly, the first voltage minimum is at Application of Eq. (2.57) with n 0 gives 3 cm 8. CHAPTER 2 So 41 Problem 2.15 A load with impedance Z L 25 j50 is to be connected to a lossless transmission line with characteristic impedance Z 0 , with Z0 chosen such that the standing-wave ratio is the smallest possible. What should Z 0 be? Solution: Since S is monotonic with (i.e., a plot of S vs. is always increasing), the value of Z0 which gives the minimum possible S also gives the minimum possible , and, for that matter, the minimum possible 2 . A necessary condition for a minimum is that its derivative be equal to zero: jXL Z0 A mathematically precise solution will also demonstrate that this point is a minimum (by calculating the second derivative, for example). Since the endpoints of the range may be local minima or maxima without the derivative being zero there, the endpoints (namely Z0 0 and Z0 ) should be checked also. Problem 2.16 A 50- lossless line terminated in a purely resistive load has a voltage standing wave ratio of 3. Find all possible values of Z L . Solution: For a purely resistive load, r ZL 50 For r , 0 5 and 1 1 05 05 15 ZL Z0 50 1 1 1 1 05 05 S 1 3 1 05 S 1 3 1 0 or . For r 0, 150 Z0 ZL 252 2 Therefore, Z0 R2 L 2 XL or 50 2 55 9 Z0 2 2 XL RL Z0 2 2 XL 2 RL Z0 RL Z0 2 2 XL 2 4RL Z0 0 jXL Z0 2 Z0 RL Z0 RL 2 2 R2 L 1 0 2e j36 0 2 XL ZL Z0 50 1 1 1 0 2e j36 0 67 0 j16 4 42 CHAPTER 2 Section 2-6: Input Impedance Problem 2.17 At an operating frequency of 300 MHz, a lossless 50- air-spaced transmission line 2.5 m in length is terminated with an impedance Z L 40 j20 . Find the input impedance. Solution: Given a lossless transmission line, Z 0 50 , f 300 MHz, l 2 5 m, 40 j20 . Since the line is air filled, up c and therefore, from Eq. and ZL (2.38), Since the line is lossless, Eq. (2.69) is valid: Problem 2.18 A lossless transmission line of electrical length l 0 35 is terminated in a load impedance as shown in Fig. 2-38 (P2.18). Find , S, and Z in . l = 0.35 Zin Z0 = 100 ZL = (60 + j30) Figure P2.18: Loaded transmission line. Solution: From Eq. (2.49a), From Eq. (2.59), S 1 1 1 1 0 307 0 307 1 89 ZL ZL Z0 Z0 60 60 j30 j30 100 100 0 307e j132 5 Zin Z0 50 ZL Z0 jZ0 tan l jZL tan l 40 j20 50 j 40 40 j20 50 50 j 40 j50 tan 2 rad/m 2 5 m j20 tan 2 rad/m 2 5 m j50 0 40 j20 j20 0 up 2 300 106 3 108 2 rad/m CHAPTER 2 From Eq. (2.63) 43 Problem 2.19 Show that the input impedance of a quarter-wavelength long lossless line terminated in a short circuit appears as an open circuit. Zin Z0 Problem 2.20 Show that at the position where the magnitude of the voltage on the line is a maximum the input impedance is purely real. which is real, provided Z0 is real. Problem 2.21 A voltage generator with vg t 5 cos 2 109 t V and internal impedance Zg 50 is connected to a 50- lossless air-spaced transmission line. The line length is 5 cm and it is terminated in a load with impedance ZL 100 j100 . Find (a) at the load. (b) Zin at the input to the transmission line. ~ (c) the input voltage Vi and input current Ii . j r 2n Z0 1 1 Z0 1 1 e jr e e jr e j r 2n Zin lmax Z0 1 1 e jr e e jr e j2lmax j2lmax Solution: From Eq. (2.56), lmax representation for , r 2n 2, so from Eq. (2.61), using polar Zin Z0 jZ0 tan 2 Z0 For l 4, l 2 4 2. With ZL 0, we have j (open circuit) Solution: ZL Z0 jZ0 tan l jZL tan l 100 j 60 j30 tan 35 100 60 j30 j100 tan 2 rad 0 2 rad 0 Zin Z0 ZL Z0 jZ0 tan l jZL tan l 35 64 8 j38 3 44 Solution: (a) From Eq. (2.49a), CHAPTER 2 (b) All formulae for Zin require knowledge of up . Since the line is an air line, up c, and from the expression for vg t we conclude 2 109 rad/s. Therefore Then, using Eq. (2.63), An alternative solution to this part involves the solution to part (a) and Eq. (2.61). (c) In phasor domain, Vg 5 V e j0 . From Eq. (2.64), and also from Eq. (2.64), Problem 2.22 A 6-m section of 150- lossless line is driven by a source with and Zg 150 . If the line, which has a relative permittivity r in a load ZL 150 j50 find (a) on the line, (b) the reflection coefficient at the load, (c) the input impedance, vg t 5 cos 8 107 t 30 (V) 2 25, is terminated Ii Vi Zin 1 4e j34 0 12 5 j12 7 78 4e j11 5 (mA) Vi 1 40e Vg Zin Zg Zin 5 50 12 5 j12 7 12 5 j12 7 j34 0 (V) 50 100 j100 j50 tan 50 j 100 j100 tan 3 3 rad rad 12 5 50 j 100 50 j100 tan rad/m 5 cm 100 j100 Zin Z0 ZL Z0 jZ0 tan l jZL tan l j50 tan 20 3 20 3 rad/m 5 cm 2 3 109 rad/s 108 m/s 20 rad/m 3 0 62e j12 7 ZL ZL Z0 Z0 100 100 j100 j100 50 50 j29 7 CHAPTER 2 (d) the input voltage Vi , (e) the time-domain input voltage vi t . Solution: 45 ~ 150 I i Zg ~ Vg + - Transmission line + Z0 = 150 + ~ VL l=6m Load z=0 IL ZL (150-j50) ~ ~ Vi Zin - Generator z = -l Zg ~ Vg + - ~ Ii + ~ Vi Figure P2.22: Circuit for Problem 2.22. Zin (a) l 0 4 6 2 4 (rad) up f up up c r 108 2 108 (m/s) 2 25 2up 2 2 108 5 m 8 107 8 107 0 4 (rad/m) 2 108 3 Vg 5e j30 V vg t 5 cos 8 107 t 30 V 46 CHAPTER 2 0 4 (d) (e) Problem 2.23 Two half-wave dipole antennas, each with impedance of 75 , are connected in parallel through a pair of transmission lines, and the combination is connected to a feed transmission line, as shown in Fig. 2.39 (P2.23(a)). All lines are 50 and lossless. (a) Calculate Zin1 , the input impedance of the antenna-terminated line, at the parallel juncture. (b) Combine Zin1 and Zin2 in parallel to obtain ZL , the effective load impedance of the feedline. (c) Calculate Zin of the feedline. Solution: (a) Zin1 ZL1 jZ0 tan l1 Z0 jZL1 tan l1 75 j50 tan 2 0 2 50 50 j75 tan 2 0 2 Z0 35 20 j8 62 vi t Vi e jt 2 2e j22 56 e jt 2 2 cos 8 107t 22 56 5e j30 0 44 e j7 44 2 2e Vi Vg Zin Zg Zin 5e j30 115 7 j27 42 150 115 7 j27 42 115 7 j27 42 5e j30 265 7 j27 42 j22 56 (V) Zin ZL jZ0 tan l Z0 jZL tan l 150 j50 j150 tan 0 4 150 150 j 150 j50 tan 0 4 Z0 115 70 j27 42 V Since this exceeds 2 (rad), we can subtract 2, which leaves a remainder l (rad). ZL Z0 150 j50 150 j50 0 16 e j80 54 . (b) ZL Z0 150 j50 150 300 j50 (c) CHAPTER 2 47 0.2 75 (Antenna) 0.3 Zin Zin1 Zin2 0.2 75 (Antenna) Figure P2.23: (a) Circuit for Problem 2.23. (b) (c) l = 0.3 Zin ' ZL Figure P2.23: (b) Equivalent circuit. Zin 50 17 60 j4 31 j50 tan 2 0 3 50 j 17 60 j4 31 tan 2 0 3 107 57 j56 7 ZL Zin1 Zin2 Zin1 Zin2 35 20 j8 62 2 2 35 20 j8 62 17 60 j4 31 48 CHAPTER 2 Section 2-7: Special Cases Problem 2.24 At an operating frequency of 300 MHz, it is desired to use a section of a lossless 50- transmission line terminated in a short circuit to construct an equivalent load with reactance X 40 . If the phase velocity of the line is 0 75c, what is the shortest possible line length that would exhibit the desired reactance at its input? Solution: On a lossless short-circuited transmission line, the input impedance is always purely sc sc imaginary; i.e., Zin jXin . Solving Eq. (2.68) for the line length, Problem 2.25 A lossless transmission line is terminated in a short circuit. How long (in wavelengths) should the line be in order for it to appear as an open circuit at its input terminals? This is evident from Figure 2.15(d). Problem 2.26 The input impedance of a 31-cm-long lossless transmission line of unknown characteristic impedance was measured at 1 MHz. With the line terminated in a short circuit, the measurement yielded an input impedance equivalent to an inductor with inductance of 0.064 H, and when the line was open circuited, the measurement yielded an input impedance equivalent to a capacitor with capacitance of 40 pF. Find Z0 of the line, the phase velocity, and the relative permittivity of the insulating material. sc Zin jL j2 Solution: Now 2 f 6 28 106 rad/s, so 106 0 064 10 6 l 2 2 n 4 n 2 j0 4 sc Solution: From Eq. (2.68), Zin Hence, jZ0 tan l. If l 2 sc n , then Zin j . for which the smallest positive solution is 8 05 cm (with n 0). l 1 1 1 tan sc Xin Z0 1 tan 8 38 rad/m 40 50 0 675 n rad 8 38 rad/m up 2 rad/cycle 0 75 300 106 cycle/s 3 108 m/s 8 38 rad/m CHAPTER 2 49 where n 0 for the plus sign and n 1 for the minus sign. For n 0, up 1 94 108 m/s 0 65c and r c up 2 1 0 652 2 4. For other values of n, up is very slow and r is unreasonably high. Problem 2.27 A 75- resistive load is preceded by a 4 section of a 50- lossless line, which itself is preceded by another 4 section of a 100- line. What is the input impedance? Solution: The input impedance of the 4 section of line closest to the load is found from Eq. (2.77): The input impedance of the line section closest to the load can be considered as the load impedance of the next section of the line. By reapplying Eq. (2.77), the next section of 4 line is taken into account: Problem 2.28 A 100-MHz FM broadcast station uses a 300- transmission line between the transmitter and a tower-mounted half-wave dipole antenna. The antenna impedance is 73 . You are asked to design a quarter-wave transformer to match the antenna to the line. (a) Determine the electrical length and characteristic impedance of the quarterwave section. (b) If the quarter-wave section is a two-wire line with d 2 5 cm, and the spacing between the wires is made of polystyrene with r 2 6, determine the physical length of the quarter-wave section and the radius of the two wire conductors. Zin 2 Z0 ZL 1002 33 33 300 Zin 2 Z0 ZL 502 75 33 33 tan j0 4 j4000 1 6 28 up l tan 1 sc oc Zin Zin 106 0 31 1 95 106 m/s 0 01 n oc and Zin 1 jC 1 j2 From Eq. (2.74), Z0 Eq. (2.75), 106 40 sc oc Zin Zin 10 12 j4000 . j0 4 j4000 40 Using 50 CHAPTER 2 Solution: (a) For a match condition, the input impedance of a load must match that of the transmission line attached to the generator. A line of electrical length 4 can be used. From Eq. (2.77), the impedance of such a line should be (b) and, from Table 2-2, Hence, ln which leads to Problem 2.29 A 50-MHz generator with Z g 50 is connected to a load ZL 50 j25 . The time-average power transferred from the generator into the load is maximum when Zg ZL where ZL is the complex conjugate of ZL . To achieve this condition without changing Z g , the effective load impedance can be modified by adding an open-circuited line in series with Z L , as shown in Fig. 2-40 (P2.29). If the line's Z0 100 , determine the shortest length of line (in wavelengths) necessary for satisfying the maximum-power-transfer condition. Solution: Since the real part of Z L is equal to Zg , our task is to find l such that the input impedance of the line is Z in j25 , thereby cancelling the imaginary part of ZL (once ZL and the input impedance the line are added in series). Hence, using Eq. (2.73), j100 cot l j25 and whose solution is a d 7 44 25 cm 7 44 3 36 mm. d 2a d 2a 2 1 7 31 1 d 2a d 2a 148 2 6 120 2 1 99 Z0 120 ln d 2a d 2a 2 1 4 up 4f c 4 r f 3 108 4 2 6 100 106 0 465 m Z0 Zin ZL 300 73 148 CHAPTER 2 51 Z 0 = 100 50 + - l ~ Vg Z L (50-j25) Figure P2.29: Transmission-line arrangement for Problem 2.29. or which leads to Since l cannot be negative, the first solution is discarded. The second solution leads to 1 816 1 816 l 0 29 2 Problem 2.30 A 50- lossless line of length l 0 375 connects a 300-MHz generator with Vg 300 V and Zg 50 to a load ZL . Determine the time-domain current through the load for: (a) ZL 50 j50 (b) ZL 50 , (c) ZL 0 (short circuit). Application of Eq. (2.63) gives: Zin Z0 50 ZL Z0 jZ0 tan l jZL tan l 50 j50 50 j 50 j50 tan 135 j50 tan 135 100 ZL ZL Z0 Z0 50 50 j50 j50 50 50 j50 100 j50 0 45 e Solution: (a) ZL 50 j50 , l 2 0 375 2 36 (rad) l 1 326 or 1 816 135 . j63 43 cot l 25 100 0 25 j50 52 CHAPTER 2 50 + Zin Generator z = -l Transmission line ~ Vg Z0 = 50 ZL (50-j50) l = 0.375 z=0 Load Zg ~ Vg + - ~ Ii + ~ Vi Zin Figure P2.30: Circuit for Problem 2.30(a). Using Eq. (2.66) gives 2 68 cos 6 108t 108 44 2 68 e j108 44 e j6 iL t 108 t (A) IL V0 1 Z0 IL e jt 150 e j135 1 50 150 e j135 (V) 0 45 e j63 43 e j135 j63 43 e j135 2 68 e 300 100 j50 50 100 j50 e jl jl 1 0 45 e j108 44 V0 Vg Zin Zg Zin 1 e (A) CHAPTER 2 (b) 53 0 (c) Section 2-8: Power Flow on Lossless Line Problem 2.31 A generator with Vg 300 V and Zg 50 is connected to a load ZL 75 through a 50- lossless line of length l 0 15. (a) Compute Zin , the input impedance of the line at the generator end. (b) Compute Ii and Vi . (c) Compute the time-average power delivered to the line, Pin 1 Vi Ii . 2 (d) Compute VL , IL , and the time-average power delivered to the load, PL 1 VL IL . How does Pin compare to PL ? Explain. 2 (e) Compute the time average power delivered by the generator, Pg , and the time average power dissipated in Zg . Is conservation of power satisfied? Solution: iL t IL V0 Zin jZ0 tan 135 jZ0 tan 135 j50 () Z0 0 300 j50 1 150 e j135 (V) j135 50 j50 e e j135 V0 150 e j135 1 1 1 6e j135 (A) Z0 50 6 cos 6 108 t 135 (A) Z0 ZL 0 1 0 iL t 3e j135 e j6 108 t 3 cos 6 108 t IL 3e j135 (A) 135 V0 300 50 1 50 50 e j135 V0 150 j135 e Z0 50 150 e j135 (V) (A) Zin 0 Z0 50 ZL 50 54 CHAPTER 2 50 + Zin Generator z = -l Transmission line ~ Vg Z0 = 50 75 l = 0.15 z=0 Load Zg ~ Vg + - ~ Ii + ~ Vi Zin Figure P2.31: Circuit for Problem 2.31. (a) (b) Vi Ii Zin 3 24 e j10 16 41 25 j16 35 Zg Zin 50 143 6 e j11 46 Ii Vg 300 41 25 j16 35 3 24 e j10 16 (A) (V) Zin Z0 50 ZL Z0 jZ0 tan l jZL tan l l 2 0 15 54 75 50 j50 tan 54 j75 tan 54 41 25 j16 35 CHAPTER 2 (c) 55 (d) PL Pin , which is as expected because the line is lossless; power input to the line ends up in the load. (e) Power delivered by generator: Power dissipated in Zg : Problem 2.32 If the two-antenna configuration shown in Fig. 2-41 (P2.32) is connected to a generator with Vg 250 V and Zg 50 , how much average power is delivered to each antenna? Solution: Since line 2 is 2 in length, the input impedance is the same as ZL1 75 . The same is true for line 3. At junction CD, we now have two 75- impedances in parallel, whose combination is 75 2 37 5 . Line 1 is 2 long. Hence at AC, input impedance of line 1 is 37.5 , and Zg Zin Ii Vg 250 50 37 5 2 86 (A) Note 1: Pg PZg Pin 478 4 W. PZg IiVZg Ii Ii Zg 1 2 1 2 1 2 Ii Zg 2 1 3 24 2 2 50 262 4 (W) Pg Vg Ii 300 3 24 e j10 16 486 cos 10 16 1 2 1 2 478 4 PL IL V0 1 Z0 1 VL IL 2 150e j54 1 02 2 4 e j54 (A) 50 1 180e j54 2 4 e j54 216 (W) 2 VL V0 1 150e j54 1 02 180e e jl jl j54 (V) V0 Vi 1 e 143 6 e j11 46 e j54 0 2 e j54 ZL ZL Z0 Z0 75 75 50 50 02 150e j54 (V) (W) Pin Vi Ii 1 2 1 143 6 e j11 46 3 24 e j10 16 2 143 6 3 24 cos 21 62 216 (W) 2 56 CHAPTER 2 /2 ZL1 = 75 (Antenna 1) 50 + 250 V B Generator Z in A /2 C Line 1 D 2 ne Li /2 Li ne 3 ZL 2 = 75 (Antenna 2) Figure P2.32: Antenna configuration for Problem 2.32. 2 This is divided equally between the two antennas. Hence, each antenna receives 153 37 76 68 (W). 2 Problem 2.33 For the circuit shown in Fig. 2-42 (P2.33), calculate the average incident power, the average reflected power, and the average power transmitted into the infinite 100- line. The 2 line is lossless and the infinitely long line is slightly lossy. (Hint: The input impedance of an infinitely long line is equal to its characteristic impedance so long as 0.) Solution: Considering the semi-infinite transmission line as equivalent to a load (since all power sent down the line is lost to the rest of the circuit), Z L Z1 100 . Since the feed line is 2 in length, Eq. (2.76) gives Z in ZL 100 and jl l 2 2 , so e 1. From Eq. (2.49a), ZL ZL Z0 Z0 100 100 50 50 1 3 Pin IiVi Ii Ii Zin 1 2 1 2 2 86 2 37 5 153 37 (W) CHAPTER 2 50 + 2V i Pav r Pav t Pav 57 /2 Z0 = 50 Z1 = 100 Figure P2.33: Line terminated in an infinite line. From Eqs. (2.84), (2.85), and (2.86), Problem 2.34 An antenna with a load impedance Z L 75 j25 is connected to a transmitter through a 50- lossless transmission line. If under matched conditions (50- load), the transmitter can deliver 20 W to the load, how much power does it deliver to the antenna? Assume Z g Z0 . t Pav Pav i Pav r Pav 10 0 mW 1 1 mW 8 9 mW r Pav i 2 Pav 1 3 2 10 mW i Pav V0 2Z0 2 1e j180 2 2 50 10 0 mW 1 1 mW 1e j180 1 (V) e jl jl 1 1 3 V0 Vg Zin Zg Zin 1 e 2 100 50 100 Also, converting the generator to a phasor gives Vg results into Eq. (2.66), 2e j0 (V). Plugging all these 1 1 58 Solution: From Eqs. (2.66) and (2.61), CHAPTER 2 Thus, in Eq. (2.86), Section 2-9: Smith Chart Problem 2.35 Use the Smith chart to find the reflection coefficient corresponding to a load impedance: (a) ZL 3Z0 , (b) ZL 2 2 j Z0 , (c) ZL 2 jZ0 , (d) ZL 0 (short circuit). Solution: Refer to Fig. P2.35. (a) Point A is zL 3 j0. (b) Point B is zL 2 j2. (c) Point C is zL 0 j2. (d) Point D is zL 0 j0. 0 5e0 0 62e 29 7 1 0e 53 1 1 0e180 0 so Pav 20 W 1 2 20 W 1 0 2772 18 46 W. ZL ZL Z0 Z0 75 75 j25 j25 50 50 0 277e j33 6 Under the matched condition, 0 and PL 20 W, so Vg 75 j25 , from Eq. (2.49a), When ZL 2 8Z0 2Z0 20 W. Pav 1 2 2 V0 2 1 2Z0 1 jl 2 2 Vg e 1 e j2l 1 e j2l Vg 2 1 8Z0 Vg e jl 1 2 Vg e jl 1 e j2l Vg e jl 1 e j2l 2 Vg Z0 1 e j2l 1 e j2l Z0 Z0 1 e j2l 1 e j2l e jl jl e jl 1 e j2l V0 Vg Zin Zg Zin 1 e CHAPTER 2 59 0.11 0.12 0.13 0.37 0.1 0.0 0.0 8 0.9 1.0 9 0.4 0.39 100 0.38 0.14 0.36 80 90 0.15 0.35 0.1 0.3 4 6 0.1 7 1.2 0.7 1.4 0 .41 110 0.8 70 0 0. 06 13 0.5 0 2.0 0 .43 0.2 1.8 .07 0. 44 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ R ,O o) ) 2 /Yo 0.4 120 (+jB CE AN PT CE S SU VE TI CI PA CA 0.6 60 1.6 0.3 3 0.3 50 0.1 2 8 19 0. 31 0. 0.4 1.0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 D A 0.2 0.1 0.4 0.6 0. 8 B 1.0 0.47 0.3 06 44 0. 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 Figure P2.35: Solution of Problem 2.35. Problem 2.36 Use the Smith chart to find the normalized load impedance corresponding to a reflection coefficient: (a) 0 5, (b) 0 5 60 , (c) 1, (d) 0 3 30 , (e) 0, (f) j. Solution: Refer to Fig. P2.36. 0.3 C 1.8 2 8 0.1 0 -5 0.5 0.3 2.0 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 5.0 0.2 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 50 0.2 1 9 20 10 4.0 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0.2 0.4 0. 4 40 0.2 0.3 0.6 3.0 0.2 0.3 0.2 1 30 0.8 1.0 9 0.22 0.28 4.0 5.0 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 20 50 -20 0.22 0.28 -30 0.2 0.2 0.3 -4 0 0. 19 0. 31 60 CHAPTER 2 0.11 0.12 0.13 0.37 0.1 0.0 0.0 8 0.9 1.0 9 0.4 0.39 100 0.38 0.14 0.36 80 90 0.15 0.35 0.1 0.3 4 6 0.1 7 1.2 F' 0.2 0.7 1.4 0 .41 110 0.8 70 0 1.8 .07 0 0. 06 13 0.5 0 2.0 .43 0. 44 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ R ,O o) ) 2 /Yo 0.4 120 (+jB CE AN PT CE S SU VE TI CI PA CA 0.6 60 1.6 0.3 3 0.3 50 0.1 2 8 19 0. 31 0. 0.4 1.0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 C' E' 0.2 A' 0.1 0.4 D' 0.6 0. 8 1.0 0.47 0.3 06 44 0. 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 Figure P2.36: Solution of Problem 2.36. Problem 2.37 On a lossless transmission line terminated in a load Z L 100 , the standing-wave ratio was measured to be 2.5. Use the Smith chart to find the two possible values of Z0 . (a) Point A is (b) Point B is (c) Point C is (d) Point D is (e) Point E is (f) Point F is 0 5 at zL 3 j0. 0 5e j60 at zL 1 j1 15. 1 at zL 0 j0. 0 3e j30 at zL 1 60 j0 53. 0 at zL 1 j0. j at zL 0 j1. 0.3 1.8 2 8 0.1 0 -5 0.5 0.3 2.0 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 5.0 0.2 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 50 0.2 1 9 20 10 4.0 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0.2 0.4 0. 4 40 0.2 0.3 0.6 3.0 0.2 0.3 0.2 1 30 0.8 1.0 9 4.0 B' 5.0 0.28 0.22 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 20 50 -20 0.22 0.28 -30 0.2 0.2 0.3 -4 0 0. 19 0. 31 CHAPTER 2 61 Solution: Refer to Fig. P2.37. S 2 5 is at point L1 and the constant SWR circle is shown. zL is real at only two places on the SWR circle, at L1, where zL S 2 5, and L2, where zL 1 S 0 4. so Z01 ZL zL1 100 2 5 40 and Z02 ZL zL2 100 0 4 250 . 0.11 0.12 0.13 0.37 0.1 0.0 0.0 8 0.9 1.0 9 0.4 0.39 100 0.38 0.14 0.36 80 90 0.15 0.35 0.1 0.3 4 6 0.1 7 1.2 0.7 1.4 0.4 0.8 1 110 70 0.0 0. 06 3 0.4 0 13 R ,O o) 0.2 1.8 7 0. ) /Yo (+jB CE AN PT CE US S VE TI CI PA CA 0.6 2 0.4 120 60 1.6 0.3 3 0.3 50 2.0 0.1 2 8 0.5 44 0.4 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ 0.6 3.0 0.8 1.0 1.0 4.0 5.0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 L2 L1 0.2 0.1 0.4 0.6 0. 8 1.0 0.47 0.3 06 44 0. 0. 0.5 2.0 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 Figure P2.37: Solution of Problem 2.37. Problem 2.38 A lossless 50- transmission line is terminated in a load with ZL 50 j25 . Use the Smith chart to find the following: (a) the reflection coefficient , (b) the standing-wave ratio, (c) the input impedance at 0 35 from the load, 0.3 1.8 2 0.1 -5 8 0 0.3 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 CA -12 0.08 PAC 0 ITI VE 0.4 RE 3 AC 0.0 TA 7 NC -1 E 30 CO M PO N EN T (-j 0. 4 5.0 0.2 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 50 0.2 1 9 0.2 1 0.22 0.28 20 10 4.0 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0.2 0.4 0. 4 0.3 0.2 0.1 19 0. 0. 31 40 0.2 0.3 0.2 30 9 20 10 20 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 50 0.28 -20 0.22 -30 0.2 0.2 0.3 -4 0 0. 19 0. 31 62 CHAPTER 2 (d) the input admittance at 0 35 from the load, (e) the shortest line length for which the input impedance is purely resistive, (f) the position of the first voltage maximum from the load. 0.1 9 0.0 0.0 8 0.9 1.0 1.2 1.4 0 2 .41 110 0.8 0.7 0 0. 06 13 0.5 0 2.0 0 .43 0.2 1.8 .07 0. 44 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ R ,O o) ) /Yo (+jB CE AN PT CE US ES V TI CI PA CA 0.6 0.4 120 1.6 1.0 Z-LOAD 0.4 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 0.2 0.1 0.4 0.6 0. 8 1.0 0.47 0.3 06 44 0. 0. 0.350 0.3 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 Figure P2.38: Solution of Problem 2.38. Solution: Refer to Fig. P2.38. The normalized impedance is at point Z-LOAD. (a) 0 24e j76 0 The angle of the reflection coefficient is read of that scale at the point r . zL 50 j25 50 1 j0 5 0.3 1.8 2 8 0.1 0 -5 0.5 2.0 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 5.0 0.2 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) Z-IN SWR 50 0.2 1 9 20 10 4.0 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0.11 0.12 0.13 0.37 0.4 0.39 100 0.38 0.14 0.36 80 90 0.15 r 0.35 0.1 0.3 4 6 0.1 7 70 60 0.3 3 0.3 50 0.1 2 8 19 0. 31 0. 0.4 0.106 0.2 0.3 0. 4 40 0.6 3.0 0.2 0.3 0.2 1 30 0.8 1.0 9 0.22 0.28 4.0 5.0 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 20 50 -20 0.22 0.28 -30 0.2 0.2 0.3 -4 0 0. 19 0. 31 CHAPTER 2 63 (b) At the point SW R: S 1 64. (c) Zin is 0 350 from the load, which is at 0 144 on the wavelengths to generator scale. So point Z-IN is at 0 144 0 350 0 494 on the WTG scale. At point Z-IN: (d) At the point on the SWR circle opposite Z-IN, (e) Traveling from the point Z-LOAD in the direction of the generator (clockwise), the SWR circle crosses the xL 0 line first at the point SWR. To travel from Z-LOAD to SWR one must travel 0 250 0 144 0 106. (Readings are on the wavelengths to generator scale.) So the shortest line length would be 0 106. (f) The voltage max occurs at point SWR. From the previous part, this occurs at 0 106. z Problem 2.39 A lossless 50- transmission line is terminated in a short circuit. Use the Smith chart to find (a) the input impedance at a distance 2 3 from the load, (b) the distance from the load at which the input admittance is Yin j0 04 S. Solution: Refer to Fig. P2.39. (a) For a short, zin 0 j0. This is point Z-SHORT and is at 0 000 on the WTG scale. Since a lossless line repeats every 2, traveling 2 3 toward the generator is equivalent to traveling 0 3 toward the generator. This point is at A : Z-IN, and (b) The admittance of a short is at point Y -SHORT and is at 0 250 on the WTG scale: which is point B : Y -IN and is at 0 324 on the WTG scale. Therefore, the line length is 0 324 0 250 0 074. Any integer half wavelengths farther is also valid. yin Yin Z0 j0 04 S 50 j2 Zin zin Z0 0 j3 08 50 j154 Yin yin Z0 1 64 j0 06 50 32 7 j1 17 mS Zin zin Z0 0 61 j0 022 50 30 5 j1 09 64 CHAPTER 2 0.11 0.12 0.13 0.37 0.1 0.0 0.0 8 0.9 1.0 9 0.4 0.39 100 0.38 0.14 0.36 80 0.300 0.15 0.35 0.1 0.3 4 6 0.1 7 90 1.2 0.7 1.4 0 .41 110 0.8 70 0 0. 06 13 0.5 0 2.0 0 .43 0.2 1.8 .07 0. 44 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ R ,O o) ) 2 /Yo 0.4 120 (+jB CE AN PT CE S SU VE TI CI PA CA 0.6 60 1.6 0.3 3 0.3 50 0.1 2 8 19 0. 31 0. 0.4 1.0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 50 Z-SHORT 0.2 0.1 0.4 0.6 0. 8 1.0 0.47 0.3 A:Z-IN 0.2 0.3 06 0. 44 0. B:Y-IN 0.3 2 0.6 1.8 1.6 0.7 1.4 0.8 0.9 1.0 1.2 Figure P2.39: Solution of Problem 2.39. Solution: Refer to Fig. P2.40. The point Z represents 1 5 point Z is at point Y , which is at 0 55 j0 26. j0 7. The reciprocal of Problem 2.40 Use the Smith chart to find yL if zL 15 j0 7. 8 0.1 0 -5 0.5 0.3 2.0 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 0.074 5.0 0.2 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 20 10 4.0 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0.2 0.4 0. 4 40 0.2 0.3 0.6 3.0 0.2 0.3 0.2 1 30 0.8 1.0 9 0.22 0.28 4.0 5.0 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 20 50 Y-SHORT -20 0.22 0.28 0.2 1 -30 0.2 9 -4 0 0. 19 0. 31 CHAPTER 2 65 0.11 0.12 0.13 0.37 0.1 0.0 0.0 8 0.9 1.0 9 0.4 0.39 100 0.38 0.14 0.36 80 90 0.15 0.35 0.1 0.3 4 6 0.1 7 1.2 0.7 1.4 0 .41 110 0.8 70 0 0. 06 13 0.5 0 2.0 0 .43 0.2 1.8 .07 0. 44 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ R ,O o) ) 2 /Yo 0.4 120 (+jB CE AN PT CE S SU VE TI CI PA CA 0.6 60 1.6 0.3 3 0.3 50 0.1 2 8 19 0. 31 0. 0.4 1.0 Y 0.4 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 0.2 0.1 0.4 0.6 8 0. 1.0 0.47 0.3 06 44 0. 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 Figure P2.40: Solution of Problem 2.40. Problem 2.41 A lossless 100- transmission line 3 8 in length is terminated in an unknown impedance. If the input impedance is Z in j2 5 , (a) use the Smith chart to find ZL . (b) What length of open-circuit line could be used to replace Z L ? Solution: Refer to Fig. P2.41. zin Zin Z0 j2 5 100 0 0 j0 025 which is at point Z-IN and is at 0 004 on the wavelengths to load scale. (a) Point Z-LOAD is 0 375 toward the load from the end of the line. Thus, on the wavelength to load scale, it is at 0 004 0 375 0 379. ZL zL Z0 0 j0 95 100 j95 0.3 1.8 2 8 0.1 0 -5 0.5 0.3 2.0 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 5.0 0.2 10 Z 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 50 0.2 1 9 20 4.0 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0. 4 40 0.2 0.3 0.6 3.0 0.2 0.3 0.2 1 30 0.8 1.0 9 0.22 0.28 4.0 5.0 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 20 50 -20 0.22 0.28 -30 0.2 0.2 0.3 -4 0 0. 19 0. 31 66 CHAPTER 2 0.11 0.12 0.13 0.37 0.1 0.0 0.0 8 0.9 1.0 9 0.4 0.39 100 0.38 0.14 0.36 80 1.2 Z-LOAD 0.2 90 0.15 0.35 0.1 0.3 4 6 0.1 7 0.7 1.4 0 .41 110 0.8 70 0 1.8 .07 0 0. 06 13 0.5 0 2.0 .43 0. 44 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ R ,O o) ) 2 /Yo 0.4 120 (+jB CE AN PT CE S SU VE TI CI PA CA 0.6 60 1.6 0.3 3 0.3 50 0.1 2 8 19 0. 31 0. 0.4 1.0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 50 0.2 0.1 0.4 0.6 0. 8 1.0 0.47 0.3 06 44 0. 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 0.246 Figure P2.41: Solution of Problem 2.41. (b) An open circuit is located at point Z-OPEN, which is at 0 250 on the j0 025 must wavelength to load scale. Therefore, an open circuited line with Z in have a length of 0 250 0 004 0 246. S S 1 1 0 29 Solution: Refer to Fig. P2.42. The SWR circle must pass through S SWR. A circle of this radius has 1 8 at point Problem 2.42 A 75- lossless line is 0 6 long. If S Smith chart to find , ZL , and Zin . 1 8 and r 60 , use the 0.3 1.8 2 8 0.1 0 -5 0.5 2.0 0.3 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 0.375 5.0 0.2 50 Z-IN RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 20 10 4.0 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0.2 0.4 0. 4 40 0.2 0.3 0.6 3.0 0.2 0.3 0.2 1 30 0.8 1.0 9 0.22 0.28 4.0 5.0 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 20 50 Z-OPEN -20 0.22 0.28 0.2 1 -30 0.2 9 0.2 0.3 -4 0 0. 19 0. 31 CHAPTER 2 67 0.11 0.12 0.13 0.37 0.1 0.0 0.0 8 0.9 1.0 9 0.4 0.39 100 0.38 0.14 0.36 80 90 0.15 0.35 0.1 0.3 4 6 0.1 7 1.2 0.7 1.4 0 .41 110 0.8 70 0 0. 06 13 0.5 0 2.0 0 .43 0.2 1.8 .07 0. 44 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ R ,O o) ) 2 /Yo 0.4 120 (+jB CE AN PT CE S SU VE TI CI PA CA 0.6 60 1.6 0.3 3 0.3 50 0.1 2 8 19 0. 31 0. 0.4 1.0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 SWR 0.2 0.1 0.4 0.6 0. 8 Z-LOAD 1.0 0.47 0.3 06 44 0. 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 0.100 Figure P2.42: Solution of Problem 2.42. 60 . The angle of the reflection The load must have a reflection coefficient with r coefficient is read off that scale at the point r . The intersection of the circle of constant and the line of constant r is at the load, point Z-LOAD, which has a value zL 1 15 j0 62. Thus, A 0 6 line is equivalent to a 0 1 line. On the WTG scale, Z-LOAD is at 0 333, so Z-IN is at 0 333 0 100 0 433 and has a value zin 0 63 j0 29 ZL zL Z0 1 15 j0 62 75 86 5 j46 6 2 0.3 3 0.1 r 7 0.3 1.8 8 0.1 0 -5 0.5 2.0 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 5.0 0.2 10 Z-IN 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 50 0.2 1 9 20 4.0 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0.2 0.4 0. 4 40 0.2 0.3 0.6 3.0 0.2 0.3 0.2 1 30 0.8 1.0 9 0.22 0.28 4.0 5.0 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 20 50 -20 0.22 0.28 -30 0.2 0.2 0.3 -4 0 0. 19 0. 31 68 CHAPTER 2 Problem 2.43 Using a slotted line on a 50- air-spaced lossless line, the following measurements were obtained: S 1 6, V max occurred only at 10 cm and 24 cm from the load. Use the Smith chart to find Z L . 0.11 0.12 0.13 0.37 0.1 0.0 9 0.9 1.0 0.4 0.39 100 0.38 0.14 0.36 80 90 0.15 0.35 0.1 0.3 4 6 0.1 7 1.2 0.6 1.6 0.357 7 0.0 3 0.4 0 13 R ,O o) 0. Yo) 0 jB/ 120 E (+ NC TA EP SC SU VE TI CI PA CA .42 0.5 0.0 0.7 8 1.4 1 0.4 110 0.8 70 60 0.3 3 0.2 1.8 06 2.0 0. 44 0.4 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ 0.6 0.8 1.0 1.0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 SWR 0.2 0.1 0.4 Z-LOAD .6 0 0. 8 1.0 0.47 0.3 06 44 0. 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 Figure P2.43: Solution of Problem 2.43. Solution: Refer to Fig. P2.43. The point SWR denotes the fact that S 1 6. This point is also the location of a voltage maximum. From the knowledge of the locations of adjacent maxima we can determine that 2 24 cm 10 cm 28 cm. 10 cm Therefore, the load is 28 cm 0 357 from the first voltage maximum, which is at 0 250 on the WTL scale. Traveling this far on the SWR circle we find point Z-LOAD 0.3 1.8 2 8 0.1 0 -5 0.5 0.3 2.0 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 0 CAP -1 AC 20 .08 ITI VE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 OM PO N EN T (-j 0. 4 5.0 0.2 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 50 0.2 1 9 20 10 4.0 3.0 50 1.0 0.8 0. 8 0.6 0.6 0.4 0.2 0.4 0. 4 0.3 Therefore Zin zin Z0 0 63 j0 29 75 47 0 j21 8 . 0.3 0.1 2 8 3.0 0. 19 0. 31 40 0.2 0.3 0.2 1 0.2 4.0 30 5.0 9 0.22 0.28 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 20 50 -20 0.22 0.28 -30 0.2 0.2 0.3 -4 0 0. 19 0. 31 CHAPTER 2 69 Problem 2.44 At an operating frequency of 5 GHz, a 50- lossless coaxial line with insulating material having a relative permittivity r 2 25 is terminated in an antenna with an impedance ZL 150 . Use the Smith chart to find Z in . The line length is 30 cm. Solution: To use the Smith chart the line length must be converted into wavelengths. Since 2 and up , Section 2-10: Impedance Matching Problem 2.45 A 50- lossless line 0 6 long is terminated in a load with ZL 50 j25 . At 0 3 from the load, a resistor with resistance R 30 is connected as shown in Fig. 2-43 (P2.45(a)). Use the Smith chart to find Z in . Zin Z0 = 50 30 Z0 = 50 ZL 0.3 0.3 ZL = (50 + j25) Figure P2.45: (a) Circuit for Problem 2.45. Hence, l 0 30 m 0 04 m 7 5. Since this is an integral number of half wavelengths, Zin ZL 150 2 2up c r f 3 2 25 108 m/s 5 109 Hz Therefore ZL zL Z0 0 82 j0 39 50 41 0 j19 5 . 0 04 m zL at 0 250 0 357 0 500 0 107 on the WTL scale, and here 0 82 j0 39 70 CHAPTER 2 0.11 0.12 0.13 0.37 0.1 0.0 9 0.9 1.0 0.4 0.39 100 0.38 0.14 0.36 80 90 0.15 0.35 0.1 0.3 4 6 0.1 7 1.2 0.6 1.6 0.300 0.08 0 .07 0 0. 06 0.7 1.4 0 .41 110 0.8 70 13 0.5 0 2.0 .43 0. 44 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ R ,O o) ) 2 /Yo 0.4 120 (+jB CE AN PT CE S SU VE TI CI PA CA 60 0.3 3 0.2 1.8 0.3 50 0.1 2 8 19 0. 31 0. 0.4 1.0 Z-LOAD 0.4 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 0.2 0.1 0.4 Y-LOAD 0.6 0. 8 Z-IN 1.0 0.47 0.3 06 44 0. 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 0.300 Figure P2.45: (b) Solution of Problem 2.45. Solution: Refer to Fig. P2.45(b). Since the 30- resistor is in parallel with the input impedance at that point, it is advantageous to convert all quantities to admittances. and is located at point Z-LOAD. The corresponding normalized load admittance is at point Y -LOAD, which is at 0 394 on the WTG scale. The input admittance of the load only at the shunt conductor is at 0 394 0 300 0 500 0 194 and is denoted by point A. It has a value of yinA 1 37 j0 45 zL ZL Z0 50 j25 50 1 j0 5 0.3 1.8 2 8 0.1 0 -5 0.5 0.3 2.0 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 5.0 0.2 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 50 0.2 1 9 20 10 4.0 3.0 1.0 0.8 0.6 0.6 Y-IN 0. 8 0.2 0.4 0. 4 40 0.2 0.3 0.6 3.0 0.2 0.3 0.2 1 30 0.8 1.0 9 0.22 0.28 4.0 5.0 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 A B 20 50 -20 0.22 0.28 -30 0.2 0.2 0.3 -4 0 0. 19 0. 31 CHAPTER 2 The shunt conductance has a normalized conductance 71 The normalized admittance of the shunt conductance in parallel with the input admittance of the load is the sum of their admittances: and is located at point B. On the WTG scale, point B is at 0 242. The input admittance of the entire circuit is at 0 242 0 300 0 500 0 042 and is denoted by point Y -IN. The corresponding normalized input impedance is at Z-IN and has a value of Thus, Problem 2.46 A 50- lossless line is to be matched to an antenna with using a shorted stub. Use the Smith chart to determine the stub length and the distance between the antenna and the stub. Solution: Refer to Fig. P2.46(a) and Fig. P2.46(b), which represent two different solutions. and is located at point Z-LOAD in both figures. Since it is advantageous to work in admittance coordinates, yL is plotted as point Y -LOAD in both figures. Y -LOAD is at 0 041 on the WTG scale. For the first solution in Fig. P2.46(a), point Y -LOAD-IN-1 represents the point at which g 1 on the SWR circle of the load. Y -LOAD-IN-1 is at 0 145 on the WTG scale, so the stub should be located at 0 145 0 041 0 104 from the load (or some multiple of a half wavelength further). At Y -LOAD-IN-1, b 0 52, so a stub with an input admittance of ystub 0 j0 52 is required. This point is Y -STUB-IN-1 and is at 0 423 on the WTG scale. The short circuit admittance zL ZL Z0 75 j20 50 ZL 75 j20 15 j0 4 Zin zin Z0 19 j1 4 50 zin 19 j1 4 95 j70 yinB g yinA 1 67 1 37 j0 45 3 04 j0 45 g 50 30 1 67 72 CHAPTER 2 0.104 0.0 0.0 8 0.11 0.12 0.13 0.37 0.1 0.4 0.9 1.0 9 0.39 100 0.38 0.14 0.36 80 90 0.15 0.35 0.1 0.3 4 6 0.1 7 1.2 0.7 1.4 0 .41 110 0.8 70 0 0. 06 13 0.5 0 2.0 0 .43 0.2 1.8 .07 0. 44 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ R ,O o) ) 2 /Yo 0.4 120 (+jB CE AN PT CE S SU VE TI CI PA CA 0.6 60 1.6 0.3 3 0.3 50 0.1 2 8 19 0. 31 0. 0.4 1.0 Y-LOAD-IN-1 Y-LOAD 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 50 0.2 Z-LOAD 0.4 0.1 0.6 0. 8 1.0 0.47 0.3 06 44 0. 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 0.173 Figure P2.46: (a) First solution to Problem 2.46. is denoted by point Y -SHT, located at 0 250. Therefore, the short stub must be 0 423 0 250 0 173 long (or some multiple of a half wavelength longer). For the second solution in Fig. P2.46(b), point Y -LOAD-IN-2 represents the point at which g 1 on the SWR circle of the load. Y -LOAD-IN-2 is at 0 355 on the WTG scale, so the stub should be located at 0 355 0 041 0 314 from the load (or some multiple of a half wavelength further). At Y -LOAD-IN-2, b 0 52, so a stub with an input admittance of ystub 0 j0 52 is required. This point is Y -STUB-IN-2 and is at 0 077 on the WTG scale. The short circuit admittance is denoted by point Y -SHT, located at 0 250. Therefore, the short stub must be 0 077 0 250 0 500 0 327 long (or some multiple of a half wavelength 0.3 1.8 2 0.3 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.2 Y-STUB-IN-1 8 0.1 0 -5 0.5 2.0 0.4 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 5.0 0.2 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 20 10 4.0 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0.2 0.4 0. 4 40 0.2 0.3 0.6 3.0 0.2 0.3 0.2 1 30 0.8 1.0 9 0.22 0.28 4.0 5.0 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 20 50 Y-SHT -20 0.22 0.28 0.2 1 -30 0.2 9 0.2 0.3 -4 0 0. 19 0. 31 CHAPTER 2 73 0.11 0.12 0.13 0.37 0.1 0.0 0.0 8 0.9 1.0 9 0.4 0.39 100 0.38 0.14 0.36 80 90 0.15 0.35 0.1 0.3 4 6 0.1 7 1.2 0.7 1.4 0 .41 110 0.8 70 0 0. 06 13 0.5 0 2.0 0 .43 Y-STUB-IN-2 0.2 1.8 .07 0. 44 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ R ,O o) ) 2 /Yo 0.4 120 (+jB CE AN PT CE S SU VE TI CI PA CA 0.6 60 1.6 0.3 3 0.3 50 0.1 2 8 19 0. 31 0. 0.4 0.314 0.2 0.3 1.0 Y-LOAD 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 50 0.2 0.2 Z-LOAD 0.4 0.1 Y-LOAD-IN-2 0.6 0. 8 1.0 0.47 0.3 06 44 0. 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 Figure P2.46: (b) Second solution to Problem 2.46. longer). Solution: Refer to Fig. P2.47(a) and Fig. P2.47(b), which represent two different solutions. ZL 100 j50 zL 2 j1 Z0 50 and is located at point Z-LOAD in both figures. Since it is advantageous to work in admittance coordinates, yL is plotted as point Y -LOAD in both figures. Y -LOAD is at 0 463 on the WTG scale. Problem 2.47 Repeat Problem 2.46 for a load with Z L 0.35 0.36 0.37 0.13 0.12 0.38 0.39 0.4 0.327 100 0.3 1.8 2 8 0.1 0 -5 0.5 0.3 2.0 3 0.1 7 -60 0.3 0.1 4 6 -70 0.15 0.14 -80 0.4 0.2 -90 0.11 -100 0.1 0.4 1 -110 0.0 9 0.4 2 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 j50 . 5.0 0.2 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 20 10 4.0 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0. 4 40 0.6 3.0 0.2 0.3 0.2 1 30 0.8 1.0 9 0.22 0.28 4.0 5.0 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 20 50 Y-SHT -20 0.22 0.28 0.2 1 -30 0.2 9 0.2 0.3 -4 0 0. 19 0. 31 74 CHAPTER 2 0.11 0.12 0.13 0.37 0.1 0.0 0.0 8 0.9 1.0 9 0.4 0.39 100 0.38 0.14 0.36 80 90 0.15 0.35 0.1 0.3 4 6 0.1 7 1.2 0.7 1.4 0 .41 110 0.8 70 0. 06 13 0.5 0 2.0 0.199 0. 44 0 0 .43 0.2 1.8 .07 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ R ,O o) ) 2 /Yo 0.4 120 (+jB CE AN PT CE S SU VE TI CI PA CA 0.6 60 1.6 0.3 3 0.3 50 0.1 2 8 19 0. 31 0. 0.4 1.0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 50 0.2 0.1 0.6 0. 8 1.0 0.47 0.3 06 44 0. 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 Figure P2.47: (a) First solution to Problem 2.47. For the first solution in Fig. P2.47(a), point Y -LOAD-IN-1 represents the point at which g 1 on the SWR circle of the load. Y -LOAD-IN-1 is at 0 162 on the WTG scale, so the stub should be located at 0 162 0 463 0 500 0 199 from the load (or some multiple of a half wavelength further). At Y -LOAD-IN-1, b 1, so a stub with an input admittance of ystub 0 j1 is required. This point is Y -STUB-IN-1 and is at 0 375 on the WTG scale. The short circuit admittance is denoted by point Y -SHT, located at 0 250. Therefore, the short stub must be 0 375 0 250 0 125 long (or some multiple of a half wavelength longer). For the second solution in Fig. P2.47(b), point Y -LOAD-IN-2 represents the point at which g 1 on the SWR circle of the load. Y -LOAD-IN-2 is at 0 338 on the 4 0.35 0.15 Y-STUB-IN-1 0.36 0.14 -80 0.3 1.8 2 8 0.1 0 -5 0.5 2.0 0.3 3 0.1 7 -60 0.3 0.1 6 -70 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 0.125 5.0 0.2 10 Y-LOAD 0.4 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 20 4.0 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0.2 0.4 0. 4 40 0.2 0.3 0.6 3.0 0.2 0.3 0.2 1 30 0.8 1.0 9 4.0 Y-LOAD-IN-1 0.22 5.0 0.28 0.2 20 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG Z-LOAD 10 0.1 20 50 Y-SHT -20 0.22 0.28 0.2 1 -30 0.2 9 0.2 0.3 -4 0 0. 19 0. 31 CHAPTER 2 75 0.11 0.12 0.13 0.37 0.1 0.0 0.0 8 0.9 1.0 9 0.4 0.39 100 0.38 0.14 0.36 0.375 0.15 0.1 0.3 4 6 0.1 7 90 1.2 80 Y-STUB-IN-2 0.35 0.7 1.4 0 .41 110 0.8 70 0 0. 06 13 0.5 0 2.0 0 .43 0.2 1.8 .07 0. 44 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ R ,O o) ) 2 /Yo 0.4 120 (+jB CE AN PT CE S SU VE TI CI PA CA 0.6 60 1.6 0.3 3 0.3 50 0.1 2 8 19 0. 31 0. 0.4 1.0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 50 0.2 0.1 0.6 0. 8 1.0 0.47 0.3 06 44 0. 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 Figure P2.47: (b) Second solution to Problem 2.47. WTG scale, so the stub should be located at 0 338 0 463 0 500 0 375 from the load (or some multiple of a half wavelength further). At Y -LOAD-IN-2, b 1, so a stub with an input admittance of ystub 0 j1 is required. This point is Y -STUB-IN-2 and is at 0 125 on the WTG scale. The short circuit admittance is denoted by point Y -SHT, located at 0 250. Therefore, the short stub must be 0 125 0 250 0 500 0 375 long (or some multiple of a half wavelength longer). Problem 2.48 Use the Smith chart to find Z in of the feed line shown in Fig. 2-44 (P2.48(a)). All lines are lossless with Z 0 50 . 0.3 2 0.3 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.2 0.375 1.8 8 0.1 0 -5 0.5 2.0 0.4 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 5.0 0.2 Y-LOAD-IN-2 4.0 10 Y-LOAD 0.4 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 20 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0.2 0.4 0. 4 40 0.2 0.3 0.6 3.0 0.2 0.3 0.2 1 30 0.8 1.0 9 0.22 0.28 4.0 5.0 0.2 20 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG Z-LOAD 10 0.1 20 50 Y-SHT -20 0.22 0.28 0.2 1 -30 0.2 9 0.2 0.3 -4 0 0. 19 0. 31 76 Z1 = (50 + j50) 0.3 CHAPTER 2 Z1 0.3 Zin 0.7 Z2 Z2 = (50 - j50) Figure P2.48: (a) Circuit of Problem 2.48. Solution: Refer to Fig. P2.48(b). and is at point Z-LOAD-1. and is at point Z-LOAD-2. Since at the junction the lines are in parallel, it is advantageous to solve the problem using admittances. y 1 is point Y -LOAD-1, which is at 0 412 on the WTG scale. y2 is point Y -LOAD-2, which is at 0 088 on the WTG scale. Traveling 0 300 from Y -LOAD-1 toward the generator one obtains the input admittance for the upper feed line, point Y -IN-1, with a value of 1 97 j1 02. Since traveling 0 700 is equivalent to traveling 0 200 on any transmission line, the input admittance for the lower line feed is found at point Y -IN-2, which has a value of 1 97 j1 02. The admittance of the two lines together is the sum of their admittances: 1 97 j1 02 1 97 j1 02 3 94 j0 and is denoted Y -JU NCT . 0 300 from Y -JU NCT toward the generator is the input admittance of the entire feed line, point Y -IN, from which Z-IN is found. Zin zin Z0 1 65 j1 79 50 82 5 j89 5 z2 Z2 Z0 50 j50 50 1 j1 z1 Z1 Z0 50 j50 50 1 j1 CHAPTER 2 77 0.11 0.12 0.13 0.37 0.1 0.0 0.0 8 0.9 1.0 9 0.4 0.39 100 0.38 0.14 0.36 80 90 0.15 0.35 0.1 0.3 4 6 0.1 7 1.2 0.7 1.4 0 .41 110 0.8 70 0. 06 13 0.5 0 2.0 0 .43 0.2 1.8 0.300 7 0.0 0. 44 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ R ,O o) ) 2 /Yo 0.4 120 (+jB CE AN PT CE S SU VE TI CI PA CA 0.6 60 1.6 0.3 3 0.3 50 0.1 2 8 19 0. 31 0. 0.4 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 Y-JUNCT 0.2 0.1 0.4 0.6 0. 8 0.47 0.3 06 44 0. 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 0.300 Figure P2.48: (b) Solution of Problem 2.48. Problem 2.49 Repeat Problem 2.48 for the case where all three transmission lines are 4 in length. Solution: Since the transmission lines are in parallel, it is advantageous to express loads in terms of admittances. In the upper branch, which is a quarter wave line, Y1 in Y02 Y1 Z1 2 Z0 0.3 1.8 2 8 0.1 0 -5 0.5 0.3 2.0 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 5.0 4.0 3.0 1.0 Y-LOAD-1 0.8 1.0 0.2 Z-IN Z-LOAD-2 -30 10 Y-IN-2 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 50 20 0. 8 Y-IN 0.6 1.0 0.6 0.4 0.2 0.4 0. 4 40 0.2 0.3 0.6 3.0 0.2 0.3 0.2 1 30 0.8 9 4.0 1.0 Y-LOAD-2 Z-LOAD-1 0.22 5.0 0.28 0.2 20 0.200 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG Y-IN-1 10 0.1 20 50 -20 0.22 0.28 0.2 1 0.2 9 0.2 0.3 -4 0 0. 19 0. 31 78 and similarly for the lower branch, CHAPTER 2 Thus, the total load at the junction is 2 Z0 Therefore, since the common transmission line is also quarter-wave, Section 2-11: Transients on Transmission Lines Problem 2.50 Generate a bounce diagram for the voltage V z t for a 1-m long lossless line characterized by Z 0 50 and up 2c 3 (where c is the velocity of light) if the line is fed by a step voltage applied at t 0 by a generator circuit with Vg 60 V and Rg 100 . The line is terminated in a load Z L 25 . Use the bounce diagram to plot V t at a point midway along the length of the line from t 0 to t 25 ns. Solution: From Eq. (2.124b), Also, The bounce diagram is shown in Fig. P2.50(a) and the plot of V t in Fig. P2.50(b). 2 T l up l 2c 3 3 3 108 5 ns V1 Vg Z0 Rg Z0 60 50 100 50 20 V L 1 3 g Rg Rg ZL ZL Z0 Z0 Z0 Z0 100 50 100 50 25 50 25 50 50 150 25 75 1 3 Zin 2 Z0 ZJCT 2 Z0 YJCT Z1 Z2 50 j50 YJCT Y1 in Y2 in Z1 Z2 50 j50 100 Y2 in Y02 Y2 Z2 2 Z0 CHAPTER 2 Voltage 79 = g = 1 3 z=0 = L = - 1 3 z = 0.5 m z=1m V1+ = 20V 5 ns -6.67 V 10 ns -2.22 V 15 ns 0.74 V 20 ns 0.25 V 25 ns -0.08 V t t Figure P2.50: (a) Bounce diagram for Problem 2.50. V(0.5 m, t) 20 V 20 V 13.34 V 11.86 V 12.10 V 11.12 V 5 10 15 20 25 t (ns) Figure P2.50: (b) Time response of voltage. 80 CHAPTER 2 Problem 2.51 Solution: Repeat Problem 2.50 for the current I on the line. Rg Rg ZL ZL Z0 Z0 Z0 Z0 From Eq. (2.124a), The bounce diagram is shown in Fig. P2.51(a) and I t in Fig. P2.51(b). = -g = - 1 3 z=0 Current z = 0.5 m = -L = 1 3 z=1m I1+ = 0.4 A 5 ns 0.133 A 10 ns -0.044 A 15 ns -0.015 A 20 ns 5 10-3 A 25 ns t Figure P2.51: (a) Bounce diagram for Problem 2.51. Rg Z0 t I1 Vg 60 100 50 04A L g 100 50 100 50 25 50 25 50 1 3 1 3 CHAPTER 2 I(0.5 m, t) 0.533 A 0.4 A 0.489 A 0.474 A 0.479 A 81 5 10 15 20 25 t (ns) Figure P2.51: (b) Time response of current. Problem 2.52 In response to a step voltage, the voltage waveform shown in Fig. 2-45 (P2.52) was observed at the sending end of a lossless transmission line with Rg 50 , Z0 50 , and r 2 25. Determine (a) the generator voltage, (b) the length of the line, and (c) the load impedance. V(0, t) 5V 3V z 0 Figure P2.52: Observed voltage at sending end. which gives Vg 2V1 10 V. V1 Solution: (a) From the figure, V1 5 V. Applying Eq. (2.124b), Vg Z0 Rg Z0 Vg Z0 Z0 Z0 Vg 2 6 s 82 (b) up CHAPTER 2 c 3 108 2 108 m/s. The first change in the waveform occurs r 2 25 6 s. But t 2l up . Hence, From Problem 2.53 In response to a step voltage, the voltage waveform shown in Fig. 2.46 (P2.53) was observed at the sending end of a shorted line with Z 0 50 and r 4. Determine Vg , Rg , and the line length. V(0, t) 12 V 3V 0 7 s 14 s 0.75 V z Figure P2.53: Observed voltage at sending end. Solution: Hence, l 525 m. 15 108 7 s 7 2l up c r 108 4 2l 10 6 s up 3 15 108 m/s ZL Z0 50 1 1 L L 1 1 04 04 21 43 V1 LV1 or L (c) Since Rg Z0 , g 0. Hence V2 to 3 V is due to V1 2 V. But 0 and the change in level from 5 V down V1 V1 2 5 04 l at t tp 2 6 10 2 6 2 108 600 m CHAPTER 2 83 is Also, Problem 2.54 Suppose the voltage waveform shown in Fig. 2-45 was observed at the sending end of a 50- transmission line in response to a step voltage introduced by a generator with Vg 15 V and an unknown series resistance R g . The line is 1 km in length, its velocity of propagation is 1 10 8 m/s, and it is terminated in a load ZL 100 . (a) Determine Rg . (b) Explain why the drop in level of V 0 t at t 6 s cannot be due to reflection from the load. (c) Determine the shunt resistance R f and the location of the fault responsible for the observed waveform. Solution: V(0, t) 5V 3V z 0 6 s Figure P2.54: Observed voltage at sending end. which gives Vg 19 2 V. V1 or 12 Vg Z0 Rg Z0 Vg 30 50 50 Rg Z0 50 1 1 g g Hence, 3 V g 12 V, or g 0 25. From Eq. (2.128), 1 1 0 25 0 25 30 V z 0t 7s V1 LV1 g LV1 From the voltage waveform, V1 12 V. At t 7s, the voltage at the sending end gV1 (because L 1 84 (a) CHAPTER 2 which gives Rg 100 and g 1 3. (b) Roundtrip time delay of pulse return from the load is which is much longer than 6 s, the instance at which V 0 t drops in level. (c) The new level of 3 V is equal to V1 plus V1 plus V2 , Problem 2.55 A generator circuit with Vg 200 V and Rg 25 was used to excite a 75- lossless line with a rectangular pulse of duration 0 4 s. The line is 200 m long, its up 2 108 m/s, and it is terminated in a load Z L 125 . (a) Synthesize the voltage pulse exciting the line as the sum of two step functions, Vg1 t and Vg2 t . (b) For each voltage step function, generate a bounce diagram for the voltage on the line. (c) Use the bounce diagrams to plot the total voltage at the sending end of the line. with Vg2 t 200U t 0 4 s Vg1 t 200U t (V) Vg t Vg1 t Vg2 t Solution: (a) pulse length 0 4 s. (V) which gives ZLf 26 92 . Since ZLf is equal to Rf and Z0 in parallel, Rf which yields f 0 3. But f ZLf ZLf Z0 Z0 03 V1 V1 V2 5 5f 5f g 3 2T 2l up 2 1 103 108 5 From Fig. 2-45, V1 5 V. Hence, 15 Rg 50 50 20 s (V) 58 33 . V1 Vg Z0 Rg Z0 CHAPTER 2 85 25 t=0 Z0 = 75 125 + 200 V 200 m Figure P2.55: (a) Circuit for Problem 2.55. V(t) Vg (t) 200 V 1 t 0.4 s -200 V Vg (t) 2 Figure P2.55: (b) Solution of part (a). (b) We will divide the problem into two parts, one for Vg1 t and another for Vg2 t and then we will use superposition to determine the solution for the sum. The solution for Vg2 t will mimic the solution for Vg1 t , except for a reversal in sign and a delay by 0 4 s. L 0 25 For Vg1 t 200U t : g Rg Rg ZL ZL Z0 Z0 Z0 Z0 25 75 25 75 125 75 125 75 05 2 T l up 200 108 1 s 86 V1 CHAPTER 2 (i) V1 0 t at sending end due to Vg1 t : = g = - 1 2 z=0 t=0 Vg (t) 1 = L = 1 4 z = 200 m V1+ = 150V 1 s 37.5V 2 s -18.75V 3 s -4.69V 4 s 2.34V 5 s 0.56V 6 s t -0.28V t Figure P2.55: (c) Bounce diagram for voltage in reaction to Vg1 t . V V1 Z0 Rg Z0 Vg ZL Rg ZL 200 75 150 V 25 75 200 125 166 67 V 25 125 CHAPTER 2 (ii) V2 0 t at sending end due to Vg2 t : = g = - 1 2 z=0 t = 0.4 s 87 -37.5V 2.4 s 18.75V 3.4 s 4.69V 4.4 s -2.34V 5.4 s 6.4 s t 0.28V t Figure P2.55: (d) Bounce diagram for voltage in reaction to Vg2 t . Vg (t) 2 = L = 1 4 z = 200 m V1+ = -150V 1.4 s -0.56V 88 (b) (i) V1 0 t at sending end due to Vg1 t : CHAPTER 2 V1 ( 0, t ) 168.75 150V 166.41 2 4 Figure P2.55: (e) V1 0 t . (ii) V2 0 t at sending end: V2 ( 0, t ) 0.4 2.4 4.4 6.4 -150V -168.75 -166.41 -167.58 -166.67 Figure P2.55: (f) V2 0 t . 167.58 166.67 t (s) 6 t (s) CHAPTER 2 89 V ( 0, t ) 150V 18.75 4 4.4 0.4 2 2.4 -2.34 0.28 t (s) 6 6.4 Figure P2.55: (g) Net voltage V 0 t . Problem 2.56 For the circuit of Problem 2.55, generate a bounce diagram for the current and plot its time history at the middle of the line. Solution: Using the values for g and L calculated in Problem 2.55, we reverse their signs when using them to construct a bounce diagram for the current. V1 Z0 V2 Z0 V ZL 150 2A 75 150 2A 75 I 1 33 A I2 I1 (iii) Net voltage V 0 t V1 0 t V2 0 t : 90 = -g = 1 2 z=0 t=0 I 1 (t) CHAPTER 2 = -L = 1 4 z = 200 m 2A 1 s -0.5A 2 s -0.25A 3 s 62.5mA 4 s 31.25mA 5 s -7.79mA 6 s t -3.90mA t Figure P2.56: (a) Bounce diagram for I1 t in reaction to Vg1 t . CHAPTER 2 = -g = 1 2 z=0 t = 0.4 s I 2 (t) 91 1 4 z = 200 m = -L = - -2A 1.4 s 0.5A 2.4 s 0.25A 3.4 s -62.5mA 4.4 s -31.25mA 5.4 s 7.79mA 6.4 s t 3.90mA t Figure P2.56: (b) Bounce diagram for current I2 t in reaction to Vg2 t . 92 (i) I1 l 2 t due to Vg1 t : CHAPTER 2 I 1 ( 100, t ) 2A 1.5 1.25 1.3125 1.3333 0.5 Figure P2.56: (c) I1 l 2 t . (ii) I2 l 2 t due to Vg2 t : I 2 ( 100, t ) 0.9 1.9 2.9 -1.5 -2A -1.25 Figure P2.56: (d) I2 l 2 t . t (s) 1.5 2.5 3.5 3.9 t (s) -1.3125 -1.3333 CHAPTER 2 93 I ( 0, t ) 2A 1.5 0.5 0.9 1.9 2.5 -0.25 -0.5 Figure P2.56: (e) Total I l 2 t . Problem 2.57 For the parallel-plate transmission line of Problem 2.3, the line parameters are given by: Find , , up , and Z0 at 1 GHz. C 172 G 0 (pF/m) L 167 (nH/m) R 1 (/m) 2.5 (iii) Net current I l 2 t I1 l 2 t I2 l 2 t : .0625 t (s) 0.5 0.5 94 CHAPTER 2 Hence, Problem 2.58 R = 600 Z0 = 300 L = 0.02 mH 31e j0 025 954e j0 05 1 2 1049e j89 95 1 1e j90 1 2 31 Z0 R G jL jC 1 2 j0 01 up 2 f 34 rad/m 0 016 Np/m 2 109 34 1 85 108 m/s 34e j89 97 34 cos 89 97 1154e j179 95 1 2 1049e j89 95 1 1e j90 1 2 j34 sin 89 97 0 016 j34 1 1049 2 e j tan 1 1049 1 1e j90 1 2 1 j1049 j1 1 1 2 j e j90 1 j2 10 9 167 10 9 0 j2 109 172 10 12 1 2 R jL G jC Solution: At 1 GHz, 2 f 2 109 rad/s. Application of (2.22) gives: CHAPTER 2 95 A 300- lossless air transmission line is connected to a complex load composed of a resistor in series with an inductor, as shown in the figure. At 5 MHz, determine: (a) , (b) S, (c) location of voltage maximum nearest to the load, and (d) location of current maximum nearest to the load. Solution: (a) (b) (c) (d) The locations of current maxima correspond to voltage minima and vice versa. Hence, the location of current maximum nearest the load is the same as location of voltage minimum nearest the load. Thus Problem 2.59 17 46 m lmin 4 2 46 15 lmax lmax 4 15 m lmax r for r 4 29 6 60 180 4 2 46 m 0 3 5 108 106 S 1 1 1 1 ZL Z0 ZL Z0 600 j628 600 j628 300 j628 900 j628 300 300 0 63e j29 6 0 63 0 63 1 67 60 m 600 j2 5 106 2 10 5 ZL R jL 600 j628 96 CHAPTER 2 RL = 75 Z0 = 50 C=? A 50- lossless transmission line is connected to a load composed of a 75- resistor in series with a capacitor of unknown capacitance. If at 10 MHz the voltage standing wave ratio on the line was measured to be 3, determine the capacitance C. Solution: Noting that: Hence Problem 2.60 A 50- lossless line is terminated in a load impedance ZL 30 j20 . 2 66 1 C 1 XC 1 107 2 41 XC 66 1 Upon substituting L we have 0 5, RL 75 , and Z0 50 , and then solving for XC , 10 2 R2 L R2 L 2 XC 2 XC 2 Z0 2 Z0 2Z0 RL 2Z0 RL 10 241 pF Z0 ZL ZL Z0 RL jXC RL jXC 2Z0 RL ZL ZL RL jXC RL jXC 2 ZL ZL ZL ZL Z0 ZL Z0 ZL ZL ZL 2 Z0 2 Z0 R2 L 2 ZL ZL Z0 Z0 ZL ZL ZL ZL Z0 Z0 Z0 Z0 2 XC ZL RL jXC where XC S S 1 1 3 3 1 1 2 4 05 1 C CHAPTER 2 97 Z0 = 50 ZL = (30 - j 20) (a) Calculate and S. (b) It has been proposed that by placing an appropriately selected resistor across the line at a distance lmax from the load (as shown in the figure below), where l max is the distance from the load of a voltage maximum, then it is possible to render Z i Z0 , thereby eliminating reflections back to the sending end. Show that the proposed approach is valid and find the value of the shunt resistance. lmax A Z0 = 50 Zi R ZL = (30 - j 20) Solution: (a) (b) We start by finding lmax , the distance of the voltage maximum nearest to the load. Using (2.56) with n 1, Applying (2.63) at l lmax 0 33, for which l 2 lmax r 4 S 2 121 180 4 2 0 33 0 33 2 07 radians, ZL Z0 ZL Z0 1 1 30 j20 50 30 j20 50 1 0 34 2 1 0 34 20 j20 80 j20 20 j20 80 j20 0 34e j121 98 the value of Zin before adding the shunt resistance is: CHAPTER 2 Thus, at the location A (at a distance lmax from the load), the input impedance is purely real. If we add a shunt resistor R in parallel such that the combination is equal to Z0 , then the new Zin at any point to the left of that location will be equal to Z 0 . Hence, we need to select R such that Problem 2.61 For the lossless transmission line circuit shown in the figure, determine the equivalent series lumped-element circuit at 400 MHz at the input to the line. The line has a characteristic impedance of 50 and the insulating layer has r 2 25. Solution: At 400 MHz, Subtracting multiples of 2, the remainder is: l 0 8 rad l or R 98 . Zin Z0 = 50 1.2 m up c 3 108 f f r 4 108 2 25 2 2 l 1 2 4 8 05 1 R 1 102 1 50 75 05m Zin ZL jZ0 tan l Z0 jZL tan l 30 j20 j50 tan 2 07 50 50 j 30 j20 tan 2 07 Z0 102 j0 CHAPTER 2 Using (2.63), 99 Zin is equivalent to a series RL circuit with R Zin L Problem 2.62 Rg + ~ Vg Z0 = 100 ZL = (50 + j 100) The circuit shown in the figure consists of a 100- lossless transmission line terminated in a load with ZL 50 j100 . If the peak value of the load voltage was measured to be VL 12 V, determine: (a) the time-average power dissipated in the load, (b) the time-average power incident on the line, and (c) the time-average power reflected by the load. 20 75 2 4 108 which is a very small inductor. or L 83 L 2 f L R 52 38 20 75 9 10 H Zin ZL Z0 75 50 50 Z0 jZ0 tan l jZL tan l j50 tan 0 8 j75 tan 0 8 52 38 j20 75 100 Solution: (a) CHAPTER 2 The time average power dissipated in the load is: 502 Hence, 1 Problem 2.63 C l1 = 3/8 B Bl Br l2 = 5/8 A Zin Z01 = 100 Z02 = 50 ZL = (75 - j 50) Use the Smith chart to determine the input impedance Z in of the two-line configuration shown in the figure. r Pav i 2 Pav (c) 0 62 2 1 0 47 0 18 W i Pav Pav 2 0 29 0 622 Pav i Pav 1 (b) 2 0 47 W 1 VL 2 RL 2 ZL 2 1 VL 2 ZL Pav 1 2 IL RL 2 2 RL 1 2 122 50 1002 0 29 W ZL ZL Z0 Z0 50 50 j100 j100 100 100 50 150 j100 j100 0 62e j82 9 CHAPTER 2 0.625 0.11 101 0.12 0.13 0.37 90 1.0 0.1 0.9 1.2 0.0 9 0.4 0.39 100 0.38 0.14 0.36 80 0.15 0.35 0.1 0.3 4 6 7 0.0 0. 06 3 0.4 0 13 R ,O o) 0. 44 ) 2 /Yo 0.4 120 (+jB CE AN PT CE S SU VE TI CI PA CA 0.6 0.5 8 0.0 1.4 1 0.4 110 0.8 70 0.7 0.1 60 1.8 1.6 0.3 7 0.1 2 19 0. 3 0.3 8 0.2 50 2.0 31 0. 0.4 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 R-- -170 EN 0.47 VEL > WA 0.0 160 60 4 .46 <-- 0 -1 0.4 4 6 0.0 IND o) UCT 0.0 5 15 jB/Y 0 IVE 5 0 E (15 0.4 RE NC 0.4 5 AC TA 5 TA 0.0 EP 0.1 NC SC SU EC VE OM I 14 40 0 CT PO -1 DU N EN IN R T (+ ,O jX o) Z /Z 0.2 X/ 1.0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 0.2 0.1 0.4 0.6 Br A 0. 8 1.0 0.47 0.3 06 44 0. 0. 0.5 2.0 0.6 1.4 0.8 0.9 1.0 1.2 Smith Chart 1 Solution: Starting at point A, namely at the load, we normalize Z L with respect to Z02 : ZL 75 j50 zL 1 5 j1 (point A on Smith chart 1) Z02 50 From point A on the Smith chart, we move on the SWR circle a distance of 5 8 to point Br , which is just to the right of point B (see figure). At B r , the normalized input impedance of line 2 is: Next, we unnormalize zin2 : Zin2 Z02 zin2 50 0 48 j0 36 zin2 0 48 j0 36 (point Br on Smith chart) j18 24 0.3 0.7 3 7 0.3 0.1 0.433 4 6 1.6 0.3 1.8 2 0.1 -5 8 0 0.1 -60 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0 .42 CAP -12 0.08 A 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 OM PO N EN T (-j 0. 4 0.308 5.0 0.2 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 50 9 20 10 4.0 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0.2 0.4 0. 4 40 0.3 0.2 0.6 3.0 1 0.2 9 0.2 30 0.3 0.8 SWR Circle 4.0 1.0 0.22 5.0 0.28 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 20 50 -20 0.22 0.28 0.2 1 -30 0.3 0.2 0.2 -4 0 0. 19 0. 31 102 0.12 CHAPTER 2 0.13 0.37 0.11 0.1 0.9 1.0 1.2 0.0 0.0 8 9 0.4 0.39 100 0.38 0.14 0.36 80 0.15 0.35 0.375 0.1 0.3 1.6 90 6 0.1 7 0.7 1.4 0.4 2 0.8 1 110 70 4 60 1.8 0.0 7 3 0.4 0 13 R ,O o) 0. 44 ) /Yo (+jB CE AN PT CE US S VE TI CI PA CA 0.6 0.4 120 0.3 3 0.2 50 0.3 0.1 2 8 0. 06 0.5 2.0 19 0. 31 0. 0.4 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 -- 0.0 0.49 GEN D LOAD < ERA OWAR 0.48 180 HS T TO 170 0 NGT R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 4 6 0.0 IND ) UCT 0.0 /Yo 5 15 0 (-jB IVE 5 0 0.4 -15 CE RE 0.4 AN 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T O (+ ), jX Zo /Z 0.2 X/ 1.0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 0.2 0.1 0.4 Bl 0.6 0. 8 1.0 0.47 0.3 06 44 0. 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 0.345 Smith Chart 2 To move along line 1, we need to normalize with respect to Z 01 . We shall call this zL1 : After drawing the SWR circle through point B , we move 3 8 towards the generator, ending up at point C on Smith chart 2. The normalized input impedance of line 1 is: which upon unnormalizing becomes: Zin 66 j125 zin 0 66 j1 25 zL1 Zin2 Z01 24 j18 100 0 24 j0 18 (point B on Smith chart 2) 0.3 1.8 2 8 0.1 0 -5 0.5 0.3 2.0 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0.4 2 C APA -12 0.08 0 CIT IVE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 4 0. 3.0 C 4.0 0.470 5.0 0.2 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 50 0.2 9 20 10 1.0 0.8 0. 8 0.6 0.6 0.4 0.2 0.4 0. 4 SWR Circle 0.6 40 0.2 0.3 0.2 3.0 0.3 0.2 1 30 0.8 9 4.0 1.0 0.22 5.0 0.28 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 20 50 -20 0.22 0.2 0.28 1 -30 0.2 0.3 -4 0 0. 19 0. 31 CHAPTER 2 103 Problem 2.64 l =? B A Z0 = 75 Z=? ZL = 25 A 25- antenna is connected to a 75- lossless transmission line. Reflections back toward the generator can be eliminated by placing a shunt impedance Z at a distance l from the load. Determine the values of Z and l. Solution: 0.250 0.11 0.12 0.13 0.37 0.1 9 0.0 0.0 8 0.9 1.0 0.4 0.39 100 0.38 0.14 0.36 80 90 0.15 0.35 0.1 0.3 4 6 0.1 7 1.2 0.7 1.4 0.4 2 0.8 1 110 70 0 06 13 0.5 0 2.0 0 0. .43 0.2 1.8 .07 0.0 --> WAVELEN 0.49 GTHS TOW ARD 0.48 AD <-- 0.0 0.49 GEN ARD LO ERA 0.48 S TOW 180 TO GTH 170 0 N R-- -17 ELE 0.47 > AV W 0.0 6 160 4 <-- 0.4 -160 0.4 .04 6 0 IND o) UCT 0.0 5 15 B/Y j 0 IVE 5 0 E (0.4 -15 RE NC 0.4 A 5 AC 5 PT TA 0.0 CE 0.1 NC US EC ES OM IV 14 40 0 CT PO -1 DU N EN IN R T (+ ,O jX o) Z /Z 0.2 X/ R ,O o) 0. ) /Yo (+jB CE AN PT CE US ES V TI CI PA CA 0.6 0.4 120 60 1.6 0.3 3 0.3 50 0.1 2 8 0. 19 44 0. 31 0.4 1.0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 0.500 5.0 10 20 A B 0.2 0.1 0.4 0.6 0. 8 1.0 0.47 0.3 06 0. 44 0. 0.6 1.6 0.7 1.4 0.8 0.9 1.0 1.2 0.3 1.8 2 8 0.1 0 -5 0.5 0.3 2.0 3 0.1 7 -60 0.3 0.1 4 6 -70 0.35 0.15 0.36 0.14 -80 0.37 0.13 0.4 0.2 -90 0.12 0.38 0.11 -100 0.39 0.1 0.4 1 -110 0.0 9 0 .42 CAP -12 0.08 AC 0 ITI VE 0.4 RE 3 AC 0.0 TA 7 NC -1 EC 30 O M PO N EN T (-j 0. 4 5.0 0.2 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 50 20 10 4.0 3.0 1.0 0.8 0. 8 0.6 0.6 0.4 0.2 0.4 0. 4 40 0.2 0.3 0.6 3.0 SWR Circle 0.2 0.3 0.2 1 30 0.8 1.0 9 0.22 0.28 4.0 5.0 0.2 20 10 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 COEFFICIENT IN 0.27 REFLECTION DEGR LE OF EES ANG 0.1 20 50 0.750 -20 0.22 0.28 0.2 1 -30 0.2 9 0.2 0.3 -4 0 0. 19 0. 31 104 The normalized load impedance is: CHAPTER 2 The Smith chart shows A and the SWR circle. The goal is to have an equivalent impedance of 75 to the left of B. That equivalent impedance is the parallel combination of Zin at B (to the right of the shunt impedance Z) and the shunt element Z. Since we need for this to be purely real, it's best to choose l such that Zin is purely real, thereby choosing Z to be simply a resistor. Adding two resistors in parallel generates a sum smaller in magnitude than either one of them. So we need for Zin to be larger than Z0 , not smaller. On the Smith chart, that point is B, at a distance l 4 from the load. At that point: which corresponds to Hence, we need y, the normalized admittance corresponding to the shunt impedance Z, to have a value that satisfies: In summary, Problem 2.65 In response to a step voltage, the voltage waveform shown in the figure below was observed at the midpoint of a lossless transmission line with Z0 50 and up 2 108 m/s. Determine: (a) the length of the line, (b) Z L , (c) Rg , and (d) Vg . Z l 4 112 5 Z z y yin y 1 1 yin 1 0 33 0 66 1 1 15 y 0 66 75 1 5 112 5 yin 0 33 zL 25 75 0 33 (point A on Smith chart) zin 3 CHAPTER 2 V (l/2 , t) 12 V 105 15 3 -3 V 9 21 t (s) Solution: (a) Since it takes 3 s to reach the middle of the line, the line length must be (b) From the voltage waveform shown in the figure, the duration of the first rectangle is 6 s, representing the time it takes the incident voltage V1 to travel from the midpoint of the line to the load and back. The fact that the voltage drops to zero at t 9 s implies that the reflected wave is exactly equal to V1 in magnitude, but opposite in polarity. That is, (c) After V1 arrives at the generator end, it encounters a reflection coefficient g . The voltage at 15 s is composed of: g From the figure, V V1 3 12 1 4. Hence, 1 4 V V1 1 1 g 1 L V V1 V1 This in turn implies that L 1, which means that the load is a short circuit: ZL 0 V2 L g V1 V1 V1 l 23 10 6 up 2 3 10 6 2 108 1200 m 106 which means that CHAPTER 2 (d) Vg 12 83 3 50 50 V1 12 Vg Z0 Rg Z0 12 Rg Z0 Z0 32 V Rg 1 1 g Z0 g 1 1 0 25 50 0 25 83 3 ...
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This note was uploaded on 09/15/2008 for the course EE 330 taught by Professor Tofighimohammad- during the Spring '08 term at Pennsylvania State University, University Park.

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