Errors in DFT Processing

7/31/2019 Errors in DFT Processing

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EECE 301

Signals & Systems

Prof. Mark Fowler

Discussion #9

Illustrating the Errors in DFT Processing DFT for Sonar Processing


2/20

Example #1

Illustrating The Errors inDFT Processing


3/20

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.


4/20

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]


5/20

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


6/20

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


7/20

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


8/20

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


9/20

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.


10/20

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


11/20

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


12/20

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


13/20

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


14/20

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


15/20

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


17/20

Example #2

Sonar Processingusing the DFT

R d /S P i i h DFT


18/20

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


19/20

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

Errors in DFT Processing

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