DSP Lecture 13-14

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Lectures 13-14 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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Discrete Fourier Transform

Discrete Fourier Transform (DFT) not same as DTFT

DFT results from sampling of DTFT (at pre-defined frequencies),

generally for an aperiodic finite duration discrete sequence

Extension of Discrete Fourier Series (DFS)

Aperiodic Discrete

Why DFT? Efficient algorithms exist for DFT computation (Fast Fourier Transform)

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be a periodic sequence with periodNso that

Discrete Fourier Series-Revisit

[ ]x n [ ] [ ]x n x n rN

12 / 2 /

0

1 1[ ]Nj N kn j N kn

k N k

x n X k e X k eN N

Fourier Series : representation of signal as sum of complex exponentials

Periodicity of discrete time exponentials implies sum is finite duration

For convenience we define : 2 /j N

NW e

Analysis equation

Synthesis equation 1

0

1[ ]

Nkn

N

k

x n X k WN

1

0

[ ]N

kn

N

n

X k x n W

DFS Pair

][~

][~ kXnxDFS

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Discrete Fourier Series-Examples

1[ ] 0r

n rNx n n rN

else

1

2 /

0

1[ ]

Nj N kn

r k

x n n rN eN

[ ] [ ]r

Y k N k rN

10

0

1[ ] [ ] 1

Nkn

N N

k

y n N k W WN

Periodic impulse train

Coefficients form periodic impulse train

Coefficients constant

Sequence constant

DUALITY

Refer to your DSP Oppenheim Book for Summary of DFS properties

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Relationship between DTFT/DFS

Periodic /Finite Duration signals Relationship between DTFT and DFS for periodic sequences

Periodic impulse train : strength proportional to DFS coefficients

2 2j

k

kX e X k

N N

Consider a finite length signal x[n] spanning from 0 to N-1

Convolve with periodic impulse train

[ ] [ ] [ ] [ ]r r

x n x n p n x n n rN x n rN

2

2 2

2 2

j j j j

k

kj

j N

k

kX e X e P e X e

N N

kX e X e

N N

2

2

kj

jNk

N

X k X e X e

Fourier Transform

DFS coeffs. correspond to sampling of

DTFT (in freq. domain) @every 2 /N

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Relationship between DTFT/DFS

Periodic /Finite Duration signals (Ex.)

1, 0 4

0 , otherwise

n

x n

42

0

sin 5 / 2

sin / 2

j j n j

nX e e e

x n represents one cycle of the following periodic signal

4 4

2 /10

10

0 0

j knkn

n n

X k W e

54 /1010

10

sin / 21

sin /101

kj k

k

kWX k e

kW

2 /10k Substitute

(4 /10)

sin / 2[ ]

sin /10j k

kX k e

k

DFS coeffs.

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Sampling the Fourier Transform

Time Domain : Periodic signal from finite duration sequence

Frequency-domain: Performing the sampling operation

Performing reverse operation : sampling of DTFT

2 /2 /

j N kj

N kX k X e X e

2 / 2 /j N k j N k

z eX k X z X e

Sampling the z-transform

1

0

1[ ] [ ]

N

knN

kx n X k W

N

Inverse DFS

*r r

x n x n n rN x n rN

Sampling of DTFT (aperiodic sequence) in frequency domain

generates a periodic repetition of the aperiodic sequence with period N

Periodicity of sequence evident from moving around unit circle

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Sampling the Fourier Transform-Time Aliasing

r

rNnxnx ][][~ One-period

Im(z)

Re(z)

12

2

Im(z)

Re(z)

DFS DTFT

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Sampling the Fourier Transform-Time Aliasing

r

rNnxnx ][][~ One-period

Im(z)

Re(z)

7

2

Im(z)

Re(z)

DFS DTFT

X

X

Aliasing in Time-Domain

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Summary

Samples of Fourier Transform of an aperiodic sequencex[n]can be

thought of as Fourier Series Coeffs. of periodic sequence obtained

by summing periodic replicas ofx[n]

Ifx[n]is finite duration and we take sufficient number of samples

of its Fourier Transform (greater than or equal to its length), thenx[n]

is recoverable from the periodic sequence

To recover or representx[n], it is therefore not necessary to knowX(ej )

at all the frequencies

Using DFS for purpose of recovering/representing a finite duration

sequence leads to DFT

][~ nx

][~ nx

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DFT-Fourier Representation ofFinite

Duration Sequences

Consider a finite length sequence x[n] of length N

Corresponding periodic sequence

To maintain duality between time and frequency

Fourier coefficients to associate with finite duration sequence x[n]

one period of

0 outside of 0 1x n n N

r

x n x n rN

mod NN

X k X k X k

mod N Nx n x n x n

No-overlap between the summation terms

[ ] 0 1[ ]

0 otherwise

x n n Nx n

One period of ][~ nx

][~

kX

[ ] 0 1[ ]

0 otherwise

X k k NX k

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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 N

else

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 DFT

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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-Ex: Rectangular Pulse N = 10

4

2 /10

0

j kn

n

X k e

2 /10 5

2 /10

1

1

j k

j k

e

e

4 /10 sin / 2

sin /10j k ke

k

Required N = 10-point DFT

DFT not same as

prev. example althoughDTFT is same

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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

Nkff sanalysis

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

frequency (scale from -0.5 to 0.5)

DSP Lecture 13-14

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