Assignment 4
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Assignment 4
λ/4 Transformer for Complex Loads
University of Ottawa
ELG 3106
November 22, 2023
Introduction
In this assignment, A λ/4 transformer (see Figure 1) is used to match a line of lossless
characteristic impedance Z
01
, with complex load Z
L
, to a feedline of lossless characteristic
impedance, Z
01
. To do that, we need to find the transformer lossless characteristic impedance Z
02
and the load line length d with the help of Smith Chart and simulation app Module 2.7 provided
in the textbook website.
Theory
The input impedance at the source end of a line of length l is given by equation 1 below:
Z
¿
=
Z
0
{
Z
L
+
j Z
0
tan
(
βl
)
Z
0
+
j Z
L
tan
(
βl
)
}
(
1
)
In this assignment, we are considering the circuit shown in Figure 1, where Z
01
= Z
in
, therefore we can use
equation 2 to equation the value of Z
02
that is derived by Z
in
= Z
0
2
/Z
L
so that we are allowed to find d
max
and d
min
to renormalize them to the original transmission line.
Z
01
Z
(
d
)
=
Z
02
2
(
2
)
Figure 1. An in-series λ/4 transformer inserted at either d
max
or d
min
Calculation and Simulation Results
Part 1
Given Z
01
= 50Ω, Z
L
= (100 - j200) Ω, and Z
in
= Z
01
using Smith Chart, z
01
= (100 – j200)/50 = 2 – j4 Ω, the point is being plotted as point A in Figure
1 below:
Figure 2. Smith Chart for d
min
From the WTG scale, since the closest voltage minimum is at 0λ (point B), d
min
= z
01
– 0λ, we can
see that d
min
is around 0.218λ, and z(d
1
) ≈ 0.096.
Therefore, by equation 2, Z(d
1
) = 4.8Ω and Z
02
= [Z
in
Z(d
1
)]
1/2
= [50(4.8)]
1/2
= 15.49Ω,
then we renormalize the line to the original transmission line by plotting the following: z(d
1
) =
0.096*Z
01
/Z
02
= 0.096 * 50/15.49 = 0.31 as point C, next we draw a SWR circle as shown in
Figure 2 and plot point D to help us to renormalize with Z
01
. Therefore: z
in
= 3.22 *15.49/50 =
0.998 ≈ 1, renormalized with Z
01
, and Γ = 0.
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