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Section 3.8 PROPERTIES OF FOURIER REPRESENTATIONS. Non-periodic (k, w ). (3.19). (3.35). (T: period). (3.20). (3.36). Periodic (k, W ). (3.10). (3.31). (3.32). (N: period). (3.11). Continuous ( w, W). Discrete [k]. Freq. property. Linearity and symmetry. E. - PowerPoint PPT Presentation
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Section 3.8
PROPERTIES OF FOURIER REPRESENTATIONSTime
propertyPeriodic
(t,n)
Nonperiodic
(t,n)
C.T.
(t)
• Fourier Series (FS) • Four Transform (FT)
D.T.
[n]
• Discrete-Time Fourier Series (DTFS)
• Discrete-Time Fourier Transform (DTFT)
TekXtx
k
tjk 2 ,][)( 0
0
,)(1
][0
0 dtetxT
kXT tjk
(T: period)
NekXnx
njkN
k
2 ,][][ 0
1
0
0
1
0
0][1
][N
n
njkenxN
kX(N: period)
dejXtx tj
)(
2
1)(
dtetxjX tj
)()(
deeXnx njj
)(
2
1][
nj
n
j enxeX
][)(
(3.19)
(3.20)
(3.10)
(3.11)
(3.35)
(3.36)
(3.31)
(3.32)
Non-periodic
(k,)
Periodic
(k,)
Freq.property
Discrete[k]
Continuous (
• Linearity and symmetry
][][][][][][
][][][
][][][)()()(
)()()(
0
0
;
;
kbYkaXkZnbynaxnz
ebYeaXeZnbynaxnz
kbYkaXkZtbytaxtz
jbYjaXjZtbytaxtz
DTFS
jjjDTFT
FS
FT
E Example 3.30, p255:
)(2
1)(
2
3)( tytxtz
Find the frequency-domain representation of z(t).
Which type of freq.-domain representation?
• FT, FS, DTFT, DTFS ?
kk
kY
kk
kjjk
ejk
dte
dtetxT
dtetxT
kX
kYkXkZ
tjktjk
T
T
tjk
x
T tjk
x
x
x
x
xx
2
4
4
28/1
8/1
2
2/
2/
0
sin1
][
sin1
sin22
1
2
1
)(1
)(1
][
][2
1][
2
3][
81
81
0
0
Periodic signals, continuous time. Thus, FS.
kk
kk
kZ FS24
2; sin2
1sin
2
3][
Symmetry:
We will develop using continuous, non-periodic signals. Results for other 2 cases may be obtained in a similar way.
a) Assume )()( real )( * txtxtx
jX
dtetx
dtetx
dtetxjX
tj
tj
tj
)(
*
**
)(
)(
)(
symmetric conjugate is )( real is )( If jXtx
Further assume
)(
real) (andeven
tx(t) xx(t)(t)xx(t),x(-t)
x(t)**
jX
dext
dtetxjX
j
tj
(why?) )( with replace
)( )(*
real. is )(even and real is )( If jXtx
dtdt
dttxdttxa
b
b
a
)()(
Further assume
)(
real) (and odd
tx(t) xx(t)(t)xx(t),x(-t)
x(t)**
jX
dtetxjX tj
)(* )(
imaginary.purely is )( odd and real is )( If jXtx
b) Assume )()(imaginary )( * txtxtx
jX
dtetxjX tj
)(* )(
symmetryeven has )( ofpart Imaginary
symmetry odd has )( ofpart Real
imaginary purely is )( If
jX
jX
tx
• Convolution: Applied to non-periodic signals.
dtxh
txthty
)()(
)()()(Let
dejXtx
jXtx
tj
FT
)()(2
1)(
)()( If
dejXjH
dejXdeh
ddeejXhty
tj
tj
jH
j
jtj
)()(
)()(
)()()(
21
)(
21
21
)()()()()()( jHjXjYthtxty FT
Conclusion: Convolution in time domain Multiplication in freq. domain.
E Example 3.31:
)(output system theFind ).2sin()(
response impulse with system a input to be )sin()(Let 1
1
tytth
ttx
t
t
)()()(
)()()(
jHjXjY
thtxty
From results of example 3.26, p264.
)sin()(||,0
||,1)(
2||,0
2||,1)()(
||,0
||,1)()(
1 ttyjY
jHth
jXtx
tFT
FT
FT
E Example 3.32: ).( Find ).(sin)()( 242 txjXtx FT
Recall that (Example 3.25, p244)
)sin(1||,0
1||,1)( 2
FT
t
ttz
).(j)(j)(j Then, ).sin()Let 2 ZZXZ(jω
).()()( Thus, tztztx
The same convolution properties hold for discrete-time, non-periodic signals.
)()()(][][][ jjjDTFT eHeXeYnhnxny
Convolution properties for periodic (DT or CT) and periodic with non-periodic signals will be discussed in Chapter 4.
• Differentiation and integration: (Section 3.11)
– Applicable to continuous functions: time (t) or frequency ( or )– FT (t, ) and DFTF ()
Differentiation in time:
followsit )(2
1)(
)(2
1)(
dejXjtxdt
d
dejXtx
tj
tj
)()( jXjtxdt
d FT
E Find FT of 0,)( atuedt
d at
ja
jtue
dt
dja
tue
FTat
FTat
)(
p243) 3.24, example class-(in 1
)(
E Find x(t) if
1||,0
1||,)(
j
jX
)sin(1
)cos(1
)()( Thus,
p246) 3.26, (example )sin(1
)()(
)()(
then,1||,0
1||,1)(Let
2t
tt
t
tzdt
dtx
tt
tzjZ
jZjjX
jZ
FT
Differentiation in frequency:
thus,)()()(
)()(
dtetxjtjXd
d
dtetxjX
tj
tj
)()(
jXd
dtjtx FT
If x(t) is periodic, frequency-domain representation is Fourier Series (FS):
][)( 0; 0 kXjktx
dt
d FS
tjk
k
ekXtx 0][)(
E Example 3.40, p275
FT. its Find .2
1)( :asgiven is pulseGaussian A 2
2t
etg
)(1
)(
Thus )(1
)(
)( )( )(
jGd
d
jjGj
jGd
d
jjGj
ttgtgdt
d
)()(
jGjGd
d
This differential equation (it has the same mathematical form as (*), and thus the functional form of G(j) is the same as that of g(t) ) has a solution given as:
2/2
)( cejG
Constant c can be determined as:
1|)()0(1)( 0 cdtetgjGdttg tj
2/22
2
2
1 Thus,
ee FTt
Integration:
– In time: applicable to FT and FS– In frequency: applicable to FT and DTFT
0 if )(1
)()()(
)()(
)()(Let
jXj
jYjXjYj
txtydt
d
dxtyt
For=0, this relationship is indeterminate. In general,
)()0()(1
)(
jXjXj
dx FTt
E Determine the Fourier transform of u(t).
1)(
)()(
FT
t
t
dtu
)(1
)(
jj
jU FT
E Problem 3.29, p279: Fund x(t), given
)()1(
1)(
j
jjjX
)()()(
1
1)(
1)(
1
11)(
tuetutx
jj
jj
jjjX
t
Review Table 3.6. Commonly used properties.
E Problem 3.22(b), p271.
22
22
2
2
2
2)(2
2
1)(
2
1)(
?)( .)(2)(
j
jtute
dt
dj
tute
jtue
jXtutedt
dtx
FTt
FTt
FTt
t
)()(
jXd
dtjtx FT
)()( jXjtxdt
d FT
• Time and frequency shift
Time shift:
0)( with Replace
)(
)()(
)()(Let
0
0
0
tjj
tj
tj
edextt
dtettx
dtetzjZ
ttxtz
)()( 0 jXejZ tj
Note: Time shift phase shift in frequency domain. Phase shift is a linear function of Magnitude spectrum does not change.
Table 3.7, p280:
][][)(][
][)()()(
00
0
00
0
;0
0
;0
0
kXennxeXennxkXettx
jXettx
njkDTFS
jnjDTFT
tjkFS
tjFT
E Example 3.41: Find Z(j)
)sin(2
)()(
p244) 3.25, (Example )sin(2
)()(
)()(
0
0
1
1 TejZtz
TjXtx
Ttxtz
TjFT
FT
E Problem 3.23(a), p282:
2
4)4(2
22
2
2
2
4
)2()4()4(
)2(
1)(
)2()()(
)()(2
1)(
)( Find .)2(
)(
j
etuet
jtute
j
jjZ
d
dtjtz
tuetzj
jZ
txj
ejX
jFTt
FTt
FT
tFT
j
Frequency shift:
??))(()()()(
FT
FT
jXjZjXtx
)(let )(2
1
))((2
1
)(2
1)(
tjtj
tj
tj
edejX
dejX
dejZtz
)()( txetz tj
Note:– Frequency shift time signal multiplied by a complex sinusoid.– Carrier modulation.
Table 3.8, p284:
][][)(][][)(
))(()(
0;
)(0
;
000
000
kkXnxeeXnxe
kkXtxejXtxe
DTFSnjk
jDTFTnj
FStjk
FTtj
E Example 3.42, p284:
Find Z(j).
||,0
||,)(
10
t
tetz
tj
))10sin((10
2))10(()()(
)sin( 2
||,0
||,1)(Let
10
jXetxtz
t
ttx
FTtj
FT
E Example 3.43, p285:
).( Find .)2()()( 3 jXtuetuedt
dtx tt
)1)(3()()( )( Thus,
1
1)()2(
)2()(3
1)()()(Let
22
222)2(
3
jj
ejejVjWjjX
jeejVetue
tuetvj
jWtuetw
j
jFTt
t
FTt
• Multiplication
?)()()( of FT theis what periodic,-non are )(),( If tztxtytxtz
READ derivation on p291!
dvdejZjvXtztxty
dejZtz
dvejvXtx
tvj
tj
jvt
)(2
)()()2(
1)()()(
)(2
1)(
)(2
1 )(
Then .first fix ,Let ddvv
dedvvjZjvXty tj
jZjX
)()(2
1
))(()(2
1
2
1)(
Inverse FT of
:][],[ signals time-discreteFor nznx
)()()()()()( 21 jZjXjYtztxty FT
)()()(][][][ 21 jjjDTFT eZeXeYnznxny *
Periodic convolution:
deZeXeZeX jjjj )()()(
*
- p296, CT, periodic signals:
m
TFS
mkZmXkZkX
kZkXkYtztxty
][][][][
][][][)()()( /2;
)()()()( Compare jZjXtztx FT
- p297, DT, periodic signals:
1
0
/2;
][][][][
][][][][][][N
m
NDTFS
mkZmXkZkX
kZkXkYnznxny
*
*
• Scaling
))/((
)(
)(let 0)(
0)(
)(
)()(
)()(Let
||1
)/(||
1
)/(1
)/(1
ajX
dex
atadex
adex
dteatx
dtetzjZ
atxtz
a
aja
aja
aja
tj
tj
E Example 3.48, p300:
2||0
2||1)(
1||0
1||1)(
t
tty
t
ttx
Find )( jY
)2sin(2
)2sin(2
22)2(2)(
)sin(2
)(
|)()(21
jXjY
jX
atxty a
E Example 3.49, p301:
)3/(1)( if )( Find
2
j
e
d
djjXtx
j
jjStuets FTt
1
1)()()( know We
– Time scaling:
– Time shift:
– Differentiation:
)3/(1)(
)3/(1))2(3(3)2(3)(
)3/(1
1
3
1)()3()(
2
2
j
e
d
djttv
j
etstztv
jjZtstz
jFT
jFT
FT
))2(3(3
))2(3(3
)2(3
)()( Thus,
)2(3
tute
tts
ttz
ttvtx
t
)2(3)( Thus, ).2())2(3( :Note )2(3 tutetxtutu t
)()(
)()(
))/(()(
00
||1
jXtjtx
jXettx
ajXatx
ddFT
tjFT
aFT
• Parseval’s relationship:
:)( signal periodic-non CT, ofEnergy tx
djXjX
ddtetxjX
dtdejXtxW
dejXtx
tj
tjx
tj
)()(2
1
)()(2
1
)(2
1)(
)(2
1)(
dttxtx
dttxWx
)()(
|)(|2
djXdttxWx
22 |)(|2
1|)(|
domain freq.in energy domain in timeEnergy b)
spectrumenergy :|)(| a) :Note 2
jX
Table 3.10, p304:
k
T
N
k
N
n
j
n
kXdttxT
kXnxN
deXnx
2
0
2
1
0
21
0
2
22
|][||)(|1
:FS
|][||][|1
:DTFS
|)(|2
1|][| :DTFT
E Example 3.50, p304:
/12
12
1][
,0
,1)sin(][
)(
)(sin Determine .
)sin(][
22
2
2
Wd
deXnxE
W
WeX
n
Wnnx
n
WnE
n
Wnnx
W
W
j
nx
jDTFT
nx
• Time-bandwidth product Compression in time domain expansion in frequency domain Bandwidth: The extent of the signal’s significant contents. It is in general a vague definition as “significant” is not mathematically defined. In practice, definitions of bandwidth include
– absolute bandwidth– x% bandwidth– first-null bandwidth. If we define
thenbandwidth, RMS :)(
)(
signalenergy an ofduration RMS :)(
)(
2/1
2
22
2/1
2
22
djX
djXB
dttx
dttxtT
w
d
2/1wd BT
• Duality )(2
)(
fjtF
jFtfFT
FT
Ejt
txjX
1
1)( if )( Find Example 3.52, p308:
)(2)(2)(
)(21
1)(Duality
)(1
1)()( know We
uefjX
fjt
jtF
jFj
tuetf
FT
FTt
jtde
dee
dejXx
jt
tj
tj
1
1
22
1
)(2
1(t)Check
0 )1(
0
E Problem 3.44, p309: )()( if )( Find ujXtx
)(2
1
2
1
2
1))((
)(
1)(
)(2)(1
)()(
)()(
tjt
ttj
tx
xj
tujtX
ujX
FT
)()0()(1
)(
jXjXj
dx FTt
Table 3.11, p311:
class!own in their stay not do FS and TFT
][
][ :FS and DTFT
][][
][][ :DTFS
1;
1/2;
/2;
D
kxeX
eXnx
kxnX
kXnx
FSjt
jDTFT
NNDTFS
NDTFS
