This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 2 y 2 )
dt
= k =0
m
k =0 So dk
b k k (c 1 u 1 + c 2 u 2 )
dt y3(t) = c1y1(t) + c2y2(t)
is the output resulting from the input
u3(t) = c1u1(t) + c2u2(t)
since y3(t) and u3 obey (1),
proving that such systems are linear. Natural Response
Can there be a nonzero output when
u(t) = 0 ?
Set u(t) = 0 in eqn. (1), and assume a
solution of the form y (t) = Bept.
n
dk y
ak k = 0
dt k =0
n
= k
=0 = Bept ak pk Bept n
k =0 ak p k = 0 The only way that this equation can be
obeyed for all t is that B = 0 or
n
ak pk = 0
k =0 This says that p must be a pole of the system.
The transfer function
m
b k sk
k
G(s) = n=0
ak sk
k =0 has n poles. Let’s denote them by p1, p2, . . . , pn
Consider
y (t) = n
Aiepit i=1 By linearity, this y (t) is also a solution.
This shows that when u(t) = 0,
y (t) = n
Aiepit i=1 is a solution of eqn. (1), no matter what
the values of the constants A1, . . . , An. Theorem 3
All solutions of eqn. (1) with u(t) = 0 are
y (t) = n
Aiepit i=1 Proof We showed above that such y ’s are
solutions. These solutions are called the
natural response.
For this type of system, the natural
response is a sum of complex exponentials,
the frequencies of which coincide with the
poles of the system. 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. Consequence
To obtain a unique solution, need to
specify the n initial conditions
dy
d2y
dn−1y
(0), 2 (0), . . . n−1 (0)
y (o),
dt
dt
dt
which then determine the values of
A1 , . . . , A n . Example ✟
✟✟
✟
❡✟
✟
❡ 2k Ω 1mF The switch is closed at t = 0,
and Vc(t) = 6V for t ≤ 0.
Find Vc(t) for t > 0. +
Vc
− dVc Vc
C
+
=0
dt
R
Vc
dVc Vc
dVc
+
=
+
=0
⇒
dt
RC
dt
2
In operator notation,
1 Vc s +
=0
2
The characteristic polynomial is
1
a( s ) = s +
2
Its roots are the system’s poles.
1
There is one pole, at s = − .
2
So all solutions are
Vc(t) = A1ep1t = A1e−1/2t
Check:
dVc Vc
1
−1/2t + 1 A e−1/2t = 0
+
= − A1 e
dt
2
2
21 The value of A1 is determined by the
initial condition
Vc(0) = 6
Vc(0) = A1e0 = A1 = 6
⇒ Vc(t) = 6e−1/2t Example ✟
✟✟
✟
❡✟
✟
❡ 1k Ω ✬✩ +
Vs = e−3t ✫✪
− The switch is closed at t = 0,
and Vc(t) = 6V for t ≤ 0.
Find Vc(t) for t > 0. 1mF +
Vc
− For t > 0,
dVc Vc − Vs
+
=0
C
dt
R
dVc
Vc
dVc
⇒
+
=
+ (Vc − Vs) = 0
dt
RC
dt
In operator nota...
View
Full
Document
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

Click to edit the document details