182
3 Superposition et le théorème de l’extra-élément
r
d
150Ω
:=
R
C
10kΩ
:=
R
E
1200Ω
:=
r
π
1kΩ
:=
R
f
2.2kΩ
:=
C
f
22nF
:=
|| x y
,
( )
x y
⋅
x y
+
:=
β 150
:=
R
1
470Ω
:=
R
2
1kΩ
:=
H
0
β R
C
⋅
R
1 r
d
+
(
)
r
π
R
E β 1
+
(
)
⋅
+
R
1 R
2
+
r
d
+
(
)
⋅
R
2 R
1 r
d
+
(
)
⋅
1
+
⋅
−
−
5.07127
=
:=
condition pour RHPZ ou LHPZ
R
n
R
f
r
π
β
−
R
E 1
1
β
+
⋅
−
985.33333Ω
τ
=
=
:
1
C
f R
n
⋅
21.67733⋅µs
=
:=
r
π
β
R
E 1
1
β
+
⋅
+
1.21467⋅kΩ
=
R
d
r
d
R
1
+
(
) || R
2 1
1
R
E r
π
+
R
E β
⋅
+
r
d
R
1
+
(
) || R
2
−
⋅
R
C 1
1
R
E r
π
+
R
E β
⋅
+
r
d
R
1
+
(
) || R
2
1
+
1
+
β
⋅
+
⋅
+
R
f
+
15.7261⋅kΩ
=
:=
τ
2
C
f R
d
⋅
345.9742⋅µs
=
:=
ω
z
1
τ
1
:=
f
z
ω
z
2π
7.342⋅kHz
=
ω
=
:
p
1
τ
2
:=
f
p
ω
p
2π
460.01968⋅Hz
=
:=
H
1 s
( ) H
0
1 s
τ
1
⋅
+
1 s
τ
2
⋅
+
⋅
:=
1
10
100 1 10
3
×
1 10
4
×
1 10
5
×
1 10
6
×
–10
0
10
20
20 log H
1 i 2
⋅ π f
k
⋅
(
) 10
,
(
)
⋅
f
k
1
10
100 1 10
3
×
1 10
4
×
1 10
5
×
1 10
6
×
100
120
140
160
180
arg H
1 i 2
⋅ π f
k
⋅
(
)
(
)
180
π
⋅
f
k
Figure 3.55 La feuille montre une réponse plate en continu, suivie d’un pôle puis
d’un zéro à une fréquence supérieure.