# Solved Problems-8

Problem-8

A load of is located at the end of a transmission line. At a distance of 0.11Î»

from the load, an adjustable stub is placed. Another 0.175Î» distance from the first stub, a second stub is placed. Using a Smith chart, determine the lengths of the two stubs.

from the load, an adjustable stub is placed. Another 0.175Î» distance from the first stub, a second stub is placed. Using a Smith chart, determine the lengths of the two stubs.

Solution

The normalised load admittance is given as,

**Normalised load admittance**

**Smith chart to determine load lengths**

**Steps of Design**

- We locate the point of load admittance on the chart, say the point
*A*. - Since the first stub is located at a distance of 0.11Î» from the load, we move clockwise a distance 0.11Î» along the rim of the chart from the point
*A*on a circle of constant radius*OA*and come to the point*B*. The*first stub is located at the point B*. The normalised admittance of this point (point without stub) as measured from chart is (0.25 â€“*j*0.6). - Also, as the distance between the stubs is 0.175Î», the
*auxiliary circle*is drawn by moving a distance of 0.175Î» counterclockwise (towards the load) as shown in the Smith chart. - This auxiliary circle intersects the constant conductance circle (with conductance 0.25) at the point
*C*. This point gives the normalised admittance with the first stub. This normalised admittance as read from the chart is, - The first stub must contribute to a susceptance of
*j*0.05 â€“ (â€“*j*0.6) =*j*0.65. The +0.65 constant susceptance circle intersects the rim of the chart at 0.09Î». - Now, we move a distance of 0.175Î» clockwise along the rim of the chart. This location cuts the unitary conductance circle at the point
*D*. Thus, we have to move from the location of stub 1 (point B) to the location of stub 2 (point D). The*second stub is located at point D*. - The normalised admittance at the point
*D*without the second stub as measured from the chart is (1 +*j*1.6). For perfect matching, the second stub must eliminate the imaginary component of this admittance, i.e., the second stub must provide a susceptance of â€“*j*1.6. This [â€“*j*1.6] circle cuts the rim of the chart at 0.34Î».- Length of first stub located at load,
*L*_{stub1}= 0.34Î» - Length of second stub,
*L*_{stub2}= 0.09Î»

- Length of first stub located at load,