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7/26/2019 Corrige 08.12
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Exercice 1 - Corrigé
1.
d
Z c
γ = jβ
0 d z
Z in
Z c, γ
ρ(z) = Z (z)/Z c − 1Z (z)/Z c + 1
.
z = d
ρ(d) = −1.
ρ(z) = ρ(0)e2γz .
ρ(0) = ρ(d)e−2γd = −e− j 2βd.
Z (z) = Z c1 + ρ(z)
1− ρ(z)
Z in = Z (0) = Z c1− e− j 2βd
1 + e− j 2βd
= j Z c tan (βd).
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0 d z
Z in
Z c, γ
ρ(d) = 1
=⇒ ρ(0) = e− j 2βd
=⇒ Z in = Z (0) = − j Z c cot(βd).
d = λ4
0 zd = λ4
C
Z in
Z c, γ
ρ(d) =
Z L
−Z c
Z L + Z c
=⇒ ρ(0) = Z L − Z cZ L + Z c
e− j 2βd = Z c − Z LZ c + Z L
,
d = λ4
βd = π2
Z in = Z (0) = Z c1 + ρ(0)
1− ρ(0) =
Z 2cZ L
.
Z in = Z cZ L+jZ c tan(βd)Z c+jZ L tan(βd)
Z L = 0, Z L →∞
βd = π
2
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2.
L = 10
1
Z c = 50 Ω
Z in = j ωL
⇐⇒
0 d z
j ωL
Z in
Z c, γ
j Z c tan(βd) = j ωL.
β = 2π/λ = 2πf/c = ω/c
βd = arctan
ωL
Z c
=⇒ d =
c
ω arctan
ωL
Z c
= 42.9
.
d < λ/4
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Exercice 2 - Corrigé
d
I 1I 1 I 2I 2
U 1U 1 U 2U 2[Z ] ⇐⇒
Z sZ s
Z p
1.
U (z) = Ae− jβz + Be jβz ,
Z cI (z) = Ae− jβz −Be jβz .
U 1 = Z sI 1 + Z p(I 1 + I 2) = (Z s + Z p)I 1 + Z pI 2,
U 2 = Z sI 2 + Z p(I 1 + I 2) = Z pI 1 + (Z s + Z p)I 2.