signals & systems ch.3 fourier transform of signals and lti system
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Signals & systems Ch.3 Fourier Transform of Signals and LTI System. Signals and systems in the Frequency domain. Fourier transform. Time [sec]. Frequency [sec -1 , Hz]. 3.1 Introduction. Orthogonal vector => orthonomal vector What is meaning of magnitude of H?. - PowerPoint PPT PresentationTRANSCRIPT
Signals & systemsCh.3 Fourier Transform of
Signals and LTI System
04/21/23
Signals and systems in the Frequency domain
04/21/23 KyungHee University 2
Time [sec]
Frequency [sec-1, Hz]
Fourier transform
3.1 Introduction
Orthogonal vector => orthonomal vector
What is meaning of magnitude of H?
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)2/1,2/1()2/1,2/1( 1 vvo
1,0,][110 1101 jiforjivvvvvvvv jioo
),(),( 1010 HHHhhh Fourier
110 vHvHh o
110 vHvHh o
110 vhHvhH o
Any vector in the 2- dimensional space can be represented by weighted sum of 2 orthonomal vectors
Fourier Transform(FT)
Inverse FT
3.1 Introduction cont’
CDMA?
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1 1 1 1( , , , )2 2 2 2ov
Orthogonal
? 1 0 1 10 1 1o ov v v v v v
[ ] , 0,1,2,3i jv v i j for i j
Any vector in the 4- dimensional space can be represented by weighted sum of 4 orthonomal vectors
Orthonormal
function?][)()(
2)cos()( 00 jidttvtv
TwheretkAtv
T jik
][)()(1
)()(2
)( **0
0 kmdttvtvT
tvtvT
whereetvT kmkm
tjkk
T
tkmj
T km dteT
dttvtvT
0)(* 1)()(
1
1
1 1 1 1( , , , )2 2 2 2
v
2
1 1 1 1( , , , )2 2 2 2
v
3
1 1 1 1( , , , )2 2 2 2
v
3.2 Complex Sinusoids and Frequency Response of LTI Systems
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cf) impulse response
.][ ][][][ )(
k
knj
k
ekhknxkhny
,)( ][][ njj
k
kjnj eeHekheny
.][)(
k
kjj ekheH
How about for complex z? (3.1) nznx ][
,)( )( )()( )( tjjtjtj ejHdehedehty
.)()(
dehjH j
How about for complex s? (3.3) stetx )(
.)()( )})(arg{( jHtjejHty
Magnitude to kill or not? Phase delay
Fourier transform
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2
1[ ] ( )
2j j nx n X e e d
( ) [ ]j j n
n
X e x n e
1
( ) ( )2
j tx t X j e d
( ) ( ) j tX j x t e dt
discrete time
Continuous time
Time domain frequency domain
1( ) ( )
2stx t X s e ds
j ( ) ( ) stX s x t e dt
Laplace transform
11[ ] ( )
2nx n X z z dz
j ( ) [ ] n
n
X z x n z
z-transform
(periodic) -
(discrete)
(discrete) -
(periodic)
3.6 DTFT: Discrete-Time Fourier Transform
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(discrete) (periodic)
deeXnx njnj )(
2
1][
n
njj enxeX ][)(
)(][ jDTFT eXnx
n
nx ][
n
nx2
][
(3.31)
(3.32)
(a-periodic) (continuous)
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Example 3.17 DTFT of an Exponential Sequence
Find the DTFT of the sequence ][][ nunx n
Solution :
n n
njnnjnj eenueX0
][)( 1,1
1)(
0
j
n
nj
ee
sincos1
1)(
jeX j
2/122/1222 )cos21(
1
)sin)cos1((
1)(
jeX
)cos1
sinarctan()}(arg{
jeX
x[n] = nu[n]. magnitude
phase
= 0.5 = 0.9
= 0.5 = 0.9
3.6 DTFT Example 3.17
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Example 3.18 DTFT of a Rectangular Pulse
Mn
Mnnx
,0
,1][ ].[nxLet Find the DTFT of
Solution : (square) (sinc)
Figure 3.30 Example 3.18. (a) Rectangular pulse in the time domain. (b) DTFT in the frequency domain.
3.6 DTFT Example 3.18
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,4,2,0,12
,4,2,0,1
1
)(
)12(
2
9
2
0
)(
Me
ee
eeeeX
j
Mjmj
M
m
mjmjM
m
Mmjj
2/2/
2/)12(2/)12(
)2/2/2/
2/)12(2/)12(2/)12(
)(
)()(
jj
MjMj
jjj
MjMjMjMjj
ee
ee
eee
eeeeeX
)2/sin(
)2/)12(sin()(
MeX j
12)2/sin(
)2/)12(sin(lim
,,4,2,0
M
M
)2/sin(
)2/)12(sin()(
MeX j
3.6 DTFT Example 3.18
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Example 3.19 Inverse DTFT of a Rectangular Spectrum
Find the inverse DTFT of
W
WeX j
,0
,1)(
Solution : (sinc) (square)
0),sin(1
2
1
2
1][
nWnnW
We
njdenx njW
W
nj
W
Wnn
xn
)sin(1
lim]0[0
)sin(1
][ Wnn
nx )/(][
Wnncsi
Wnx
Figure 3.31 (a) Rectangular pulse in the frequency domain. (b) Inverse DTFT in the time domain.
3.6 DTFT Example 3.19
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Example 3.20 DTFT of the Unit ImpulseFind the DTFT of ][][ nnx
Solution : (impulse) - (DC) 1][)(
nj
n
j eneX
1][ DTFTn
Example 3.21 Find the inverse DTFT of a Unit Impulse Spectrum.
Solution : (impulse train) (impulse train)
denx nj)(
2
1][
k
j keX )2()(
),(2
1 DTFT
3.6 DTFT Example 3.20-21
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Example 3.23 Multipath Channel : Frequency Response
]1[][][ naxnxny
Solution : jj aeeH 1)( })arg{(1 ajea })arg{sin(})arg{cos(1 aajaa
2/122/1222 }))arg{cos(21(}))arg{(sin}))arg{cos(1(()( aaaaaaaeH j
1,1
1)(
aae
eHj
jinv
( arg{ })
1 1( )
1 1 cos( arg{ }) sin( arg{ })inv j
j aH e
a e a a j a a
)(
1
}))arg{cos(21(
1)(
2/12
j
jinv
eHaaaeH
|)(| jeH (a) a = 0.5ej2/3. (b) a = 0.9ej2/3.
|)(| jinv eH (a) a = 0.5ej2/3. (b) a = 0.9ej2/3.
3.6 DTFT Example 3.23
3.7 CTFT
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(continuous aperiodic) (continuous aperiodic)
Inverse CTFT (3.35)
dwejwXtx jwt)(
2
1)(
CTFT (3.26)
dtetxjwX jwt)()(
)()( jwXtx CTFT
Condition for existence of Fourier transform:
dttx
2)(
3.7 CTFT Example 3.24
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Example 3.24 FT of a Real Decaying ExponentialFind the FT of )()( tuetx at
Solution : 0,0
adte at Therefore, FT not exists.
jwae
jwa
dtedtetuejwXa
tjwa
tjwajwtat
11
)()(,0
0
)(
0
)(
22
1)(
wajwX
)/arctan()}(arg{ awjwX
LPF or HPF? Cut-off from 3dB point?
)()( tuetx at
3.7 CTFT Example 3.25
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Example 3.25 FT of a Rectangular Pulse
0
00
,0
,1)(
Tt
TtTtx Find the FT of x(t).
Solution : (square) (sinc)
0),sin(21
)()( 0
0
0
0
0
wwTw
ejw
dtedtetxjwXT
T
jwtT
T
jwtjwt
000
2)sin(2
lim,0 TwTw
wForw
Example 3.25. (a) Rectangular pulse. (b) FT.
)sin(2
)( 0wTw
jwX )/(2)( 00 wTincsTjwX
w
wTjwX
)sin(2)( 0
0/)sin(,
0/)sin(,0)}(arg{
0
0
wwT
wwTjwX
3.7 CTFT Example 3.25
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Example 3.26 Inverse FT of an Ideal Low Pass Filter!!Fine the inverse FT of the rectangular spectrum depicted in Fig.3.42(a) and given by
Ww
WwWjwX
,0
,1)(
Solution : (sinc) -- (square)
)()()sin(1
)(
,/)sin(1
lim,0
0),sin(1
2
1
2
1)(
0
Wtncsi
WtxorWt
ttx
WWtt
twhen
tWtt
etj
dwetx
t
W
W
jwtW
W
jwt
3.7 CTFT Example 3.27-28
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Example 3.27 FT of the Unit Impulse
)()( ttx
Solution : (impulse) - (DC)
1)()( dtetjwX jwt 1)( FTt
Example 3.28 Inverse FT of an Impulse Spectrum
Find the inverse FT of )(2)( wjwX
Solution : (DC) (impulse)
1)(22
1)(
dwewtx jwt
)(21 wFT
3.7 CTFT Example 3.29
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Example 3.29 Digital Communication SignalsRectangular (wideband)
2/,0
2/)(
0
0,
Tt
TtAtx
r
r
Separation between KBS and SBS. Narrow band
2/,0
2/)),/2cos(1)(2/()(
0
00
Tt
TtTtAtx
c
c
Figure 3.44 Pulse shapes used in BPSK communications. (a) Rectangular pulse. (b) Raised cosine pulse.
3.7 CTFT Example 3.29
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Figure 3.45 BPSK (a) rectangular pulse shapes
(b) raised-cosine pulse shapes.
Solution : 22/
2/
2
0
0
0
1r
T
T rr AdtAT
P
crc
T
T
c
T
T cc
AAA
dtTtTtT
A
dtTtAT
P
3
8
8
3
)]/4cos(2/12/1/2cos(21[4
))/2cos(1)(4/(1
2
2/
2/ 000
2
2/
2/
20
2
0
0
0
0
0
the same power constraints
w
wTjwX r
)sin(2)( 2/0
f
fTjfX r
)sin()( 0'
2/
2/ 0
0
0
))/2cos(1(3
8
2
1)(
T
T
jwtc dteTtjwX
2/
2/
)/2(2/
2/
)/2(2/
2/
0
0
00
0
00
0 3
2
2
1
3
2
2
1
3
2)(
T
T
tTwjT
T
tTwjT
T
jwtc dtedtedtejwX
)2/sin(
2 02/
2/
0
0
Tdte
T
T
tj
0
00
0
000
/2
)2/)/2sin((
3
2
/2
)2/)2sin((
3
2)2/sin(
3
22)(
Tw
TTw
Tw
TTw
w
wTjwX c
)/1(
))/1(sin(
3
25.0
)/1(
))/1(sin(
3
25.0
)sin(
3
2)(
0
00
0
000'
Tf
TTf
Tf
TTf
t
fTjfX c
3.7 CTFT Example 3.29
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rectangular pulse.
One sinc
Raised cosine pulse 3 sinc’s
The narrower main lobe, the narrower bandwidth. But, the more error rate as shown in the time domain
Figure 3.47 sum of three frequency-shifted sinc functions.
Fourier transform
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2
1[ ] ( )
2j j nx n X e e d
( ) [ ]j j n
n
X e x n e
1( ) ( )
2j tx t X j e d
( ) ( ) j tX j x t e dt
Discrete time
Continuous time
Time domain frequency domain
3.9.1 Linearity Property
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)()()(][][][
)()()()()()(
jjjDTFT
FT
ebYeaXeZnbynaxnz
jwbYjwaXjwZtbytaxtz
)(2
1)(
3
2)( tytxtz
)(tx
)(ty
)2/sin(2
1)4/sin(
2
3][)(
)2/sin())/(1(][)(
)4/sin())/(1(][)(
2;
2;
2;
kk
kk
kZtz
kkkYty
kkkXtx
FS
FS
FS
3.9.1 Symmetry Properties
Real and Imaginary Signals
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dtetx
dtetxjwX
jwt
jwt
)(
)()((3.37)
(real x(t)=x*(t)) (conjugate symmetric)
dtetxjwX twj )()()( )()( jwXjwX (3.38)
3.9.2 Symmetry Properties of FT
EVEN/ODD SIGNALS
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(even) (real)(odd) (pure
imaginary)
)(txe )}(Re{ jX
)(txo )}(Im{ jXj
For even x(t), .* jwXdexdtetxjwX jwtjw
real
3.10 Convolution Property
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(convolution) (multiplication)
dtxhtxthty *
.2
1
dwejwXtx tjw
.
2
1
dwejwXtx jwt
But given
ddweejwXhty jwjwt
2
1
change the order of integration
.
2
1dwejwXdeh jwtjw
,
2
1dwejwXjwHty jwt
40.3.* jwHjwXjwYtxthty FT
3.10 Convolution Property Example 3.31
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Example 3.31 Convolution problem in the frequency domain
tttx sin1Input to a system with impulse response .2sin1 ttth
Find the output .* thtxty
Solution:
w
wjwXtx FT
,0
,1 .
2,0
2,1
w
wjwHth FT
,* jwHjwXjwYthtxty FT
,,0
,1
w
wjwY .sin1 ttty
3.10 Convolution Property Example 3.32
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Example 3.32 Find inverse FT’S by the convolution property
Use the convolution property to find x(t), where
.sin4 2
2w
wjwXtx FT
.sin2
ww
jwZ ,1,0
1,1jwZ
t
ttz FT
Ex 3.32 (p. 261). (a) Rectangular z(t). (b) txtztz )(*)(
3.10.2 Filtering
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Continuous time Discrete time(periodic with 2π
LPF
HPF
BPF
Figure 3.53 (p. 263)
Frequency dependent gain (power spectrum) .log20log20 jeHorjwH
,222
jwXjwHjwY kill or not (magnitude)
3.10 Convolution Property Example 3.34
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Example 3.34 Identifying h(t) from x(t) and y(t)
The output of an LTI system in response to an input is .Find frequency response and the impulse response of this system.
tuetx t2 tuety t
Solution:
2
1
jwjwX .
1
1
jwjwY But
jwX
jwYjwH
.1
11
1
1
1
1
1
2
jwjwjw
jw
jw
jwjwH .tuetth t
2
1
s
sX .1
1
s
sY But note sX
sYsH
.1
11
1
1
1
1
1
2
sss
s
s
ssH .tuetth t
as
tue Lat
1
3.10 Convolution Property Example 3.35
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EXAMPLE 3.35 Equalization(inverse) of multipath channel
jwYjwHjwX inv , jjinvj eYeHeXor
Consider again the problem addressed in Example 2.13. In this problem, a distorted received signal y[n] is expressed in terms of a transmitted signal x[n] as .1,1 anaxnxny .
,0
1,
0,1
otherwise
na
n
nh
.* nnhnh inv ,1 jjinv eHeH
Then .1
j
jinv
eHeH
.1 jjDTFT aeeHnh .1
1
j
jinv
aeeH
.nuanh ninv
3.11 Differentiation and Integration Properties
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.
2
1dwejwXtx jwt
dwjwejwXtx
dt
d jwt
21 .jwjwXFT
EXAMPLE 3.37 The differentiation property implies that
.jwa
jwtue
dt
d FTat
. tuaetetuaetuedt
d atatatat
jwa
jw
jwa
atue
dt
d FTat
1
dssesXtx
dt
d st
21 .ssXL
.sa
stue
dt
d Lat
sa
s
sa
atue
dt
d Lat
1
3.11 Differentiation and Integration Properties
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예제 한 두개
3.11.2 DIFFERENTIATION IN FREQUENCY
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,
dtetxjwX jwt Differentiate w.r.t. ω, ,
dtetjtxjwXdw
d jwt
Then, .jwXdw
dtjtx FT
Example 3.40 FT of a Gaussian pulse
Use the differentiation-in-time and differentiation-in-frequency properties for the FT of the Gaussian pulse, defined by and depicted in Fig. 3.60.
22
21 tetg
48.3.2 22
ttgettgdt
d t
Figure 3.60 (p. 275) Gaussian pulse g(t).
,)( jwjwGttgtgdt
d FT jwGdw
dtjtg FTand
jwGdw
d
jttg FT 1 .jwwGjwG
dw
d
.22wcejwG Then (But, c=?) .1210 22
dtejG t
.21 22 22 wFTt ee
Laplace transform and z transform
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as
tue Lat
1
2)(
1
astute Lat
11
1][
znu zn
21
1
)1(][
z
znun zn
3.11.3 Integration
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;t
dxty 52.3.1
jwXjw
jwY
53.3.01
wjXjwXjw
dx FTt
.t
dtu .1w
jwjwUtu FT Ex) Prove
54.3.sgn2
1
2
1ttu Note where a=0)()( tuetu at
.
0,1
0,0
0,1
sgn
t
t
t
t We know .21 wFT
.2sgn ttdt
d .2jwjwSgn
Fig. a step fn. as the sum of a constant and a signum fn.
0,0
0,2
)(
jjS
jtu FT 1
since linear
Differentiation and Integration Properties
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Common Differentiation and Integration Properties.
jDTFT
FTt
FT
FT
eXd
dnjnx
jXjXj
dx
jXd
dtjtx
jXjtxdt
d
01
3.12.1 Time-Shift Property
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dtettxetzjZ tjtj 0
jXedexedex tjjtjtj 000
)](arg[]arg[ 0
0 jXtjXe tj
Table 3.7 Time-Shift Properties of Fourier Representations
jnjDTFT
tjFT
eXennx
jXettx
0
0
0
0
Fourier transform of time-shifted z(t) = x(t-t0)
Note that x(t-t0) = x(t) * δ(-t0) and 0)( 0tjFT ett
3.12 Time-and Frequency-Shift Properties
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Example)
Figure 3.62 1Ttxtz jXejZ Tj 1
0sin2
TjX
TejZ Tj
sin2
1
3.12.2 Frequency-Shift Property
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dejX
dejZtz
tj
tj
2
12
1
txedejXe
dejXtz
tjtjtj
tj
2
12
1
jXettx tjFT 00
Recall
Table 3.8 Frequency-Shift Properties
00 0
00 0
;0
;0
FTj t
FSjk t
jDTFTj n
DTFSjk n
e x t X j
e x t X k k
e x n X e
e x n X k k
3.12.2 Frequency-Shift Property
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Example 3.42 FT by Using the Frequency-Shift Property
t
tetz
tj
,0
,10
Solution: We may express as the product of a complex sinusoid and a rectangular pulse
tz tje 10
t
ttx
,0
,1 )(10 txetz tj
sin2 jXtx FT
10sin10
210)(10
jXtxetz FTtj
3.12 Shift Properties Ex. 3.43
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Example 3.43 Using Multiple Properties to Find an FT
23 tuetuedt
dtx tt
Sol) Let and tuetw t3 2 tuetv t
Then we may write tvtwdt
dtx
By the convolution and differentiation properties
jVjWjjX
The transform pair
ja
tue FTat
1
jjW
3
1
222 tueetv t
j
eejV
j
1
22
jj
ejejX
j
31
22
3.12 Shift Properties Ex. 3.43
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Example 3.43 Using Multiple Properties to Find an FT
23 tuetuedt
dtx tt
Sol) Let and tuetw t3 2 tuetv t
Then we may write tvtwdt
dtx
By the convolution and differentiation properties
X s s W s V s
The transform pair
1Laplaceate u ta s
1
3W s
s
222 tueetv t
2
2
1 3
sseX s e
s s
2
2
1
seV j e
s
s
3.13 Inverse FT: Partial-Fraction Expansions
3.13.1 Inverse FT by using
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1FTate u ta s
1 01
1 1 0
MM
N NN
B sb s b s bX s
s a s a s a A s
0
M Nk
kk
B sX s f s
A s
Nkfordk ,,1N roots,
0
1
Mk
kk
N
kk
b sX j
s d
partial fraction
1
Nx
k k
CX s
s d
10sdte u t for d
s d
1 1
k
N Nd t s k
kk k k
Cx t C e u t X s
s d
3.13 Inverse FT: Partial-Fraction Expansions
3.13.1 Inverse FT by using
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ja
tue FTat
1
jA
jB
ajajaj
bjbjbjX N
NN
MM
01
11
01
NM
k
kk jA
jBjfjX
0
~
Let then ,jv 0011
1 avavav N
NN
Nkfordk ,,1N roots,
N
kk
M
k
kk
dj
jbjX
1
0
partial fraction
N
k k
x
dj
CjX
1
01
dfordj
tue FTdt
N
k
N
k k
kFTtdk dj
CjXtueCtx k
1 1
Inverse FT: Partial-Fraction Expansions
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1Laplaceate u ts a
'( ) ( )Laplacex t sX s 2 ( ) 3 '( ) ''( ) 2 ( ) 3 '( )y t y t y t x t x t 22 ( ) 3 ( ) ( ) 2 ( ) 3 ( )Laplace Y s sY s s Y s X s sX s
2
( ) 2 3( )
( ) 2 3 1 2
Y s s A BH s
X s s s s s
11
( 1)( )( 1)
2ss
B sH s s A A
s
2( ) ( ) ( )t th t Ae Be u t
( )x t( )h t
( ) ( )* ( )y t x t h t( ) ( ) ( )Y s X s H s( )X s
3.13.2 Inverse DTFT
3.13.2
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11
)1(1
01
jNjN
NjN
jMjMj
eee
eeeX
N
k
jk
jNjN
NjN edeee
11
11 11
012
21
1
NNNNN vvvv
N
kj
k
kj
ed
CeX
1 1
jk
DTFTnk ed
nud1
1where
Then
N
k
nkk nudCnx
1
Inverse FT: Partial-Fraction Expansions
04/21/23 KyungHee University 48
1
1[ ]
1Zna u n
az
1[ ] ( )Z
ox n n z X z 2 [ ] 3 [ 1] [ 2] 2 [ ] 3 [ 1]y n y n y n x n x n 1 2 12 ( ) 3 ( ) ( ) 2 ( ) 3 ( )Z transform Y z z Y z z Y z X z z X z
1
1 2 1 1
( ) 2 3( )
( ) 2 3 1 1 2
Y z z A BH z
X z z z z z
1
111
1
(1 )( )(1 )
1 2zz
B zH z z A A
z
[ ] ( 2 ) [ ]nh n A B u n
[ ]x n[ ]h n
[ ] [ ]* [ ]y n x n h n( ) ( ) ( )Y z X z H z( )X z
3.13.2 Inverse DTFT by z-transform
3.13.2
04/21/23 KyungHee University 49
1
1 0( 1) 1
1 1 1
MM
N NN N
z zX z
z z z
1 1 11 1
1
1 1N
NNN N k
k
z z z d z
11 1
Nk
k k
CX z
d z
1
1
1n z
kk
d u nd z
where
Then
N
k
nkk nudCnx
1
3.13 Inverse FT Example 3.45
04/21/23 KyungHee University 50
Example 3.45 Inversion by Partial-Fraction Expansion
2
6
1
6
11
56
5
jj
j
j
ee
eeX
Solution:
jjjj
j
e
C
e
C
ee
e
3
11
2
11
6
1
6
11
56
5
21
2
Using the method of residues described in Appendix B, We obtain
4
3
11
56
5
6
1
6
11
56
5
2
11
22
21
jj e
j
j
e
jj
j
j
e
e
ee
eeC
1
3
11
56
5
6
1
6
11
56
5
3
11
33
22
jj e
j
j
e
jj
j
j
e
e
ee
eeCAnd
Hence, nununx nn 3/12/14 2011.5.4
3.13 Inverse FT Example 3.45
04/21/23 KyungHee University 51
Example 3.45 Inversion by Partial-Fraction Expansion
1
1 2
55
61 1
16 6
zX z
z z
Solution:1
1 2
1 2 1 1
55
61 1 1 1
1 1 16 6 2 3
z C C
z z z z
Using the method of residues described in Appendix B, We obtain
1 1
1 1
11
1 2 1
2 2
5 55 51 6 61 4
1 1 12 1 16 6 3z z
z zC z
z z z
1 1
1 1
12
1 2 1
3 3
5 55 51 6 61 1
1 1 13 1 16 6 3z z
z zC z
z z z
And
Hence, nununx nn 3/12/14 2011.5.4
3.14 Multiplication (modulation) Property
04/21/23 KyungHee University 52
Given and
dvejvXtx jvt
21
dejZtz tj
2
1
dvdejZjvXtztxty tvj
22
1)()(
Change of variable to obtain
dedvvjZjvXty tj
2
1
2
1
jZjXjYtztxty FT 2
1
dvvjZjvXjZjX Where
jjjDTFT eZeXeYnznxny21
(3.56)
(3.57)
denotes periodic convolution.
Here, and are -periodic. jeX jeZ 2
deZeXeZeX jjjj
Modulation property
23年 4月 21日 MediaLab , Kyunghee University
53
jZjXjYtztxty
nconvolutiotionmultiplica
jZjXjYtztxty
ionmultilicatnconvolutio
FT
FT
FT
FT
2
1
2
1*
3.14 Modulation Property Ex 3.46
04/21/23 KyungHee University 54
Example 3.46 Truncating the sinc function
Sol) truncated by
2sin
1 n
nnh
otherwise
Mnnw
,0
,1
otherwise
Mnnhnht
,0
, jt
DTFTt eHnh
1( )
2jj j
tH e H e W e d
2sin
1 n
nnh
2/,0
2/,1jeH
2/sin
2/12sin
MeW j
/2
/2
1
2F d
2/,0
2/,
j
jjeW
eWeHF
Figure 3.66 The effect of Truncating the impulse response of a discrete-time system. (a) Frequency response of ideal system. (b) for near zero. (c) for slightly greater than (d) Frequency response of system with truncated impulse response.
F F
2/
3.15 Scaling Properties
04/21/23 KyungHee University 55
dteatxetzjZ tjtj
0,/1
0,/1
/
/
adexa
adexajZ
aj
aj
dexajZ aj //1
ajXaatxtz FT //1 (3.60)
3.15 Scaling Properties Example 3.48
04/21/23 KyungHee University 56
Example 3.48 SCALING A RECTANGULAR PULSE
Let the rectangular pulse
1,0
1,1)(
t
ttx
2,0
2,1)
2()(
t
ttxty
Solution :
).sin(2
)( ww
jwX 2/1).2/()( atxty
).2sin(2
)2sin()2
2(2
)2(2)(
ww
ww
wiXjwY
3.15 Scaling Properties Example 3.49
04/21/23 KyungHee University 57
Example 3.49 Multiple FT Properties for x(t) when
}.)3/(1
{)(2
wj
e
dw
djjwX
wj
Solution)jw
jwStuetsFT
t
1
1)()()(
)}.3/({)( 2 jwSedw
djjwX wj
we define )3/()( jwSjwY ).(3)3(3)3(3)( 33 tuetuetsty tt
Now we define )()( 2 jwYejwW wj
).2(3)2()( )2(3 tuetytw t
Finally, since ),.()( jwWdw
djjwX
).2(3)()( )2(3 tutettwtx t
3.15 Scaling Properties Example 3.49
04/21/23 KyungHee University 58
Example 3.49 Multiple FT Properties for x(t) when
.13/
)(2
s
e
ds
dsX
s
Solution)s
sStuetsL
t
1
1)()()(
)}.3/({)( 2 sSedw
dssX s
we define )3/()( sSsY ).(3)3(3)3(3)( 33 tuetuetsty tt
Now we define )()( 2 sYesW s
).2(3)2()( )2(3 tuetytw t
Finally, since ).()( sWds
dsX
).2(3)()( )2(3 tutettwtx t
3.16 Parseval’s Relationships
04/21/23 KyungHee University 59
? |)(|?|)(| 22 에너지쓴주자가바이올린꽝언제 jwXtx
.|)(|2
1|)(|
)()(2
1
.])(2
1)[(
.)(2
1)(
).()(|)(|
,|)(|
22
*
*
**
*2
2
dwjwXdttx
thatconcludeSo
dwjwXjwXW
dtdwejwXtxW
dwejwXtx
txtxtxthatNote
dttxW
x
jwtx
jwt
x
Representation
Parseval Relation
FT
DTFT
deXnx
dwjwXdwjwX
j
n
22
22
|)(|2
1|][|
|)(|2
1|)(|
Table 3.10 Parseval Relationships for the Four Fourier Representations
3.16 Parseval’s Relationships Example 3.50
04/21/23 KyungHee University 60
Example 3.50 Calculate the energy in a signal
n
WnnxLet
)sin(
][ 2
22 2
sin ( )E = | [ ]| .
n n
Wnx n
n
Use the Parseval’s theorem
Solution)
.|)(|
2
1 2 deXE j
./12
1
WdEW
W
||,0
||,1{)(][
W
WeXnx jDTFT