dtft, dft, fft.pdf

Connexions module: m10247 1

Discrete-Time Fourier Transform

(DTFT)

Don Johnson

This work is produced by The Connexions Project and licensed under the

Creative Commons Attribution License

Abstract

Discussion of Discrete-time Fourier Transforms. Topics include comparison with analog transforms

and discussion of Parseval's theorem.

The Fourier transform of the discrete-time signal s (n) is dened to be

S(ei2pif

)=

n=

(s (n) e(i2pifn)

)(1)

Frequency here has no units. As should be expected, this denition is linear, with the transform of a

sum of signals equaling the sum of their transforms. Real-valued signals have conjugate-symmetric spectra:

S(e(i2pif)

)= S (ej2pif ).Exercise 1 (Solution on p. 7.)

A special property of the discrete-time Fourier transform is that it is periodic with period one:

S(ei2pi(f+1)

)= S

(ei2pif

). Derive this property from the denition of the DTFT.

Because of this periodicity, we need only plot the spectrum over one period to understand completely the

spectrum's structure; typically, we plot the spectrum over the frequency range

[ ( 12) , 12]. When the signalis real-valued, we can further simplify our plotting chores by showing the spectrum only over

[0, 12]; the

spectrum at negative frequencies can be derived from positive-frequency spectral values.

When we obtain the discrete-time signal via sampling an analog signal, the Nyquist frequency

1

corre-

sponds to the discrete-time frequency

12 . To show this, note that a sinusoid having a frequency equal to the

Nyquist frequency

12Tshas a sampled waveform that equals

cos(

2pi1

2TsnTs

)= cos (pin) = (1)n

The exponential in the DTFT at frequency

12 equals e

( i2pin2 ) = e(ipin) = (1)n, meaning that discrete-timefrequency equals analog frequency multiplied by the sampling interval

fD = fATs (2)

Version 2.31: Jul 6, 2009 5:25 pm GMT-5

http://creativecommons.org/licenses/by/1.0

1

"The Sampling Theorem"

http://cnx.org/content/m10247/2.31/


fD and fA represent discrete-time and analog frequency variables, respectively. The aliasing gure2

pro-

vides another way of deriving this result. As the duration of each pulse in the periodic sampling signal

pTs (t) narrows, the amplitudes of the signal's spectral repetitions, which are governed by the Fourier seriescoecients

3

of pTs (t), become increasingly equal. Examination of the periodic pulse signal4

reveals that as

decreases, the value of c0, the largest Fourier coecient, decreases to zero: |c0| = ATs . Thus, to maintaina mathematically viable Sampling Theorem, the amplitude A must increase as 1 , becoming innitely largeas the pulse duration decreases. Practical systems use a small value of , say 0.1 Ts and use ampliers torescale the signal. Thus, the sampled signal's spectrum becomes periodic with period

1Ts. Thus, the Nyquist

frequency

12Tscorresponds to the frequency

12 .

Example 1

Let's compute the discrete-time Fourier transform of the exponentially decaying sequence s (n) =anu (n), where u (n) is the unit-step sequence. Simply plugging the signal's expression into theFourier transform formula,

S(ei2pif

)=

n=

(anu (n) e(i2pifn)

)=

n=0

((ae(i2pif)

)n) (3)This sum is a special case of the geometric series.

n=0

(n) = , || < 1 :(

11

)(4)

Thus, as long as |a| < 1, we have our Fourier transform.

S(ei2pif

)=

11 ae(i2pif) (5)

Using Euler's relation, we can express the magnitude and phase of this spectrum.

|S (ei2pif) | = 1(1 acos (2pif))2 + a2sin2 (2pif)(6)

(S(ei2pif

))=

(tan1

(asin (2pif)

1 acos (2pif)))(7)

No matter what value of a we choose, the above formulae clearly demonstrate the periodic natureof the spectra of discrete-time signals. Figure 1 (Spectrum of exponential signal) shows indeed that

the spectrum is a periodic function. We need only consider the spectrum between ( 12) and 12 tounambiguously dene it. When a > 0, we have a lowpass spectrumthe spectrum diminishes asfrequency increases from 0 to

12with increasing a leading to a greater low frequency content; for

a < 0, we have a highpass spectrum (Figure 2 (Spectra of exponential signals)).

2

"The Sampling Theorem", Figure 2: aliasing

3

"Complex Fourier Series", (10)

4

"Complex Fourier Series", Figure 1



Spectrum of exponential signal

-2 -1 0 1 2

1

2

f

|S(ej2pif)|

-2 -1 1 2

-45

45

f

S(ej2pif)

Figure 1: The spectrum of the exponential signal (a = 0.5) is shown over the frequency range [-2, 2],clearly demonstrating the periodicity of all discrete-time spectra. The angle has units of degrees.



Spectra of exponential signals

f

a = 0.9

a = 0.5

a = 0.5

Spec

tral M

agni

tude

(dB)

-10

0

10

20

0.5

a = 0.9a = 0.5

a = 0.5

Angl

e (de

grees

)

f

-90

-45

0

45

90

0.5

Figure 2: The spectra of several exponential signals are shown. What is the apparent relationship

between the spectra for a = 0.5 and a = 0.5?

Example 2

Analogous to the analog pulse signal, let's nd the spectrum of the length-N pulse sequence.

s (n) =

1 if 0 n N 10 otherwise (8)The Fourier transform of this sequence has the form of a truncated geometric series.

S(ei2pif

)=N1n=0

(e(i2pifn)

)(9)

For the so-called nite geometric series, we know that

N+n01n=n0

(n) = n01 N1 (10)

for all values of .

Exercise 2 (Solution on p. 7.)

Derive this formula for the nite geometric series sum. The "trick" is to consider the dierence

between the series' sum and the sum of the series multiplied by .

Applying this result yields (Figure 3 (Spectrum of length-ten pulse).)

S(ei2pif

)= 1e

(i2pifN)1e(i2pif)

= e(ipif(N1)) sin(pifN)sin(pif)(11)



The ratio of sine functions has the generic form of

sin(Nx)sin(x) , which is known as the discrete-time sinc

functiondsinc (x). Thus, our transform can be concisely expressed as S(ei2pif

)= e(ipif(N1))dsinc (pif).The discrete-time pulse's spectrum contains many ripples, the number of which increase with N , the pulse'sduration.

Spectrum of length-ten pulse

Figure 3: The spectrum of a length-ten pulse is shown. Can you explain the rather complicated

appearance of the phase?

The inverse discrete-time Fourier transform is easily derived from the following relationship:

12

( 12 )e(i2pifm)ei2pifndf =

1 if m = n0 if m 6= n= (m n)(12)

Therefore, we nd that 12

( 12 )S(ei2pif

)ei2pifndf =

12

( 12 )m

(s (m) e(i2pifm)ei2pifn

)df

=m

(s (m)

12

( 12 )e((i2pif))(mn)df

)= s (n)

(13)

The Fourier transform pairs in discrete-time are

S(ei2pif

)=n=

(s (n) e(i2pifn)

)s (n) =

12

( 12 )S(ei2pif

)ei2pifndf(14)



The properties of the discrete-time Fourier transform mirror those of the analog Fourier transform. The

DTFT properties table

5

shows similarities and dierences. One important common property is Parseval's

Theorem. n=

((|s (n) |)2

)= 1

2

( 12 )

(|S (ei2pif) |)2df (15)To show this important property, we simply substitute the Fourier transform expression into the frequency-

domain expression for power. 12

( 12 )(|S (ei2pif) |)2df = 12( 12 ) (n (s (n) e(i2pifn)))m (s (n)ei2pifm) df

=

(n,m)

(s (n) s (n)

12

( 12 )ei2pif(mn)df

)(16)

Using the orthogonality relation (12), the integral equals (m n), where (n) is the unit sample6. Thus,the double sum collapses into a single sum because nonzero values occur only when n = m, giving Parseval'sTheorem as a result. We term

n

(s2 (n)

)the energy in the discrete-time signal s (n) in spite of the fact thatdiscrete-time signals don't consume (or produce for that matter) energy. This terminology is a carry-over

from the analog world.


Suppose we obtained our discrete-time signal from values of the product s (t) pTs (t), where theduration of the component pulses in pTs (t) is . How is the discrete-time signal energy related tothe total energy contained in s (t)? Assume the signal is bandlimited and that the sampling ratewas chosen appropriate to the Sampling Theorem's conditions.

5

"Discrete-Time Fourier Transform Properties"

6

"Discrete-Time Signals and Systems", Figure 2: Unit sample



Solutions to Exercises in this Module

Solution to Exercise 1 (p. 1)

S(ei2pi(f+1)

)=

n=

(s (n) e(i2pi(f+1)n)

)=

n=

(e(i2pin)s (n) e(i2pifn)

)=

n=

(s (n) e(i2pifn)

)= S

(ei2pif

) (17)Solution to Exercise 2 (p. 4)

N+n01n=n0

(n)N+n01n=n0

(n) = N+n0 n0

which, after manipulation, yields the geometric sum formula.


If the sampling frequency exceeds the Nyquist frequency, the spectrum of the samples equals the analog

spectrum, but over the normalized analog frequency fT . Thus, the energy in the sampled signal equals theoriginal signal's energy multiplied by T .



Discrete Fourier Transform

Don Johnson



Abstract

The Fourier transform can be computed in discrete-time despite the complications caused by a nite

signal and continuous frequency.

The discrete-time Fourier transform (and the continuous-time transform as well) can be evaluated when

we have an analytic expression for the signal. Suppose we just have a signal, such as the speech signal used

in the previous chapter. You might be curious; how did we compute a spectrogram such as the one shown

in the speech signal example

1

? The big dierence between the continuous-time and discrete-time worlds is

that we can exactly calculate spectra in discrete-time. For analog-signal spectra, use must build special

devices, which turn out in most cases to consist of A/D converters and discrete-time computations. Certainly

discrete-time spectral analysis is more exible than in continuous-time.

The formula for the DTFT

2

is a sum, which conceptually can be easily computed save for two issues.

Signal duration. The sum extends over the signal's duration, which must be nite to compute thesignal's spectrum. It is exceedingly dicult to store an innite-length signal in any case, so we'll

assume that the signal extends over [0, N 1]. Continuous frequency. Subtler than the signal duration issue is the fact that the frequency variableis continuous: It may only need to span one period, like

[ 12 , 12] or [0, 1], but the DTFT formula as itstands requires evaluating the spectra at all frequencies within a period. Let's compute the spectrum at

a few frequencies; the most obvious ones are the equally spaced ones k, k {k, . . . ,K 1} : (f = kK ).We thus dene the discrete Fourier transform (DFT) to be

Discrete Fourier transform

k, k {k, . . . ,K 1} :(S (k) =

N1n=0

S (n) e(i)2pinkK

)(1)

Here, S (k) is shorthand for S(ei2pi

kK

).

Version 2.5: May 9, 2005 7:41 pm GMT-5


1

"Analyzing the Spectrum of Speech", Figure 1: spectrogram

2

"Discrete-Time Fourier Transform(DTFT)", (1) : Fourier Transform



The Fast Fourier Transform (FFT)

Justin Romberg



Abstract

This module describes the fast Fourier Transform (FFT).

1 Introduction

The Fast Fourier Transform (FFT) is an ecient O(NlogN) algorithm for calculating DFTs The FFT

1

exploits

symmetries in the W matrix to take a "divide and conquer" approach. We will rst discuss deriving theactual FFT algorithm, some of its implications for the DFT, and a speed comparison to drive home the

importance of this powerful algorithm.

2 Deriving the FFT

To derive the FFT, we assume that the signal's duration is a power of two: N = 2l . Consider what happensto the even-numbered and odd-numbered elements of the sequence in the DFT calculation.

S (k) = s (0) + s (2) e(i)2pi2kN + + s (N 2) e(i) 2pi(N2)kN +

s (1) e(i)2pikN + s (3) e(i)

2pi(2+1)kN + + s (N 1) e(i) 2pi(N2+1)kN =

s (0) + s (2) e(i) 2pikN

2 + + s (N 2) e(i)2pi(N2 1)k

N2 +s (1) + s (3) e(i) 2pikN2 + + s (N 1) e(i) 2pi(N2 1)kN2

e(i2pik)N

(1)

Each term in square brackets has the form of a

N2 -length DFT. The rst one is a DFT of the even-

numbered elements, and the second of the odd-numbered elements. The rst DFT is combined with the

second multiplied by the complex exponential e(i2pik)

N. The half-length transforms are each evaluated at

frequency indices k {0, . . . , N 1} . Normally, the number of frequency indices in a DFT calculation rangebetween zero and the transform length minus one. The computational advantage of the FFT comes from

recognizing the periodic nature of the discrete Fourier transform. The FFT simply reuses the computations

made in the half-length transforms and combines them through additions and the multiplication by e(i2pik)

N

Version 2.7: Jul 29, 2010 2:46 pm GMT-5


1

"Fast Fourier Transform (FFT)"



, which is not periodic over

N2 , to rewrite the length-N DFT. Figure 1 (Length-8 DFT decomposition)

illustrates this decomposition. As it stands, we now compute two length-

N2 transforms (complexity 2O

(N2

4

)), multiply one of them by the complex exponential (complexity O (N) ), and add the results (complexityO (N) ). At this point, the total complexity is still dominated by the half-length DFT calculations, but theproportionality coecient has been reduced.

Now for the fun. Because N = 2l , each of the half-length transforms can be reduced to two quarter-lengthtransforms, each of these to two eighth-length ones, etc. This decomposition continues until we are left with

length-2 transforms. This transform is quite simple, involving only additions. Thus, the rst stage of the

FFT has

N2 length-2 transforms (see the bottom part of Figure 1 (Length-8 DFT decomposition)). Pairs of

these transforms are combined by adding one to the other multiplied by a complex exponential. Each pair

requires 4 additions and 4 multiplications, giving a total number of computations equaling 8N4 =N2 . This

number of computations does not change from stage to stage. Because the number of stages, the number of

times the length can be divided by two, equals log2N , the complexity of the FFT is O (N logN) .

Length-8 DFT decomposition

(a)

(b)

Figure 1: The initial decomposition of a length-8 DFT into the terms using even- and odd-indexed

inputs marks the rst phase of developing the FFT algorithm. When these half-length transforms are

successively decomposed, we are left with the diagram shown in the bottom panel that depicts the

length-8 FFT computation.

Doing an example will make computational savings more obvious. Let's look at the details of a length-8

DFT. As shown on Figure 1 (Length-8 DFT decomposition), we rst decompose the DFT into two length-



4 DFTs, with the outputs added and subtracted together in pairs. Considering Figure 1 (Length-8 DFT

decomposition) as the frequency index goes from 0 through 7, we recycle values from the length-4 DFTs

into the nal calculation because of the periodicity of the DFT output. Examining how pairs of outputs are

collected together, we create the basic computational element known as a buttery (Figure 2 (Buttery)).

Buttery

Figure 2: The basic computational element of the fast Fourier transform is the buttery. It takes two

complex numbers, represented by a and b, and forms the quantities shown. Each buttery requires one

complex multiplication and two complex additions.

By considering together the computations involving common output frequencies from the two half-length

DFTs, we see that the two complex multiplies are related to each other, and we can reduce our computational

work even further. By further decomposing the length-4 DFTs into two length-2 DFTs and combining their

outputs, we arrive at the diagram summarizing the length-8 fast Fourier transform (Figure 1 (Length-8 DFT

decomposition)). Although most of the complex multiplies are quite simple (multiplying by e(ipi) meansnegating real and imaginary parts), let's count those for purposes of evaluating the complexity as full complex

multiplies. We have

N2 = 4 complex multiplies and 2N = 16 additions for each stage and log2N = 3 stages,making the number of basic computations

3N2 log2N as predicted.Exercise 1 (Solution on p. 6.)

Note that the ordering of the input sequence in the two parts of Figure 1 (Length-8 DFT decom-

position) aren't quite the same. Why not? How is the ordering determined?

2.1 FFT and the DFT

We now have a way of computing the spectrum for an arbitrary signal: The Discrete Fourier Transform

(DFT)

2

computes the spectrum at N equally spaced frequencies from a length- N sequence. An issue thatnever arises in analog "computation," like that performed by a circuit, is how much work it takes to perform

the signal processing operation such as ltering. In computation, this consideration translates to the number

of basic computational steps required to perform the needed processing. The number of steps, known as

the complexity, becomes equivalent to how long the computation takes (how long must we wait for an

answer). Complexity is not so much tied to specic computers or programming languages but to how many

steps are required on any computer. Thus, a procedure's stated complexity says that the time taken will be

proportional to some function of the amount of data used in the computation and the amount demanded.

For example, consider the formula for the discrete Fourier transform. For each frequency we chose, we

must multiply each signal value by a complex number and add together the results. For a real-valued signal,

2

"Discrete Fourier Transform", (1) : Discrete Fourier transform



each real-times-complex multiplication requires two real multiplications, meaning we have 2N multiplicationsto perform. To add the results together, we must keep the real and imaginary parts separate. Adding Nnumbers requires N 1 additions. Consequently, each frequency requires 2N + 2 (N 1) = 4N 2 basiccomputational steps. As we have N frequencies, the total number of computations is N (4N 2).In complexity calculations, we only worry about what happens as the data lengths increase, and take the

dominant termhere the 4N2 termas reecting how much work is involved in making the computation.As multiplicative constants don't matter since we are making a "proportional to" evaluation, we nd the

DFT is an O(N2)computational procedure. This notation is read "order N -squared". Thus, if we doublethe length of the data, we would expect that the computation time to approximately quadruple.


In making the complexity evaluation for the DFT, we assumed the data to be real. Three ques-

tions emerge. First of all, the spectra of such signals have conjugate symmetry, meaning that

negative frequency components (k =[N2 + 1, ..., N + 1

]in the DFT

3

) can be computed from the

corresponding positive frequency components. Does this symmetry change the DFT's complexity?

Secondly, suppose the data are complex-valued; what is the DFT's complexity now?

Finally, a less important but interesting question is suppose we want K frequency values insteadof N ; now what is the complexity?

3 Speed Comparison

How much better is O(NlogN) than O( N2)?

Figure 3: This gure shows how much slower the computation time of an O(NlogN) process grows.

N 10 100 1000 106 109

N2 100 104 106 1012 1018

N logN 1 200 3000 6 106 9 109

Table 1

Say you have a 1 MFLOP machine (a million "oating point" operations per second). Let N = 1million =106.An O( N2) algorithm takes 1012 ors 106 seconds ' 11.5 days.An O( N logN) algorithm takes 6 106 Flors 6 seconds.3

"Discrete Fourier Transform", (1) : Discrete Fourier transform



note: N = 1million is not unreasonable.

Example 1

3 megapixel digital camera spits out 3106 numbers for each picture. So for two N point sequencesf [n] and h [n]. If computing (f [n]~ h [n]) directly: O( N2) operations.taking FFTs O(NlogN)

multiplying FFTs O(N)

inverse FFTs O(NlogN).

the total complexity is O(NlogN).

4 Conclusion

Other "fast" algorithms have been discovered, most of which make use of how many common factors the

transform length N has. In number theory, the number of prime factors a given integer has measures how

composite it is. The numbers 16 and 81 are highly composite (equaling 24 and 34 respectively), thenumber 18 is less so ( 2132 ), and 17 not at all (it's prime). In over thirty years of Fourier transformalgorithm development, the original Cooley-Tukey algorithm is far and away the most frequently used. It

is so computationally ecient that power-of-two transform lengths are frequently used regardless of what

the actual length of the data. It is even well established that the FFT, alongside the digital computer, were

almost completely responsible for the "explosion" of DSP in the 60's.



Solutions to Exercises in this Module


The upper panel has not used the FFT algorithm to compute the length-4 DFTs while the lower one has.

The ordering is determined by the algorithm.


When the signal is real-valued, we may only need half the spectral values, but the complexity remains

unchanged. If the data are complex-valued, which demands retaining all frequency values, the complexity is

again the same. When only K frequencies are needed, the complexity is O (KN).


dtft, dft, fft.pdf

Documents