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Description
Sujets
Informations
Publié par | bankexam |
Publié le | 10 novembre 2009 |
Nombre de lectures | 255 |
Langue | Français |
Extrait
U
U(t) = 0 t < 0
U(t) = 1 t> 0
R ]−∞;0[
e
e(t) = 4[U(t)−U(t−2)].
e
E e
E(p)
s
′4s (t)+s(t) = e(t) s(0) = 0.
s S
1 −2p
S(p) = 1− .
1
p p+
4
a b
1 a b
= + .
11 p
p+p p+
44
f F
1 1 1 1
−2p −2pF(p)
1 1p p
p+ p+
4 4
f(t) U(t)
s(t) t
s(t) = 0 t < 0
t
−
4s(t) = 4−4 06 t < 2
t 1 −
4 2s(t) = 4 −1 t> 2
s [0;2]
lims(t)
t→2
t<2
s [2;+∞[
lim s(t)
t→+∞
s
E ]0;3[
f R
E×t 06 t < 1
(3−E)t+2E−3 16 t < 2f(t) = . 5 3 26 t6
2
E = 2
5
f(t) [0;1[ [1;2[ 2;
2
f [−5;10]
E
S f
+∞
2nπ 2nπ
S(t) = a + a cos t +b sin t0 n n
5 5
n=1
E +3
f a = 20
5
b nn
n
1 2nπ 5 2nπ 25 2nπ
tcos t t = sin + cos −1 .
2 25 2nπ 5 4n π 50
5
2
22nπ 2nπ
f(t)cos t t f(t)cos t t
5 51 2
n
5
2 2nπ 25 2nπ 4nπ
f(t)cos t t = (2E−3)cos +(3−E)cos −E .
2 25 4n π 5 50
n
5 2nπ 4nπ
a = (2E−3)cos +(3−E)cos −E .n 2 2n π 5 5
n u nn
2nπ 2nπ
u (t) = a cos t +b sin t tn n n
5 5
u (t) t5
E E0
E u (t) t3
−210 E0
u (t) = u (t) = 03 5