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Example: Determining
Transmission Line Length
A load terminating at transmission line has a normalized
impedance z L′ = 2.0 + j 2.0 . What should the length of
transmission line be in order for its input impedance to be:
a) purely real (i.e., xin = 0 )?
b)
have a real (resistive) part equal to one (i.e., rin = 1.0 )?
Solution:
a) Find z L′ = 2.0 + j 2.0 on your Smith Chart, and then rotate
clockwise until you “bump into” the contour x = 0 (recall this is
contour lies on the Γr axis!).
When you reach the x = 0 contour—stop! Lift your pencil and
note that the impedance value of this location is purely real
(after all, x = 0 !).
Now, measure the rotation angle that was required to move
clockwise from z L′ = 2.0 + j 2.0 to an impedance on the x = 0
contour—this angle is equal to 2 β !
You can now solve for , or alternatively use the electrical
length scale surrounding the Smith Chart.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
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One more important point—there are two possible solutions!
Solution 1:
2β = 30
= 0.042λ
z L′ = 2 + j 2
Γ (z )
′
zin = 4.2 + j 0
x =0
Jim Stiles
The Univ. of Kansas
Dept. of EECS
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Solution 2:
z L′ = 2 + j 2
′
zin = 0.24 + j 0
x =0
Γ (z )
2β = 210
= 0.292λ
Jim Stiles
The Univ. of Kansas
Dept. of EECS
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b) Find z L′ = 2.0 + j 2.0 on your Smith Chart, and then rotate
clockwise until you “bump into” the circle r = 1 (recall this circle
intersects the center point or the Smith Chart!).
When you reach the r = 1 circle—stop! Lift your pencil and note
that the impedance value of this location has a real value equal
to one (after all, r = 1 !).
Now, measure the rotation angle that was required to move
clockwise from z L′ = 2.0 + j 2.0 to an impedance on the r = 1
circle—this angle is equal to 2β !
You can now solve for , or alternatively use the electrical
length scale surrounding the Smith Chart.
Again, we find that there are two solutions!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
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Solution 1:
z L′ = 2 + j 2
Γ (z )
r =1
′
zin = 1.0 − j 1.6
2β = 82
= 0.114λ
Jim Stiles
The Univ. of Kansas
Dept. of EECS
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Solution 2:
′
zin = 1.0 + j 1.6
Γ (z )
z L′ = 2 + j 2
r =1
2β = 339
= 0.471λ
Jim Stiles
The Univ. of Kansas
Dept. of EECS
7.
2/16/2010
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′
Q: Hey! For part b), the solutions resulted in zin = 1 − j 1.6 and
′
zin = 1 + j 1.6 --the imaginary parts are equal but opposite! Is
this just a coincidence?
A: Hardly! Remember, the two impedance solutions must result
in the same magnitude for Γ --for this example we find
Γ ( z ) = 0.625 .
Thus, for impedances where r =1 (i.e., z ′ = 1 + j x ):
Γ=
jx
z ′ − 1 (1 + jx ) − 1
=
=
z ′ + 1 (1 + jx ) + 1 2 + j x
and therefore:
2
Γ =
jx
2
2+ j x
2
x2
=
4 +x2
Meaning:
x =
2
4 Γ
2
1− Γ
2
of which there are two equal by opposite solutions!
x = ±
2 Γ
1− Γ
2
Which for this example gives us our solutions x = ±1.6 .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
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