2/9/2010
Zin Calculations using the Smith Chart.doc
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Zin Calculations using
the Smith Chart
z 0′ = 1
zin′
z L′
A
z = −A
z = 0
The normalized input impedance zin′ of a transmission line length
A , when terminated in normalized load z L′ , can be determined as:
zin′ =
=
Zin
Z0
1
Z0
⎛ Z L + j Z 0 tan β A ⎞
⎟
⎝ Z 0 + j Z L tan β A ⎠
Z0 ⎜
Z L Z 0 + j tan β A
1 + j Z L Z 0 tan β A
z ′ + j tan β A
= L
1 + j z L′ tan β A
=
Q: Evaluating this unattractive expression
looks not the least bit pleasant. Isn’t there a
less disagreeable method to determine zin′ ?
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/9/2010
Zin Calculations using the Smith Chart.doc
2/7
A: Yes there is! Instead, we could determine this normalized
input impedance by following these three steps:
1. Convert z L′ to ΓL , using the equation:
ZL − Z0
ZL + Z0
Z Z −1
= L 0
ZL Z0 + 1
z ′ −1
= L
z L′ + 1
ΓL =
2. Convert ΓL to Γin , using the equation:
Γin = ΓL e − j 2 β A
3. Convert Γin to zin′ , using the equation:
zin′ =
Zin 1 + Γin
=
Z 0 1 − Γin
Q: But performing these three
calculations would be even more
difficult than the single step
you described earlier. What
short of dimwit would ever use
(or recommend) this approach?
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/9/2010
Zin Calculations using the Smith Chart.doc
3/7
A: The benefit in this last approach is that each of the three
steps can be executed using a Smith Chart—no complex
calculations are required!
1. Convert z L′ to Γ L
Find the point z L′ from the impedance mappings on your
Smith Chart. Place you pencil at that point—you have now
located the correct Γ L on your complex Γ plane!
For example, say z L′ = 0.6 − j 1.4 . We find on the Smith
Chart the circle for r =0.6 and the circle for x =-1.4. The
intersection of these two circles is the point on the
complex Γ plane corresponding to normalized impedance
z L′ = 0.6 − j 1.4 .
This point is a distance of 0.685 units from the origin, and
is located at angle of –65 degrees. Thus the value of ΓL is:
Γ L = 0.685 e − j 65
D
2. Convert Γ L to Γin
Since we have correctly located the point ΓL on the
complex Γ plane, we merely need to rotate that point
clockwise around a circle ( Γ = 0.685 ) by an angle 2 β A .
When we stop, we are located at the point on the complex
Γ plane where Γ = Γin !
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/9/2010
Zin Calculations using the Smith Chart.doc
4/7
For example, if the length of the transmission line
terminated in z L′ = 0.6 − j 1.4 is A = 0.307 λ , we should
rotate around the Smith Chart a total of 2β A = 1.228π
radians, or 221D . We are now at the point on the complex
Γ plane:
D
Γ = 0.685 e + j 74
This is the value of Γin !
3. Convert Γin to zin′
When you get finished rotating, and your pencil is located
at the point Γ = Γin , simply lift your pencil and determine
the values r and x to which the point corresponds!
For example, we can determine directly from the Smith
D
Chart that the point Γin = 0.685 e + j 74 is located at the
intersection of circles r = 0.5 and x =1.2. In other words:
zin′ = 0.5 + j 1.2
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/9/2010
Zin Calculations using the Smith Chart.doc
5/7
Step 1
Γ = 0.685
Γ L = 0.685 e − j 65
D
θ Γ = −65D
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/9/2010
Zin Calculations using the Smith Chart.doc
6/7
Step 2
A 2 = 0.147 λ
Γin = 0.685 e + j 74
D
Γ = 0.685
ΓL = 0.685 e − j 65
D
A 1 = 0.16λ
A = A 1 + A 2 = 0.160λ + 0.147 λ = 0.307 λ
2β A = 221D
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/9/2010
Zin Calculations using the Smith Chart.doc
7/7
Step 3
zin′ = 0.5 + j 1.2
Jim Stiles
The Univ. of Kansas
Dept. of EECS