Problem 238 A lossless 50 \u2126 transmission line is terminated in a load with Z L

# Problem 238 a lossless 50 ω transmission line is

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Problem 2.38 A lossless 50- transmission line is terminated in a load with Z L 50 j 25 . Use the Smith chart to find the following: (a) the reflection coefficient Γ , (b) the standing-wave ratio, (c) the input impedance at 0 35 λ from the load, Full file at
62 CHAPTER 2 (d) the input admittance at 0 35 λ 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 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 0.7 0.7 0.7 0.8 0.8 0.8 0.9 0.9 0.9 1.0 1.0 1.0 1.2 1.2 1.2 1.4 1.4 1.4 1.6 1.6 1.6 1.8 1.8 1.8 2.0 2.0 2.0 3.0 3.0 3.0 4.0 4.0 4.0 5.0 5.0 5.0 10 10 10 20 20 20 50 50 50 0.2 0.2 0.2 0.2 0.4 0.4 0.4 0.4 0.6 0.6 0.6 0.6 0.8 0.8 0.8 0.8 1.0 1.0 1.0 1.0 20 -20 30 -30 40 -40 50 -50 60 -60 70 -70 80 -80 90 -90 100 -100 110 -110 120 -120 130 -130 140 -140 150 -150 160 -160 170 -170 180 ± 0.04 0.04 0.05 0.05 0.06 0.06 0.07 0.07 0.08 0.08 0.09 0.09 0.1 0.1 0.11 0.11 0.12 0.12 0.13 0.13 0.14 0.14 0.15 0.15 0.16 0.16 0.17 0.17 0.18 0.18 0.19 0.19 0.2 0.2 0.21 0.21 0.22 0.22 0.23 0.23 0.24 0.24 0.25 0.25 0.26 0.26 0.27 0.27 0.28 0.28 0.29 0.29 0.3 0.3 0.31 0.31 0.32 0.32 0.33 0.33 0.34 0.34 0.35 0.35 0.36 0.36 0.37 0.37 0.38 0.38 0.39 0.39 0.4 0.4 0.41 0.41 0.42 0.42 0.43 0.43 0.44 0.44 0.45 0.45 0.46 0.46 0.47 0.47 0.48 0.48 0.49 0.49 0.0 0.0 A N G L E O F R E F L E C T I O N C O E F F I C I E N T I N D E G R E E S > W A V E L E N G T HS T O W A R D G E N E R A T O R > < W A V E L E N G T H S T O W A R D L O A D < I N D U C T I V E R E A C T A N C E C O M P O N E N T ( + j X / Z o ) , O R CA P A C I T I V E S U S C E P T A N C E ( + j B / Y o ) C A P A C I T I V E R E A C T A N C E CO M P O N E N T ( - j X / Z o ) , O R I N D U C T I V E S U S C E P T A N C E ( - j B / Y o ) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) Z-LOAD SWR Z-IN θ r 0.350λ 0.106λ Figure P2.38: Solution of Problem 2.38. Solution: Refer to Fig. P2.38. The normalized impedance z L 50 j 25 50 1 j 0 5 is at point Z-LOAD . (a) Γ 0 24 e j 76 0 The angle of the reflection coefficient is read of that scale at the point θ r . Full file at
CHAPTER 2 63 (b) At the point SWR : S 1 64. (c) Z in is 0 350 λ from the load, which is at 0 144 λ on the wavelengths to generator scale. So point Z - IN is at 0 144 λ 0 350 λ 0 494 λ on the WTG scale. At point Z-IN : Z in z in Z 0 0 61 j 0 022 50 30 5 j 1 09 (d) At the point on the SWR circle opposite Z-IN , Y in y in Z 0 1 64 j 0 06 50 32 7 j 1 17 mS (e) Traveling from the point Z-LOAD in the direction of the generator (clockwise), the SWR circle crosses the x L 0 line first at the point SWR . To travel from Z-LOAD to SWR one must travel 0 250 λ 0 144 λ 0 106 λ . (Readings are on the wavelengths to generator scale.) So the shortest line length would be 0 106 λ . (f) The voltage max occurs at point SWR . From the previous part, this occurs at z 0 106 λ . 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 λ from the load, (b) the distance from the load at which the input admittance is Y in j 0 04 S.

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