decimation in frequency -...

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FFTFFTDecimation in FrequencyDecimation in Frequency

Spring

2009© Ammar Abu-Hudrouss -Islamic

University Gaza

Slide ٢Digital Signal Processing

Decimation in Frequency

1

01,.....,2,1,0)()(

N

n

knN NkWnxkX

12/

0

1

2/)()()(

N

n

N

Nn

knN

knN WnxWnxkX

12/

0

12/

0

)2/()2/()()(N

n

N

n

NnkN

knN WNnxWnxkX

NjN eW /2

The discrete Fourier transform can be found using

Where N = 2, 4, 8, 16,… and

X (k ) can be expressed as

12/

0

12/

0

2/ )2/()()(N

n

N

n

knN

kNN

knN WNnxWWnxkX

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Slide ٣Digital Signal Processing


kkNNkkNN eeW )1(/)2/(2)2/(

12/

0

2)2/()()2(N

n

mnNWNnxnxmX

12/

0

12/

0

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

n

N

n

knN

kknN WNnxWnxkX

But

Then

If k = 2m or an even number

12/

02/)()2(

N

n

mnNWnamX

Slide ٤Digital Signal Processing


mnN

NmnjNmnjmnN WeeW 2/

)2//(2/222

Noting That

)2/()()( Nnxnxna

12/

0

2)2/()()12(N

n

mnN

nNWWNnxnxmX

)2/()()( Nnxnxnb

12/

02/)()12(

N

n

mnN

nN WWnbmX

If k = 2m+1 (odd number) and using the same method

Then X(2m) is N/2-point DFT for a(n)

X(2m+1) is N/2-point DFT for nNWnb )(

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Slide ٥Digital Signal Processing


28W

18W

38W

N/2-pointDFT

X(0)

X(2)

X(4)

X(6)

a(0)

a(1)

a(2)

a(3)

N/2-pointDFT

X(1)

X(3)

X(5)

X(7)

b(0)

b(1)

b(2)

b(3)

x(0)

x(1)

x(2)

x(3)

x(4)

x(5)

x(6)

x(7)

08W

-1

-1

-1

-1

Slide ٦Digital Signal Processing


28W

18W

38W

X(0)

X(4)

X(2)

X(6)

x(0)

x(1)

x(2)

x(3)

X(1)

X(5)

X(3)

X(7)

x(4)

x(5)

x(6)

x(7)

08W

N/4 point DFT

N/4 point DFT

N/4 point DFT

N/4 point DFT

28

14 WW

08

04 WW

-1

-1

-1

-1

-1

-1-1

-1

28

14 WW

08

04 WW

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Slide ٧Digital Signal Processing


X(0)

X(4)

X(2)

X(6)

X(1)

X(5)

X(3)

X(7)0

8W

08W

08W

08W

-1

-1

-1

-1

28W

18W

38W

x(0)

x(1)

x(2)

x(3)

x(4)

x(5)

x(6)

x(7)

08W

08W

28W

08W

28W

-1

-1

-1

-1

-1

-1-1

-1

Slide ٨Digital Signal Processing


Using the previous algorithm , the complex multiplications needed is only 12. While using the normal DFT would require 64 complex multiplications

In general

Complex multiplication of DFT is: N2

Complex multiplication of FFT is (N/2) log2(N)

If N = 1024

Complex multiplication of DFT is: 1,048,576

Complex multiplication of FFT is: 5,120

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Slide ٩Digital Signal Processing


Index mapping for Fast Fourier Transform

Input Data index n

Index Bits Reversal Bits

Output data indexk

0 000 000 01 001 100 42 010 010 23 011 110 64 100 001 15 101 101 56 110 011 37 111 111 7

Slide ١٠Digital Signal Processing


Example Given a sequence x(n) where x(0) = 1, x(1) = 2, x(2) = 3, x(3) = 4 and x(n) = 0 elsewhere ,find DFT for the first four points

solution

X(0)

X(2)

X(1)

X(3)

x(0) =1

x(1) =2

x(2) =3

x(3) =4

104 W

jW 14 10

4 W

104 W -26

-2

-2

4 10

-2+2j

-2-2j-1

-1

-1

-1

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Slide ١١Digital Signal Processing

Inverse Fourier Transform

1

01,.....,2,1,0)(1)(

N

k

knN NnWkX

Nnx

1

0

~

1,.....,2,1,0)(1)(N

k

knN NnWkX

Nnx

The inverse discrete Fourier transform can be found using

Which can be expressed as

where knN

knN WW

~

We can see that the difference between the inverse discrete Fourier and forward Fourier transform is the twiddled factor and the division by 1/N

is called the twiddled factor

Slide ١٢Digital Signal Processing


~2

8W

~1

8W

~3

8W

x(0)

x(4)

x(2)

x(6)

X(0)

X(1)

X(2)

X(3)

x(1)

x(5)

x(3)

x(7)

X(4)

X(5)

X(6)

X(7)

~0

8W

~0

8W

~2

8W

~0

8W

~2

8W~

08W

~0

8W

~0

8W

~0

8W

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

1/N

1/N

1/N

1/N

1/N

1/N

1/N

1/N

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Slide ١٣Digital Signal Processing


Example Given a sequence X(n) given in the previous example. Find the IFFT using decimation in frequency method

solution

x(0) = 1

x(2) = 3

x(1) = 2

x(3) = 4

X(0) =10

X(2) =-2

X(1) = -2+2j

1~

04 W

jW ~

14 1

~0

4 W

1~

04 W 12-4

12

j4

8 4

8

16X(3) = -2-2j

-1

-1

1/4

1/4

1/4

1/4-1

-1

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