8/18/2019 Theorie Circuits2
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a = ar + 7 a
i , = ,
r 7 ,
i
q = e γ/2 t ! !ar + 7 a
i # ! cos!ω*t# + 7 sin!ω*t## + !,
r + 7 ,
i #! cos!ω*t# 7 sin!ω*t## #
q = e γ/2 t ! !ar + ,
r # !cos!ω*t# + !,
i a
i # !sin!ω*t## + 7 !!a
r ,
r #!sin!ω*t# + !a
i + ,
i #cos!ω*t###
q est r%el donc la partie ia8inaire est nulle
donc ar ,r = 0 et ai + ,i = 0. donc ar = ,r et ai = ,i donc , =
a9 ! aleur
con7u8u%e de a #q = e%/! t ) a e 4 '# t + a5 e
4'# t)n prenant la partie r%elle. on o,tient 3
q = C uc = e%/! t )6 cos)'# t + 7
sin)'# t
i = e %/! t ))7'# %6/! cos)'# t )6'# + %7/!
sin)'# t
Conditions initiales ' t = 0 ( = ) =$ q0 = C) et i = 0C) =
'
0 = :ω* γ'/2 donc :ω* = γ'/2 = γC)/)2ω* q = C- e %/!
t )cos)'# t + %/)!'# sin)'# t
uc = - e %/! t )cos)'# t + %/)!'# sin)'# ti = C- e %/!
t )'# + %/)2'# sin)'# t
1.8 "olution apériodique critique % = !' .
ω* = ω
0 γ/2 = 0 ou γ = 2ω
0
solution oscillante 3 q = C) e γ/2 t !cos!ω* t# +
γ/2/ ω*
sin!ω* t##
On ;ait tendre ω* ers 0 donc γ/2 sin!ω* t#/ ω*
tend ers γ/2 ω*t/
ω* = γ/2 t
La solution 8%n%rale est donc
q = )6 + 7t e(p)'t
i = ))7 '6 '7t e(p)'t
Conditions initiales 3 ' t = 0 ( = ) =$ q0 = C) et i = 0
q = C- )1 + ' t e(p)' tuc= - )1 + ' t e(p)' ti = C-
'! t e(p)' t
1.9 "olution non amortie.
sin! ω0
t + ? # aec ω0
= !1/LC#1/2 et @0
= 2A/ω0
= 2A!LC#1/2
uc = >
sin! ω
0t + ? # = >
/C sin! ω
0t + ? # = (
sin! ω
0t + ? #
i = dq/dt = >
ω0
cos! ω0
t + ? # = C (
ω
0 cos! ω
0t + ? # = B
cos! ω
0t + ? #
Conditions initiales 3 ' t = 0. i = B0 et u
c = (
0
uc = (
sin? = (
0
i = C (
ω
0 cos? = B
0
tan? = sin? /cos? = Cω0
(0/B
0
(
= (0/sin?
)5eple 3 ' t = 0. i = 0 et uc
= ). on o,tient uc
= ) sin!ω0
t + A/2 # = ) cos!ω0
t# .
solution que l*on retroue en ;aisant γ = 0 dans la solution
pseudop%riodique correspondant
au5 es conditions initiales 3 uc = ) e γ/2 t !cos!ω* t# +
γ/2 ω* sin!ω* t##
