# DSP 3rd LEC15 63c77ab7b7f1f0f39e023c1962f5123a

```FFT
Decimation in Time
1
Decimation in Time
In this method we split x (n) into the even indexed x (2m) and the odd indexed x
(2m + 1) each N/2 long. DFT can be expressed as:
X (k ) 
N / 2 1

x(2m)W
m 0
X (k ) 
2 mk
N
N / 2 1

m 0
x(2m)W
mk
N /2

N / 2 1
W
 x(2m  1)W
m 0
k ( 2 m 1)
N
N / 2 1
k
N
 x(2m  1)W
m 0
km
N /2
K = 0,1,2,3,4,……………..,N-1
We have used the fact
WN2 mk  e  j 2 mk2 / N  e  j 2mk /( N / 2)  WNmk/ 2
(1)
Decimation in Time
Define new functions as
G (K) is a DFT for the even samples of x (n), or
G (k ) 
N / 2 1
 x(2m)W
m 0
k  0,1,2,...., N / 2  1
mk
N /2
H (K) is a DFT for the odd samples of x(n), or
H (k ) 
N / 2 1
 x(2m  1)W
m 0
km
N /2
k  0,1,2,3,..., N / 2  1
Then equation (1) for the first N/2 points can be expressed as
X (k )  G (k )  WNk H (k )
k  0,1,2,...., N / 2  1
Decimation in Time
But
WN( N / 2 k )  WNk
G (k )  G ( N / 2  k ),
k  0,1,2,..., N / 2  1
H (k )  H (k  N / 2)
k  0,1,2,...., N / 2  1
Then
X (k  N / 2)  G (k )  WNk H (k )
k  N / 2, N / 2  1,...., N
This can be expressed in the following diagram
Decimation in Time
G(0)
x(0)
x(2)
x(4)
G(1)
4-point
DFT
x(7)
G(2)
H(0)
x(1)
x(5)
X(1)
X(2)
G(3)
x(6)
x(3)
X(0)
4-point
DFT
X(3)
W80
H(1)
W81
H(2)
W82
H(3)
W83
-1
X(4)
X(5)
-1
-1
-1
X(6)
X(7)
Decimation in Time
x(0)
x(4)
x(2)
x(6)
x(1)
x(5)
x(3)
x(7)
X(0)
2-point
DFT
2-point
DFT
X(1)
W80
X(2)
-1
W82
X(3)
-1
W80
2-point
DFT
2-point
DFT
-1
W81
X(5)
-1
W82
W80
-1
W83
W82
-1
X(4)
-1
-1
X(6)
X(7)
Decimation in Time
X(0)
x(0)
x(4)
W80
-1
W80
x(2)
x(6)
X(1)
W82
W80
W80
-1
W81
W80
-1
x(3)
x(7)
X(3)
-1
-1
x(1)
x(5)
X(2)
-1
-1
W80
W82
-1
X(5)
-1
W82
W80
-1
W83
X(4)
-1
-1
X(6)
X(7)
Decimation in Time
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
4
x(0)=1
x(2)=3
W40  1
-2
-1
6
x(1)=2
x(3)=4
X(0 )= 10
W40  1
-2
-1
X(1)= -2+2j
W40  1
-1
W41   j
-1
X(2)= -2
X(3)= -2-2j
Inverse Fourier Transform
The inverse discrete Fourier can be calculated using the same
method but after changing the variable WN and multiplying the
result by 1/N
Example Given a sequence X(n) given in the previous example.
Find the IFFT using decimation in time method
Solution
X(0) =10
8
4
12
8
1/4
x(1) = 2
12
1/4
x(2) = 3
16
1/4
x(3) = 4
1/4
x(0) = 1
~
X(2) =-2
W40  1
-1
X(1) = -2+2j
~
X(3) = -2-2j
~
-4
j4
W40  1
-1
W40  1
-1
~
W41  j
-1
```