EKT 241: ELECTROMAGNETIC THEORY
SEM 2 2009/2010
TUTORIAL 4 - ANSWER
1. A 300-Ω lossless air transmission line is connected to a complex load composed
of a resistor in series with an inductor as shown in Figure 1. At 5 MHz, determine:
a) reflection coefficient, Γ
b) voltage standing wave ratio, VSWR
c) location of voltage maximum nearest to the load
Figure 1
Solution
(b)
VSWR 
1 |  |
 4.4
1 |  |
2. At an operating frequency of 300 MHz, a lossless 50-Ω air spaced transmission
line 2.5 m in length is terminated with an impedance ZL = (40 + j20)Ω. Find the
input impedance.
Solution
v  f
3  10 8  (300  10 6 )
  1m
For transmission line length, l = nλ/2,
Then, the input impedance becomes,
where n is an integer;
Zin = ZL.
Transmission line length = 2.5m = 5 x λ/2
Hence, Zin = ZL = (40+ j20) Ω.
3. A lossless 50-Ω transmission line is terminated in a load Z L  (50  j 25). Use
the Smith chart to find the following:
a) the reflection coefficient, Γ
b) the voltage standing wave ratio, VSWR
c) the input impedance 0.35λ from the load, Zin
d) the position of the first voltage maximum from the load
Solution
The normalized load impedance,
(50  j 25)
50
 (1  j 0.5)
zL 
a) By measuring the distance between the center of Smith chart and the point where
zL present on the Smith chart using the reflection coefficient scale marked with *;
we get the magnitude of the reflection coefficient, |Γ| = 0.245.
The phase angle of the reflection coefficient, θr = 76°.
Hence, reflection coefficient,
Γ = 0.245 ej76°.
b) By measuring the distance between the center of Smith chart and the point where
zL present on the Smith chart using the SWR scale marked with # (measure from the
middle of scale);
VSWR (ratio) = 1.65.
c) The load impedance on the WTG = 0.1442λ. Hence, moving in clockwise direction
on the WTG scale for 0.35λ, the normalized input impedance zin must lie on the line
where WTG = 0.1442 λ +0.35 λ = 0.4942λ.
The intersection point of this line and the constant SWR circle of 1.65, we get the
point zin.
Hence, zin = (0.6 –j0.025)Ω.
Zin = zin x Z0 = (30 – j1.25) Ω
d) Point where first voltage maximum is plotted on the Smith chart with Pmax.
Hence, the position on the transmission line with first voltage maximum;
lmax = 0.25λ – 0.1442λ = 0.1058λ. (rotation in clockwise direction, WTG scale)