Formule simboliche per il trasporto di Impedenza e Ammettenza

Trasporto Simbolico di impedenza
Z0
ZC = RC + jXC
Z IN
L
Z IN = Z 0 ⋅
= Z0 ⋅
Z C + j ⋅ Z 0 ⋅ tan (β ⋅ l )
Z + j ⋅ Z0 ⋅T
= Z0 ⋅ C
=
Z 0 + j ⋅ Z C ⋅ tan (β ⋅ l )
Z0 + j ⋅ ZC ⋅T
(RC
+ j ⋅ X C ) + j ⋅ Z0 ⋅T
R + j ⋅ (X C + Z0 ⋅ T )
= Z0 ⋅ C
=
(Z 0 − X C ⋅ T ) + j ⋅ RC ⋅ T
Z 0 + j ⋅ (RC + j ⋅ X C ) ⋅ T
= ( Razionalizzando) = Z 0 ⋅
= Z0 ⋅
[RC + j ⋅ ( X C + Z 0 ⋅ T )] ⋅ [(Z 0 − X C ⋅ T ) − j ⋅ RC ⋅ T ] =
(Z 0 − X C ⋅ T )2 + (RC ⋅ T )2
RC ⋅ Z 0 − RC ⋅ X C ⋅ T + RC ⋅ X C ⋅ T + RC ⋅ Z 0 ⋅ T 2
+
Z 02 − 2 ⋅ Z 0 ⋅ X C ⋅ T + X C2 ⋅ T 2 + RC2 ⋅ T 2
[
]
j ⋅ X C ⋅ Z 0 − X C2 ⋅ T + Z 02 ⋅ T − X C ⋅ Z 0 ⋅ T 2 − RC2 ⋅ T
+
=
Z 02 − 2 ⋅ Z 0 ⋅ X C ⋅ T + X C2 ⋅ T 2 + RC2 ⋅ T 2
= Z0 ⋅
(
[
)
(
Z 02 + Z C
2
Im(Z IN ) = Z 0 ⋅
(
)
Z 0 ⋅ RC ⋅ T 2 + 1
Z + ZC
2
0
(
⋅T 2 − 2 ⋅ Z0 ⋅ X C ⋅T
Separando quindi parte reale e parte immaginaria otteniamo:
Re(Z IN ) = Z 0 ⋅
)
Z 0 ⋅ RC ⋅ T 2 + 1 + j ⋅ Z 0 ⋅ X C ⋅ 1 − T 2 + Z 02 − Z C
2
⋅T − 2 ⋅ Z0 ⋅ X C ⋅T
2
(
)
(
Z 0 ⋅ X C ⋅ 1 − T 2 + Z 02 − Z C
Z 02 + Z C
2
2
)⋅ T
⋅T 2 − 2 ⋅ Z0 ⋅ X C ⋅T
2
)⋅ T ]
Per l’ammettenza di ingresso valgono formule analoghe: basta sostituire rispettivamente Z0 con Y0,
RC con GC e XC con BC, dove si ricorda che:
Y0 =
1
Z0
YC =
1
1
=
= GC + jBC
Z C RC + jX C
⎡
⎤
1
⎡
⎤
1
GC = Re[YC ] = Re ⎢
⎥ , BC = Im[YC ] = Im ⎢
⎥
⎣ RC + jX C ⎦
⎣ RC + jX C ⎦
Æ
Da cui l’espressione di YIN:
YIN = Y0 ⋅
(
[
)
(
)
(
Y0 ⋅ GC ⋅ T 2 + 1 + j ⋅ Y0 ⋅ BC ⋅ 1 − T 2 + Y02 − YC
2
)⋅ T ]
Y + YC ⋅ T − 2 ⋅ Y0 ⋅ BC ⋅ T
2
2
0
2
Separando quindi parte reale e parte immaginaria otteniamo:
Re(YIN ) = Y0 ⋅
Im(YIN ) = Y0 ⋅
(
)
Y0 ⋅ GC ⋅ T 2 + 1
Y + YC ⋅ T − 2 ⋅ Y0 ⋅ BC ⋅ T
2
0
2
2
(
)
(
Y0 ⋅ BC ⋅ 1 − T 2 + Y02 − YC
2
)⋅ T
Y02 + YC ⋅ T 2 − 2 ⋅ Y0 ⋅ BC ⋅ T
2