Question

A load ZL = 80− j100 is located at z = 0 on a lossless 50- line.
The operating frequency is 200 MHz and the wavelength on the line is 2 m.
(a) If the line is 0.8 m in length, use the Smith chart to find the input impedance.
(b) What is s? (c) What is the distance from the load to the nearest voltage
maximum? (d) What is the distance from the input to the nearest point at which
the remainder of the line could be replaced by a pure resistance?

Answer 1

The answers of the above question are as following

a)We have to move from Point 'p' by '0.4λ' In clockwise direction where Smith-circle cuts at Q.

therefore, in = Q= 1.6 + jz

so   in = Zo x in = 50(1.6 + jz) = 80 + j100 Ω

b)So if we normalize Zl = l = (Zl/Zo) = ((80 - j100)/50) it will come to 1.6 - jz

we can plot this as point p in the chart .

So if we take m as center and mp as radius , then draw Smith chart .This S circle cuts r = 4.6 circle on right side of

the chart.So the asnwer is r = 4.6.

c)Now if we move from p upto Smith circle it cuts at 0(opencircuit load) in clockwise direction. so dmax will be equal to Arc po = 0.45 λ.

d)Now if we move from load (point p) and then upto point T(min impudence Zmin) Where impudence if real

Arc pt = dmin= 0.2λ

so the distance from source to minima will be

= l- dmin = 0.4λ - 0.2 λ

= 0.2λ