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1
Signals and Systems Lecture 7
•Convergence of CTFS•Properties of CTFS
2
Chapter 3 Fourier Series
§3.4 Convergence (收敛) of the Fourier Series
1. Approximation( 近似性 )
tjkk
N
NkN eatx 0ˆ
txtxte NN ——Error
1 NEN
dtetxT
aa tjk
Tkk0
1ˆ 2
dtteE NTN
2
EN 最小
0 N NE
dtteteT NN
If T dttx 2|)(| , then the series is convergent. ( xN(t) x(t) )
tjkk
k
eatx 0
0
00
1 jk tk Ta x t e dt
T
3
ka
Chapter 3 Fourier Series
2. Dirichlet Conditions:
Condition 1 dttx
T
dtetxT
a tjk
Tk0
1
dttx
T T
1
1T , 1t0 , /1 ttx
4
Chapter 3 Fourier Series
Condition 2.
In any finite interval , is of bounded variation. tx
1T , 1t0 , /2sin ttx
5
Chapter 3 Fourier Series
Condition 3.
In any finite interval , there are only a finite number
of discontinuities.
6
The Dirichlet Conditions (cont.)
• Dirichlet conditions are met for most of the signals we will encounter in the real world. Then– The Fourier series = x(t) at points where x(t) is continuous– The Fourier series = “midpoint” at points of discontinuity
• Still, convergence has some interesting characteristics:
• As N → ∞, exhibits Gibbs’ phenomenon at points of discontinuity.
N
Nk
tjkN eatx 0)(
)(txN
Chapter 3 Fourier Series
7
Chapter 3 Fourier Series
Gibbs Phenomenon:
Figure 3.9
Any continuity:
xN(t1) x(t1)
Vicinity of discontinuity:
ripples
peak amplitude does not seem to decrease
Discontinuity:
overshoot 9%
8
Chapter 3 Fourier Series
§3.5 Properties of Continuous-Time Fourier Series
§3.5.1 Linearity
k kx t a y t b FS FS
k k kz t Ax t By t c Aa Bb FS
§3.5.2 Time Shifting
kx t aFS
0 00
jk tkx t t a e FS
x(t) and y(t) may have the same period T.
9
Chapter 3 Fourier Series
§3.5.3 Time Reversal
kx t aFS kx t a FS
x t x t k ka a
x t x t k ka a
10
§3.5.6 Conjugation and Conjugate Symmetry(共轭及共轭对称性)
kk atxatx FSFS
Chapter 3 Fourier Series
kk aa txtx kk aa
or
txtx k ka a k ka a
11
Example (more symmetry properties )3.42 (P262)
kk atxatx FSFS
Chapter 3 Fourier Series
real even tx ka real even
real odd tx kaPurely imaginary
odd
[x(t)real ] }{ txEv }Re{ ka
[x(t)real ] }{ txOd }Im{ kaj
12
Chapter 3 Fourier Series
§3.5.4 Time Scaling
tjkk
k
eatx 0
0jk a t
kk
x at a e
0kak 0akak
§3.5.5 Multiplication (相乘)
kk btyatx FSFS
tytx FSk m k m
m
h a b
Convolution Sum
The Fourier series representation has changed!
13
Average Powerof kth harmonic
Average Power of tx
Chapter 3 Fourier Series
§3.5.7 Parseval’s Relation (帕兹瓦尔关系式)
221k
kT
adttxT
§3.5.8 Differential Property
katx FS
knn ajktx 0
FS
kajktx 0FS
14
Example (Proof Multiplication and Parseral’s Relation )3.46 (P264)
Chapter 3 Fourier Series
k
tjkkectytxtz 0)(
dtetytxT
dtetzT
cT
tjk
T
tjkk
0
0
0
0 )(11
00
dtetyeaT T
tjk
n
tjnk
0
00 )(1
0
nT
tnkjn dtety
Ta
0
0)(
0
)(1
nnknba
15
Example ( Continue )
Chapter 3 Fourier Series
k
tjkkectxtxtz 0)(*
nnnk
nnnkk aaaac **
k
tjk
nnnk eaatx 0*2
dteaaT
dttxT T
n
tjn
knnkT
0
0
0
*
0
2
0
11
dteeaT
ak
T
tjktknj
nnkn
0
00)(
0
* 1
dtetxT
ak
T
tjkk
0
0)(1
0
*
k
kk
kk aaa2*
16
Chapter 3 Fourier Series
Example 3.6 tg
-2 -1 0 1 2 t
2
1
2
1
Figure of Example 3.5
2/ t T 0
t 1
1
1
T
Ttx
1 tx
-T -T/2 –T1 0 T1 T/2 T t
Based on Property of linear and time-shifting, we may get
g(t)=x(t-1)-1/2
dk=bk+ck
2
jk
kk eab
02/1
00
kfor
kforck
17
Chapter 3 Fourier Series
1 kctx FS
-4 -2 0 2 4 t
Example 3.7
Figure of Example 3.6-2 -1 0 1 2 t
2
1
2
1 ketg FS
dt
tdgtx
)(
kk ejkc 0Differentiation Property
2
20
T
18
Chapter 3 Fourier Series
Example 3.9
21
210)( tjtj eaeaatx
According to Fact 3
2
20
T
According to Fact 2
According to Fact 1 *11 aa
Synthesis Equation
Symmetry Property
}Re{2
)(2
10
2*1
210
tj
tjtj
eaa
eaeaatx
So
19
Chapter 3 Fourier Series
According to Fact 4 Time-ReversalTime-Shifting
kkFS abtxty )()(
Example 3.9 Continue
kjk
kFS aebtxty 0)1()(
So, ck=a-k => z(t)=x(-t)
=x(-(t-1))
bk=e-jkπ/2ck => y(t)=z(t-1)
Because bk is odd, b0=0, b1=-b-1
20
Chapter 3 Fourier Series
According to Fact 5
2
1)(
4
1
)1(4
1)(
4
1
2
1
2
14
2
4
2
4
2
bbdtty
dttxdttx
Example 3.9 Continue
Parseval’s Relation
So, b1=-b-1 =±j/2
Because bk=e-jkπ/2a-k, so, a0=0, a1=b-1ejπ/2=jb-1
then, x(t)=±cos(πt/2)
21
Chapter 3 Fourier Series
kk btyatxFSFS
kkk BbAactBytAxtz FS
00
FS
0tjk
keattx
kk aa txtx kk aa
or
txtx k ka a k ka a
22
Readlist
• Signals and Systems:– 3.6~3.7
• Question: – Calculation of DTFS.
23
Problem Set
• 3.5 P251
• 3.8 P252 Reference Example 3.9(P210)
• 3.40 P261 Reference Table3.1(P206)