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# hw4 - Figure P236 Solution of Problem 2.36 1 5.0 5.I J ﬂ...

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Unformatted text preview: Figure P236: Solution of Problem 2.36. 1 5 . .0 5 .I. J . ﬂ m .+ .1 0110 :J =.J =0 . + +0.72”. 3 a0m++ __ m __oo 10 how 5.61 _I_.__‘ mam—21m. 55.13am 00 _ 00.] ===== ._ FFFFF .Bkﬁﬁk. wwamy t t. .mmmmm PPPP )))) hawk“ ﬁeoﬂaut: .2 «ﬁn... _. mu... . 509 = (so + j25) Q = 1 + JO 5 The angle of the reﬂection coefficient is read of that scale at Figure P238: Solution of Problem 2.38. Solution: Refer to Fig. P238. The normalized impedance ZL 00. .6 D7; A.“ mg 20. __ .. .mrm W).m. m8 (0 .m .m ’ 48 - CHAPTERZ (b) At the point SWR: S = 1.64. to (c) Zn. is 0.3501 from the load, which is at 0.144;» on the wavelengths scale. So point Z-IN is at 0.1441+ 0.35% = 0,4941 on the WTG scale- Z-IN: At point Zin = zina) = (0.61 - j0.022) x 50 S2 = (30.5 — j1.09) Q. (d) At the point on the SWR circle opposite Z-IN, yg _ (1.64+j0.06) zo " 5052 (e) Traveling from the point Z-LOAD in the direction of the generator (clockwise), the SWR circle crosses the x1, = 0 line ﬁrst at the point SWR. To travel from Z-LOAD to SWR one must travel 0.2507L— 0.1441 = 0.1067». (Readings are on the wavelengths '- 2 or scale.) So the shortest line length would be 0.106).. voltage max occurs at point SWR. From the previous part, this occurs at 2L. Yin = = (32.7 + jl.17) mS. 45),..1,Z}_::éla {mung}, .o/xmaes is 1 GHz- ‘ ﬂ.___.._.;__._ Llézé _ pué-gxwzww Tb; MES/Jonah»? (‘rxﬂAt I‘M/7ZCIAAC? (‘5. *' (+77?” ~5/J5657 \ 1 j - _ ﬂ ..... E \ \ 5 ‘ ‘L. if V 3’ , mod/Mo ,2. 7" ima { E» ‘ m c 2,“) 22,704 5 3;”; : 3-0/2“ -0 SO J : ,————————~ 106 .7 . Vﬂ’é 2 S + TO K [6% Z. L 31 : Wmﬂ 7 (I. £662 3 ,44401/ i 37‘ 5‘0 (00 Mm (WNW 031me Lﬁbﬂ‘e [mJL- [Wk % 6N9“ ’73, . ...
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