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EECE 301
Signals & Systems
Prof. Mark Fowler
Discussion #9
Illustrating the Errors in DFT Processing DFT for Sonar Processing
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Example #1
Illustrating The Errors inDFT Processing
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Illustrating the Errors in DFT processing
This example does a nice job of showing the relationships between:
the CTFT,
the DTFT of the infinite-duration signal,
the DTFT of the finite-duration collected samples,
and the DFT computed from those samples.
However, it lacks any real illustration of why we do DFT processing in practice.
There are many practical applications of the DFT and well look at one in the nextexample.
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ADC
x[0]
x[1]
x[2]
x[N-1]
DFTprocessing
X[0]
X[1]
X[2]
memory array
Inside Computersensor
Recall the processing setup:
)(tx ][nx
(Note: no anti-aliasing filter shown but we should have one!)
)(XCTFT
(theory)
)(XDTFT(theory)
Zero-
padding to
lengthNzp
We study the
theory of thesePractical
computed DFT
to understand what this shows us
DTFTN(theory)
)(NX
0
0
memory array
X[Nzp-3]X[Nzp-2]
X[Nzp-1]
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Lets imagine we have the following CT Signal: 0)()( >=
bfortuetxbt
)(tx
1
If we samplex(t) at the rate ofFs samples/second That is, sample every T= 1/Fssec we get the DT Signal coming out of the ADC is:
)(|)(][ nTxtxnx nTt == =For this example we get:
[ ]( ) ][][
][)(][
nuanue
nuetuenx
nnbT
bTn
nTt
bt
=
==
==
Note: |a| < 1
From our FT Table we find the FT ofx(t) is:
bfjfX
bjX
+=
+=
2
1)(
1)(
CTFT Result(Theory)
A
t
Now analyze what we will get from the DFT processing for this signal
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Now imagine that in theory we have all of the samplesx[n] - < n < at the ADCoutput.
Then, in theory the DTFT of this signal is found using the DTFT table to be:
Now, in reality we can collect onlyN< samples in our computer:
elsewhere)0][Assume""(
)1(0,][
=
=
nx
Nnanx
n
n
n
Necessary to connect the DFT result to the theoretical
results wed like to see.
The DTFT of this collected finite-duration is easily found by hand:
( )
= j
Nj
Nae
aeX1
1)(C
=
jae
X1
1)(
DTFT Result(Theory)
For |a| < 1 which we have because:
0,0& >>= Tbea bTB
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Note that we think of this as follows:
[ ]nwnxnx NN ][][ =
..and DTFT theory tells us that
)()()()(2
1)( ==
NNN WXdWXX
A form of convolution (DT Freq. Domain Convolution)
and this convolution has a smearing effect.
Finally, the DFT of the zero-padded collected samples is
0...
0
]1[
...
]1[
]0[
Nx
x
x
=][nxzp Total ofNzppoints
[ ]
=
=1
0
2
][zp
zp
N
n
N
knj
zpzp nxkX
(The only part of this
example wed really do)
D
[ ]
=otherwise
NnwN
,0
1,,2,1,0,1 K
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Our theory tells us that the zero-padded DFT is nothing more than points on
DTFTN:)(][
kNzpXkX =
Now run the m-file called DFT_Relations.m for different Fs ,N, &Nap values
where 1...,,2,1,02
== NkN
k
zp
k
Spacing between DFT
points is 2/Nzp
Increasing the amountof zero-padding gives
closer spacing
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Results from DFT_Relations.m
Plot #1: shows CTFT computed using:
bfj
fX
+
=
2
1)( A
Notice that this is not ideally bandlimited, but is essentially bandlimited.
Therefore we expect some aliasing when we sample.
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Plot #2: shows DTFT computed using:
=j
ae
X
1
1)((For plotting)
(CTFT rescaled to and then shiftedby multiples of 2)
Our theory says that:
=
+
= ksF
k
XTX
2
21
)((For analysis)
So we should see replicas in X ( ) and we do!
We plot TX() to undo the 1/T here
B
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We also plot the CTFT against =
sFf 2
=
jae
X1
1)(DTFT:
bFj
X
bfj
fX
s+
=+
=
)2/(2
1)(
2
1)(
Plot #2:
Recall the two
equivalent axes:
f
sF sF02
sF
2
sF
2 2 0
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The theory in
says well see significant aliasing inX() unless Fs is high enough
The first error visible in plot #2
=
+=
k
sFk
XT
X
2
21)(
DTFT CTFT
Nj
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Plot #3 shows DTFTN computed using
=
j
Nj
Nae
aeX
1
)(1)(
The second error visible in plot #3
C
DTFT DTFTN
This leakage error is less significant as we
increase N, the number of collected samples
We see thatXN() shows signs of the smearing due to: )()()( = NN WXX
Also called leakage error
1 2N knj
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Plot #4 shows DFT computed using: =
=1
0
][][zp
zp
N
n
N
zpzp enxkX
It is plotted vs.
zp
kN
k2= but with the right half moved down to lie
between -& 0 rad/sample
For comparison we also plotXN()
DTFTN
DFT
Note: We show an artificially small number of DFT points here
D
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Theory says Xzp[k] points should lie on top ofXN() notX() !!
We see that this is true
IfNzp is too small (i.e.Nzp = N) then there arent enough DFT points onXN()to allow us to see the real underlying shape ofXN()
This is Grid Error and it is less significant whenNzp is large.
The third error visible in plot #4
DTFTN
DFT
sensorSummary of Results:
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)(tx
CTFT
ADC
Inside Computer
sensorSummary of Results:
x[0]
x[1]
x[2]
x[N-1]
DFT
processing
X[0]
X[1]
X[2]
memory array
Zero-
padding tolengthNzp
0
0
memory array
X[Nzp-3]
X[Nzp-2]
X[Nzp-1]
][nx
DTFT
DTFTN DFT
Aliasing
Error
Smearing
/Leakage
Error
Grid
Error
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Example #2
Sonar Processingusing the DFT
R d /S P i i h DFT
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Radar/Sonar Processing using the DFT
Imagine a stationary sonar and moving target
x
y
Radar/Sonar
targetComponent of velocity along line of sight
v
v
sV
Tx = Transmit
Rx = Receive
Say we transmit a sinusoidal pulse:
= else
TttfA
tx
oo
TX
,0
0),2cos(
)(
Physics tells us (Doppler effect) that the reflected signal received will be:
+
+=
else
Tttc
VffA
txosoo
RX
,0
0,2cos)(
Doppler shift in Hz
(for radars, this is generally in the kHz range)
(for sonar, this is in the 100s of Hz range)
(c speed of propagation 331m/sfor sound in air)
O CTFT th t ll th t th CTFT f th T i l ill b
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Our CTFT theory tells us that the CTFT of the Tx signal will be:
)( fXtx
of of
f
effBeffB
abs. value of
shifted sinc
Assume thatfo is large enough that this decays to a
negligible level byf 0 Hz and byf Beff Center fo between 0 andBeff Fs 2Beff
Also CTFT theory tells us that the CTFT of the Rx signal will be:
)( fXtx
rso
o f
C
Vff =
C
Vfff soor +=
f
effB
Peak is shifted
fromfo by the
doppler shift
If we knowfo and we can find where
this peak is then we can find Vs:)/(1 sm
f
fcV
o
rs
=
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Anti-aliasing
filterADC
DFT
processor
peak
searchGet Vs
Computer: DT processingHydrophone
sensor
CT
signal
CT
signal
DTsignal
DFTvalues
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