Problem Solutions
2.38. A lossless 50-W line terminated in ZL = 50+j25W Find the reflection coefficient
at the load.
Answer: Normalize the load impedance (1+.5) and find reflection from the radius
(using lower Reflect. Coeff for E scales) and the angle tracing a line
from the center of Chart to the point of zL(=1+.5).2.40. Use Smith Chart to find yL if zL is 1.5-j.7W.
Answer: yload = .55+j.260 (Hint: Move half-a circle which is the same
as a quarter-wave long)2.42 A 75 W lossless line is .6l(=1 whole turn + .1l) long, a standing
wave ratio of 1.8 was measured and qr = -60o.
Use the Smith Chart to determine the load impedance|G|, ZL and Zin.
Answer: Draw the circle of s=1.8, it has a radius of .29 so |G|= .29,
ZL is where the circle intersecs the line between the center of the Smith chart
and the angle of -60. ZL= (1.15-j.62)(75) = 86.5-j46.6).To find Zin, move .1l fromZL toward the generator.
Zin=.63-j.29 * 75 = 476-j21.8W.
2.43 Using a slotted line on a 50 W lossless , a standing
wave ratio of 1.6 was measured and the first two voltage maxima nearest to the load at 10cm and 24cm. Use the Smith Chart to determine the load impedance ZL.
Answer: Draw the circle of s=1.6 in the Smith chart.
The point of s=1.6 is also the location of a voltage maximum. From the knowledge
of the locations of adjacent maxima we can determine that l=2(24-10)=28 cm.
Therefore, the load is 10/28l= .357. from the first maximum, which is
at .25l on the scale of toward the load. The point will be at .25 +.357 -.5(one whole circle)
= .107l. Which will end up in ZL = .82-j.39 or 41-j19.5 ohms.2.44 At an operating frequency of 5GHz, a 50 ohm lossless coax line with er=2.25 terminated in Zl=75. Find Zin. The line is 30cm.
Answer: With the f and er find l= 4cm. Line length 30/4= 7.5 l .which is 15 whole turns
around the chart. Therefore you endup in the same place. Zin=Zl=75 ohms!
