CHAPTER

Fourier Transform Pairs

11

For every time domain waveform there is a corresponding frequency domain waveform, and viceversa. For example, a rectangular pulse in the time domain coincides with a sinc function [i.e.,sin(x)/x] in the frequency domain. Duality provides that the reverse is also true; a rectangularpulse in the frequency domain matches a sinc function in the time domain. Waveforms that
correspond to each other in this manner are called Fourier transform pairs. Several common
pairs are presented in this chapter.

Delta Function Pairs
For discrete signals, the delta function is a simple waveform, and has anequally simple Fourier transform pair. Figure 11-1a shows a delta function in
the time domain, with its frequency spectrum in (b) and (c). The magnitude
is a constant value, while the phase is entirely zero. As discussed in the lastchapter, this can be understood by using the expansion/compression property.
When the time domain is compressed until it becomes an impulse, the frequency
domain is expanded until it becomes a constant value.
In (d) and (g), the time domain waveform is shifted four and eight samples tothe right, respectively. As expected from the properties in the last chapter,
shifting the time domain waveform does not affect the magnitude, but adds a
linear component to the phase. The phase signals in this figure have not been
unwrapped, and thus extend only from -B to B. Also notice that the horizontal
axes in the frequency domain run from -0.5 to 0.5. That is, they show thenegative frequencies in the spectrum, as well as the positive ones. The
negative frequencies are redundant information, but they are often included in
DSP graphs and you should become accustomed to seeing them.
Figure 11-2 presents the same information as Fig. 11-1, but with thefrequency domain in rectangular form. There are two lessons to be learned
here. First, compare the polar and rectangular representations of the

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The Scientist and Engineer's Guide to Digital Signal Processing

Time Domain

Frequency Domain

2

2

6

b. Magnitude

a. Impulse at x[0]40Phase (radians)

Amplitude

Amplitude

1
1

0

-1

c. Phase

2
0
-2
-4

-2
-0.5

-1
0

16

32

48

Sample number

64
63

2

0

Frequency

2

4

Amplitude

0Phase (radians)

1
1

0

0.5

0

0.5

0

0.5

Frequency

6e. Magnitude

d. Impulse at x[4]Amplitude

-6
-0.5

0.5

0

-1

f. Phase

2
0
-2
-4

-2
-0.5

-1
0

16

32

48

Sample number

64
63

2

0

Frequency

2

4

Amplitude

0Phase (radians)

11Frequency

6h. Magnitude

g. Impulse at x[8]Amplitude

-6
-0.5

0.5

0

-1

i. Phase

2
0
-2
-4

-1
0

16

32

48

Sample number

64
63

-2
-0.5

0

Frequency

0.5

-6
-0.5

Frequency

FIGURE 11-1
Delta function pairs in polar form. An impulse in the time domain corresponds to a
constant magnitude and a linear phase in the frequency domain.

frequency domains. As is usually the case, the polar form is much easier tounderstand; the magnitude is nothing more than a constant, while the phase isa straight line. In comparison, the real and imaginary parts are sinusoidal
oscillations that are difficult to attach a meaning to.
The second interesting feature in Fig. 11-2 is the duality of the DFT. In theconventional view, each sample in the DFT's frequency domain corresponds
to a sinusoid in the time domain. However, the reverse of this is also true,
each sample in the time domain corresponds to sinusoids in the frequency
domain. Including the negative frequencies in these graphs allows theduality property to be more symmetrical. For instance, Figs. (d), (e), andChapter 11- Fourier Transform Pairs

Time Domain

211

Frequency Domain

2

2

2

b. Real Parta. Impulse at x[0]c. Imaginary part

0

1

Amplitude

Amplitude

Amplitude

1
1

0

-1

-2
-0.5

-1
0

16

32

48

Sample number

64
63

2

-1

0

Frequency

-2
-0.5

0.5

2

1

Amplitude

Amplitude

Amplitude

0

0

-1

-2
-0.5

-1
16

32

48

Sample number

64
63

2

0

Frequency

-2
-0.5

0.5

0

i. Imaginary part

0

-1

-1
16

32

48

Sample number

64
63

-2
-0.5

0.5

1

Amplitude

Amplitude

Amplitude

10Frequency

2h. Real Part
1

0

0

-1

2

g. Impulse at x[8]0.5f. Imaginary part1
1

0Frequency

2

e. Real Part

d. Impulse at x[4]

0

0

0

-1

0

Frequency

0.5

-2
-0.5

0

Frequency

0.5

FIGURE 11-2
Delta function pairs in rectangular form. Each sample in the time domain results in a cosine wave in the real part,and a negative sine wave in the imaginary part of the frequency domain.

(f) show that an impulse at sample number four in the time domain results infour cycles of a cosine wave in the real part of the frequency spectrum, andfour cycles of a negative sine wave in the imaginary part. As you recall, an
impulse at sample number four in the real part of the frequency spectrum
results in four cycles of a cosine wave in the time domain. Likewise, animpulse at sample number four in the imaginary part of the frequency spectrum
results in four cycles of a negative sine wave being added to the time domain
wave.
As mentioned in Chapter 8, this can be used as another way to calculate theDFT (besides correlating the time domain with sinusoids). Each sample in thetime domain results in a cosine wave being added to the real part of the

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The Scientist and Engineer's Guide to Digital Signal Processing

frequency domain, and a negative sine wave being added to the imaginary part.
The amplitude of each sinusoid is given by the amplitude of the time domain
sample. The frequency of each sinusoid is provided by the sample number of
the time domain point. The algorithm involves: (1) stepping through each time
domain sample, (2) calculating the sine and cosine waves that correspond toeach sample, and (3) adding up all of the contributing sinusoids. The resulting
program is nearly identical to the correlation method (Table 8-2), except thatthe outer and inner loops are exchanged.

The Sinc Function
Figure 11-4 illustrates a common transform pair: the rectangular pulse and thesinc function (pronounced “sink”). The sinc function is defined as:sinc(a) ' sin(Ba)/ (Ba) , however, it is common to see the vague statement: "thesinc function is of the general form: sin(x)/x ." In other words, the sinc is a sine
wave that decays in amplitude as 1/x. In (a), the rectangular pulse issymmetrically centered on sample zero, making one-half of the pulse on theright of the graph and the other one-half on the left. This appears to the DFT
as a single pulse because of the time domain periodicity. The DFT of thissignal is shown in (b) and (c), with the unwrapped version in (d) and (e).
First look at the unwrapped spectrum, (d) and (e). The unwrapped
magnitude is an oscillation that decreases in amplitude with increasing
frequency. The phase is composed of all zeros, as you should expect fora time domain signal that is symmetrical around sample number zero. We
are using the term unwrapped magnitude to indicate that it can have both
positive and negative values. By definition, the magnitude must always be
positive. This is shown in (b) and (c) where the magnitude is made all
positive by introducing a phase shift of B at all frequencies where theunwrapped magnitude is negative in (d).
In (f), the signal is shifted so that it appears as one contiguous pulse, but is nolonger centered on sample number zero. While this doesn't change themagnitude of the frequency domain, it does add a linear component to thephase, making it a jumbled mess. What does the frequency spectrum look likeas real and imaginary parts ? Too confusing to even worry about.
An N point time domain signal that contains a unity amplitude rectangular pulse
M points wide, has a DFT frequency spectrum given by:
EQUATION 11-1
DFT spectrum of a rectangular pulse. In thisequation, N is the number of points in thetime domain signal, all of which have a valueof zero, except M adjacent points that have avalue of one. The frequency spectrum iscontained in X[k] , where k runs from 0 to
N/2. To avoid the division by zero, use
X[0] ' M . The sine function uses radians,
not degrees. This equation takes into
account that the signal is aliased.

sin(BkM / N )Mag X [ k] ' /0/000 sin(Bk /N ) 00

Chapter 11- Fourier Transform Pairs

Time Domain

213

Frequency Domain

2

20

6

a. Rectangular pulse

b. Magnitude

0Phase (radians)

Amplitude

Amplitude

1

10
5
0

-1
32

64

96

Sample number

128
127

0.1

0.2

0.3

Frequency

0.4

0Phase (radians)

Amplitude

10
5

0.1

0.2

0.3

Frequency

0.4

0.2

0.5

0.3

0.4

0.5

Frequency

h. Phase

4

Phase (radians)

Amplitude

Amplitude

0.1

6

15
10
5
0

2
0
-2
-4
-6

-5

Sample number

0.4

0
-2

0

g. Magnitude

128
127

0.3

2

0.5

20

f. Rectangular pulse

96

0.5

-6
0

64

0.4

-4

-5

32

0.3

e. Phase

4

0

0

0.2

Frequency

6

15

-1

0.1

or

d. Unwrapped Magnitude

0

-2

0.5

20

1

0

-6
0

2

2

-4

-5
0

c. Phase

4

15

0

0.1

0.2

0.3

Frequency

0.4

0.5

0

0.1

0.2

Frequency

FIGURE 11-3
DFT of a rectangular pulse. A rectangular pulse in one domain corresponds to a sincfunction in the other domain.

Alternatively, the DTFT can be used to express the frequency spectrum as afraction of the sampling rate, f:
EQUATION 11-2
Equation 11-1 rewritten in terms of thesampling frequency. The parameter, f , isthe fraction of the sampling rate, running
continiously from 0 to 0.5. To avoid thedivision by zero, use Mag X(0) ' M .

sin(B f M )Mag X ( f ) ' /0/
00 sin(B f ) 000

In other words, Eq. 11-1 provides N/2 % 1 samples in the frequency spectrum,
while Eq. 11-2 provides the continuous curve that the samples lie on. These

The Scientist and Engineer's Guide to Digital Signal Processing

equations only provide the magnitude. The phase is determined solely by theleft-right positioning of the time domain waveform, as discussed in the lastchapter.
Notice in Fig. 11-3b that the amplitude of the oscillation does not decay tozero before a frequency of 0.5 is reached. As you should suspect, thewaveform continues into the next period where it is aliased. This changes
the shape of the frequency domain, an effect that is included in Eqs. 11-1
and 11-2.
It is often important to understand what the frequency spectrum looks like whenaliasing isn't present. This is because discrete signals are often used torepresent or model continuous signals, and continuous signals don't alias. To
remove the aliasing in Eqs. 11-1 and 11-2, change the denominators fromsin (Bk M/ N ) to BkM /N , and sin (Bf ) to Bf , respectively. Figure 11-4 shows
the significance of this. The quantity Bf can only run from 0 to 1.5708, since f
can only run from 0 to 0.5. Over this range there isn't much difference
between sin (Bf ) and Bf . At zero frequency they have the same value, andat a frequency of 0.5 there is only about a 36% difference. Withoutaliasing, the curve in Fig. 11-3b would show a slightly lower amplitude
near the right side of the graph, and no change near the left side.
When the frequency spectrum of the rectangular pulse is not aliased(because the time domain signal is continuous, or because you are ignoring
the aliasing), it is of the general form: sin (x)/ x , i.e., a sinc function. For
continuous signals, the rectangular pulse and the sinc function are Fouriertransform pairs. For discrete signals this is only an approximation, with theerror being due to aliasing.
The sinc function has an annoying problem at x ' 0 , where sin (x)/x becomeszero divided by zero. This is not a difficult mathematical problem; as x
becomes very small, sin (x) approaches the value of x (see Fig. 11-4).

1.6
1.4

FIGURE 11-4
Comparing x and sin(x). The functions: y(x) ' x ,and y(x) ' sin(x) are similar for small values of x,
and only differ by about 36% at 1.57 (B /2). This
describes how aliasing distorts the frequency
spectrum of the rectangular pulse from a puresinc function.

y(x) = x

1.2

y(x)

214

1
0.8

y(x) = sin(x)

0.6
0.4
0.2
0
0.0

0.2

0.4 0.6

0.8

x

1.0 1.2

1.4

1.6

Chapter 11- Fourier Transform Pairs

215

This turns the sinc function into x/x , which has a value of one. In other words,
as x becomes smaller and smaller, the value of sinc (x) approaches one, which
includes sinc (0) ' 1 . Now try to tell your computer this! All it sees is adivision by zero, causing it to complain and stop your program. The important
point to remember is that your program must include special handling at x ' 0
when calculating the sinc function.
A key trait of the sinc function is the location of the zero crossings. Theseoccur at frequencies where an integer number of the sinusoid's cycles fit
evenly into the rectangular pulse. For example, if the rectangular pulse is
20 points wide, the first zero in the frequency domain is at the frequency
that makes one complete cycle in 20 points. The second zero is at thefrequency that makes two complete cycles in 20 points, etc. This can beunderstood by remembering how the DFT is calculated by correlation. The
amplitude of a frequency component is found by multiplying the time
domain signal by a sinusoid and adding up the resulting samples. If thetime domain waveform is a rectangular pulse of unity amplitude, this is thesame as adding the sinusoid's samples that are within the rectangular pulse.
If this summation occurs over an integral number of the sinusoid's cycles,
the result will be zero.
The sinc function is widely used in DSP because it is the Fourier transform pairof a very simple waveform, the rectangular pulse. For example, the sincfunction is used in spectral analysis, as discussed in Chapter 9. Consider the
analysis of an infinitely long discrete signal. Since the DFT can only workwith finite length signals, N samples are selected to represent the longer signal.
The key here is that "selecting N samples from a longer signal" is the same as
multiplying the longer signal by a rectangular pulse. The ones in therectangular pulse retain the corresponding samples, while the zeros eliminate
them. How does this affect the frequency spectrum of the signal? Multiplying
the time domain by a rectangular pulse results in the frequency domain being
convolved with a sinc function. This reduces the frequency spectrum's
resolution, as previously shown in Fig. 9-5a.

Other Transform Pairs
Figure 11-5 (a) and (b) show the duality of the above: a rectangular pulse in
the frequency domain corresponds to a sinc function (plus aliasing) in the time
domain. Including the effects of aliasing, the time domain signal is given by:

EQUATION 11-3
Inverse DFT of the rectangular pulse. In thefrequency domain, the pulse has anamplitude of one, and runs from sample
number 0 through sample number M-1. The
parameter N is the length of the DFT, andx[i] is the time domain signal with i runningfrom 0 to N-1. To avoid the division byzero, use x[0] ' (2M& 1)/ N .

x [i ] '1 sin(2B i (M & 1/2) / N )Nsin(B i / N )216

The Scientist and Engineer's Guide to Digital Signal Processing

To eliminate the effects of aliasing from this equation, imagine that thefrequency domain is so finely sampled that it turns into a continuous curve.
This makes the time domain infinitely long with no periodicity. The DTFT is
the Fourier transform to use here, resulting in the time domain signal being
given by the relation:
EQUATION 11-4
Inverse DTFT of the rectangular pulse. In
the frequency domain, the pulse has anamplitude of one, and runs from zerofrequency to the cutoff frequency, fc , a value
between 0 and 0.5. The time domain signal isheld in x [i ] with i running from 0 to N-1. To
avoid the division by zero, use x[0] ' 2fc .

x[i ] 'sin(2B f c i )iBThis equation is very important in DSP, because the rectangular pulse in thefrequency domain is the perfect low-pass filter. Therefore, the sinc function
described by this equation is the filter kernel for the perfect low-pass filter.
This is the basis for a very useful class of digital filters called the windowedsinc filters, described in Chapter 15.
Figures (c) & (d) show that a triangular pulse in the time domain coincides
with a sinc function squared (plus aliasing) in the frequency domain. Thistransform pair isn't as important as the reason it is true. A 2M & 1 point
triangle in the time domain can be formed by convolving an M point
rectangular pulse with itself. Since convolution in the time domain results in
multiplication in the frequency domain, convolving a waveform with itself willsquare the frequency spectrum.
Is there a waveform that is its own Fourier Transform? The answer is yes, andthere is only one: the Gaussian. Figure (e) shows a Gaussian curve, and (f)shows the corresponding frequency spectrum, also a Gaussian curve. This
relationship is only true if you ignore aliasing. The relationship between thestandard deviation of the time domain and frequency domain is given by:
2BFf ' 1/Ft . While only one side of a Gaussian is shown in (f), the negative
frequencies in the spectrum complete the full curve, with the center ofsymmetry at zero frequency.
Figure (g) shows what can be called a Gaussian burst. It is formed bymultiplying a sine wave by a Gaussian. For example, (g) is a sine wave
multiplied by the same Gaussian shown in (e). The corresponding frequency
domain is a Gaussian centered somewhere other than zero frequency. As
before, this transform pair is not as important as the reason it is true. Since
the time domain signal is the multiplication of two signals, the frequency
domain will be the convolution of the two frequency spectra. The frequency
spectrum of the sine wave is a delta function centered at the frequency of thesine wave. The frequency spectrum of a Gaussian is a Gaussian centered atzero frequency. Convolving the two produces a Gaussian centered at thefrequency of the sine wave. This should look familiar; it is identical to theprocedure of amplitude modulation described in the last chapter.

Chapter 11- Fourier Transform Pairs

Time Domain

Frequency Domain
153

a. Sincb. Rectangular pulse
10

Amplitude

2

Amplitude

217

1

0

5

0

-1

-5
0

16

32

48

64

80

Sample number

96

112

128
127

0

2

0.1

0.2

0.3

0.4

0.5

0.2

0.3

0.4

0.5

0.2

0.3

0.4

0.5

0.2

0.3

0.4

0.5

Frequency

15c. Triangle

d. Sinc squared

1

Amplitude

Amplitude

10

0

5

0

-1

-5
0

16

32

48

64

80

Sample number

96

112

128
127

0

2

0.1

Frequency

15

e. Gaussianf. Gaussian

1

Amplitude

Amplitude

10

0

5

0

-1

-5
0

16

32

48

64

80

Sample number

96

112

128
127

0

3

Frequency

15

g. Gaussian burst

h. Gaussian

2

10

Amplitude

Amplitude

0.1

1
0

5

0

-1

-5

-2
0

16

32

48

64

80

Sample number

96

112

128
127

0

FIGURE 11-5
Common transform pairs.

0.1Frequency

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The Scientist and Engineer's Guide to Digital Signal Processing

Gibbs EffectFigure 11-6 shows a time domain signal being synthesized from sinusoids. The
signal being reconstructed is shown in the last graph, (h). Since this signal is1024 points long, there will be 513 individual frequencies needed for acomplete reconstruction. Figures (a) through (g) show what the reconstructedsignal looks like if only some of these frequencies are used. For example, (f)shows a reconstructed signal using frequencies 0 through 100. This signal was
created by taking the DFT of the signal in (h), setting frequencies 101 through512 to a value of zero, and then using the Inverse DFT to find the resulting
time domain signal.
As more frequencies are added to the reconstruction, the signal becomes closer
to the final solution. The interesting thing is how the final solution isapproached at the edges in the signal. There are three sharp edges in (h). Two
are the edges of the rectangular pulse. The third is between sample numbers
1023 and 0, since the DFT views the time domain as periodic. When only
some of the frequencies are used in the reconstruction, each edge shows
overshoot and ringing (decaying oscillations). This overshoot and ringing is
known as the Gibbs effect, after the mathematical physicist Josiah Gibbs,
who explained the phenomenon in 1899.
Look closely at the overshoot in (e), (f), and (g). As more sinusoids areadded, the width of the overshoot decreases; however, the amplitude of theovershoot remains about the same, roughly 9 percent. With discrete signals
this is not a problem; the overshoot is eliminated when the last frequency isadded. However, the reconstruction of continuous signals cannot be explainedso easily. An infinite number of sinusoids must be added to synthesize acontinuous signal. The problem is, the amplitude of the overshoot does notdecrease as the number of sinusoids approaches infinity, it stays about the same
9%. Given this situation (and other arguments), it is reasonable to question if
a summation of continuous sinusoids can reconstruct an edge. Remember thesquabble between Lagrange and Fourier?
The critical factor in resolving this puzzle is that the width of the overshootbecomes smaller as more sinusoids are included. The overshoot is still present
with an infinite number of sinusoids, but it has zero width. Exactly at thediscontinuity the value of the reconstructed signal converges to the midpoint of
the step. As shown by Gibbs, the summation converges to the signal in thesense that the error between the two has zero energy.
Problems related to the Gibbs effect are frequently encountered in DSP. For
example, a low-pass filter is a truncation of the higher frequencies, resulting
in overshoot and ringing at the edges in the time domain. Another common
procedure is to truncate the ends of a time domain signal to prevent them fromextending into neighboring periods. By duality, this distorts the edges in thefrequency domain. These issues will resurface in future chapters on filter
design.

Chapter 11- Fourier Transform Pairs
22

b. Frequencies: 0 & 1

1

Amplitude

Amplitude

a. Frequencies: 0

0

-1

1

0

-1
0

256

512

Sample number

768

1024
1023

0

2

1

Amplitude

Amplitude

512

Sample number

768

1024
1023

768

1024
1023

768

1024
1023

768

1024
1023

d. Frequencies: 0 to 10

0

-1

1

0

-1
0

256

512

Sample number

768

1024
1023

0

2

256

512

Sample number

2

e. Frequencies: 0 to 30

f. Frequencies: 0 to 100

1

Amplitude

Amplitude256

2

c. Frequencies: 0 to 3

0

-1

1

0

-1
0

256

512

Sample number

768

1024
1023

0

2

256

512

Sample number

2

g. Frequencies: 0 to 300

h. Frequencies: 0 to 512

1

Amplitude

Amplitude

219

0

-1

1

0

-1
0

256

512

Sample number

768

1024
1023

0

FIGURE 11-6.
The Gibbs effect.

256

512

Sample number

220

The Scientist and Engineer's Guide to Digital Signal Processing

Harmonics
If a signal is periodic with frequency f, the only frequencies composing thesignal are integer multiples of f, i.e., f, 2f, 3f, 4f, etc. These frequencies arecalled harmonics. The first harmonic is f, the second harmonic is 2f, thethird harmonic is 3f, and so forth. The first harmonic (i.e., f) is also givena special name, the fundamental frequency. Figure 11-7 shows anTime Domain

Frequency Domain

2700

a. Sine wave
1

500

Amplitude

Amplitude

b. Fundamental

600

0

400
300
200

-1

100
-2

0
0

128

256

384

512

640

Sample number

768

896

1024
1023

0

0.04

0.06

0.08

0.1

0.06

0.08

0.1

0.06

0.08

0.1

Frequency

700

2c. Asymmetrical distortiond. Fundamental plus
even and odd harmonics

600

1

500

Amplitude

Amplitude

0.02

0

400
300
200

-1

100
0

-2
0

128

256

384

512

640

Sample number

768

896

1024
1023

0

0.04

Frequency

700

2e. Symmetrical distortion

f. Fundamental plus
odd harmonics

600

1

500

Amplitude

Amplitude

0.02

0

400
300
200

-1

100
0

-2
0

128

256

384

512

640

Sample number

768

896

1024
1023

0

0.02

0.04

Frequency

FIGURE 11-7
Example of harmonics. Asymmetrical distortion, shown in (c), results in even and odd harmonics,(d), while symmetrical distortion, shown in (e), produces only even harmonics, (f).

Chapter 11- Fourier Transform Pairs

Time Domain

Frequency Domain

2700

a. Distorted sine wave

b. Frequency spectrum

600
500

Amplitude

1

0

-1

400
300
200
100
0
-100

-2
0

20

40

60

Sample number

80

100

0

0.1

0.2

0.3

Frequency

0.4

0.5

100
105

FIGURE 11-8
Harmonic aliasing. Figures (a) and (b) show
a distorted sine wave and its frequency
spectrum, respectively. Harmonics with afrequency greater than 0.5 will become
aliased to a frequency between 0 and 0.5.
Figure (c) displays the same frequency
spectrum on a logarithmic scale, revealing
many aliased peaks with very low amplitude.

c. Frequency spectrum (Log scale)

804
10
603
10
402
10

Amplitude

Amplitude

221

201
10

1000
-20-1
10
-40-2
10
-60-3
10
-80-4
10
-100
10-5
0

0.1

0.2

0.3

Frequency

0.4

0.5

example. Figure (a) is a pure sine wave, and (b) is its DFT, a single peak.
In (c), the sine wave has been distorted by poking in the tops of the peaks.
Figure (d) shows the result of this distortion in the frequency domain.
Because the distorted signal is periodic with the same frequency as theoriginal sine wave, the frequency domain is composed of the original peakplus harmonics. Harmonics can be of any amplitude; however, they usually
become smaller as they increase in frequency. As with any signal, sharpedges result in higher frequencies. For example, consider a common TTL
logic gate generating a 1 kHz square wave. The edges rise in a few
nanoseconds, resulting in harmonics being generated to nearly 100 MHz,the ten-thousandth harmonic!
Figure (e) demonstrates a subtlety of harmonic analysis. If the signal issymmetrical around a horizontal axis, i.e., the top lobes are mirror images ofthe bottom lobes, all of the even harmonics will have a value of zero. As
shown in (f), the only frequencies contained in the signal are the fundamental,
the third harmonic, the fifth harmonic, etc.
All continuous periodic signals can be represented as a summation ofharmonics, just as described. Discrete periodic signals have a problem that
disrupts this simple relation. As you might have guessed, the problem isaliasing. Figure 11-8a shows a sine wave distorted in the same manner as
before, by poking in the tops of the peaks. This waveform looks much less

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regular and smooth than in the previous example because the sine wave is
at a much higher frequency, resulting in fewer samples per cycle. Figure(b) shows the frequency spectrum of this signal. As you would expect, you
can identify the fundamental and harmonics. This example shows that
harmonics can extend to frequencies greater than 0.5 of the sampling
frequency, and will be aliased to frequencies somewhere between 0 and 0.5.
You don't notice them in (b) because their amplitudes are too low. Figure(c) shows the frequency spectrum plotted on a logarithmic scale to reveal
these low amplitude aliased peaks. At first glance, this spectrum looks likerandom noise. It isn't; this is a result of the many harmonics overlappingas they are aliased.
It is important to understand that this example involves distorting a signal
after it has been digitally represented. If this distortion occurred in an
analog signal, you would remove the offending harmonics with an antialias
filter before digitization. Harmonic aliasing is only a problem whennonlinear operations are performed directly on a discrete signal. Eventhen, the amplitude of these aliased harmonics is often low enough that they
can be ignored.
The concept of harmonics is also useful for another reason: it explains why the
DFT views the time and frequency domains as periodic. In the frequency
domain, an N point DFT consists of N/2+1 equally spaced frequencies. You
can view the frequencies between these samples as (1) having a value of zero,or (2) not existing. Either way they don't contribute to the synthesis of the time
domain signal. In other words, a discrete frequency spectrum consists ofharmonics, rather than a continuous range of frequencies. This requires thetime domain to be periodic with a frequency equal to the lowest sinusoid in thefrequency domain, i.e., the fundamental frequency. Neglecting the DC value,
the lowest frequency represented in the frequency domain makes one complete
cycle every N samples, resulting in the time domain being periodic with aperiod of N. In other words, if one domain is discrete, the other domain mustbe periodic, and vice versa. This holds for all four members of the Fouriertransform family. Since the DFT views both domains as discrete, it must also
view both domains as periodic. The samples in each domain represent
harmonics of the periodicity of the opposite domain.

Chirp Signals
Chirp signals are an ingenious way of handling a practical problem in echo
location systems, such as radar and sonar. Figure 11-9 shows the frequency
response of the chirp system. The magnitude has a constant value of one, while
the phase is a parabola:

EQUATION 11-7
Phase of the chirp system.

Phase X [k ] ' "k % $k 2

Chapter 11- Fourier Transform Pairs
2

223

50

a. Chirp magnitude

b. Chirp phase
Phase (radians)

Amplitude

0
1

0

-50
-100

-150

-200

-1
0

0.1

0.2

0.3

0.4

Frequency

0.5

0

0.1

0.2

0.3

Frequency

0.4

0.5

FIGURE 11-9
Frequency response of the chirp system. The magnitude is a constant, while the phase is a parabola.

The parameter " introduces a linear slope in the phase, that is, it simply shifts
the impulse response left or right as desired. The parameter $ controls thecurvature of the phase. These two parameters must be chosen such that thephase at frequency 0.5 (i.e. k = N/2) is a multiple of 2B. Remember, whenever
the phase is directly manipulated, frequency 0 and 0.5 must both have a phaseof zero (or a multiple of 2B, which is the same thing).
Figure 11-10 shows an impulse entering a chirp system, and the impulse
response exiting the system. The impulse response is an oscillatory burst thatstarts at a low frequency and changes to a high frequency as time progresses.
This is called a chirp signal for a very simple reason: it sounds like the chirpof a bird when played through a speaker.
The key feature of the chirp system is that it is completely reversible. If you
run the chirp signal through an antichirp system, the signal is again made into
an impulse. This requires the antichirp system to have a magnitude of one,and the opposite phase of the chirp system. As discussed in the last

1.5

1.5

b. Impulse Response(chirp signal)a. Impulse
1

Chirp
System0.5

0Amplitude

Amplitude

1

0.5

0

-0.5

-0.5
0

20

40

60

80

Sample number

100

120

0

20

40

60

80

Sample number

FIGURE 11-10
The chirp system. The impulse response of a chirp system is a chirp signal.

100

120

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The Scientist and Engineer's Guide to Digital Signal Processing

chapter, this means that the impulse response of the antichirp system is foundby preforming a left-for-right flip of the chirp system's impulse response.
Interesting, but what is it good for?
Consider how a radar system operates. A short burst of radio frequency energy
is emitted from a directional antenna. Aircraft and other objects reflect someof this energy back to a radio receiver located next to the transmitter. Since
radio waves travel at a constant rate, the elapsed time between the transmitted
and received signals provides the distance to the target. This brings up thefirst requirement for the pulse: it needs to be as short as possible. For
example, a 1 microsecond pulse provides a radio burst about 300 meters long.
This means that the distance information we obtain with the system will havea resolution of about this same length. If we want better distance resolution,
we need a shorter pulse.
The second requirement is obvious: if we want to detect objects farther away,
you need more energy in your pulse. Unfortunately, more energy and shorter
pulse are conflicting requirements. The electrical power needed to provide apulse is equal to the energy of the pulse divided by the pulse length. Requiring
both more energy and a shorter pulse makes electrical power handling alimiting factor in the system. The output stage of a radio transmitter can only
handle so much power without destroying itself.
Chirp signals provide a way of breaking this limitation. Before the impulse
reaches the final stage of the radio transmitter, it is passed through a chirp
system. Instead of bouncing an impulse off the target aircraft, a chirp signal
is used. After the chirp echo is received, the signal is passed through anantichirp system, restoring the signal to an impulse. This allows the portionsof the system that measure distance to see short pulses, while the power
handling circuits see long duration signals. This type of waveshaping is afundamental part of modern radar systems.