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 = 2π k , 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 - j2π kn (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.
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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,