Post on 21-Dec-2015
CHAPTER 4APPLICATIONS OF FOURIER REPRESENTATIONS TO
MIXED SIGNAL CLASSES
jeX
kX
jX
kX
DTFT:periodic-non
][DTFS:periodic :DT
)(FT:periodic-non
][FS:periodic :CT
• What about the Fourier representation of a mixture of a) periodic and non-periodic signals b) CT and DT signals?
(periodic)
assuch )()( tyHtx Examples: )()()cos( 0 tytht
][sampler)( nytx
• We will go through: a) FT of periodic signals, which we have used FS:
TekXtx
k
tjk 2 ,][)( 0
0
dtetxT
kXtxT tjkFS 0
; 00 )(1
][)(
We can take FT of x(t).
b) Convolution and multiplication with mixture of periodic and non-periodic signals.
c) Fourier transform of discrete-time signals.
FT of periodic signals
Chapter 3: for CT periodic signals, FS representations.What happens if we take FT of periodic signals?
(*) ][)( 0
k
tjkekXtx
FS representation of periodic signal x(t):
)(21 :shift Freq.
)(21 :1 of FT
00
ke FTtjk
FT
Take FT of equation (*)
kk
FTtjk kkXekXtx )(][2][)( 00
Note:a) FT of a periodic signal is a series of impulses spaced by the fundamental frequency 0.b) The k-th impulse has strength 2X[k].c) FT of x(t)=cos(0t) can be obtained by replacing
versaor vise ),(2 with 00 ke tjk
FS and FT representation of a periodic continuous-time signal.
E Example 4.1, p343:
)cos()( of ransformFourier t Find 0ttx
1,0
1,2/1)cos( 0 k
kt FS
)()()cos( Thus, 000 FTt
E Example 4.2, p344:
n
nTttp )()( train impulseunit theof FT theDetermine
p(t) is periodic with fundamental period T, fundamental frequency 0. FS coefficients:
k
T
T
tjk
kT
jP
kT
dtetT
kP
)(2
)(
,1
)(1
][
0
2/
2/
0
Relating DTFT to DTFS
N-periodic signal x[n] has DTFS expression
00
1
0
,),(2
)( ][][
0
0
kke
ekXnx
DTFTnjk
N
k
njk
Extending to any interval:
m
DTFTnjk mke )2(2 00
This, DTFT of x[n] given in (*) is expressed as:
m
N
k
jDTFTN
k
njk mkkXeXekXnx )2(][2][][ 0
1
0
1
0
0
Since X[k] is N periodic and N0=2, we have
k
jDTFTN
k
njk kkXeXekXnx )(][2][][ 0
1
0
0
Note:a) DTFS DTFT:
b) DTFT DTFS:
Also, replace sum intervals from 0~N-1 for DTFS to - ~ for DTFT
2by scale then )( 00 ke njk
)21/(by scale then )( 00 njkek
E Problem 4.3(c), p347:
. x[n]of DTFT theFind ].10[][
k
knnx
Fundamental period? :DTFS e Us/5.,10 0 N
kenxkXn
njk
,
10
1][
10
1][
9
0
5
Use note a) last slide:
n
DTFT knx 510
12][
Question: if we take inverse DTFS of X[k], we get
9
0
9
0
5
5
10
1
][][
n
njkn
njk
e
ekXnx
n
knnx ].10[][
expression original theequal toseemnot does which
Exercise: use Matlab to verify.
Convolution and multiplication with mixture of periodicand non-periodic signals
)(th)(tx )(ty ][nh][nx ][ny
For periodic inputs:
FT Use
)()()(periodic-nonperiodic
thtxty
DTFT Use
][][][periodic-nonperiodic
nhnxny
1) Convolution of periodic and non-periodic signals
tscoefficien FS :][
,][2)()(
)()()()()()(
0
kX
kkXjXtx
jHjXjYthtxty
k
FT
FT
k
k
FT
kkXjkH
jHkkXjYthtxty
00
0
][)(2
)(][2)()()()(
E Problem 4.4(a), p350: LTI system
has an impulse response
)(th)(tx )(ty
).(output system thedemermine toFT theuse
),4sin()cos(1)( signalinput For
ty
tttx
)./()sin()4cos(2)( tttth
2
4
2
4 )(
)sin()( 44
rectrectjH
eet
tth tjtj
2
)sin()( rect
t
tth FT
1
)4()4()()()(2 )( j jX
Because h(t) is an ideal bandpass filter with a bandwidth 2 centered at 4the Fourier transform of the output signal is thus
)4()4( )( j jY
which has a time-domain expression given as: )4sin()( tty
k
jkDTFT kkXeHnhnxny 0
periodic
][2][][][ 0
For discrete-time signals:
2) Multiplication of periodic and non-periodic signals
k
FT
kkXjG
jXjGjY
txtgty
0
21
periodic
][)(
)()()(
)()()(
k
FT
k
tjk
kkXjX
ekXtx
0][2)(
][)(
:Note0
Carrying out the convolution yields:
k
FT kGkXjYtxtgty 0][)()()()(
DT case:
1
0
)(
periodic
0][)(][][][N
k
kjjDTFT eZkXeYnznxny
E Problem 4.7, p357(b):
Consider the LTI system and input signal spectrum X(ej) depicted by the figure below. Determine an expression for Y(ej), the DTFT of the output y[n] assuming that z[n]=2cos(n/2).
..,0
1,1][
:][ of tscoefficien FS . with periodic :][
][][][][
20
wo
kkZ
nznz
nznxnxny
Thus,
)()(22 jjjj eXeXeXeY
E Example 4.6, p353: AM Radio
(a) Simplified AM radio transmitter & receiver.(b) Spectrum of message signal. Analyze the system in the frequency domain.
Signals in the AM transmitter and receiver. (a) Transmitted signal r(t) and spectrum R(j). (b) Spectrum of q(t) in the receiver. (c) Spectrum of receiver output y(t).
))(())(()(
)cos()()(
21
21
cc
FTc
jMjMjR
ttmtr
In the receiver, r(t) is multiplied by the identical cosine used in the transmitter to obtain:
))2(()())2((
))(())(()(
)cos()()(
41
21
41
21
21
cc
cc
FTc
jMjMjM
jRjRjG
ttrtg
After low-pass filtering:
)()( 21 jMjY