Discrete Fourier Transform (DFT)

Recall the DTFT: X(ω) =

∞ X

x(n)e−jωn.

n=−∞

DTFT is not suitable for DSP applications because • In DSP, we are able to compute the spectrum only at specific discrete values of ω, • Any signal in any DSP application can be measured only in a finite number of points. A finite signal measured at N points:  n < 0,  0, y(n), 0 ≤ n ≤ (N − 1), x(n) =  0, n ≥ N, where y(n) are the measurements taken at N points. EE 524, Fall 2004, # 5

1

Sample the spectrum X(ω) in frequency so that X(k) = X(k∆ω), X(k) =

N −1 X

∆ω = −j2π kn N

x(n)e

2π N

=⇒

DFT.

n=0

The inverse DFT is given by: N −1 kn 1 X x(n) = X(k)ej2π N . N k=0

1 x(n) = N =

N −1 X

(N −1 X

k=0

m=0

N −1 X

( x(m)

m=0

|

EE 524, Fall 2004, # 5

) −j2π km N

x(m)e

1 N

N −1 X

−j2π

e

j2π kn N

e

k(m−n) N

) = x(n).

k=0

{z

δ(m−n)

}

2

The DFT pair:

X(k) =

N −1 X

kn

x(n)e−j2π N

analysis

n=0

x(n) =

N −1 1 X j2π kn X(k)e N N

synthesis.

k=0

Alternative formulation:

X(k) =

N −1 X

x(n)W kn

2π

←− W = e−j N

n=0

x(n) =

N −1 1 X X(k)W −kn. N k=0

EE 524, Fall 2004, # 5

3

EE 524, Fall 2004, # 5

4

Periodicity of DFT Spectrum

X(k + N ) =

N −1 X

−j2π

x(n)e

(k+N )n N

n=0 N −1 X

=

! −j2π kn N

x(n)e

e−j2πn

n=0

= X(k)e−j2πn = X(k) =⇒ the DFT spectrum is periodic with period N (which is expected, since the DTFT spectrum is periodic as well, but with period 2π). Example: DFT of a rectangular pulse:  x(n) =

X(k) =

1, 0 ≤ n ≤ (N − 1), 0, otherwise.

N −1 X

kn

e−j2π N = N δ(k) =⇒

n=0

the rectangular pulse is “interpreted” by the DFT as a spectral line at frequency ω = 0.

EE 524, Fall 2004, # 5

5

DFT and DTFT of a rectangular pulse (N=5)

EE 524, Fall 2004, # 5

6

Zero Padding

What happens with the DFT of this rectangular pulse if we increase N by zero padding: {y(n)} = {x(0), . . . , x(M − 1), 0, . . . , 0} }, | 0,{z N −M positions where x(0) = · · · = x(M − 1) = 1. Hence, DFT is

Y (k) =

N −1 X

−j2π kn N

y(n)e

=

M −1 X

n=0

=

EE 524, Fall 2004, # 5

sin(π kM N ) sin(π Nk )

−j2π kn N

y(n)e

n=0 −jπ

e

k(M −1) N

.

7

DFT and DTFT of a Rectangular Pulse with Zero Padding (N = 10, M = 5)

Remarks: • Zero padding of analyzed sequence “approximating” its DTFT better,

results

in

• Zero padding cannot improve the resolution of spectral components, because the resolution is “proportional” to 1/M rather than 1/N , • Zero padding is very important for fast DFT implementation (FFT). EE 524, Fall 2004, # 5

8

Matrix Formulation of DFT

Introduce the N × 1 vectors 

x(0)  x(1) x= ..  x(N − 1)

  , 





X(0)   X(1)  . X= ..  X(N − 1)

and the N × N matrix     W=  

0

0

0

W W W W0 W1 W2 W0 W2 W4 .. .. .. W 0 W N −1 W 2(N −1)

0

··· W · · · W N −1 · · · W 2(N −1) .. .. (N −1)2 ··· W

    .  

DFT in a matrix form: X = Wx. Result: Inverse DFT is given by x= EE 524, Fall 2004, # 5

1 H W X, N 9

which follows easily by checking W H W = WW H = N I, where I denotes the identity matrix. Hermitian transpose: xH = (xT )∗ = [x(1)∗, x(2)∗, . . . , x(N )∗]. Also, “∗” denotes complex conjugation. Frequency Interval/Resolution: DFT’s frequency resolution Fres ∼

1 NT

[Hz]

and covered frequency interval ∆F = N ∆Fres

1 = = Fs T

[Hz].

Frequency resolution is determined only by the length of the observation interval, whereas the frequency interval is determined by the length of sampling interval. Thus • Increase sampling rate =⇒ expand frequency interval, • Increase observation time =⇒ improve frequency resolution. Question: Does zero padding alter the frequency resolution? EE 524, Fall 2004, # 5

10

Answer: No, because resolution is determined by the length of observation interval, and zero padding does not increase this length.

Example (DFT Resolution): Two complex exponentials with two close frequencies F1 = 10 Hz and F2 = 12 Hz sampled with the sampling interval T = 0.02 seconds. Consider various data lengths N = 10, 15, 30, 100 with zero padding to 512 points.

DFT with N = 10 and zero padding to 512 points. Not resolved: F2 − F1 = 2 Hz < 1/(N T ) = 5 Hz. EE 524, Fall 2004, # 5

11

DFT with N = 15 and zero padding to 512 points. Not resolved: F2 − F1 = 2 Hz < 1/(N T ) ≈ 3.3 Hz.

DFT with N = 30 and zero padding to 512 points. Resolved: F2 − F1 = 2 Hz > 1/(N T ) ≈ 1.7 Hz. EE 524, Fall 2004, # 5

12

DFT with N = 100 and zero padding to 512 points. Resolved: F2 − F1 = 2 Hz > 1/(N T ) = 0.5 Hz.

EE 524, Fall 2004, # 5

13

DFT Interpretation Using Discrete Fourier Series

Construct a periodic sequence by periodic repetition of x(n) every N samples: {e x(n)} = {. . . , x(0), . . . , x(N − 1), x(0), . . . , x(N − 1), . . .} | {z } | {z } {x(n)}

{x(n)}

The discrete version of the Fourier Series can be written as x e(n) =

X

j2π kn N

Xk e

k

1 X e 1 X e j2π kn = X(k)e N = X(k)W −kn, N N k

k

e where X(k) = N Xk . Note that, for integer values of m, we have W

−kn

j2π kn N

=e

j2π

=e

(k+mN )n N

= W −(k+mN )n.

As a result, the summation in the Discrete Fourier Series (DFS) should contain only N terms: N −1 1 X e j2π kn x e(n) = X(k)e N N

DFS.

k=0

EE 524, Fall 2004, # 5

14

Inverse DFS The DFS coefficients are given by

e X(k) =

N −1 X

kn

x e(n)e−j2π N

inverse DFS.

n=0

Proof. N −1 X

−j2π kn N

x e(n)e

n=0

  N −1  N −1  X 1 X e j2π pn −j2π kn N N X(p)e = e N  n=0 p=0 N −1 X

=

( e X(p)

p=0

1 N

N −1 X

j2π

e

(p−k)n N

) e = X(k).

n=0

|

{z

δ(p−k)

}

2 The DFS coefficients are given by

e X(k) =

N −1 X

kn

x e(n)e−j2π N

analysis,

n=0

x e(n) =

N −1 1 X e j2π kn X(k)e N N

synthesis.

k=0

EE 524, Fall 2004, # 5

15

• DFS and DFT pairs are identical, except that − DFT is applied to finite sequence x(n), − DFS is applied to periodic sequence x e(n). • Conventional (continuous-time) FS vs. DFS − CFS represents a continuous periodic signal using an infinite number of complex exponentials, whereas − DFS represents a discrete periodic signal using a finite number of complex exponentials.

EE 524, Fall 2004, # 5

16

DFT: Properties

Linearity Circular shift of a sequence: if X(k) = DFT {x(n)} then km

X(k)e−j2π N = DFT {x((n − m) mod N )} Also if x(n) = DFT −1{X(k)} then km

x((n − m) mod N ) = DFT −1{X(k)e−j2π N } where the operation mod N denotes the periodic extension x e(n) of the signal x(n): x e(n) = x(n mod N ).

EE 524, Fall 2004, # 5

17

DFT: Circular Shift

N −1 X

x((n − m)modN )W kn

n=0

= W km

N −1 X

x((n − m)modN )W k(n−m)

n=0 EE 524, Fall 2004, # 5

18

= W km

N −1 X

x((n − m)modN )W k(n−m)modN

n=0

= W kmX(k), where we use the facts that W k(lmodN ) = W kl and that the order of summation in DFT does not change its result. Similarly, if X(k) = DFT {x(n)}, then j2π mn N

X((k − m)modN ) = DFT {x(n)e

}.

DFT: Parseval’s Theorem

N −1 X

x(n)y ∗(n) =

n=0

N −1 1 X X(k)Y∗(k) N k=0

Using the matrix formulation of the DFT, we obtain yH x = = EE 524, Fall 2004, # 5



H 



1 H 1 H W Y W Y N N 1 H 1 H H Y W W } X = Y X. | {z N2 N NI 19

DFT: Circular Convolution If X(k) = DFT {x(n)} and Y (k) = DFT {y(n)}, then X(k)Y (k) = DFT {{x(n)} ~ {y(n)}} Here, ~ stands for circular convolution defined by {x(n)} ~ {y(n)} =

N −1 X

x(m)y((n − m) mod N ).

m=0

DFT {{x(n)} ~ {y(n)}} N −1 h i X PN −1 kn = x(m)y((n − m) mod N ) W m=0 | {z } n=0 {x(n)}~{y(n)}

=

N −1 h X

PN −1 n=0

m=0 |

= Y (k)

N −1 X

|m=0

EE 524, Fall 2004, # 5

i

y((n − m) mod N )W kn x(m) {z } Y (k)W km

x(m)W km = X(k)Y (k). {z

X(k)

}

20