DISCRETE FOURIER TRANSFORM

1. Introduction

The sampled discrete-time fourier transform (DTFT) of a finite length, discrete-time signal is

known as the discrete Fourier transform (DFT). The DFT contains a finite number of samples

equal to the number of samples N in the given signal. Computationally efficient algorithms for

implementing the DFT go by the generic name of fast Fourier transforms (FFTs). This chapter

describes the DFT and its properties, and its relationship to DTFT.

2. Definition of DFT and its Inverse

Lest us consider a discrete time signal x (n) having a finite duration, say in the range 0 ≤ n ≤N-1.

The DTFT of the signal is

N-1

X ( ω) = Σ x (n)e-jwn (1)

n-0

Let us sample X using a total of N equally spaced samples in the range : ω ∈(0,2π), so the

sampling interval is 2π That is, we sample X(ω) using the frequencies.

N

ω= ωk = 2πk , 0 ≤ k ≤N-1. N-1

Thus X (k) = Σ x (n)e-jwn (2)

n-0

N-1

X (k) = Σ x (n)e-

j2πkn (2)

n-0

N

The result is, by definition the DFT.

That is ,

Equation (0.2) is known as N-point DFT analysis equation. Fig 0.1 shows the Fourier transform

of a discrete – time signal and its DFT samples.

x(w)

0 π 2π →w

Fig.1 Sampling of X(w) to get x(k)

While working with DFT, it is customary to introduce a complex quantity

WN = e-j2π /N

Also, it is very common to represent the DFT operation

N=1

X(k) = DFT ( x(n)) = Σ x(n) WNkn, 0≤n≤N-1

n=0

The complex quantity Wn is periodic with a period equal to N. That is,

WNa+N = e-j+2π/N(a+N) = e-j2π /N n = WN

a where a is any integer.

Figs. 0.2(a) and (b) shows the sequence for 0≤n≤N-1 in the z-plane for N being even and

odd respectively.

6 5

5 7 4 6

4 0 0

3 1 3 1

2 2

(a) (b)

Fig.2 The Sequence for even N (b) The sequence for odd N.

The sequence WkN

N for 0 ≤ n ≤ N-1 lies on a circle of unit radius in the complex plane

and the phases are equally spaced, beginning at zero.

The formula given in the lemma to follow is a useful tool in deriving and analyzing

various DFT oriented results.

2.1. Lemma

N-1

Σ Wkn = N δ (k) = { N, k = 0 (3)

n-0 N 0, k ≠ 0

Proof :

N-1

Σ an

= 1 - aN

: a≠ 1

n-0 1 - a

We know that

Applying the above result to the left side of equation (3.3), we get

N-1

Σ (WkN) n = 1- Wk

NN = 1- e-j2π kN

N

n= 0 1- Wk

NN 1- e-j2π k

NN : k ≠ 0

= 1 - 1

1- e- j2π kN

N

= 0, k ≠ 0

when k = 0, the left side of equation (3.3) becomes

N-1 N-1

Σ WN0xn = Σ 1 = N

n= 0 n =0

N-1 N, k = 0

Hence, we may write Σ WN

0xn = 0, k ≠ 0

n=0

= N δ (k), 0≤ k≤ N-1

2.2 Inverse DFT

The DFT values (X(k), 0≤ k≤ N-1), uniquely define the sequence x(n) through the inverse DFT

formula (IDFT) :

N-1

x (n) = IDFT (X(k) = 1 Σ X(k) WN-kn , 0≤ k≤ N-1

N k=0

The above equation is known as the synthesis equation.

N-1 N-1 N-1

Proof : 1 Σ X(k) WN-kn = 1 Σ [Σ x(m) WN

km] = WN-kn

N k=0 N k=0 m=0

N-1 N-1 N-1

= 1 Σ x(m) [Σ WN

-(n-m)k]

N k=0 m=0

It can be shown that

N-1

Σ WN(n-m)k = N , n = m

0, n ≠ m

Hence,

N-1

1 Σ x(m) Nδ (n-m)

N

= 1 x Nx (m) ( sifting property)

N m=n

= x(n)

2.3 Periodicity of X (k) and x (n)

The N-point DFT and N-point IDFT are implicit period N. Even though x (n) and X (k) are

sequences of length – N each, they can be shown to be periodic with a period N because the

exponentials WN±kn in the defining equations of DFT and IDFT are periodic with a period N. For

this reason, x (n) and X (k) are called implicit periodic sequences. We reiterate the fact that for

finite length sequences in DFT and IDFT analysis periodicity means implicit periodicity. This

can be proved as follows :

N-1

X (k) = Σ x(n) WNkn

N-p=0

⇒ X (k+N) = Σ x(n) WN(k+N)n

Since, WNNn = e-j2πnNn = 1, we get

N-1

X (k+N) = Σ x(n) WN-kn

n=0

= X (k)

N-1

Similarly, x (n) Σ X (k) WN-kn

k=0

N-1

⇒ x (n+N) = 1 Σ X(k) WN-k(n+N)

N k=0

N-1

= 1 Σ X(k) WN-kn WN

-kn

N k=0

Since, WN-kn = e-j2π/N kN = e-j2πk = 1, we get

N-1

x(n+N) = 1 Σ X(k) WN-kn

= x (n)

Since, DFT and its inverse are both periodic with period N, it is sufficient to compute the

results for one period (0 to N-1). We want to emphasize that both x (n) and X (k) have a starting

index of zero.

A very important implication of x (n), being periodic is, if we wish to find DFT of a

periodic signal, we extract one period of the periodic signal and then compute its DFT.

Example 1 Compute the 8 – point DFt of the sequence x (n) given below :

x (n) = (1,1,1,1,0,0,0,0)

Solution

The complex basis functions, W for 0 n 7 lie on a circle of unit radius as shown in Fig. Ex.3

W86

W85 W87

W84 1.0 Re(z)

W83 W81

W82

Fig. 3 Sequence W8

0 for 0≤ n ≤ 8.

Since N = 8 we get W = e –j2π/8

Thus,

W80 = 1

W81 = e -jπ/4 = 1 – j 1

√2 √2

W82 = e -jπ/2 = – j

W83 = e -j3π/4 = 1 – j 1

√2 √2

W84 = -W8

0 = -1

By definition, the DFT of x (n) is

X ( k) = DFTI (x (n))

= W8kn

= 1+1 x W8k + W8

-2k + W83k .

= 1 + W8k + W8

2k + W83k k = 0, 1…7

X(0) = 1+1+1+1 = 4

X(1) = 1+ W81 + W8

2 + W83 = 1 – j2.414

X(2) = 1+ W82 + W8

4+ W86 = 0

X(3) = 1+ W83 + W8

6+ W81 = 1 – j0.414

X(4) = 1 + W84 + W8

0+ W84 = 0

X(5) = 1+ W85 + W8

2 + W87 = 1+j0.414

X(6) = 1+ W86 + W8

4 + W82 = 0.

X(7) = 1+ W87 + W8

6 + W85 = 1+j2.414

Please note the periodic property : WNa = WN

a+N where a is any integer.

Example 2 : Compute the DFT of the sequence defined by x (n) = (-1) n for

a. = N= 3

b. N = 4,

c. N odd,

d. N even.

Solution

X (k) = DFT (x-n)

N-1

= Σ (-1)n WNnk

n-0

N-1

Σ (-1)n [WNk] n

n=0

= 1 – (1)N for WNk ≠-1

1+ WNk

a. N = 3

X(k) = 2 = 2 0≤ k ≤ 2

1+ WNk 1+ cos (2kπ/3) – j sin ((2πk/3)

b. N = 4 X (k) = 0 for W4k ≠-1 or k≠ 2

With k = 2 we get

N-1

X(2) Σ (-1)n W42n

n=0

= 1 - W42 + W4

4 - W46

= 1 – (-1) + (-1)2– (-1)2 = 4

Hence, X (k) = 48(k-2)

c. We know that

W42n – e-j2π / N k

If N = 2k, we get WNk= -1.

Since N is odd no k exists. This means to say that WNk ≠ -1 for all k from 0 to N-1.

Therefore,

X(k) = 2 0≤ k ≤ N-1

1+ WNk

d. N even WNk = - 1, if k = N/2.

X ( k) = 0 for k ≠ N/2

With k = 2, we get

N-1

And x (N/2) = Σ [- WNk] n

n=0

N-1

=Σ [1] = N

n=0

Hence X (k) = N δ (k-N/2)

Example 3. Compute the inverse DFT of the sequence,

X(k) = (2,1+j,0,1 –j)

Solution

x (n) = IDFT (X(k))

N-1

∆ 1 Σ x(n) WN-kn , 0 ≤ n ≤N -1

N n=0

Please note that :

WN-kn = [ WN

kn]*

Since, N = 4, we get

N-1

x (n) = 1 Σ X(k) W4-kn , 0 ≤ n ≤3

4 n=0

= 1 [X(0) W4-0xn + X(1) W4

-n + X (2) W4-2n + X (3) W4

-3n]

4

= 1 [2 + (1+j) W4-n +0 + (1-j) + X (3) W4

-3n]

4

Hence, x (0) = 1 [2 + (1+j) +(1-j)] = 1

4

x (1) = 1 [2 + (1+j) W4-1 +(1-j) W4

-3] = 1

4

x (2) = 1 [2 + (1+j) W4-2 +(1-j) W4

-6] = 1

4

Because of periodicity, W4-6 = W4

-2

Hence, x (2) = 1 [2 + (1+j) (-1) + (1-j) (-1)] = 0

4

x (3) = 1 [2 + (1+j) W4-3 +(1-j) W4

-9] = 1

4

Because of periodicity, W4-9 = W4

-5 = W4-1

Hence, x (3) = 1 [2 + (1+j) W4-3 +(1-j) W4

-1]

4

= 1 [2 + (1+j) (-j) (-j) + (1-j) ] = 1

4

Hence, x(n) = (1,0,0,1)

3. Matrix Relation for Computing DFT

The defining relation for DFT of a finite length sequence x (n) is

N-1

X (k) = Σ x(n) Wnkn , 0.1, …… N-1

n=0

Let us evaluate X (k) for different values of k in the range (0, N-1) as given below :

X (0) = WN0 x (0) + WN

0 x (1) + …… + WN0 x (N-1)

X (1) = WN0 x (0) + WN

1 x (1) + …… + WN(N-1) x (N-1)

X (2) = WN0 x (0) + WN

2 x (1) + …… + WN2(N-1) x (N-1)

X (N-1) = WN0 x (0) + WN

(N-1)x (1) + ….. =+ WN(N-1) (N-1) x (N-1)

Putting the N DFT equations in N unknowns in the matrix form, we get

X = WNx

Here X and x are (N x 1) matrices, and Wn is an (N x N) square matrix called the DFT matrix.

The full matrix form is described by

X(0) WN0 WN

0 WN0 … WN

0 x(0)

X(1) WN0 WN

1 WN2 … WN

(N-1) x(1)

X(2) = WN0 WN

2 WN4 … WN

kn x(2)

: : : : : :

X(N-1) WN0 WN

N-1 WN2(N-1)… WN

(N-1)(N-1) x(N-1)

The elements WNkn of WN are called complex basis functions or twiddle factors.

Example : Compute the 4-point DFT of the sequence, x(n) = (1,2,1,0).

Solution

With N = 4, W4 = e-j2 π /4 = -j.

We know that

X = WN x

X(0) W40 W4

0 W40 W4

0 x(0)

X(1) W40 W4

1 W42 W4

3 x(1)

X(2) = W40 W4

2 W44 W4

6 x(2)

X(3) W40 W4

3 W46 W4

0 x(3)

Exploiting the periodic property WN0 = WN

n+N where a is any integer the above matrix relation

becomes.

X(0) W40 W4

0 W40 W4

0 x(0)

X(1) W40 W4

1 W42 W43 x(1)

X(2) = W40 W4

2 W40 W4

2 x(2)

X(3) W40 W4

3 W42 W4

1 x(3)

X(0) 1 1 1 1 1

X(1) 1 -j -1 j 2

X(2) = 1 -1 1 -1 1

X(3) 1 j -1 -j 0

Hence,

X(k) = (4, -j2, 0, j2)

4. Matrix Relation for Computing IDFT

We know that x = WN1X

Premultiplying both the sides of the above equation by we get

WN-1-X = WN

-1 WNx

WN-1X = x

Or x = WN-1X

In the above equation (.8) x = WN-1 is called IDFT matrix.

The defining equation for finding IDFT of a sequence X (k) is

N-1

X (n) = 1 Σ X(k) W , 0≤ n ≤ N-1

N k=0

N-1

= 1 X Σ (k) [WNkn ]*

N k=0

The first set of N IDFT equation in N unknowns may be expressed in the matrix form as

x = 1 W*NX

N

Where W*N denotes the complex conjugate of WN. Comparision of equation (3.8) and (3.9)

leads us to conclude that

WN-1 = 1/N W*N

This very important result shows that W-1N requires only conjugation of Wn multiplied by 1/N.

an obvious computational advantage. The matrix relations (.7) and (.9) together define DFT as a

linear transformation.

5. Using the DFT to Find the IDFT

We know that

N-1

x* (n) = 1 Σ X(k) WN-kn , n= 0,1 …. N-1

N k=0

Taking complex conjugates on both the sides of the above equation, we get

N-1

x*(n) = 1 Σ X(k) WN-kn *

N k=0

N-1

⇒ x*(n) = 1 Σ X*(k) WN-kn * (10)

N k =0

The right – hand side of equation (3.10) is recognized as the DFT of X* (k), so we can rewrite

equation (3.10) as follows :

x*(n) = 1 DFT (X*(k))

N

Taking complex conjugates on both the sides of equation (11), we get

x*(n) = 1 [DFT (X*(k))]

N

The above results suggests the DFT algorithm itself can be used to find IDFT. In practice, this is

indeed what is done.

6. Properties of DFT

In the following section, we shall discuss some of the important properties of the DFt. They are

strikingly similar to other frequency domain transforms, but must always be used in keeping with

implied periodicity for both DFT and IDFT in time and frequency domains.

6.1 Linearity

DFT (ax1(n) + bx2(n)j = aX1(k) + bX2(k), k = 0, N =1

If X1(k) and X2(k) are the DFTs of the sequence x1 (n) and x2 (n), respectively, both of lengths

N.

Proof :

N-1

We know that DFT [x(n)] = Σ x(n) WNkn

n=0

Letting x (n) = ax1(n) + bx2(n) we get

N-1

DFT = ax1(n) + bx2(n) = Σ (ax1(n) +bx2 (n)] WNkn

n=0

N-1

= a Σ x1(n) WNkn + b Σ x1(n) WN

kn

n=0

= aX1 (k) + bX2(k), 0 ≤ k≤ N-1

Sometimes we represent the linearity property as given below

DFT

ax1 (n) + bx2(n) aX1 (k) + bX2(k)

Example : Find the 4- point DFT of the sequence

X(n) = cos ( π n) + sin ( π n)

4 4

Use linearity property.

Solution

Given N = 4,

WN = e j2π /n ⇒ W = e jπ /2

We know that, W40 = 1

Hence, W40 = 1

W41 = e = -j

W43 = e = -1

x1 (n) = cos (π /4 n )

Let

x2 (n) = sin (π /4 n )

and

Then, the values of x (n) and x2 (n) for 0 <n <3 tabulated below :

N x1 (n) = cos (π /4 n ) x2 (n) = sin (π /4 n )

0 1 0

1 1 1

√2 √2

2 0 1

3 -1 1

√2 √2

X1 (k) = DFT (x1 (n))

3

∆ Σ x1 (n) W4kn, k = 0,1,2,3

n=0

⇒ X1 (k) = 1 + 1 W4k +0 – 1 W4

3k

√2 √2

Hence, X1 (0) = 1 + 1 – 1 = 1

√2 √2

X1 (1) = 1 + 1 W41 – 1 W4

3 = 1 –j1.414

√2 √2

X1 (2) = 1 + 1 W42 – 1 W4

6

√2 √2

= 1 + 1 W42 – 1 W2

4 = 1

√2 √2

X1 (3) = 1 + 1 W43 – 1 W4

9

√2 √2

= 1 + 1 W43 – 1 W4

1

√2 √2

Similarly, X2 (k) = DFT (x2 (n))

3

∆ Σ x2 (n) W4kn

n=0 ⇒ X2 (k) = 1 W4

k + W42k + 1 W4

3k

√2 √2

Hence, X2 (0) = 1 + 1 + 1 = 2.414

√2 √2

X2 (1) = 1 W41 + W4

2 + 1 W43 = -1

√2 √2

X2 (2) = 1 W42 + W4

0 + 1 W49

√2 √2

= 1 W43 + W4

6 + 1 W49 = -0.414

√2 √2

X2 (3) = 1 W43 + W4

6 + 1 W49

√2 √2

= 1 W43 + W4

2 + 1 W41 = -1

√2 √2

Finally, applying the linearity property, we get

X (k) = DFT ( x1(n) + x2(n))

= X1(k) + X2(k)

= ( X1(0) + X2(0), X1(1) + X2(1). X1(2), X2(2), X1(3) + X2 (3) )

= (3.414. – j1.414.0.586. j1.414)

↑ k=0

It may be noted that the arrow, ↑explicitly represents the position index of k = 0 or n = 0 of a

given sequence. The absence of this arrow also implicitly means that the first element in a

sequence always has the index k = 0 or n = 0.

Example Compute DFT (x(n)) of the sequence given below using the linearity property.

x (n) = cosh an, 0 ≤ n ≤ N-1

Solution

Given x(n) = cosh an, 0 ≤ n ≤ N-1

Then the N point DFT of the sequence x (n) is

X (k) = DFT [x(n)] = DFT ( cosh an)

= DFT 1 ean + 1 e-an

2 2

Applying linearity property, we get

X (k) = 1 DFT [e an ] + 1 DFT [e -an ], 0 ≤ n ≤ N-1

2 2

We know from Example 3.5, that

DFT (b N ) = b N -1 , 0 ≤ k ≤ N-1

b WNk -1

Hence, X (k) = 1 e a(N) - 1 + e a(N) - 1

2 e a(N) WNk -1 e -a WN

k - 1

= WN-kn ( e a(N-1) + e -a(N-1) - e -a + e –a ] - e –aN- e –aN +2

2[1- WNk (ea – ea ) + WN

k ]

= 1 – cosh Na + WNk [cosh ( N-1)a – cosh a] , 0 ≤ k ≤ N-1

1- 2 WNk cosha + WN

k

6.2 Circular time shift

If DFT [x(n)] = X (k).

Then DFt [x(n-m))] = X(k), 0 ≤ n ≤ N-1

Proof :

N=1

x (n) = 1 Σ [ X (k) WN-kn

N k=0

N=1

x (n-m) = 1 Σ [ X (k) WNk(n-m)

N k=0

Since, the time shift is circular, we can write the above equation as

N=1

x (n-m) = 1 Σ [ X (k) WNkm ] WN

-kn

N k=0

⇒ x (n-m)N = IDFt [ X(k) WNkm]

or DFT [x(n-m) N ] = WNkm X (k)

In terms of the transform pair, we can write the above equation is

DFT

x (n-m)N ↔ WNkm X (k)

Example Find the 4- point DFT of the sequence, x(n) = (1, -1, 1, -1) Also, using time shift

property, find the DFT of the sequence, y(n) = x (n-2)4.

Solution

Given N = 4

We know that

W40= 1, W4

1= -j

W42 = -1, W4

3 = j

Hence, X (k) = DFT (x (n) )

3

= Σ x(n) W4kn , 0 ≤ k≤ 3

n=0

= 1 W40k x – 1 x W4

k +1 x W42k - 1 x W4

3k

= 1- W41+W4

2k - W43k

X(0) = 1 -1 +1 -1 = 0

X(1) = 1 - W41+ W4

2- W43 = 0

X(2) = 1 - W42 + W4

4 - W46

= 1- W42+ W4

0 - W42 = 4

X(3) = 1 - W43 + W4

6- W49

= 1- W43 + W4

2- W41= 0

Given y (n) = x(n-2) 4

Applying circular time shift property, we get

Y(k) = W42k X (k), k = 0,1,2,3

Y(0) = W40 X(0) = 0

Y(1) = W42 X (1) = 0

Y(2) = W44 = W4

0 x(2) = 4

Y(3) = W46x (3) = W4

2 x (3) = 0

Hence, Y(k) = (0,0,4,0)

↑ k-0

Example Suppose x(n) is a sequence defined on 0 -7 only as ( 0,1,2,3,4,5,6,7),

a. Illustrate x(n-2) is

b. If DFT (x(n)) = X (k), what is the DFT (x (n-2)s)

Solution

a. Given

To generate x (n-2) move the last 2 samples of x (n) to the beginning.

That is, x (n-2) = (6,7,0,1,2,3,4,5)

s(n-2)g

7

5

6 4

2 3

1

0 1 2 3 4 5 6 7 n

It should be noted that x (n-2) is implicity periodic with a period N = 8.

b. Let y(n)=x(n-2) 8

Applying circular time shift property, we get

Y(k) = W32k X(k)

Example : Let X (k) donate a 6-point DFT of a length – 6 real sequence, x(n). The sequence is

shown in Fig. Ex 3.17, without computing the IDFT, determine the length -6 sequence, y(n)

whose 6-point DFT is given by, Y (k) = W32k X(k)

1 2 3 x(n)

0

4 5

-1

Fig. Sequence x(n)

Solution

We may write

W32k = e-j

2π/3nx2k

= e-j2π/3nx2k

Hence, W32k = W6

4k

It is given in the problem that

Y(k) = W3 j2k X(k)

Y(k) = W64k X(k)

We know that DFT (x(n=m) N = WNmk X(k)

IDFT WNmk X(k) = x(n-m) N

Hence, y(n) = x(n-4)6

Since, x(n) = (1,-1,2,3,0,0)

We get x(n-4) is by moving the last 4 samples of x(n) to the beginning

y(n) = x(n-4) 6

= (2,3,0,0,1-1)

Circular frequency shift ( Multiplication by exponential in time-domain)

If DFT (x(n)) = X (k), then DFT = X (k-1) N.

Proof :

N-1

X (k) = DFT (x (n)) = Σ x(n) WNkn, 0 ≤ k ≤ N-1

n=0

N-1

⇒ X(k-1) = Σx(n) WN(k-1)n

Since, the shift is frequency is circular, we may write the above equation as

N-1

X(k-1)8 = Σ { x(n) WN-ln } WN

kn

n=0

Hence, DFT {x(n) WN-ln } = X (k=1)N

Example Compute the 4-point DFT of the sequence x (n) = (1,0,1,0), Also, find y (n) if Y (k) =

X (k-2) 4.

Solution

Given N = 4.

Also W40 = 1, W4

l =-j, W42 =-1, W4

3 =j,

The DFT of the sequence, x (n) is

3

X(k) = Σ x(n) W4kn , 0 ≤ k≤ 3

= 1x W40k + 0+1 x W4

2k =0

= 1 +W42k

X(0) = 1+1= 2

X(1) = 1+W24

= 0

X(2) = 1+W04

= 2

X(3) = 1+W24

= 0

X(0) = 1+1= 2

X(k) = X(k-2))4

Given Y(k) = X(k-2) 4

We know that, DFT (WN-ln x (n)) = X(k-l)N

That is, y(n) = WN-ln x (n)

DFT Y(k) = X(k-1)N

Hence, y(n) = W4-2n x (n)

⇒ y(0) = W4-0 x (0) = 1

y(1) = W4-2 x (1) = 0

y (2) = W4-4 x (2)

= W4-0 x (2) = 1x 1 =1

y(3) = W4-6 x (3) = W4

-2 x (3) = 0

That is, y(n) = (1,0,1,0)

↑ n=0

Circular convolution

Unlike DFT convolution in DFT in circular consider two sequence x(n) and y(n) the

circular convolution of x(n) and y(n) in given by

Let f (n) = x (n) x y(n)

N-1

x(n+N) = 1 Σ x(n-m) Nh(m) 0 ≤ n ≤ N-1

m=0

N-1

= Σ x(m) h(n-m) n

n=0

Point to be noted here in that x(n) and y(n) should be of same length

Example :

Let x (n) = 1,1,1 y (n) = 1,-2,2

Retain x (n) as it is and circularly fold y (n) i.e. y(n) = 1,2,-2.

N x(m) y(n-m)N f(n)

0 1,1,1 1,2,-2 1 x 1+1x2+1x-2 = 1

1 1,1,1 -2,1,2 1 x -2 + 1x1 + 1x2 = 1

2 1,1,1 2,-2,1 1 x 2, +1x-2 + 1x1 = 1

∴h (n) = 1,1,1

Summary

N-1

1) x (k) = Σ x(n) e-jwn

n=0

N-1

= Σ x(n) Wnkn

n=0

N-1

2) x (n) = 1 Σ x(n) Wnkn 0≤ n≤ N-1

N k=0

3) Periodicity of Wn

kn

W6

8

W5

8 W

7

8

W4

8 W

0

8 =W

4

6 W

α

N = W

N

α+N

-W7

8= W

3

8 W

1

8 = -W

5

8

W2

8

= W8

8

4) DFT { ax1

(n) + bx2

(n) } = ax1

(k) + bx2

(k)

5) DFT {x(n-m)N

} = WN

mk x (k)

6) DFT {Wn

-lnx(n) } = x(k-l)

N

7) X (k) = Xy (N-k)

8) DFT {x(N-n)} = x(N-k)

9) DFT { xe (n) } = 1 DFT {x(n) + 1 DFT { x(-n)N)}

2

= 1 x (k) + 1 x (-k) N

2 2

10) DFT { x1 (n) x2(n) } = 1 x 1(k) x2 (k)

N