SME
Engineering Electromagnetics · Problem type
Input impedance
Use this for “A 50 Ω line λ/8 long ends in a 100 + j50 Ω load. Find its input impedance.”
Diagram
|Zᵢₙ|
=
√(
50
2
+
(−50)
2
)
=
70.71
Ω.
θ = −45°: Zᵢₙ = 70.71∠−45° Ω.
The current leads the voltage by 45°.
This is an example. Type your own values to start a new problem.
Z₀
Line impedance
example · Ω
ℓ ÷ λ
Length in wavelengths
example
βℓ
Electrical length
calculated · °
t
Tangent of βℓ
calculated
Re ZL
Load impedance, real part
example · Ω
Im ZL
Load impedance, imaginary part
example · Ω
Re Zᵢₙ
Input impedance, real part
worked out (not typed) · Ω
Im Zᵢₙ
Input impedance, imaginary part
worked out (not typed) · Ω
Formulas
βℓ
=
360°
×
(
ℓ
λ
)
45 = 360° × 0.125
t
=
tan(βℓ)
1 = tan(45°)
Re
Zᵢₙ
=
Re(
Z₀
((Re
ZL
+
jIm
ZL)
+
jZ₀
×
t
)
÷
(
Z₀
+
j(Re
ZL
+
jIm
ZL)
×
t
))
50 = Re(50((100 + j50) + j50 × 1) ÷ (50 + j(100 + j50) × 1))
Im
Zᵢₙ
=
Im(
Z₀
((Re
ZL
+
jIm
ZL)
+
jZ₀
×
t
)
÷
(
Z₀
+
j(Re
ZL
+
jIm
ZL)
×
t
))
−50 = Im(50((100 + j50) + j50 × 1) ÷ (50 + j(100 + j50) × 1))
Assumptions
•
j² = −1, as circuits texts write it; an impedance is R + jX in ohms.
•
A lossless line: Zᵢₙ repeats every half wavelength, so ℓ ÷ λ runs from 0 to 0.5.
•
βℓ = 360° × ℓ ÷ λ is the electrical length; λ/8 makes tan βℓ = 1.
•
At ℓ = λ/4, tan βℓ has no bound and Zᵢₙ = Z₀² ÷ ZL, as the quarter-wave page uses.
Step-by-step
WE KNOW
Z₀ = 50 Ω, ℓ ÷ λ = 0.125, Re ZL = 100 Ω, Im ZL = 50 Ω
FIND
electrical length (βℓ), tangent of βℓ (t), input impedance, real part (Re Zᵢₙ), input impedance, imaginary part (Im Zᵢₙ)
Step 1 · Find electrical length (βℓ)
Use βℓ = 360° × (ℓ ÷ λ)
One wavelength of line turns the phase through 360°.
βℓ = 360° × 0.125
βℓ = 45°
Step 2 · Find tangent of βℓ (t)
Use t = tan(βℓ)
The tangent of the electrical length.
t
=
tan(45°)
t
=
1
Step 3 · Find input impedance, real part (Re Zᵢₙ)
Use Re Zᵢₙ = Re(Z₀((Re ZL + jIm ZL) + jZ₀ × t) ÷ (Z₀ + j(Re ZL + jIm ZL) × t))
Work out the top and the bottom, then divide: multiply both by the bottom’s conjugate.
Re Zᵢₙ = Re(50((100 + j50) + j50 × 1) ÷ (50 + j(100 + j50) × 1))
(100 + j50) + (j50) = (100 + 0) + j(50 + 50) = 100 + j100
(50)(100 + j100) = 5,000 + j5,000
(j1)(100
+
j50)
=
j100
+
j
2
50
= −50 + j100
50 + (−50 + j100) = (50 + (−50)) + j(0 + 100) = j100
5,000 + j5,000
j100
=
(5,000
+
j5,000)
−j100
(j100)(−j100)
=
500,000 − j500,000
0
2
+
100
2
=
500,000 − j500,000
10,000
= 50 − j50
Re Zᵢₙ = 50 Ω
Step 4 · Find input impedance, imaginary part (Im Zᵢₙ)
Use Im Zᵢₙ = Im(Z₀((Re ZL + jIm ZL) + jZ₀ × t) ÷ (Z₀ + j(Re ZL + jIm ZL) × t))
The imaginary part of the same Zᵢₙ.
Im Zᵢₙ = Im(50((100 + j50) + j50 × 1) ÷ (50 + j(100 + j50) × 1))
Im Zᵢₙ = −50 Ω → Zᵢₙ = 50 − j50 Ω = 70.7107∠−45° Ω
Check
Put the numbers back in. Both sides should match.
45 = 360° × 0.125 ✓
1 = tan(45°) ✓
50 = Re(50((100 + j50) + j50 × 1) ÷ (50 + j(100 + j50) × 1)) ✓
−50 = Im(50((100 + j50) + j50 × 1) ÷ (50 + j(100 + j50) × 1)) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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