Problem: A lossless 50-Ω transmission line is terminated in a load with ZL = (50 + j25) Ω. Use the Smith chart to find the following: (a) The reflection coefficient Γ. (b) The standing-wave ratio. (c) The input impedance at 0.35λ from the load. (d) The input admittance at 0.35λ from the load. (e) The shortest line length for which the input impedance is purely resistive. (f) The position of the first voltage maximum from the load.

 (a). ΓL = 0.246 < 75°

(b). S =  1.7

(c). Zin =  (30-j)λ

(d). jreal = Arc Po = 0.105λ

(e). jmax = jreal = 0.105λ

Explanation:

attached is a document to help in understanding.

So we will begin with a step by step analysis of the problem.

from the diagram we have that  ZL = (50 + j25) Ω.

where ZL = ZL / Z₀ = 50 + j25 / 50 = 1 + j0.5

so we mark this on the chart as point 'P'

(a) ΓL = mP/m 'P' < Θ L = 1.7/6.9 < 75°

        ΓL = 0.246 < 75°

(b) This s-circle 's' is given thus s = r = 1.7 on the RHS of the chart

       S =  1.7

(c) we are to calculate the input impedance;

ζin = Q = 0.6 - j0.02

therefore Zin = Z₀ζin = 50(0.6 - j0.02) = (30-j)λ

Zin = (30-j)λ

(d) here we are taking R as the diameter opposite of Q on the s=circle

   so R = γin = 1.7 + j0.02

         yin = yo (γin) = (1.7+j0.02) / 50 = (34 + j0.4)ms

          yin = (34 + j0.4)ms

(e) move from 'p' on s-circle to 'o'

where maximum impedance = Znxl = Zos

which gives jreal =  Arc Po = 0.105λ

(f) jmax = jreal = 0.105λ

cheers i hope this helps

Ithink the answer is c but then again i'm probably wrong

Answer(a)   the matererials that are in contact, the two materials and the nature of their surfaces/// (b)/// the force pushing the two surfaces together, pushing the surfaces together causes the more of the asperities to come together and increases the surface area in contact with each other