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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_.__‘
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hawk“ ﬁeoﬂaut:
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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.
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A.“
mg
20. __ ..
.mrm
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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 ZIN is at 0.1441+ 0.35% = 0,4941 on the WTG scale
ZIN: 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 ZIN, yg _ (1.64+j0.06)
zo " 5052 (e) Traveling from the point ZLOAD in the direction of the generator (clockwise),
the SWR circle crosses the x1, = 0 line ﬁrst at the point SWR. To travel from ZLOAD
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
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 Spring '08
 Pham

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