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⇒ Vc (s + 1) = Vs
The transfer function is
Vc
1
= G (s ) =
Vs
s+1
The characteristic polynomial is
a(s) = (s + 1)
so the natural response is
Vcn = A1e−t
The forced response is
−3t = − 1 e−3t
Vcf = G(−3)e 2 The total response is
−t − 1 e−3t
Vc = Vcn + Vcf = A1e 2 The initial condition is
Vc(0) = 6
1
⇒ Vc(0) = A1 − = 6
2
13
1
⇒ A1 = + 6 =
2
2
13 −t 1 −3t
Vc = e − e
2
2 Check: dVc
+ Vc = Vs
dt
13 −t 3 −3t
13 −t 1 −3t
− e +e
+
e −e
2
2
2
2
= e−3t = Vs Example
Solve
du
d2y
dy
+ 4u
+ 3 + 2y =
2
dt
dt
dt
when u(t) = 2e−3t and the initial
dy (0)
conditions are
= −1 and y (0) = 1.
dt
In operator notation,
s2 + 3s + 2 y = (s + 4) u so the transfer function is
y
s+4
= G (s ) = 2
u
s + 3s + 2
s+4
=
(s + 1)(s + 2)
The poles are s = −1 and s = −2.
The natural response is
A1e−t + A2e−2 The forced response is
−3t = −3 + 4 2e−3t = e−3t
G(−3)2e
9−9+2
The total response is then
y (t) = A1e−t + A2e−2 + e−3t
The values of A1 and A2 are determined
from the initial conditions.
y (0) = A1 + A2 + 1 = 1
dy (t)
= −A1e−t − 2A2e−2 − 3e−3t
dt
dy (0)
= −A1 − 2A2 − 3 = −1
dt
⇒ A1 + A2 = 0
⇒ A1 + 2A2 = −2
Solving these two equations gives
A1 = 2,
The solution is A2 = −2 y (t) = 2e−t − 2e−2 + e−3t Theorem 1
If u(t) = Aept, then one solution of eqn (1)
is y (t) = G(p)u(t).
Theorem 2
Systems described by
n
m
dk y dk u
ak k =
bk k
dt
dt
k =0 (1) k =0 are linear. Theorem 3
All solutions of eqn. (1) with u(t) = 0 are
y (t) = n
i=1 Aiepit Theorem 4
Suppose that u(t) is nonzero, and that
y1(t) is any one solution of (1), and let
y2(t) = n
Aiepit i=1 Then y1 + y2 is also a solution, for all
values of A1, . . . , An. dn−1y
dny
dy
an n + an−1 n−1 + . . . + a1
+ a0 y
dt
dt
dt
dm−1u
du
dmu
+ b0 u
= bn m +bm−1 m−1 +. . .+b1
dt
dt
dt...
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This note was uploaded on 01/22/2014 for the course ENG 282IN taught by Professor Haines during the Spring '12 term at Pima CC.
 Spring '12
 Haines
 Volt

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