DSP Lecture 16-17

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Lectures 16-17 EE-802 ADSP SEECS-NUST

EE 802-Advanced Digital SignalProcessing

Dr. Amir A. Khan

Office : A-218, SEECS

9085-2162; [email protected]

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Lecture Outline

Discrete Fourier Transform (DFT)

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DFT-Analysis and Synthesis Equations

12 /

0

1[ ]

Nj N kn

kx n X k eN

12 /

0[ ]

Nj N kn

nX k x n e

DFS Pair

12 /

0

[ ] 0 1

0

Nj N kn

n

x n e k NX k

else

1

2 /

0

10 1

[ ]0

Nj N kn

k

X k e n Nx n

Nelse

PERIODIC

ONE PERIOD ONLY

DFT

x n X k

1

0

[ ] ( ) , 0 1N

kn

N

n

X k x n W k N

1

0

1[ ] ( ) , 0 1

Nkn

N

k

x n X k W n NN

Analysis equation Synthesis equation

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DFT-Ex: Rectangular Pulse

Minimum Value of N ?

Minimum Value ofN = 5

42 /5

0

2

2 /5

1

1

5 0, 5, 10,...0

j k n

n

j k

j k

X k e

e

e

kelse

DFS

Required N = 5-point DFT

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DFT- Frequency Interpretation

1

0

[ ] ( ) , 0 1N

kn

Nn

X k x n W k N

DFT determines the spectral content of input at Nequally spaced frequency points What are the exact frequencies in the DFT spectrum?

depends on the number of DFT points N and

on the sampling frequencyfs, at which the original signal was sampled

X[k] : frequency content at discrete radial frequency: (rad/sample)

or cycles/sample

In real frequency terms (Hz) :

kN

k

2

N

kfk

N

kff sanalysis

In case your sampling frequency is not specified or not known, we talk of normalized

frequency (scale from -0.5 to 0.5)

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DFT- Example

-1.5

-1

-0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7n

)4

320002sin(5.0)10002sin()(

tttx

Sample it with fs= 8000 Hz and take 8 samplesx[n]

0 1 2 3 4 5 6 70

0.5

1

1.5

2

2.5

3

3.5

4

k

|X

[k]|

Plug back in manually the real freq. axis

X[1] corresponds to 1x8000/8 = 1000 Hz

X[2] corresponds to 2x8000/8 = 2000 Hz

What are X[6] and X[7]?

-4000 -3000 -2000 -1000 0 1000 2000 30000

0.5

1

1.5

2

2.5

3

3.5

4

f(Hz)

|X(f)|

Apply 8-point DFT

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DFT Leakage

N = 64 (samples of input signal)

Falls in DFT Bin # k =3

NO DFT Bin # k = 3.4

Leaks into multiple DFT Bins

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DFT Leakage (The reality)

-4000 -3000 -2000 -1000 0 1000 2000 3000 4000

0

5

10

15

20

25

X= 875

Y= 15.9248

X= 750

Y= 24.4898

f (Hz)

)8002sin()( ttx Sample it with fs= 8000 Hz and take 64 samplesApply 64-point DFT

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DFT Leakage Work Around

Windowing (rectangular) is always taking place

For sinusoids with integer number of cycles in analysis period,

leakage does not occur

Unfortunately, for practical signals leakage will always be an issue

So we need to ensure better leakage management

One solution is to use better windows with reduced side lobes

Hamming, Hanning, Triangular, etc.

Cost paid would be the reduced resolution of different frequencies

Self-Exercises Further ReadingUnderstanding DSP (Lyons)

Chapter 3 : DFT (Sections 3.2, 3.5, 3.8, 3.9, 3.13, 3.17)

Practice on MATLAB at least with sinusoids and different

windows to clarify your concepts

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Properties of DFT-Linearity

1 1

2 2

1 2 1 2

DFT

DFT

DFT

x n X kx n X k

ax n bx n aX k bX k

N1 = length (x1) N2 = length (x2)

length (x3) =N3 = max(N1, N2)

x3 denote the sum sequence

For DFT ofx3 to meaningful, compute DFTs ofx1 andx2 on lengthN3

DFT of greater length than actual samples can be computed by zero

padding

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Properties of DFT-Circular Shift

2 /0 n N-1

DFT

j k N mDFT

N

x n X kx n m X k e

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Circular Convolution as LinearConvolution with Aliasing

[ ] ( )

DTFT j

x n X e

2

[ ] ( )

kj

N

X k X e

[ ] [ ]r

x n x n rN

Extracting one period of to form the finite-length sequence[ ]X k

2

( ) 0 1[ ]

0 otherwise

kj

NX e k NX k

[ ] 0 1[ ]

0 otherwisep

x n n Nx n

3 1 2j j j

X e X e X e

Linear Convolution:

2 /3 3 , 0 1j k NX k X e k N

2 / 2 / 3 1 2 , 0 1j k N j k NX k X e X e k N

3 1 2

3

3

[ ] [ ] [ ]

[ ] 0 1[ ]

0 otherwise

IDFT

rp

X k X k X k

x n rN n Nx n

may suffer from the time-aliasing distortion

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Circular Convolution as LinearConvolution with Aliasing (Contd.)

3 1 2

3

3

[ ] [ ] [ ]

[ ] 0 1[ ]

0 otherwise

IDFT

rp

X k X k X k

x n rN n Nx n

may suffer from the time-aliasing distortion

If , then x1

[n] =x1p

[n], i.e., no aliasing distortion in x1

[n],

likewise no distortion in x2[n]

PLN and

What about x3[n]? Possible Distortion unless Nis large enough

Nshould be greater than or equal to the length ofx3[n]

1 PLNIn this case circular convolution is same as performing the linear convolution and

hence no aliasing distortion is present

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Ex 2: Circular Convolution as LinearConvolution with Aliasing

Indexing problems in Book Fig. 8.18

Oppenheim

If N = L = P, all of sequence values of

circular convolution will be different from

linear convolution

-1 0

+

+

=

+ over interval 0 to N-1

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Ex : Circular Convolution as LinearConvolution with Aliasing

L > P

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Ex : Circular Convolution as LinearConvolution with Aliasing (Contd.)

FirstP-1 values in linear andcircular convolution are different

20 Pn

NextL-P+1 values in linear andcircular convolution are same

11 LnP

No values in circular convolutionfor

Ln Though present in linear convolution

Ill i f Ci l C l i

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Illustration of Circular Convolution asLinear Convolution with Aliasing

Wrapping around of circular convolution

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Linear FIR Filtering using DFT-filtering oflong data sequences

Assume that input sequence is very long and LTI system is FIR filter

Wait until all sequence has been acquired (too lengthy computation of

DFT)

or do Block convolution : Break up the input sequence into blocks,

process and restack

Block Convolution Methods (using DFT)

Method 1

Overlap Add

Method 2 Overlap Save

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Overlap Add Method

Partition the long input sequence into non-overlapping sections of

shorter length (L)

Consider the FIR filterh[n] of length P

Required operation

Perform the above operations using DFTs and IDFT

0

[ ] [ ]rr

x n x n rL

[ ] 0 1

[ ]0 otherwise

r

x n rL n Lx n

where

nhnxny

nhrLnxnyr

r

0

nhnxny rr

0r

r rLnyny

Linear filtering of total input

Linear filtering of each

input section

Total output = Addition of individual

outputs (separated by L samples)

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Overlap Add Method

Linear convolution = circular convolution if N = L+P-1

zero-pad filter (L-1 zeros) and input (P-1 zeros)

Take N-point DFTs of input chunk and filter

Multiply the DFTs to get an output chunks DFT

Take IDFT to obtain one chunk of output

Repeat for next input chunk of L samplesAdd (overlapping) output sections to obtain overall output

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Overlap Save Method

Partition input into overlapping sections (length L) & overlap (P-1) samples

Use N = L point DFTs and IDFTs (to obtain output chunks of length L)

time aliasing will occur as N = L+P-1 is not satisfied

discard starting P-1 points because they are incorrect (aliased)

Retain the correct data of lengthL-(P-1) points & patch them together (save)

DSP Lecture 16-17

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