application of combined fourier series transform...
TRANSCRIPT
Department of Electrical and Computer Engineering
Application of Combined Fourier Series Transform
(Sampling Theorem)
t
k
skTttx )()(
sT
t
)(tm t
)(tms
][kX
)( fM
)( fM s
s
sT
f1
Sampling Frequency
Department of Electrical and Computer Engineering
tsT
][kX
k
skTttx )()( We need Fourier Series
))2sin(][)2cos(][(]0[)(1
tkfkXtkfkXXtx sss
k
k
c
s
T
s
Tt
ts Tdtt
Tdttx
TX
ss 1)(
1)(
1]0[
0
0
0
0)2sin()(2
][0
0
dttkftxT
kXsTt
t
s
s
s
Department of Electrical and Computer Engineering
dttkftxT
kXsTt
t
s
s
c
0
0
)2cos()(2
][
ss
T
s
s TTdttkft
T
s 2)0cos(
2)2cos()(
2
0
))2sin(][)2cos(][(]0[)(1
tkfkXtkfkXXtx sss
k
k
c
Let’s go back to Fourier Series representation
k
k
s
ss
tkfTT
tx1
)2cos(21
)(
)2cos(][]0[1
tkfkXX s
k
k
c
Department of Electrical and Computer Engineering
t
)(tm
f
)( fM
BB
Fourier Transform
B is the Bandwidth of m(t)
What about m(t)? – need Fourier Transform
Department of Electrical and Computer Engineering
k
k
s
ss
tkfTT
tx1
)2cos(21
)(
k
k
s
ss
tkftmT
tmT
txtm1
)2cos()(2
)(1
)()(
We now have x(t) as frequency representation via Fourier Series
And m(t) as Frequency representation via Fourier Transform
)()( fMtm
Let’s multiply m(t) with x(t)
What is the frequency representation of m(t)x(t)?
Department of Electrical and Computer Engineering
k
k
s
ss
tkftmT
tmT
txtm1
)2cos()(2
)(1
)()(
k
k
s
ss
tkftmT
tmT
FtxtmF1
)]2cos()(2
)(1
[)]()([
We need to take Fourier Transform
))]()((2
1[
2)(
1)(
1
s
k
k
s
ss
s kffMkffMT
fMT
fM
)]()([1
)(1
1
s
k
k
s
ss
kffMkffMT
fMT
Department of Electrical and Computer Engineering
f
)( fM
BB
)]()([1
)(1
)()]()([1
s
k
k
s
ss
s kffMkffMT
fMT
fMtxtmF
f
sT
1
BB
)( fM s
sf sf2 sf3 sf4sf
sf2sf3sf4
Department of Electrical and Computer Engineering
f
sT
1
BB
)( fM s
sf sf2 sf3 sf4sf
sf2sf3sf4
f
sT
1
BBsf sf2
sf3 sf4
sfsf2
sf3
sf4
f
sT
1
BBsf sf2 sf3 sf4
sfsf2sf3
sf4
Bfs 2
Bfs 2
Bfs 2
Department of Electrical and Computer Engineering
f
sT
1
BB
)( fM s
sf sf2 sf3 sf4sf
sf2sf3sf4
f
sT
1
BBsf sf2
sf3 sf4
sfsf2
sf3
sf4
f
sT
1
BBsf sf2 sf3 sf4
sfsf2sf3
sf4
Bfs 2
Bfs 2
Bfs 2
Sampling Theorem
Department of Electrical and Computer Engineering
Sampling frequency should be at least equal to or greater
than twice the bandwidth of the message signal for
successful recovery of the signal from it’s samples
Bfs 2
Sampling Theorem
f
sT
1
BBsf sf2 sf3 sf4
sfsf2sf3
sf4
Lowpass Filter
bandwidth of lowpass filter = B Hz
Department of Electrical and Computer Engineering
A practical problem with Sampling
t
k
skTttx )()(
sT
t
)(tm t
)(tms
][kX
)( fM
)( fM s
s
sT
f1
Sampling Frequency
Not practical to generate!
Department of Electrical and Computer Engineering
Let’s Try Square Wave
t
)(tx
sT
t
)(tm t
)(tms
][kX
)( fM
)( fM s
s
sT
f1
Sampling Frequency
Problem Resolved1
Department of Electrical and Computer Engineering
t
)(tx
sT
][kX We need Fourier Series
)2
sin(4
][k
kkX c
2
1]0[ X
1
0][ kX s
We know
))2sin(][)2cos(][(]0[)(1
tkfkXtkfkXXtx sss
k
k
c
and
)2cos()2
sin(4
2
1)(
1
tkfk
ktx s
k
k
Department of Electrical and Computer Engineering
k
k
stkftmk
ktmtxtm
1
)2cos()()2
sin(4
)(2
1)()(
We now have x(t) as frequency representation via Fourier Series
And m(t) as Frequency representation via Fourier Transform
)()( fMtm
Let’s multiply m(t) with x(t)
What is the frequency representation of m(t)x(t)?
)2cos()2
sin(4
2
1)(
1
tkfk
ktx s
k
k
Department of Electrical and Computer Engineering
k
k
stkftmk
ktmtxtm
1
)2cos()()2
sin(4
)(2
1)()(
We need to take Fourier Transform
])2cos()()2
sin(4
)(2
1[)]()([
1
k
k
stkftmk
ktmFtxtmF
))]()((2
1)[
2sin(
4)(
2
1)(
1
s
k
k
ss kffMkffMk
kfMfM
))]()()[(2
sin(2
)(2
1
1
s
k
k
s kffMkffMk
kfM
Department of Electrical and Computer Engineering
f
)( fM
BB
)]()(sin([4
)(2
1)()]()([
1
s
k
k
ss kffMkffMk
fMfMtxtmF
f
2
1
BB
)( fM s
sf sf2 sf3 sf4sf
sf2sf3sf4
Department of Electrical and Computer Engineering
f
)( fM
BB
)]()(sin([4
)(2
1)()]()([
1
s
k
k
ss kffMkffMk
fMfMtxtmF
f
2
1
BB
)( fM s
sf sf2 sf3 sf4sf
sf2sf3sf4
Lowpass Filter
Sampling Theorem still works with practical samples
Department of Electrical and Computer Engineering
Distortion and Linear systems
)(*)()( thtxty
)(ty)(tx )(th
)(th - Impulse Response of a Linear Time Invariant System
)()()( HXY )()( Hth
Department of Electrical and Computer Engineering
Distortionless System
)()( dttkxty
dtjekXY
)()(
dtjkeH
)(
)()()( HXY we know
kH )(
dtH )(
Department of Electrical and Computer Engineering
Filters
centerfcenterf
Bandpass Filter
f
)( fH
offcutf offcutf
Lowpass Filter
f
offcutf offcutf
Highpass Filter
f
Cut-off Frequency
Cut-off Frequency
Center Frequency
Bandwidth
B
)( fH
)( fH
Department of Electrical and Computer Engineering
Lowpass and Highpass combined to generate Bandpass Filter
centerfcenterf
Highpass Filter
f
)( fH
LPcf ,
Lowpass Filter
f
HPcLPc ffB ,,
)( fH
HPcf ,LPcf ,HPcf ,
B 2
,, HPcLPc
center
fff
Department of Electrical and Computer Engineering
Lowpass and Highpass combined to generate Bandstop Filter
stopbandcenterf ,stopbandcenterf ,
Highpass Filter
f
)( fH
LPcf ,
Lowpass Filter
f
LPcHPcstopband ffB ,,
)( fH
HPcf ,LPcf ,HPcf ,
stopbandB 2
,,
,
HPcLPc
stopbandcenter
fff
Department of Electrical and Computer Engineering
How a signal should look like after passing through a filter
Department of Electrical and Computer Engineering
Practical Lowpass Filter
)(tx C
R
)(ty
)(ti
RCjH
1
1)(
Basic Lowpass Filter
Department of Electrical and Computer Engineering
R=1, 000 Ohm
R=10,000 Ohm
R=1, 000 Ohm
R=10,000 Ohm
C=1 mF
C=1 mF
Ma
gn
itu
de
Ph
ase
(ra
d)
Frequency in rads/sec
sec1mRC
sec10mRC
Department of Electrical and Computer Engineering
)(tx C
R
)(ty
)(ti
RCjH
1
1)(
Single Stage Lowpass Filter
C
R
)(ti
C
R
)(ti
Two Stage Lowpass Filter
)1
1)(
1
1()(
RCjRCjH
Department of Electrical and Computer Engineering
R=1000 Ohm
C=1 mF
R=1000 Ohm
C=1 mF
Single Stage
Dual Stage
Single Stage
Dual Stage
Ma
gn
itu
de
Ph
ase
(ra
d)
Frequency in rads/sec
Department of Electrical and Computer Engineering
Practical Bandpass Filter
)(tx C
R
)(ty
)(ti
)/1()/()(
/)(
2 LCRCjj
RCjH
Basic Bandpass Filter
L
)/1(0 LC
Department of Electrical and Computer Engineering
Magnitude and Phase Response of BPF
Department of Electrical and Computer Engineering
D=12
1
][nx ][ny
][]1[2
1][ nxnyny
Discrete Time Filters
][2
1][ nunh
n
Impulse Response
Fe
FH2
2
11
1)(
Transfer Function by taking DTFT of h[n]
Department of Electrical and Computer Engineering
CT vs. DT Lowpass Filters
)(tx C
R
)(ty
)()()(
txtydt
tdyRC
Continuous Time
)(ti
)()()(
tthdt
tdhRC
D=12
1
][nx ][ny
][]1[2
1][ nxnyny
Discrete Time
][]1[2
1][ nnhnh
)()( / tueth RCt ][2
1][ nunh
n
Department of Electrical and Computer Engineering
][2
1][ nunh
n
)()( / tueth RCt
t n
F
)(FH
11f
)( fH
CT vs. DT Lowpass Filters … cont.
Time Response
Frequency Response
Department of Electrical and Computer Engineering
F
)(FH
11f
)( fH
F
)(FH
11f
)( fH
CT vs. DT Lowpass Filters … cont.
Magnitude Response
Phase Response