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信号与系统(英文版)第三版 卡门
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2010-11-17
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课件是英文的,作者【美】Edward W .Kamen 和Bonnie S. Heck著 里面还有一些附件最其中的知识点进行讲解!
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EECE 301
Signals & Systems
Prof. Mark Fowler
Discussion #3a
• Review of Differential Equations
Differential Equations Review
Differential Equations like this are Linear and Time Invariant:
)(
)(
...
)(
)(...
)()(
010
1
1
1
tfb
dt
tdf
b
dt
tfd
btya
dt
tyd
a
dt
tyd
a
m
m
m
n
n
n
n
n
n
+++=+++
−
−
−
-coefficients are constants ⇒ TI
-No nonlinear terms ⇒ Linear
.,
)()(
),(,
)()(
),( etc
dt
tyd
dt
tyd
ty
dt
tyd
dt
tyd
tf
p
p
k
k
n
p
p
k
k
n
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
Examples of Nonlinear Terms:
In the following we will BRIEFLY review the basics of solving Linear, Constant
Coefficient Differential Equations under the Homogeneous
Condition
“Homogeneous” means the “forcing function” is zero
That means we are finding the “zero-input response” that occurs due to the effect
of the initial coniditions.
⇒ Write D.E. like this:
(
)
(
)
)(...)(...
)(
01
)(
01
1
1
tfbDbDbtyaDaDaD
DP
m
m
DQ
n
n
n
4443444214444434444421
ΔΔ
==
−
−
+++=++++
⇒.. EqDiff
)()()()( tfDPtyDQ
=
m is the highest-order derivative
on the “input” side
n is the highest-order derivative
on the “output” side
We will assume: m ≤ n
)(
)(
tyD
dt
tyd
k
k
k
≡
Use “operational notation”:
Due to linearity: Total Response = Zero-Input Response + Zero-State Response
Z-I Response: found assuming the input f(t) = 0 but with given IC’s
Z-S Response: found assuming IC’s = 0 but with given f(t) applied
()
00)(...
0)()(:..
01
1
1
>∀=++++⇒
=⇒
−
−
ttyaDaDaD
tyDQED
zi
n
n
n
zi
(▲)
numberscomplex possibly areand
)(Consider
0
λ
λ
c
cety
t
=
“linear combination” of y
zi
(t) & its derivatives must be = 0
Can we find c and λ such that y
0
(t) qualifies as a homogeneous solution?
Finding the Zero-Input Response (Homogeneous Solution)
Assume f(t) = 0
Put y
0
(t) into (▲) and use result for the derivative of an exponential:
0)...(
01
1
1
=++++
−
−
tn
n
n
eaaac
λ
λλλ
must = 0
solutiona is
solutiona is
solutiona is
2
1
2
1
t
n
t
t
n
ec
ec
ec
λ
λ
λ
M
Then, choose c
1
, c
2
,…,c
n
to satisfy the given IC’s
t
n
tt
zi
n
ecececty
λ
λλ
+++= ...)(:Solution I-Z
21
21
tn
n
tn
e
dt
ed
λ
λ
λ
=
Characteristic polynomial
Q(
λ
) has at most n unique roots
(can be complex)
))...()(()(
21 n
Q
λ
λ
λ
λ
λ
λ
λ
−
−
−
=⇒
So…any linear combination
is also a solution to (▲)
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- Star_dsd2014-09-11这个是不是只有前一部分呀?
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