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Solved Problems-8

Problem-8
A load of Description: Description: 101724.png is located at the end of a transmission line. At a distance of 0.11λ
from the load, an adjustable stub is placed. Another 0.175λ 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,
 
Description: Description: 101751.png
 
Description: Description: 101737.png
Normalised load admittance
 
Description: Description: 101744.png
Smith chart to determine load lengths
 
Steps of Design
  1. We locate the point of load admittance on the chart, say the point A.
  2. Since the first stub is located at a distance of 0.11λ from the load, we move clockwise a distance 0.11λ 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 – j0.6).
  3. Also, as the distance between the stubs is 0.175λ, the auxiliary circle is drawn by moving a distance of 0.175λ counterclockwise (towards the load) as shown in the Smith chart.
  4. 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,
     
    Description: Description: 101762.png
  5. The first stub must contribute to a susceptance of j0.05 – (– j0.6) = j0.65. The +0.65 constant susceptance circle intersects the rim of the chart at 0.09λ.
     
    So, the length of the first stub as measured from the short-circuit admittance point (extreme left hand point on the chart) is given as,
     
    Description: Description: 101774.png
  6. Now, we move a distance of 0.175λ 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.
  7. The normalised admittance at the point D without the second stub as measured from the chart is (1 + j1.6). For perfect matching, the second stub must eliminate the imaginary component of this admittance, i.e., the second stub must provide a susceptance of – j1.6. This [– j1.6] circle cuts the rim of the chart at 0.34λ.
     
    So, the length of the second stub as measured from the short-circuit admittance point (extreme left-hand point on the chart) is given as,
     
    Description: Description: 101782.png
     
    Hence, the lengths of the two stubs are as given:
    • Length of first stub located at load, Lstub1 = 0.34λ
    • Length of second stub, Lstub2 = 0.09λ




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