Chapter 5_dsp1_co thuc_full, dsp1, co thuc, xu li tin hieu, bkdn, dsp

CHAPTER 5:

DISCRETE FOURIER TRANSFORM (DFT)

Lesson #19: DTFT of DT periodic signals

Lesson #20: DFT and Inverse DFT

Lesson #21: DFT properties

Lesson #22: Fast Fourier Transform (FFT)

Lesson #23: Applications of DFT/FFT

Lesson #24: Using of DFT/FFT

Lecture #19

DTFT of periodic signals

1. Review of Fourier Transforms (FT)

2. Fourier Series expansions

3. DTFT of periodic signals

FT of CT periodic signals

Tdtetx

Taeatx

tjk

T

k

tjk

k

k

2;)(

1;)( 0

00

Discrete and aperiodic in frequency domain

Continuous and periodic in time domain

FT of CT aperiodic signals

Continuous and aperiodic in time domain

1

Continuous and aperiodic in frequency domain

FT of DT aperiodic signals

Discrete and aperiodic in time domain

Continuous and periodic in frequency with period 2π

Recapitulation

Can guess?

Lecture #19





Nn

njk

k

njk

Nk

k enxN

aeanx 00 ][1

;][ N

20

dtetxT

aeatxtjk

T

k

tjk

k

k00 )(

1;)(

• CT periodic signals with period T:

• DT periodic signals with period N:

Note: finite sums over an interval length of one periodic N

n)Nk(jn

N

2)Nk(jn

N

2jk

njk 00 eeee

T

20

Fourier Series expansion

Given a DT periodic signals with period N:

Example of Fourier Series expansion

[ ] [ ]k

p n n kN

N

1e]n[p

N

1a

1N

0n

N/n2jk

k

. . . -2N -N 0 N 2N . . . . .

1

0

2

][N

k

nN

jk

keanp

Lecture #19





• CT periodic signals with period T:

• DT periodic signals with period N:


)k(a2)(Xea)t(x0

kk

Ftjk

kk

0

)k(a2)(X]n[x0

kk

F

. . . -2N -N 0 N 2N . . . . . . . . n

Example of calculating DTFT of

periodic signals

kk0k

N

2k

N

2ka2)(P

2π/N 2π/N 2π/N 2π/N 2π/N

. . . -4π/N -2π/N 0 2π/N 4π/N . . . . Ω

Height = 1

Spacing = N

Height = 2π/N

Spacing = 2π/N

Another approach to get DTFT of

periodic signals

0

[ ] 0 1[ ]

0 otherwise

x n n Nx n

0 N-1

x(n) is periodic signal; x0(n) is a part of x(n) that is repeated

Another approach (cont)

0

[ ] 0 1[ ]

0 otherwise

x n n Nx n

0 0 0[ ] [ ] [ ] [ ] [ ] [ ]k k k

x n x n kN x n n kN x n n kN

p(n) in previous example

0 0[ ] [ ] [ ] ( ) ( ) ( )F

x n x n p n X P X

Another approach (cont)

0 0[ ] [ ] [ ] ( ) ( ) ( )F

x n x n p n X P X

k

k

Nk

NkX

N

Nk

NXX

)2

()2

(2

)2

(2

)()(

0

0

It has N distinct values at k = 0, 1, …, N-1

Inverse DTFT

1

0

2

0

2

0

0

2

0

0

2

0

1

)2

(1

)2

()2

(1

)2

()2

(2

2

1

)(2

1][)(

N

k

N

nkj

k

nj

nj

k

nj

eN

kXN

deN

kN

kXN

deN

kN

kXN

deXnxXDFT

This is obtained from this property of the impulse:

elsewhere

btatfdttttf

b

a0

)()()(

00

0

Summary

Procedure to calculate DTFT of

periodic signals

Step 1:

Start with x0(n) – one period of x(n), with zero everywhere else

Step 2:

Find the DTFT X0(Ω) of the signal x0[n] above

Step 3:

Find X0(Ω) at N equally spacing frequency points X0(k2π/N)

Step 4:

Obtain the DTFT of x(n): k N

kN

kXN

X )2

()2

(2

)( 0

Example of calculating DTFT of

periodic signals

33

0

00 21)()( jj

n

nj eeenxX

k = 0 X0(0) = 4; k = 1 X0(1) = 1+j

k = 2 X0(2) = -2; k = 3 X0(3) = 1-j

0 1 2 3

Example (cont)

Example (cont)

a) Draw the magnitude plot of the FT of a CT sinusoid x(t), that

has a frequency of 24 Hz

b) Sample this signal x(t) at 40 Hz. Draw the magnitude plot of

the FT of the sampled signal

c) Reconstruct a CT signal from the samples above with a LPF, it

will look like a CT sinusoid of what frequency?

d) Consider DT signal that is obtained by the sampling x(t) at a

sampling frequency of 40 Hz is x(n). Is x(n) a periodic signal?

If yes, what is its period?

e) Specify the DTFT of the DT signal x(n).

DTFT and DFS example

a) Draw the magnitude plot of the FT of a CT sinusoid x(t), that has

a frequency of 24 Hz

The FT of a sinusoidal has nonzero values only at certain points

DTFT and DFS examples

-24 24 f [Hz]

-48π 48π ω [rad/s]

b) Sample this signal x(t) at 40 Hz. Draw the magnitude plot of the

FT of the sampled signal

Sampled signal FT magnitude = copies of unsampled signal FT

magnitude; each copy is centered at k40(Hz) or k80π(rad/s)


-64 -56 -24 -16 0 16 24 40 56 64 f[Hz]

-128π -48π -32π 32π 48π 80π 112π

ω[rad/s]

fs = 40Hz fs = 40Hz

8Hz

c) Reconstruct a CT signal from the samples above with a LPF, it will

look like a CT sinusoid of what frequency?

The unit circle represents the scale of frequency range for a

signal sampled at 40Hz the reconstructed signal looks like

16Hz.


20

24

16

0 16 24 40 56 64 f[Hz]

32π 48π 80π 112π ω[rad/s]

8Hz

d) Consider DT signal that is obtained by the sampling x(t) at a

sampling frequency of 40 Hz is x(n). Is x(n) a periodic signal? If

yes, what is its period?

The unsampled signal: x(t) = cos(2π.24t)

The sampled signal:

x[n] = cos(2π.24.(1/40)t) = cos(6π/5)n = cos (4π/5)n

The sampled signal is periodic with N = 5


e) Specify the DTFT of the DT signal x(n).


Lecture #20

DFT and inverse DFT

1. DFT and inverse DFT formulas

2. DFT and inverse DFT examples

3. Frequency resolution of the DFT

DFT to the rescue!

Both CTFT and DTFT produce continuous function of

frequency can’t calculate an infinite continuum of frequencies

using a computer

Most real-world data is not in the form like anu(n)

DFT can be used as a FT approximation that can calculate a finite

set of discrete-frequency spectrum values from a finite set of

discrete-time samples of an analog signal

Could we calculate the frequency spectrum of a signal

using a digital computer with CTFT/DTFT?

Building the DFT formula

“Window” x(n) is like multiplying

the signal by the finite length

rectangular window

Discrete time

signal x(n)

Continuous time

signal x(t)

Discrete time

signal x0(n)

Finite length

sample

window

Building the DFT formula (cont)

Discrete time

signal x(n)

Continuous time

signal x(t)

Discrete time

signal x0(n)

Finite length

sample

window

DTFT

1

0

000 ][][)(N

n

nj

n

nj enxenxX

Building the DFT formula (cont)

Discrete time

signal x(n)

Continuous time

signal x(t)

Discrete time

signal x0(n)

Finite length

sample

window

DTFT

Sample

at N

values

around

the unit

circle

Discrete Fourier Transform DFT X(k)

Discrete + periodic with period N

X0(Ω)

Continuous + periodic

with period 2π

Notation conventions

Notation conventions (cont)

DFT and inverse DFT formulas

Notation

You only have to store N points

N

2j

N eW

Lecture #20

DFT and inverse DFT




DFT and IDFT examples

Ex.1. Find the DFT of x(n) = 1, n = 0, 1, 2, …, (N-1)

k = 0 X(k) = X(0) = N

k ≠ 0 X(k) = 0


Ex.2. Given y(n) = δ(n-2) and N = 8, find Y(k)


Ex.3. Find the IDFT of X(k) = 1, k = 0, 1, …, 7.


Ex.4. Given x(n) = δ(n) + 2δ(n-1) +3δ(n-2) + δ(n-3) and

N = 4. Find X(k).


Ex.5. Given X(k) = 2δ(k) + 2δ(k-2) and N = 4. Find x(n).

DFT in matrix forms

)1)(1()1(21

)1(21

1

0

1

1

1111

]1[

]1[

]0[

]1[

]1[

]0[

:'

1,...,1,0][][

NN

N

N

N

N

N

N

NNN

N

NN

N

n

kn

N

WWW

WWWW

NX

X

X

X

Nx

x

x

x

definesLet

NkWnxkX

The N-point DFT may be expressed in matrix form as:

XN = WNxN

IDFT in matrix forms

)1)(1()1(21

)1(21

1

0

1

1

1111

]1[

]1[

]0[

]1[

]1[

]0[

:'

1,...,1,0][1

][

NN

N

N

N

N

N

N

NNN

N

NN

N

k

kn

N

WWW

WWWW

NX

X

X

X

Nx

x

x

x

definesLet

NnWkXN

nx

The N-point DFT may be expressed in matrix form as:

matrixidentityI

NIWW

XWN

XWx

N

NNN

NNNNN

:

1

*

*1

Example of calculation of DFT in matrix form

Find DFT of the signal

x[n] =

δ[n-1] + 2δ[n-2] + 3δ[n-3]

j

jxWXThen

jj

jj

WWW

WWW

WWWW

x

definesLet

22

2

22

6

.

11

1111

11

1111

1

1

1

1111

3

2

1

0

:'

444

9

4

6

4

3

4

6

4

4

4

2

4

3

4

2

4

1

4

4

4

Lecture #20

DFT and inverse DFT




Discrete frequency spectrum computed from DFT has the

spacing between frequency samples of:

Frequency resolution of the DFT

N

ff s

N

2

The choice of N determines the resolution of the frequency

spectrum, or vice-versa

To obtain the adequate resolution:

1. Increasing of the duration of data input to the DFT

2. Zero padding, which adds no new information, but effects a better

interpolator

Examples of N = 5 and N = 15

Examples of N = 15 and N = 30

Lecture #21

DFT properties

1. Periodicity and Linearity

2. Circular time shift

3. Circular frequency shift

4. Circular convolution

5. Multiplication

6. Parseval’s theorem

DFT properties

x(n)

Finite length

xp(n)

Periodic

N periodic extension

truncate


truncate

X(k)

Finite length Xp(k)

Periodic

DFT

DFT

-1

Xp(Ω)

Periodic

DTFT

DTFT-1

DFT properties

x(n)

Finite length

xp(n)

Periodic


truncate


truncate

X(k)

Finite length Xp(k)

Periodic

DFT

DFT

-1

Xp(Ω)

Periodic

DTFT

DTFT-1

Periodicity

x[n+N] = x[n] for all n

X[k+N] = X[k] for all k

If x[n] and X[k] are an N-point DFT pair, then

Linearity

Note: The length of x1[n] is same with the length of x2[n]

][][][][ 22112211 kXakXanxanxaDFT

Proof:

Infer from the definition formula of DFT

Lecture #21

DFT properties





5. Multiplication


Circular time shift property

]k[XW]mn[x kmDFT

Proof:

Infer from the relation between DFT and DTFT

][][)(

:2

)(][

2

kXWkXeXe

Nk

Xemnx

kmm

Njk

mj

mjDTFT

Circular

time

shift

Circular shift by m is the same as a shift by m modulo N

x(n)

xp(n)

xp(n+2)

x(n+2)

Lecture #21

DFT properties





5. Multiplication


Circular frequency shift property

][][ ln lkXWnxDFT

Proof:

Similar to the circular time shift property

Lecture #21

DFT properties





5. Multiplication


Circular convolution property

The non-zero length of x1(n) and x2(n) can be no longer

than N

The shift operation is circular shift

The flip operation is circular flip

1

0

212121 ][][][][][].[N

p

DFT

pnxpxnxnxkXkX

The circular convolution is not the linear convolution in chapter 2

Direct method to calculate circular

convolution

1. Draw a circle with N values of x(n) with N equally spaced angles

in a counterclockwise direction.

2. Draw a smaller radius circle with N values of h(n) with equally

spaced angles in a clockwise direction. Superimpose the centers of 2

circles, and have h(0) in front of x(0).

3. Calculate y(0) by multiplying the corresponding values on each

radial line, and then adding the products.

4. Find succeeding values of y(n) in the same way after rotating the

inner disk counterclockwise through the angle 2πk/N

Example to calculate circular convolution

Evaluate the circular convolution, y(n) of 2 signals:

x1(n) = [ 1 2 3 4 ]; x2(n) = [ 0 1 2 3 ]

1

2

3

4 0

1

2

3

y(0) = 16

1

2

3

4 1

2

3

0

y(1) = 18

Example (cont)

Evaluate the circular convolution, y(n) of 2 signals:

x1(n) = [ 1 2 3 4 ]; x2(n) = [ 0 1 2 3 ]

1

2

3

4 2

3

0

1

y(2) = 16

1

2

3

4 3

0

1

2

y(3) = 10

Another method to calculate circular convolution

x1(n)

x2(n)

X1(k)

X2(k)

DFT

DFT

IDFT y(n)

Ex. x1(n) = [ 1 2 3 4 ]; x2(n) = [ 0 1 2 3 ]

X1(k) = [ 10, -2+j2, -2, -2-j2 ];

X2(k) = [ 6, -2+j2, -2, -2-j2 ];

Y(k) = X1(k).X2(k) = [ 60, -j8, 4, j8 ]

y(n) = [ 16, 18, 16, 10 ]

Lecture #21

DFT properties





5. Multiplication


Multiplication

The non-zero length of X1(k) and X2(k) can be no longer

than N



1

0

212121 ][][1

][][1

][].[N

l

DFT

lkXlXN

kXkXN

nxnx

Lecture #21

DFT properties





5. Multiplication


Parseval’s theorem

The general form:

Special case:

1

0

1

0

][*][1

][*][N

k

N

n

kYkXN

nynx

1

0

21

0

2 |][|1

|][|N

k

N

n

kXN

nx

Parseval’s theorem example

Given x(n) = δ(n) + 2δ(n-1) +3δ(n-2) + δ(n-3) and N = 4.

1

0

2222222

22221

0

2

154

60)121127(

4

1|][|

1

151321|][|

N

k

N

n

kXN

nx

Multiple choice questions

Ex1:

It is desired to build a sampling system which takes a

recorded orchestra music signal s(t) and converts it to a

discretely-sampled form. In order to experimentally determine

the bandwidth (in frequency) of this signal the best transform

would be

a. The Fourier transform of a continuous function or a discrete

sequence

b. The Z-transform

c. The Fourier series

d. The discrete Fourier transform


Ex2:

A mathematical model of human speech puts out a perfectly

periodic signal for steady vowel sounds. In order to accurately

determine the harmonics of the fundamental frequency for

this mathematical model, the best transform should be


sequence

b. The Z-transform




Ex3:

A student wishes to convolve the continuous-time mathematical

functions g(t) and h(t). She/he uniformly samples these

functions, and uses a computer to get, at best, an approximate

result. Another student claims to be able to get an exact result

via a transform. She/he must be thinking of which transform


sequence

b. The Z-transform




Ex4:

A student wishes to convolve the discrete-time mathematical

functions g(n) and h(n). She/he claims to be able to get an

exact result via one of the transforms below. She/he must be

thinking of which transforms


sequence

b. The Z-transform



Lecture #22

Fast Fourier Transform (FFT)

1. What is FFT?

2. The decomposition-in-time Fast Fourier

Transform algorithm (DIT-FFT)

3. The decomposition-in-frequency Fast

Fourier Transform algorithm (DIF-FFT)

Using some "tricks" and choosing the appropriate N value, the DFT

can be implemented in a very computationally efficient way – this is

the Fast Fourier Transform (FFT) - which is the basis of lots of

contemporary DSP hardware

N

2j

N eW1Nn0W]k[XN

1]n[x

1Nk0W]n[x]k[X

1N

0n

nk

N

1N

0n

nk

N

Recall DFT and IDFT definition

Nth complex root of unity:

What does it take to compute DFT or IDFT?

N2 complex multiplications for all N points of X(k) or x[n]

FFT: the algorithm to optimize the computational process by

breaking the N-point DFT into smaller DFTs

Ex.: N is a power of 2: split N-point DFT into two N/2-point DFTs,

then split each of these into two of length N/4, etc., until we

have N/2 subsequences of length 2

Cooley and Tukey (1965):


tionsmultiplicacomplex 2

log2

2N

NN

Some properties of can be exploited

Other useful properties:

Wnk

N

k andn in y periodicit ,

symmetry conjugatecomplex ,

)()(

*)(

WWW

WWWnNk

N

Nnk

N

kn

N

kn

N

nNk

N

knN

WW

WW

eWWWW

knN

kn

N

kn

N

kn

Njnkn

N

nN

N

kn

N

nN

k

N

2

2

2)

2(

oddn if ,

evenn if ,

FFT (cont)

N Co

mp

lex m

ult

iplicati

on

nu

mb

ers

FFT

Comparison of DFT and FFT efficiency

Comparison of DFT and FFT efficiency

Number of Complex multiplications Complex multiplications Speed improvement

points N in direct DFT in FFT factor

4

8

16

32

64

128

256

512

1024

16

64

256

1,024

4,096

16,384

65,536

262,144

1,048,576

4

12

32

80

192

448

1,024

2,304

5,120

4.0

5.3

8.0

12.8

21.3

36.6

64.0

113.8

204.8

Lecture #22


1. What is FFT?

2. The decomposition-in-time Fast

Fourier Transform algorithm (DIT-FFT)



G(k) is N/2 points DFT of the even numbered data: x(0),

x(2), x(4), …., x(N-2).

H(k) is the N/2 points DFT of the odd numbered data:

x(1), x(3), …, x(N-1).

even odd

[ ] [ ] [ ]kn kn

n n

X k x n W x n W

DIT-FFT with N as a 2-radix number

1N,...,1,0k ),k(H)k(G)k(X Wk

N

We divide X(k) into 2 parts:

2 21 1

2 (2 1)

0 0

[ ] [2 ] [2 1]

N N

mk k m

m m

X k x m W x m W

2 21 1

2 2

0 0

[2 ]( ) [2 1]( )

N N

mk k mk

m m

x m W W x m W=

2/N

)2/N/(2j2N/2j2

NWeeW

G(k) and H(k) are of length N/2; X(k) is of length N

G(k)=G(k+N/2) and H(k)=H(k+N/2)

)()(][][)(

1

0

2/

1

0

2/

22

kHkGWmhWWmgkX Wk

Nm

mk

N

k

N

m

mk

N

NN

DIT-FFT (cont)

Note:

4

k

848]k[HW]k[G]k[X

]3[HW]3[G]7[X]3[HW]3[G]3[X

]2[HW]2[G]6[X]2[HW]2[G]2[X

]1[HW]1[G]5[X]1[HW]1[G]1[X

]0[HW]0[G]4[X]0[HW]0[G]0[X

7

8

3

8

6

8

2

8

5

8

1

8

4

8

0

8

DIT-FFT of length N = 8

DFT

N = 4

DFT

N = 4

# complex multiplications are reduced: 82 = 64 2x(4)2 + 8 = 40

Signal

flow

graph

for

8-point

FFT

2424 ][][][ kHWkGkX k


]1[]1[]3[

]0[]0[]2[

]1[]1[]1[

]0[]0[]0[

3

4

2

4

1

4

0

4

HWGX

HWGX

HWGX

HWGX

]1[]1[]3[ 3

4 HWGX

]1[]1[]1[ 1

4 HWGX

]0[]0[]2[ 2

4 HWGX

]0[]0[]0[ 0

4 HWGX

DFT

N = 2

DFT

N = 2

W0

x[0]

x[2]

x[1]

x[3]

W1

W2

W3

# multiplications are reduced: 42 = 16 2x(2)2 + 4 = 12


# multiplications are

reduced: 22 = 4 0;

just 2 additions!!! x[1] - 1

x[0] X[0]

X[1]

]1[]0[]1[]0[]1[

]1[]0[]1[]0[]0[

1,10,][][

1.11.0

0.10.0

2

21

0

xxWxWxX

xxWxWxX

eWkWnxkXj

n

nk

Butterfly diagram

DFT

N = 4

DFT

N = 4

First step: split 8-point DFT into two 4-point DFTs

Example

of

8-point

FFT

0W

0W

W4

W2

W0

W0

W6

W6

W2

W4

DFT N = 2

DFT N = 2

DFT N = 2

DFT N = 2

G[0]

G[1]

G[2]

G[3]

H[0]

H[1]

H[2]

H[3]

Example

of

8-point

FFT

(cont)

Second step: split each of 4-point DFT into two 2-point DFTs

Example

of

8-point

FFT

(cont) 0W

0W

W4

W2

W0

W0

W6

W6

W2

W4

Third step: combine all signal flow graphs of step 1 and step 2

W0

W1

W2

W3

W4

W5

W6

W7

There are 3 = log28 stages; 4 butterfly diagrams in each stage

Wr+N/2

Wr+N/2 = -Wr

Wr - 1

Butterfly diagram

Now the overall computation is reduced to:

tionsmultiplicacomplex 2

log2

2N

NN

The

overall

signal

flow

graph

for

8-point

DIT-FFT

Wr - 1

In-place computation

a

b

A

B

Each butterfly takes (a,b) to produce (A,B) no need to save

(a,b) can store (A,B) in the same locations as (a,b)

We need 2N store registers to store the results at each stage and

these registers are also used throughout the computation of the

N-point DFT

The order of the input: bit-reversed order

The order of the output: natural order

Lecture #22


1. What is FFT?

2. The decomposition-in-time Fast Fourier

Transform algorithm (DIT-FFT)



Rewrite the second half:

DIF-FFT with N as a 2-radix number

1)2/(

0

1)2/(

0

1

)2/( 2

2)( 22

N

n

kn

N

k

N

N

n

nk

N

N

Nn

kn

N WN

nxWWN

nxWnxNN

1,...,1,0 ,)()()(1)2/(

0

1

)2/(

NkWnxWnxkXN

n

N

Nn

kn

N

kn

N

kjNNkjk

N eeWN

)1()2/)(/2(2

We divide X(k) in half – first half and second half

Since:

then

1,...,1,0 ,)2/()1()()(1)2/(

0

NkWNnxnxkXN

n

kn

N

k

DIF-FFT (cont)

12

,...,1,0 ,)2/()()12(

12

,...,1,0 ,)2/()()2(

1)2/(

0

2/

1)2/(

0

2/

NkWWNnxnxkX

NkWNnxnxkX

N

n

kn

N

n

N

N

n

kn

N

2

2/

2

NN WW

Decompose X(k) into two frequency sequences, even and odd:

Since:

then,

12

,...,1,0 ,)2/()()12(

12

,...,1,0 ,)2/()()2(

1)2/(

0

)12(

1)2/(

0

2

NkWNnxnxkX

NkWNnxnxkX

N

n

nk

N

N

n

kn

N

DIF-FFT butterfly

x[n]

x[n+N/2]

x[n] + x[n+N/2]

(x[n] - x[n+N/2])WNn

-1 WN

n

DIF-FFT of length N = 4

-1

-1

-1

-1 W41

W40

x[0] + x[2]

x[1] + x[3]

(x[0] - x[2])W40

(x[1] - x[3])W41

x[0]

x[1]

x[2]

x[3]

X[0]

X[2]

X[1]

X[3]

Lecture #23

DFT applications

1. Approximation of the Fourier Transform

of analog signals

2. Linear convolution

Approximation of Fourier Transform

Using DFT as a discrete-frequency approximation of the DTFT

and also an approximation of the CTFT

Step1. Determine the resolution ΔΩ = 2π/N required for the DFT to

be useful for its intended purpose determine N (N often = 2n)

Step 2. Determine the sampling frequency required to sample the

analog signal so as to avoid aliasing ωs ≥ 2ωM

Step 3. Accumulate N samples of the analog signal over N.T

seconds (T = 2π/ωs)

Step 4. Calculate DFT directly or using FFT algorithm

0 0.5 1 1.5 2 2.5 30

0.5

1

n

f[n]

0 1 2 3 4 50

1

2

3

omega

mag

nitu

de

Example

f(t) = rect[(t-1)/2]

F(ω) = 2sinc(ω)e-jω

Ts = 1s

%Error of sample 1 = 50%

Example

0 5 10 150

0.5

1

n

f[n]

0 5 10 15 20 25 300

1

2

3

omega

mag

nitu

de

f(t) = rect[(t-1)/2]


Ts = 0.2s


0 5 10 15 20 25 30 350

0.5

1

n

f[n]

0 10 20 30 40 50 60 700

1

2

3

omega

mag

nitu

de

Example

f(t) = rect[(t-1)/2]


Ts = 0.1s


Windowing

Source of error: truncation or “windowing” of the periodic

extension of the DT sequence implied in the DFT development

Windowing = multiplying the periodic extension of x[n] by a

rectangular function with duration N.T DFT involves a convolution

DFT = sampled(DTFT*sinc)

This multiplication causes spectrum-leakage distortion

To reduce the affect of spectrum-leakage distortion:

- Increase the sampling frequency

- Increase the number of samples

- Choose an appropriate windowing function (Hamming, Hanning)

Lecture #23

DFT applications

1. Approximation of the Fourier Transform of

analog signals

2. Linear convolution

Recall linear convolution

N1: the non-zero length of x1(n); N2: the non-zero

length of x2(n); Ny = N1 + N2 -1

The shift operation is the regular shift

The flip operation is the regular flip

p

pnxpxnxnxny ][][][*][][ 2121

Recall circular convolution

The non-zero length of x1(n) and x2(n) can be no longer

than N



1

0

2121 ][][][][][N

p

pnxpxnxnxny

Calculation of the linear convolution

The circular convolution of 2 sequences of length N1 and N2

can be made equal to the linear convolution of 2 sequences

by zero padding both sequences so that they both consists

of N1+N2-1 samples.

x1(n)

N1 samples

x2(n)

N2 samples

x’1(n)

x’2(n)

Zero padding

(N2-1) zeros IDFT

x1(n)*x2(n) Zero padding

(N1-1) zeros

DFT

DFT

Example of calculation of the linear convolution

x1(n) = [ 1 2 3 4 ]; x2(n) = [ 0 1 2 3 ]

x’1(n) = [ 1 2 3 4 0 0 0 ]; x’2(n) = [ 0 1 2 3 0 0 0 ]

X’1(k) = [ 10, -2.0245-j6.2240, 0.3460+j2.4791, 0.1784-j2.4220, 0.1784+j2.4220, 0.3460-j2.4791, -2.0245-j6.2240 ];

X’2(k) = [ 6, -2.5245-j4.0333, -0.1540+j2.2383, -0.3216-j1.7950, -0.3216+j1.7950, -0.1540-j2.2383, -2.5245+j4.0333];

Y’(k) = [ 60, -19.9928+j23.8775, -5.6024+j0.3927, -5.8342-j0.8644, -4.4049+j0.4585, -5.6024-j0.3927, -19.9928 +j23.8775 ]

IDFTY’(k) = y’(n) = [ 0 1 4 10 16 17 12 ]

Lecture #24

Using DFT/FFT

The response, y(n) of a filter with the impulse response, h(n)

to an input, x(n) can be calculated by using the DFT

Limitations: All sample values of the signal must be

accumulated before the process begins

Great memory

Long time delay

Not suitable for long duration input

Solution: block filtering

Filtering

Divide long input signal, x(n) into non-overlapped blocks,

xb(n) of appropriate length for FFT calculation.

Convolve each block xb(n) and h(n) to get the output,

yb(n)

Overlap-add outputs, yb(n) together to form the output

signal, y(n)

Overlap-add technique in block filtering

Example of overlap-add technique

h(n)

x(n)

Example of overlap-add technique

x0(n)

x1(n)

x2(n)

Example (cont)

y0(n)

y1(n)

y2(n)

Prob.1 Prob. 5

12.10 12.18

Prob. 2 Prob. 6

12.11 12.22

Prob. 3 Prob. 7

12.12 12.24

Prob. 4 Prob. 8

12.15 12.30

HW

Prob.9

Given signals below and their DFT-5

HW

]n[]n[s)c(

]1n[]n[x)b(

]4n[4]3n[3]2n[2]1n[]n[x)a(

2

1

1. Find y[n] so that Y[k] = X1[k].X2[k]

2. Does x3[n] exist, if S[k] = X1[k].X3[k]?

Prob.10

Prove the Parseval’s theorem

HW

1

0

1

0

][*][1

][*][N

k

N

n

kYkXN

nynx

Lecture #20

DFT applications

1. Approximation of the Fourier Transform

of analog signals

2. Convolution

3. Filtering

4. Correlation

5. ESD estimation

Windowing

0 10 20 30 40 50 60 700

0.5

1

n

f[n]

0 5 10 15 20 25 30 350

1

2

3

omega

mag

nitu

de

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Chapter 5_dsp1_co thuc_full, dsp1, co thuc, xu li tin hieu, bkdn, dsp