Corrige 08.12

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Exercice 1 - Corrigé 1. d  Z c  γ  =  j β 0  d z Z in Z c , γ ρ(z) =  Z (z)/ Z c 1 Z (z)/ Z c  + 1 . z  =  d ρ(d) = 1. ρ(z) = ρ(0)e 2γz . ρ(0) = ρ(d)e 2γd = e  j 2βd . Z (z) = Z c 1 + ρ(z) 1 ρ(z) Z in  =  Z (0) = Z c 1 e  j 2βd 1 + e  j2βd  = j Z c  ta n (βd).

Transcript of 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.

U (z = 0) = U 1, I (z = 0) = I 1,

U (z = d) = U 2, I (z = d) = −I 2.

z = 0

z = d

U 1 = A + B,

Z cI 1 = A −B,

U 2 = A/E + BE ,

Z cI 2 =−

A/E + BE ,

E = e jβd.

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A

B

U 1 =

Z c

j tan(βd) I 1 +

Z c

j sin(βd) I 2,

U 2 = Z c

j sin(βd)I 1 +

Z c j tan(βd)

I 2.

Z s + Z p = Z c

j tan(βd),

Z p = Z c

j sin(βd),

Z s = Z c

j tan(βd)− Z c

j sin(βd) = j Z c tan

βd

2

.

2.

d λ

sin(βd) ≈ βd

tan

βd

2

≈ βd

2 .

Z c =

L/C β = ω√ LC

Z s = j ωLd

2,

L := Ld

2

,

Z p = 1

j ωC d,

C := C d

.

⇐⇒

d λ

Z c, γ

Ld/2 Ld/2

C d