12 Signals and Systems 05 LTI systems and convolution

dy (t ) ay (t ) bx (t ) dt
If y(0)=y0, and
x(t ) Ke u(t )
bt
,
K at K bt y(t ) y0 e , t 0 e a b a b
y(t ) y0e
零输入响应？
at
, t<0



h(t τ)
t>0
0 t
x(t ) * h(t ) x( ) h(t ) d
=?
25
x x(t) ( ) t 0

h(t τ)
0 t
if t>0
a
x(t ) * h(t ) x( ) h(t ) d

e 0 t , t 0 x( )h(t ) 0 others
conditions （i.e. initial condition）
N
yp(h) is a particular solution of the following equation:
M d k y(t ) d j x(t ) ak bj k j dt dt k 0 j 0
5
17
The impulse response
(t )
h(t ) T [ (t )]
If the response in an LTI system to impulse is h(t) , What is the response to an arbitrary signal x(t)?
dyh (t ) a dt yh (t )
ln[yn (t )] at C
yh (t ) Be
at
（ 3）
7
yp(t) is a particular solution to the equation, and it can be guessed as:
y p (t ) Ae
Signals and Systems --Lecture 5
Y. C. Zeng Sep. 19th, 2014
2
第二章
连续时间系统的时域分析
2.1 Systems are described by linear
differential equations
d y(t ) d x(t ) ak bj k j dt dt k 0 j 0
where ak、bj are real constants.
N
k
M
j
An example:
R x (t) i(t) C
y(t)
dy (t ) y (t ) 1 x(t ) dt RC RC dy (t ) ay (t ) bx (t ) dt
4
Solution of the above equation can be set as：
3T<t y(t)=0

T
t 2T T
h(t )d
t-2T
t
34
x( )
1
t<0
y(t)=0

T
0
0<t<T
h(t τ)
1 2 y (t ) t 2
T<t<2T

-2T
0
3T
t
1 2 y (t ) Tt T 2 2T<t<3T
1 2 3 2 y (t ) t Tt T 2 2
h( )
0
h(t τ)

0

29
-2T
x( )
1

T
0
h(t τ)
-2T
0

30
x( )
1

T
The fourth step
0
h(t τ)
-2T
0
T


t<0 =0
31
x(t ) * h(t ) x( ) h(t ) d
x( )
1
x( )

(t )
t
x(t)
0
0
14
To sample continuous-time signals
x(t) x(t)
x(t0 )
t
t 0

t0

0 t0
to express x(t0) by using unit impulse


x(t ) (t t0 )dt x(t0 ) (t t0 )dt x(t0 )

T

0
h(t τ)
h(t )d
0
-2T
0 T

t
(t )d
0
T
1 2 Tt T 2
33
T<t<2T
x( )
1
x( )

T
1

T
0
0
h(t τ)
h(t τ)
2T<t<3T
-2T
0

T 2T -2T

0
T 2T
1 2 3 2 (t )d t Tt T t 2T 2 2
y(t ) yh (t ) y p (t )
yh(t) is a solution of the following homogeneous differential equation: N k
d y (t ) ak 0 k dt k 0
yh(t)’s exact form depends on N auxiliary (辅助)

And x(t)
x(t ) x( ) (t )d

15
对一般信号 x(t ) ，可以将其分成很多 宽度的区 段，用一个阶梯信号 x (t ) 近似表示x(t ) 。当 0 x(t ) 时，有 x (t ) x(t )
x (t )
x ( k )
0
dy (t ) ay (t ) x (t ) dt
where a is a constant
If y(0)=y0, and
（ 1）
x(t ) Ke u(t )
bt
,
to find y(t)
先找齐次方程的通解：
dy h (t ) ay h (t ) 0 （2） dt
6
dy h (t ) ay h (t ) 0 dt
Abe
bt
bt
Take it into the original equation:
aAe
bt
Ke , t 0
bt
A K /(a b)
K bt （ 4） y p (t ) e , t 0 a b K bt at So y (t ) Be e , t0 a b
an LTI system
h(t )
x(t )
an LTI system
y(t ) ?
18
To analyze as following：
(t )
x(t) an LTI system y(t)= T [x(t)]
h(t )
y(t)=?
variable t
T
T and integral are all linear operators Time-invariant T [ (t )] h(t ) T [ (t )] ?



x( ) (t )d

19



x( ) T (t ) d x( ) h(t ) d


The result：
(t )
an LTI system
h(t )
y(t)
x(t)



x( ) h(t ) d
k (k 1)
t
1/ 引用 (t ) ，即： (t ) 0
1 则有： (t ) 0 0t otherwise
0t otherwise
16
第k个矩形可表示为： x(k ) (t k ) 这些矩形叠加起来就成为阶梯形信号 x (t ) ，
即： x (t )
k
x(k)


(t k )
当 0 时， k
d

(t k ) (t )
于是： x(t )
x (t ) x(t )



x( ) (t )d
表明：任何连续时间信号 x(t )都可以被分解成移位 加权的单位冲激信号的线性组合。
1 0 t T x(t ) other t 0 t 0 t 2T h(t ) other t 0
x(t)
1
u (t) h (t) t
T
t 0
2T
28
0
The first step
x( )
1
h( )

T
0
0
2T

The second step
The third step
T
1

T
0
0
h(t τ)
h(t τ)
Байду номын сангаас
-2T

0 T

-2T
0 T

t t y(t ) x( )h(t )d h(t )d 0 t 1 2 (t )d t 0 2
0<t<T
32
x( )
1

T
y(t ) x( )h(t )d
an LTI system
x(t ) e u(t )
0 x(t)
at
y(t)=?
(t )
t 0 y(t) t 0 0
u(t) t
?
t
22
x(t) t
u (t) h (t) t
0 0 How to calculate the convolution of x(t) and h(t) ?
y(t ) x(t ) * h(t ) x( ) h(t ) d
20
Define a new operation—卷积(convolution)
x(t ) * h(t ) x( ) h(t ) d


Notice: variables in the expression
t
21
For example
(t )
h(t ) u (t )
If K=0, then x(t)=0, but y(t) isn’t equal to 0:
bt
y(t ) y0e
at
, t<0
If y(0)=0, then , when x(t)=0, so that y(t)=0
11
几个概念：
• 响应（response） 输出是系统对输入的响应 • 零输入响应（zero input response） 输入x(t)=0时，系统的响应（输出） • 零状态响应（zero state response） 系统初始状态（条件）全为0时，系统的响 应（输出） 12


x x(t) ( )
t 0
h( )
0



23
h( )
0
h(t τ)
0 t
x x(t) ( ) t 0


=0
t<0
h(t τ)
t
t t t t tt tt0 tttt

x(t ) * h(t ) x( ) h(t ) d

24
x x(t) ( ) t 0


t 0
e
a
d
t
1 a e a
0
1 at 1 e a


26
In Conclusion:
1 at y (t ) 1 e u (t ) a
1 a
y(t)


t 0
27
Another problem: To calculate the convolution of two signals as shown as following
零状态响应？
13
2.2 Continuous-Time Linear TimeInvariant Systems
• How does the LTI system response an input signals
Impulse response: （冲激响应）
We recall something about impulse(Dirac function)
9
t<0, x(t)=0, so
y(t ) Be
B y0
at
consider the initial condition: y(0)=y0
y(t ) y0e
at
10
Conclusion：
x(t ) Ke u(t )
K at K bt y(t ) y0 e , t 0 e a b a b
8
y (t ) Be
at
K bt e , t 0 a b
consider the initial condition: y(0)=y0
therefore
K B y0 a b
K at K bt y(t ) y0 e , t 0 e a b a b