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Audio &
DS
P Lab.
Fourier Transform of Signals
訊號的傅立葉轉換
Lecture 3-6
2
Audio &
DS
P Lab.
Fourier Representations
DTFS: x[n]: discrete-time and periodic signal
X[k]: discrete and periodic spectrum
FS: x(t): continuous-time and periodic signal
X[k]: discrete-time and non-periodic spectrum
DTFT: x[n]: discrete-time and non-periodic signal
X(ejΩ): continuous and periodic spectrum
FT: x(t): continuous-time and non-periodic signal
X(jω): continuous-time and non-periodic spectrum
3
Audio &
DS
P Lab.
Linearity and Symmetry
Linear Properties: (想一想線性特性的用處？)FT
][][][
)()()(
][][][
)()()(
kbYkaXkZ
ebYeaXeZ
kbYkaXkZ
jbYjaXjZ
jjj
+=
+=
+=
+=
ΩΩΩ
ωωω
][][][
][][][
)()()(
)()()(
nbynaxnz
nbynaxnz
tbytaxtz
tbytaxtz
+=
+=
+=
+=
FS
DTFT
DTFS
4
Audio &
DS
P Lab.
Representation of the periodic signal z(t) as a weighted sum of periodic square waves:
z(t) = (3/2)x(t) + (1/2)y(t).
(a) z(t). (b) x(t). (c) y(t).
Z[k] = (3/2)X[k] +(1/2)Y[k]
X[k] = ?
Y[k] = ?
5
Audio &
DS
P Lab.
z(t) = (3/2)x(t) + (1/2)y(t).
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Audio &
DS
P Lab. 利用線性性質求取下述訊號傅立葉表示法
Problem 3.16:
)(3)(2)( 2 tuetuetx tt −− −=
Solution:
?2
31
22
131
12
)(3)(2)( 2
=+
−+
=
+
−
+
=
⋅−⋅= −−
ωω
ωω
ω
jj
jj
tueFTtueFTjX tt
請繼續完成…
7
Audio &
DS
P Lab.
Symmetry Properties: Real Signals
x(t) 是實數)()( * txtx =Q
)(
)()(
)()()(
)(
**
*
ω
ω
ωω
ωω
jX
dtetxdtetx
dtetxdtetxjX
tjtj
tjtj
−=
==
=
=
∫∫
∫∫∞+
∞−
−−∞+
∞−
∞+
∞−
∞+
∞−
−
8
Audio &
DS
P Lab.
)()( * txtx =Q )()(* ωω jXjX −=x(t) 是實數
X (jω)為複數共軛對稱 (complex-conjugate symmetric)
X (jω) = ReX(jω) + j ImX(jω),
X∗(jω) = ReX(jω) - j ImX(jω),
X (-jω) = ReX(-jω) + j ImX(-jω),
Re X(jω) = Re X(-jω)
Im X(jω) = -Im X(-jω)
x(t) FT後實數部份是頻率偶函數(Even function)
x(t) FT後虛數部份是頻率奇函數(Odd function)
Q
振幅頻譜是偶對稱 相位頻譜是奇對稱
9
Audio &
DS
P Lab.
Symmetry Properties: Imaginary Signals
)()( * txtx =−Q x(t) 是虛數
)()(
)()(
)()()(
)(
**
*
ω
ω
ω
ωω
ωω
jXdtetx
dtetxdtetx
dtetxdtetxjX
tj
tjtj
tjtj
−−=−=
−=−=
=
=
∫∫∫
∫∫
∞+
∞−
−−
∞+
∞−
∞+
∞−
∞+
∞−
∞+
∞−
−
10
Audio &
DS
P Lab.
)()(* ωω jXjX −−=)()( * txtx =−Q x(t) 是虛數
複數共軛對稱 (complex-conjugate symmetric)
X(jω) = ReX(jω) + j ImX(jω),
X∗(jω) = ReX(jω) - j ImX(jω),
-X(-jω) = -ReX(-jω) - j ImX(-jω),
ReX(jω) = -ReX(-jω)
ImX(jω) = ImX(-jω)
x(t) FT後實數部份是頻率奇函數(Odd function)
x(t) FT後虛數部份是頻率偶函數(Even function)
Q
振幅頻譜是偶對稱 相位頻譜是奇對稱
11
Audio &
DS
P Lab.
x(t) 是實數
FT後實數部份是頻率偶函數(Even function)
FT後虛數部份是頻率奇函數(Odd function)
振幅頻譜是偶對稱 相位頻譜是奇對稱
x(t) 是虛數
FT後實數部份是頻率奇函數(Odd function)
FT後虛數部份是頻率偶函數(Even function)
振幅頻譜是偶對稱 相位頻譜是奇對稱
12
Audio &
DS
P Lab.
Symmetry Property 應用範例
若實數輸入訊號： ( )2
cos)(00
0
tjjtjj eAeeAetAtxωφωφ
φω−− +
=−=
)( ωjH系統振幅響應：
系統相位響應： [ ])(arg ωjH
)(2
)(2
)( 0000 ωω ωφωφ jHeeAjHeeAt tjjtjj −⋅+⋅= −−系統輸出： y
重新整理系統輸出：
[ ] [ ] )(arg
0
)(arg0
00
00
)(2
)(2
)(
ωωφ
ωωφ
ω
ω
jHjtjj
jHjtjj
ejHeeA
ejHeeAty
−−
−
−⋅+
⋅=
13
Audio &
DS
P Lab.
0ω0ω−ω
φjeA −
2φjeA
2
)( ωjX
)( 0ωjH)( 0ωjH −××
)( ωjH
ω
請補上“相位響應”後，繪出“輸出頻譜” )( ωjY
14
Audio &
DS
P Lab.
系統振幅響應是偶對稱：
系統相位響應是奇對稱：
)()( ωω jHjH −=
[ ] [ ])(arg)(arg ωω jHjH −−=
整理系統輸出：
[ ] [ ] [ ] [ ] [ ] [ ]
[ ] [ ]( ) [ ]( ))(argcos)(
2/)(
)(2
)(2
)(2
)(2
)(2
)(
000
)(arg())(arg(0
)(arg)(arg0
)(arg0
)(arg0
)(arg0
)(arg0
0000
0000
0000
0000
ωφωω
ω
ω
ωω
ωω
ωφωωφω
ωωφωωφ
ωωφωωφ
ωωφωωφ
jHtjHA
eejHA
eeeeeejHA
ejHeeAejHeeA
ejHeeAejHeeAty
jHtjjHtj
jHjtjjjHjtjj
jHjtjjjHjtjj
jHjtjjjHjtjj
+−=
+=
+=
−+=
−+=
+−−+−
−−−
−−−
−−−
15
Audio &
DS
P Lab.
A sinusoidal input to an LTI system:It results in a sinusoidal output of the same frequency.The amplitude and phase are modified by the system’s frequency response.
)cos( 0 φω −tA
0ωφ
[ ]( ))(argcos)()( 000 ωφωω jHtjHAty +−=
)( 0ωjHA
)( 0ωjHA− [ ]0
0 )(argω
ωφ jH−
16
Audio &
DS
P Lab.
Symmetry Properties: Even Signals
Even function : x(-t) = x(t)
Real Function: x*(t) = x(t)
Even and Real Function: x(t)* = x(-t)
)(
)(
,)()(
)()()(
)(
**
*
ω
ττ
τ
ω
ωτ
ωω
ωω
jX
dex
tletdtetxdtetx
dtetxdtetxjX
j
tjtj
tjtj
=
=
−=−=−=
=
=
∫∫∫
∫∫
∞+
∞−
−
∞+
∞−
−−∞+
∞−
∞+
∞−
∞+
∞−
−
.)( realisjX ω
17
Audio &
DS
P Lab.
Symmetry Properties: Odd Signals
Odd function : x(-t) = -x(t)
Real Function: x*(t) = x(t)
Odd and Real Function: x(t)* = - x(-t)
)(
)(
,)()(
)()()(
)(
**
*
ω
ττ
τ
ω
ωτ
ωω
ωω
jX
dex
tletdtetxdtetx
dtetxdtetxjX
j
tjtj
tjtj
−=
−=
−=−−=−−=
=
=
∫∫∫
∫∫
∞+
∞−
−
∞+
∞−
−−∞+
∞−
∞+
∞−
∞+
∞−
−
.)( imaginaryisjX ω
18
Audio &
DS
P Lab.
補充資料: Odd Signals
Odd function : x(-t) = -x(t)
Imaginary Function: x*(t) = -x(t)
Odd and Imaginary Function: x(t)* = x(-t)
)(
)(
,)()(
)()()(
)(
**
*
ω
ττ
τ
ω
ωτ
ωω
ωω
jX
dex
tletdtetxdtetx
dtetxdtetxjX
j
tjtj
tjtj
=
=
−=−=−=
=
=
∫∫∫
∫∫
∞+
∞−
−
∞+
∞−
−−∞+
∞−
∞+
∞−
∞+
∞−
−
.)( realisjX ω
19
Audio &
DS
P Lab.
補充資料: Even Signals
Even function : x(-t) = x(t)
Imaginary Function: x*(t) = -x(t)
Even and Imaginary Function: x(t)* = -x(-t)
)(
)(
,)()(
)()()(
)(
**
*
ω
ττ
τ
ω
ωτ
ωω
ωω
jX
dex
tletdtetxdtetx
dtetxdtetxjX
j
tjtj
tjtj
−=
−=
−=−−=−−=
=
=
∫∫∫
∫∫
∞+
∞−
−
∞+
∞−
−−∞+
∞−
∞+
∞−
∞+
∞−
−
.)( imaginaryisjX ω
20
Audio &
DS
P Lab.
Problem 3.17. 判定(a)及(b) 所對應時域訊號是實、複數以及是偶、奇函數?
ω
ω
( )ωjX
( ) ωjXarg
21
Audio &
DS
P Lab.
(a) The Magnitude Response is even, and the Phase Response is odd functions.
The x(t) is a real and odd function.
(b) The Magnitude Response is even, and the Phase Response is even functions.
The x[n] is a real and even function.
時域中(實/奇)訊號 偶振幅奇相位/頻域虛數 /相位 ±π/2 變化
時域中(實/偶)訊號 偶振幅偶相位/頻域實數 相位 0 和 π變化
22
Audio &
DS
P Lab.
Convolution Property 褶積性質
)()()()(*)()( ωωω jXjHjYtxthty ⋅==FT
)()()(][*][][ ΩΩΩ ⋅== jjj eXeHeYnxnhny DTFT
23
Audio &
DS
P Lab.
Convolution of Non-periodic Signal非週期訊號的褶積
)()()(
)()()(*)()(
ωωω
τττ
jXjHjYFT
dtxhtxthty
⋅=
−== ∫+∞
∞−
c
where
[ ] ωωπ
ω dejYty tj)(21)( ∫
+∞
∞−=
24
Audio &
DS
P Lab.
[ ] [ ]
[ ] ωωπ
ωωωπ
ωωωπ
ωωττπ
τωωπ
τ
ωωπ
ωωπ
τ
ωωπ
τττ
ω
ωω
ωτω
ωτω
ωτωτω
ω
dejY
dejXjHdejXjH
dejXdeh
ddeejXhty
deejXdejXtx
dejXtx
dtxhtxthty
tj
tjtj
tjj
tjj
tjjtj
tj
)(21
)()(21)()(
21
)()(21
])(21[)()(
)(21)(
21)(
)(21)(
)()()(*)()(
)(
∫
∫∫
∫∫
∫∫
∫∫
∫
∫
∞+
∞−
∞+
∞−
∞+
∞−
−∞+
∞−
∞+
∞−
−∞+
∞−
∞+
∞−
−∞+
∞−
−∞+
∞−
∞+
∞−
+∞
∞−
=
⋅==
=
=
==−∴
=
−==
Q
摺積 vs. 乘積
25
Audio &
DS
P Lab.
Example 3.31:x(t) = (1/πt) sin(πt), h(t) = (1/πt) sin(2πt),
?)(*)()( == thtxty回想
( )
>
<<−=↔=
WWW
jXWtt
txFT
ωω
ωπ ,0
,1)(sin1)(
ω
( )ωjX
( )ttty ππ sin)/1()( =Solution:
26
Audio &
DS
P Lab.
x(t) = (1/πt) sin(πt),
( )
>
<<−=↔=
πωπωπ
ωππ ,0
,1)(sin1)( jXt
ttx
ω
( )ωjX
ππ−
1
313 1
)(tx
27
Audio &
DS
P Lab.
h(t) = (1/πt) sin(2πt),
( )
>
<<−=↔=
πωπωπ
ωππ 2,0
22,1)(2sin1)( jHt
tth
ω
( )ωjH
π2π2−
1
21
23
21
23
)(th
28
Audio &
DS
P Lab.
)()()()(*)()( ωωω jYjHjXthtxty =⋅↔=
ω
( )ωjX
ππ−
1
313 1
)(tx
ω
( )ωjH
π2π2−
1
21
23
21
23
)(th
ω
( )ωjY
ππ−
1
313 1
)(ty
×
⇓
∗
⇓
29
Audio &
DS
P Lab.
Example 3.32: X(jω) = (4/ω2) sin2(ω) ; x(t) = ?
Hint: X(jω) = (2/ω) sin(ω) • (2/ω) sin(ω)
求出 x(t) = ?
Example 3.32.(a) Rectangular pulse z(t). (b) Convolution of z(t) with itself gives x(t).
