Question 2.3: SMITH CHART OPERATIONS USING ADMITTANCES A load of ZL = 100 ...
SMITH CHART OPERATIONS USING ADMITTANCES A load of \ Z_{L}=100+j50\Omega terminates a \ 50\Omega line. What are the load admittance and input admittance if the line is \ 50\lambda long?
The normalized load impedance is \ z_{L}=2+j1. A standard Smith chart can be used for this problem by initially considering it as an impedance chart and plotting \ z_{L} and the SWR circle. Conversion to admittance can be accomplished with a \ \lambda/4 rotation of \ z_{L} (easily obtained by drawing a straight line through \ z_{L} and the center of the chart to intersect the other side of the SWR circle). The chart can now be considered as an admittance chart, and the input admittance can be found
by rotating \ 0.15\lambda from y_{L} .
Alternatively, we can use the combined \ z_{y}chart of Figure 2.12, where conversion between impedance and admittance is accomplished merely by reading the appropriate scales. Plotting \ z_{L} on the impedance scales and reading the admittance scales at this same point gives \ y_{L}=0.40-j0.20 . The actual load admittance is then
\ Y_{L}=y_{L}Y_{0}=\frac{y_{L}}{Z_{0}} =0.0080-j0.0040 S.Then, on the WTG scale, the load admittance is seen to have a reference position of \ 0.214\lambda . Moving \ 0.15\lambda past this point brings us to \ 0.36 \lambda. A radial line at this point on the WTG scale intersects the SWR circle at an admittance of \ y=0.61+j0.66. The actual input admittance is then \ Y = 0.0122 + j0.0132 S.
