Chapter 3 Fourier Series Representation of Periodic Signals
第3章周期信号的傅里叶级数表示
Main content ：
1.The Frequency Analysis of Periodic
Siganl(周期信号的频域分析）
2.The Frequency Analysis of LTI（LTI系统的频域分析）
3.Properties of Fourier Series（傅立叶级数的性质）
3.0 Introduction(引言)
?The basis for time domain(chapter2)
1)Signal can be represented as linear combination of shift impulses。
2)System is LTI。
?Periodic Singal can be represented as linear combination of complex exponentials.
3.1 Historical Perspective (历史的回顾)
1、The concept of using trigonometric sums to describe periodic phenomena goes back to Babylonians
2、Euler examined the motion of Vibrating string is a linear combination of a few normal mode in 1748.
3.1 Historical Perspective (cont)
3、Largange criticized the use of trigonometric series to examine vibrating string in 1759.
4、Fourier claimed that any periodic signal could be represented by harmonically related sinusoids in1807.
some story about Fourier
?Born in France in 1768
?Fourier claimed that any periodic
signal could be represented by
harmonically related sinusoids in
1807
?Due to Lagrange …s objection his 1768—1830
paper never appeared
?His paper appeared in “The
Analytical Theory of Heat” in 1822
?Dirichlet provide precise conditions
in 1829
傅里叶的两个最重要的贡献——
?“周期信号都可以表示为成谐波关系的正弦信号的加权和”——傅里叶的第一个主要论点?“非周期信号都可以用正弦信号的加权积分来表示”——傅里叶的第二个主要论点
3.2 The Response of LTI Systems to Complex Exponentials(LTI 系统对复指数信号的响应
)
st
e
n
z
()
h n ()h t st
e
()
y t n
z
()
y n
continuios time
discrete time
Using Time domain method ，
()
()()()()s t st
s st
y t e
h d e
h e d H s e
ττ
ττττ∞
∞
---∞
-∞
===??
()
()()()()n k n
k
n
k k y n z
h k z
h k z
H z z
∞
∞
--=-∞
=-∞
=
==∑∑
Eigenvalue
Gain is called “Eigenvalue”
Eigenfunction in-> Same function out with gain Eigenfunction
discrete time
()
h n ()
h t st
e
()st
H s e
n
z
()n
H z Z
continuious time
Eigenfunction
Eigenvalue
()()st
H s h t e dt
∞
--∞=?()()n
k H z h n z
∞
-=-∞
=
∑The usefulness of decomposition in term of eigenfuction is important for the analysis of LTI systems . t
s k
k k k e
s H a t y ∑=)()(t
s k
k k e
a t x ∑=)(If ：
n
k
k
k Z
a n x ∑=)(n k
k
k k Z
Z H a n y ∑=)()(
complex exponential signal 、are eigenfuction
of LTI systems
、are eignevalue.st
e n
z ()H s ()H z Conclusion:
How broad a class of signals could be represented as a linear combination of complex exponentials?
qustion
Example 1 ( 3.1)：
? a LTI systems y(t)=x(t-3) , now the input x(t)=cos(4t)+cos(7t), detemin y(t)?s
s e
d e
s H 3)3()(--+∞
∞-=-=?ττδτ
y(t)= 1/2e -j12e j4t + 1/2e j12e -j4t + 1/2e -j21e j7t + 1/2e j21e -j7t
=cos[4(t-3)}+cos[7(t-3)]
x(t)= 1/2e j4t +1/2e -j4t +1/2e j7t +1/2e -j7t
The set of harmonically related complex exponentials
0(){}jk t
k t e
ωΦ=0,1,2,
k =±
±Each of these signals has a fundamental frequency that is
multiple of ω0,each is periodic with period 0
2T π
ω=
3.3 Fourier Series Representation of Continuous-Time Periodic Signals(连续时间周期信号的傅里叶级数表示)
3.3.1. Linear Combinations of Harmonically Related
Complex exponentials
Thus , is also periodic,the form is referred to as the Fourier series representation
这表明用傅里叶级数可以表示连续时间周期信号，即: 连续时间周期信号可以分解成无数多个复指数谐波分量。
0(),0,1,2
jk t
k k x t a e k ω∞
=-∞
==±±∑Example 2：0
()cos x t t ω=001122
j t j t
e e ωω-=+112
a ±=
Example 3 ：00
()cos 2cos3x t t t
ωω=+0000331[]2j t j t j t j t e e e e ωωωω--=+++11
2
a ±=31a ±=
Some alternative form for the Fourier series 0000*
()jk t jk t jk t jk t k k k k k k k k x t a e a e a e a e
ωωωω∞
∞
∞
∞
-*
**
-=-∞=-∞
=-∞=-∞??====????∑∑∑∑or
k k
a a
*
-∴=*k
k
a a -=)
()(t x t x *
=Suppose x(t) is real ,then is expressed in polar form as k
j k k a A e θ
=k a 0001
()
()
01
()k
k k j jk t
j k t j k t k
k
k k k k x t A e
e
a A e
A e
θωωθωθ∞
-∞
++=-∞
=-∞
==
=+
+∑∑∑
Some alternative form for the Fourier series (CONT)
0001[]
k
k
jk t
j jk t
j k k k a A e
e
A e
e ωθωθ-∞
--==++∑*k
k
j j k
k
k k a a A e
A e
θθ----=∴=Q thus ：k k
A A -=k k
θθ-=-Conclusion: is even ,and k a k θis odd
0001
()[]
k
k
jk t
j jk t
j k k k x t a A e
e
A e
e
ωθωθ-∞
--=∴=++∑001
2cos()
k k k a A k t ωθ∞
==++∑——trigonometric functions form
is expressed in rectangular form as k k k
a B jC =+k a 001
01
()()()jk t
jk t
k
k k k k k x t a B
jC e
B j
C e
ωω-∞
=-∞
==+
+++∑∑0001
()()jk t
jk t
k k k k k a B jC e
B j
C e
ωω∞
---=??=++++??
∑*
k
k
a a -=Q k k k k
B j
C B jC --∴-=+thus k k
B B -=k k
C C -=-Conclusion: the real part of is even ,
the imaginary part of is odd k a k
a
0001
()()()jk t
jk t
k k k k k x t a B jC e B jC e
ωω∞
-=??=+++-?
?
∑[]
0001
2cos sin k k k a B k t C k t ωω∞
==+-∑——trigonometric functions form
(another form)
3.3.2. Determination of the Fourier Series
Representation of a continuous-time Periodic Signal Assuming periodic signal x(t) can be represented with the Fourier series
0(),
jk t
k k x t a e
ω∞
=-∞
=
∑00
2T πω=
00()()jn t
j k n t
k
k x t e
a e
ωω∞
--=-∞
=
∑0
00()0
()T T jn t
j k n t
k
k x t e dt a e
dt
ωω∞
--=-∞
=
∑?
?
00
0()000
cos()sin()T T T j k n t
e
dt k n tdt j k n tdt
ωωω-=-+-?
??{
00,,
T =
k n ≠k n
=0
00
0()T jn t
n x t e
dt a T ω-∴=?consequently
000
01()T jn t
n a x t e dt T ω-=?Notice : the integration can be over any
interval of length T
01()jk t
k a x t e
dt
ω-=
?
01()T a x t dt
T =
?
a 0is simply the average value of x(t) over one period
1
T 0
T -t
()
x t ????
????The spectrum of periodic square wave
Example4 (3.5) ：
1
1||1,()|/2
0,
t T x t T t T
<
Chapter 3 Fourier Series Representation of Periodic Signals 第3章周期信号的傅里叶级数表示
Main content ： 1.The Frequency Analysis of Periodic Siganl(周期信号的频域分析） 2.The Frequency Analysis of LTI（LTI系统的频域分析） 3.Properties of Fourier Series（傅立叶级数的性质）
3.0 Introduction(引言) ?The basis for time domain(chapter2) 1)Signal can be represented as linear combination of shift impulses。 2)System is LTI。 ?Periodic Singal can be represented as linear combination of complex exponentials.
3.1 Historical Perspective (历史的回顾) 1、The concept of using trigonometric sums to describe periodic phenomena goes back to Babylonians 2、Euler examined the motion of Vibrating string is a linear combination of a few normal mode in 1748.
3-9 求图题3-9所示各信号的傅里叶变换。 解： ()()()() ()()() 1 222 j j j j a j 1Sa e e 12 b j 1j e T F E F T T ττττ---=?=-=--ωωωωωωωωω 3-10 试求下列信号的频谱函数。 ()()()()()()()()sgn()()()() t t f t e t f t t G t f t t f t e t εδε () -=--=-+=-=312234j212122113 4 2 解：() ()()()()()()j j e F F e Sa j ωωπδωω -+-=-=++3 121j 4 2j 223 ωωω ()()()()()() F F j πδ ==-+ - 34113 j j 4 j 22ωωωω ω 3-11 利用傅里叶变换的对称性求下列信号的频谱函数。 (1)) 2(π) 2(π2sin )(1--= t t t f (2)()()f t G t =22 解：()()()()()()F G e F Sa ω-==j2 124π1 j 2 j 2ωωωω 3-12 已知信号f (t )的频谱函数F (j ?)如下，求信号f (t )的表达式。 ()()();()()()(). 0001 j 3 j F F δεε =-=+--ωωωωωωωω 解：()()()()( ).000j 11 3 Sa 2ππ t f t e f t t == ωωω △3-13 利用傅立叶变换的微积分性质求图所示信号的频谱函数F (j ?)。 解：()[()cos()] 2 j 2j F Sa =-ωωωω 3-15 已知f (t )* f '(t )=(1-t )e -t ε(t )，求信号f (t )。 解：()()e t f t t ε-=± (b)