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CHAPTER
10 Fourier Transform Properties
The time and frequency domains are alternative ways of representing signals. The Fourier
transform is the mathematical relationship between these two representations. If a signal is
modified in one domain, it will also be changed in the other domain, although usually not in the
same way. For example, it was shown in the last chapter that convolving time domain signals
results in their frequency spectra being multiplied. Other mathematical operations, such as
addition, scaling and shifting, also have a matching operation in the opposite domain. These
relationships are called properties of the Fourier Transform, how a mathematical change in one
domain results in a mathematical change in the other domain.
Linearity of the Fourier Transform
The Fourier Transform is linear, that is, it possesses the properties of
homogeneity and additivity. This is true for all four members of the Fourier
transform family (Fourier transform, Fourier Series, DFT, and DTFT).
Figure 10-1 provides an example of how homogeneity is a property of the
Fourier transform. Figure (a) shows an arbitrary time domain signal, with the
corresponding frequency spectrum shown in (b). We will call these two
signals: and , respectively. Homogeneity means that a change in x[ ] X[ ]
amplitude in one domain produces an identical change in amplitude in the other
domain. This should make intuitive sense: when the amplitude of a time
domain waveform is changed, the amplitude of the sine and cosine waves
making up that waveform must also change by an equal amount.
In mathematical form, if and are a Fourier Transform pair, then x[ ] X[ ] k x[ ]
and are also a Fourier Transform pair, for any constant k. If the kX[ ]
frequency domain is represented in rectangular notation, means that both kX[ ]
the real part and the imaginary part are multiplied by k. If the frequency
domain is represented in polar notation, means that the magnitude is kX[ ]
multiplied by k, while the phase remains unchanged.
The Scientist and Engineer's Guide to Digital Signal Processing 186
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c. k x[ ]
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a. x[ ]
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b. X[ ]
Frequency
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0
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d. k X[ ]
Amplitude Amplitude
Amplitude Amplitude
Frequency Domain Time Domain
FIGURE 10-1
Homogeneity of the Fourier transform. If the amplitude is changed in one domain, it is changed by
the same amount in the other domain. In other words, scaling in one domain corresponds to scaling
in the other domain.
F.T.
F.T.
Additivity of the Fourier transform means that addition in one domain
corresponds to addition in the other domain. An example of this is shown
in Fig. 10-2. In this illustration, (a) and (b) are signals in the time domain
called and , respectively. Adding these signals produces a third x1[ ] x2[ ]
time domain signal called , shown in (c). Each of these three signals x3[ ]
has a frequency spectrum consisting of a real and an imaginary part, shown
in (d) through (i). Since the two time domain signals add to produce the
third time domain signal, the two corresponding spectra add to produce the
third spectrum. Frequency spectra are added in rectangular notation by
adding the real parts to the real parts and the imaginary parts to the
imaginary parts. If: , then: x1[n] % x2[n] ' x3[n] ReX1[f ] % ReX2[ f ] ' ReX3[ f ]
and . Think of this in terms of cosine and sine ImX1[f ] % ImX2[ f ] ' ImX3[f ]
waves. All the cosine waves add (the real parts) and all the sine waves add
(the imaginary parts) with no interaction between the two.
Frequency spectra in polar form cannot be directly added; they must be
converted into rectangular notation, added, and then reconverted back to
Chapter 10- Fourier Transform Properties 187
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e. Re X2[ ]
Frequency
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h. Im X2[ ]
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f. Re X3[ ]
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c. x3[ ]
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Frequency
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g. Im X1[ ]
Amplitude
Amplitude
Amplitude
Amplitude
Amplitude
Amplitude
Amplitude
Amplitude
Amplitude
FIGURE 10-2
Additivity of the Fourier transform. Adding two or more signals in one domain results in the
corresponding signals being added in the other domain. In this illustration, the time domain signals
in (a) and (b) are added to produce the signal in (c). This results in the corresponding real and
imaginary parts of the frequency spectra being added.
Frequency Domain Time Domain
F.T.
F.T.
F.T.
+ + +
= = =
polar form. This can also be understood in terms of how sinusoids behave.
Imagine adding two sinusoids having the same frequency, but with different
amplitudes ( and ) and phases ( and ). If the two phases happen to A1 A2 N1 N2
be same ( ), the amplitudes will add ( ) when the sinusoids are N1
' N2 A1
%A2
added. However, if the two phases happen to be exactly opposite ( ), N1
' &N2
the amplitudes will subtract ( ) when the sinusoids are added. The point A1
&A2
is, when sinusoids (or spectra) are in polar form, they cannot be added by
simply adding the magnitudes and phases.
The Scientist and Engineer's Guide to Digital Signal Processing 188
In spite of being linear, the Fourier transform is not shift invariant. In other
words, a shift in the time domain does not correspond to a shift in the
frequency domain. This is the topic of the next section.
Characteristics of the Phase
In mathematical form: if & , then a shift in the x[n] : Mag X[f ] Phase X[f ]
time domain results in: & , (where x[n%s] : Mag X[f ] Phase X[f ] % 2Bs f f
is expressed as a fraction of the sampling rate, running between 0 and 0.5). In
words, a shift of s samples in the time domain leaves the magnitude unchanged,
but adds a linear term to the phase, . Let's look at an example of how 2Bs f
this works.
Figure 10-3 shows how the phase is affected when the time domain waveform
is shifted to the left or right. The magnitude has not been included in this
illustration because it isn't interesting; it is not changed by the time domain
shift. In Figs. (a) through (d), the waveform is gradually shifted from having
the peak centered on sample 128, to having it centered on sample 0. This
sequence of graphs takes into account that the DFT views the time domain as
circular; when portions of the waveform exit to the right, they reappear on the
left.
The time domain waveform in Fig. 10-3 is symmetrical around a vertical
axis, that is, the left and right sides are mirror images of each other. As
mentioned in Chapter 7, signals with this type of symmetry are called linear
phase, because the phase of their frequency spectrum is a straight line.
Likewise, signals that don't have this left-right symmetry are called
nonlinear phase, and have phases that are something other than a straight
line. Figures (e) through (h) show the phase of the signals in (a) through
(d). As described in Chapter 7, these phase signals are unwrapped,
allowing them to appear without the discontinuities associated with keeping
the value between B and -B.
When the time domain waveform is shifted to the right, the phase remains a
straight line, but experiences a decrease in slope. When the time domain is
shifted to the left, there is an increase in the slope. This is the main property
you need to remember from this section; a shift in the time domain corresponds
to changing the slope of the phase.
Figures (b) and (f) display a unique case where the phase is entirely zero. This
occurs when the time domain signal is symmetrical around sample zero. At first
glance, this symmetry may not be obvious in (b); it may appear that the signal
is symmetrical around sample 256 (i.e., N/2) instead. Remember that the DFT
views the time domain as circular, with sample zero inherently connected to
sample N-1. Any signal that is symmetrical around sample zero will also be
symmetrical around sample N/2, and vice versa. When using members of the
Fourier Transform family that do not view the time domain as periodic (such
as the DTFT), the symmetry must be around sample zero to produces a zero
phase.
Chapter 10- Fourier Transform Properties 189
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g.
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Frequency
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f.
Phase (radians) Phase (radians) Phase (radians)
FIGURE 10-3
Phase changes resulting from a time domain shift.
Time Domain Frequency Domain
Amplitude Amplitude Amplitude Amplitude
Phase (radians)
The Scientist and Engineer's Guide to Digital Signal Processing 190
Sample number
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a. A low frequency
1 sample shift
= 1/32 cycle
Sample number
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b. 1/2 of sampling frequency
1 sample shift
= 1/2 cycle
FIGURE 10-4
The relationship between samples and phase. Figures (a) and (b) show low and high frequency sinusoids,
respectively. In (a), a one sample shift is equal to 1/32 of a cycle. In (b), a one sample shift is equal to 1/2 of a
cycle. This is why a shift in the waveform changes the phase more at high frequencies than at low frequencies.
Amplitude
Amplitude
Figures (d) and (h) shows something of a riddle. First imagine that (d) was
formed by shifting the waveform in (c) slightly more to the right. This means
that the phase in (h) would have a slightly more negative slope than in (g).
This phase is shown as line 1. Next, imagine that (d) was formed by starting
with (a) and shifting it to the left. In this case, the phase should have a
slightly more positive slope than (e), as is illustrated by line 2. Lastly, notice
that (d) is symmetrical around sample N/2, and should therefore have a zero
phase, as illustrated by line 3. Which of these three phases is correct? They
all are, depending on how the B and 2B phase ambiguities (discussed in Chapter
8) are arranged. For instance, every sample in line 2 differs from the
corresponding sample in line 1 by an integer multiple of 2B, making them
equal. To relate line 3 to lines 1 and 2, the B ambiguities must also be taken
into account.
To understand why the phase behaves as it does, imagine shifting a waveform
by one sample to the right. This means that all of the sinusoids that compose
the waveform must also be shifted by one sample to the right. Figure 10-4
shows two sinusoids that might be a part of the waveform. In (a), the sine
wave has a very low frequency, and a one sample shift is only a small fraction
of a full cycle. In (b), the sinusoid has a frequency of one-half of the sampling
rate, the highest frequency that can exist in sampled data. A one sample shift
at this frequency is equal to an entire 1/2 cycle, or B radians. That is, when a
shift is expressed in terms of a phase change, it becomes proportional to the
frequency of the sinusoid being shifted.
For example, consider a waveform that is symmetrical around sample zero,
and therefore has a zero phase. Figure 10-5a shows how the phase of this
signal changes when it is shifted left or right. At the highest frequency,
one-half of the sampling rate, the phase increases by B for each one sample
shift to the left, and decreases by B for each one sample shift to the right.
At zero frequency there is no phase shift, and all of the frequencies between
follow in a straight line.
Chapter 10- Fourier Transform Properties 191
Frequency
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-50
0
50
100
150
-14
-7
0
15B
10B
5B
0
-5B
b.
number of samples
shifted in time domain
FIGURE 10-5
Phases resulting from time domain shifting. For each sample that a time domain signal is shifted in the positive
direction (i.e., to the right), the phase at frequency 0.5 will decrease by B radians. For each sample shifted in the
negative direction (i.e., to the left), the phase at frequency 0.5 will increase by B radians. Figure (a) shows this for
a linear phase (a straight line), while (b) is an example using a nonlinear phase.
Frequency
0 0.1 0.2 0.3 0.4 0.5
-30
-20
-10
0
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shifted in time domain
3B
2B
B
0
-B
-2B
-3B
a.
0
Phase (radians)
Phase (radians)
All of the examples we have used so far are linear phase. Figure 10-5b shows
that nonlinear phase signals react to shifting in the same way. In this example
the nonlinear phase is a straight line with two rectangular pulses. When the
time domain is shifted, these nonlinear features are simply superimposed on the
changing slope.
What happens in the real and imaginary parts when the time domain
waveform is shifted? Recall that frequency domain signals in rectangular
notation are nearly impossible for humans to understand. The real and
imaginary parts typically look like random oscillations with no apparent
pattern. When the time domain signal is shifted, the wiggly patterns of the
real and imaginary parts become even more oscillatory and difficult to
interpret. Don't waste your time trying to understand these signals, or how
they are changed by time domain shifting.
Figure 10-6 is an interesting demonstration of what information is contained in
the phase, and what information is contained in the magnitude. The waveform
in (a) has two very distinct features: a rising edge at sample number 55, and
a falling edge at sample number 110. Edges are very important when
information is encoded in the shape of a waveform. An edge indicates when
something happens, dividing whatever is on the left from whatever is on the
right. It is time domain encoded information in its purest form. To begin the
demonstration, the DFT is taken of the signal in (a), and the frequency
spectrum converted into polar notation. To find the signal in (b), the phase is
replaced with random numbers between -B and B, and the inverse DFT used to
reconstruct the time domain waveform. In other words, (b) is based only on the
information contained in the magnitude. In a similar manner, (c) is found by
replacing the magnitude with small random numbers before using the inverse
DFT. This makes the reconstruction of (c) based solely on the information
contained in the phase.
The Scientist and Engineer's Guide to Digital Signal Processing 192
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a. Original signal
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b. Reconstructed from the magnitude
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c. Reconstructed from the phase
Amplitude Amplitude
Amplitude
FIGURE 10-6
Information contained in the phase. Figure (a)
shows a pulse-like waveform. The signal in (b)
is created by taking the DFT of (a), replacing the
phase with random numbers, and taking the
Inverse DFT. The signal in (c) is found by
taking the DFT of (a), replacing the magnitude
with random numbers, and taking the Inverse
DFT. The location of the edges is retained in
(c), but not in (b). This shows that the phase
contains information on the location of events in
the time domain signal.
The result? The locations of the edges are clearly present in (c), but totally
absent in (b). This is because an edge is formed when many sinusoids rise at
the same location, possible only when their phases are coordinated. In short,
much of the information about the shape of the time domain waveform is
contained in the phase, rather than the magnitude. This can be contrasted with
signals that have their information encoded in the frequency domain, such as
audio signals. The magnitude is most important for these signals, with the
phase playing only a minor role. In later chapters we will see that this type
of understanding provides strategies for designing filters and other methods of
processing signals. Understanding how information is represented in signals
is always the first step in successful DSP.
Why does left-right symmetry correspond to a zero (or linear) phase? Figure
10-7 provides the answer. Such a signal can be decomposed into a left half
and a right half, as shown in (a), (b) and (c). The sample at the center of
symmetry (zero in this case) is divided equally between the left and right
halves, allowing the two sides to be perfect mirror images of each other. The
magnitudes of these two halves will be identical, as shown in (e) and (f), while
the phases will be opposite in sign, as in (h) and (i). Two important concepts
fall out of this. First, every signal that is symmetrical between the left and
right will have a linear phase because the nonlinear phase of the left half
exactly cancels the nonlinear phase of the right half.
Chapter 10- Fourier Transform Properties 193
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63
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a. x[ ]
63
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e. Mag X1[ ]
Frequency
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h. Phase X1[ ]
Frequency
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Frequency
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i. Phase X2[ ]
Frequency
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Frequency
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g. Phase X[ ]
Sample number
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63
Frequency Domain Time Domain
Decompose
Amplitude
Amplitude
Amplitude
Amplitude
Amplitude
Amplitude
Amplitude
Amplitude
Amplitude
FIGURE 10-7
Phase characteristics of left-right symmetry. A signal with left-right symmetry, shown in (a), can be
decomposed into a right half, (b), and a left half, (c). The magnitudes of the two halves are identical, (e) and
(f), while the phases are the negative of each other, (h) and (i).
Second, imagine flipping (b) such that it becomes (c). This left-right flip in the
time domain does nothing to the magnitude, but changes the sign of every point
in the phase. Likewise, changing the sign of the phase flips the time domain
signal left-for-right. If the signals are continuous, the flip is around zero. If
the signals are discrete, the flip is around sample zero and sample N/2,
simultaneously.
Changing the sign of the phase is a common enough operation that it is given
its own name and symbol. The name is complex conjugation, and it is
The Scientist and Engineer's Guide to Digital Signal Processing 194
represented by placing a star to the upper-right of the variable. For example,
if consists of and , then is called the X[f ] Mag X[f ] Phase X[f ] X t[f ]
complex conjugate and is composed of and . In Mag X[f ] &Phase X[f ]
rectangular notation, the complex conjugate is found by leaving the real part
alone, and changing the sign of the imaginary part. In mathematical terms, if X[f ]
is composed of and , then is made up of and ReX[ f ] ImX[ f ] X t[f ] ReX[ f ]
. &ImX[ f ]
Here are several examples of how the complex conjugate is used in DSP. If
has a Fourier transform of , then has a Fourier transform of x[n] X[f ] x [&n]
. In words, flipping the time domain left-for-right corresponds to X t[f ]
changing the sign of the phase. As another example, recall from Chapter 7 that
correlation can be performed as a convolution. This is done by flipping one
of the signals left-for-right. In mathematical form, is convolution, a[n] t b[n ]
while is correlation. In the frequency domain these operations a[n] t b[&n ]
correspond to and , respectively. As the last A[f ] × B[ f ] A[f ] × B t[f ]
example, consider an arbitrary signal, , and its frequency spectrum, . x[n] X[f ]
The frequency spectrum can be changed to zero phase by multiplying it by its
complex conjugate, that is, . In words, whatever phase X[ f ] × X t[ f ] X[f ]
happens to have will be canceled by adding its opposite (remember, when
frequency spectra are multiplied, their phases are added). In the time domain,
this means that (a signal convolved with a left-right flipped x [n ] t x [&n ]
version of itself) will have left-right symmetry around sample zero, regardless
of what is. x [n ]
To many engineers and mathematicians, this kind of manipulation is DSP. If
you want to be able to communicate with this group, get used to using their
language.
Periodic Nature of the DFT
Unlike the other three Fourier Transforms, the DFT views both the time domain
and the frequency domain as periodic. This can be confusing and inconvenient
since most of the signals used in DSP are not periodic. Nevertheless, if you
want to use the DFT, you must conform with the DFT's view of the world.
Figure 10-8 shows two different interpretations of the time domain signal. First,
look at the upper signal, the time domain viewed as N points. This represents
how digital signals are typically acquired in scientific experiments and
engineering applications. For instance, these 128 samples might have been
acquired by sampling some parameter at regular intervals of time. Sample 0
is distinct and separate from sample 127 because they were acquired at
different times. From the way this signal was formed, there is no reason to
think that the samples on the left of the signal are even related to the samples
on the right.
Unfortunately, the DFT doesn't see things this way. As shown in the lower
figure, the DFT views these 128 points to be a single period of an infinitely
long periodic signal. This means that the left side of the acquired signal is
Chapter 10- Fourier Transform Properties 195
Sample number
-384 -256 -128 0 128 256 384
-2
-1
0
1
2
Sample number
0 32 64 96 128
-2
-1
0
1
2
127
FIGURE 10-8
Periodicity of the DFT's time domain signal. The time domain can be viewed as N samples in length, shown
in the upper figure, or as an infinitely long periodic signal, shown in the lower figure.
The time domain
viewed as N points
The time domain
viewed as periodic
Amplitude
Amplitude
connected to the right side of a duplicate signal. Likewise, the right side of the
acquired signal is connected to the left side of an identical period. This can
also be thought of as the right side of the acquired signal wrapping around and
connecting to its left side. In this view, sample 127 occurs next to sample 0,
just as sample 43 occurs next to sample 44. This is referred to as being
circular, and is identical to viewing the signal as being periodic.
The most serious consequence of time domain periodicity is time domain
aliasing. To illustrate this, suppose we take a time domain signal and pass
it through the DFT to find its frequency spectrum. We could immediately
pass this frequency spectrum through an Inverse DFT to reconstruct the
original time domain signal, but the entire procedure wouldn't be very
interesting. Instead, we will modify the frequency spectrum in some manner
before using the Inverse DFT. For instance, selected frequencies might be
deleted, changed in amplitude or phase, shifted around, etc. These are the
kinds of things routinely done in DSP. Unfortunately, these changes in the
frequency domain can create a time domain signal that is too long to fit into
The Scientist and Engineer's Guide to Digital Signal Processing 196
a single period. This forces the signal to spill over from one period into the
adjacent periods. When the time domain is viewed as circular, portions of
the signal that overflow on the right suddenly seem to reappear on the left
side of the signal, and vice versa. That is, the overflowing portions of the
signal alias themselves to a new location in the time domain. If this new
location happens to already contain an existing signal, the whole mess adds,
resulting in a loss of information. Circular convolution resulting from
frequency domain multiplication (discussed in Chapter 9), is an excellent
example of this type of aliasing.
Periodicity in the frequency domain behaves in much the same way, but is
more complicated. Figure 10-9 shows an example. The upper figures show
the magnitude and phase of the frequency spectrum, viewed as being composed
of samples spread between 0 and 0.5 of the sampling rate. This is the N/2 % 1
simplest way of viewing the frequency spectrum, but it doesn't explain many
of the DFT's properties.
The lower two figures show how the DFT views this frequency spectrum as
being periodic. The key feature is that the frequency spectrum between 0 and
0.5 appears to have a mirror image of frequencies that run between 0 and -0.5.
This mirror image of negative frequencies is slightly different for the
magnitude and the phase signals. In the magnitude, the signal is flipped leftfor-
right. In the phase, the signal is flipped left-for-right, and changed in sign.
As you recall, these two types of symmetry are given names: the magnitude is
said to be an even signal (it has even symmetry), while the phase is said to
be an odd signal (it has odd symmetry). If the frequency spectrum is
converted into the real and imaginary parts, the real part will always be even,
while the imaginary part will always be odd.
Taking these negative frequencies into account, the DFT views the frequency
domain as periodic, with a period of 1.0 times the sampling rate, such as -0.5
to 0.5, or 0 to 1.0. In terms of sample numbers, this makes the length of the
frequency domain period equal to N, the same as in the time domain.
The periodicity of the frequency domain makes it susceptible to frequency
domain aliasing, completely analogous to the previously described time
domain aliasing. Imagine a time domain signal that corresponds to some
frequency spectrum. If the time domain signal is modified, it is obvious that
the frequency spectrum will also be changed. If the modified frequency
spectrum cannot fit in the space provided, it will push into the adjacent periods.
Just as before, this aliasing causes two problems: frequencies aren't where they
should be, and overlapping frequencies from different periods add, destroying
information.
Frequency domain aliasing is more difficult to understand than time domain
aliasing, since the periodic pattern is more complicated in the frequency
domain. Consider a single frequency that is being forced to move from 0.01
to 0.49 in the frequency domain. The corresponding negative frequency is
therefore moving from -0.01 to -0.49. When the positive frequency moves
Chapter 10- Fourier Transform Properties 197
Frequency
-1.5 -1 -0.5 0 0.5 1 1.5
-1
0
1
2
3
W W W
Frequency
0 0.1 0.2 0.3 0.4 0.5
-1
0
1
2
3
Magnitude
FIGURE 10-9
Periodicity of the DFT's frequency domain. The frequency domain can be viewed as running from 0 to 0.5 of
the sampling rate (upper two figures), or an infinity long periodic signal with every other 0 to 0.5 segment
flipped left-for-right (lower two figures).
Frequency
0 0.1 0.2 0.3 0.4 0.5
-4
-3
-2
-1
0
1
2
3
4
Phase
Frequency
-1.5 -1 -0.5 0 0.5 1 1.5
-4
-3
-2
-1
0
1
2
3
4
The frequency domain
viewed as periodic
The frequency domain
viewed as 0 to 0.5 of
the sampling rate
Amplitude Phase (radians)
Amplitude Phase
The Scientist and Engineer's Guide to Digital Signal Processing 198
Time 0 N-1
signal exiting
to the right
reappears
on the left
Frequency 0 0.5
signal exiting
to the right
reappears
on the right
FIGURE 10-10
Examples of aliasing in the time and frequency domains, when only a single period is considered. In the time
domain, shown in (a), portions of the signal that exits to the right, reappear on the left. In the frequency
domain, (b), portions of the signal that exit to the right, reappear on the right as if they had been folded over.
a. Time domain aliasing b. Frequency domain aliasing
across the 0.5 barrier, the negative frequency is pushed across the -0.5
barrier. Since the frequency domain is periodic, these same events are
occurring in the other periods, such as between 0.5 and 1.5. A clone of the
positive frequency is crossing frequency 1.5 from left to right, while a clone
of the negative frequency is crossing 0.5 from right to left. Now imagine
what this looks like if you can only see the frequency band of 0 to 0.5. It
appears that a frequency leaving to the right, reappears on the right, but
moving in the opposite direction.
Figure 10-10 illustrates how aliasing appears in the time and frequency
domains when only a single period is viewed. As shown in (a), if one end of
a time domain signal is too long to fit inside a single period, the protruding end
will be cut off and pasted onto the other side. In comparison, (b) shows that
when a frequency domain signal overflows the period, the protruding end is
folded over. Regardless of where the aliased segment ends up, it adds to
whatever signal is already there, destroying information.
Let's take a closer look at these strange things called negative frequencies.
Are they just some bizarre artifact of the mathematics, or do they have a
real world meaning? Figure 10-11 shows what they are about. Figure (a)
is a discrete signal composed of 32 samples. Imagine that you are given
the task of finding the frequency spectrum that corresponds to these 32
points. To make your job easier, you are told that these points represent a
discrete cosine wave. In other words, you must find the frequency and
phase shift ( and 2) such that matches the given f x[n] ' cos(2Bnf /N % 2)
samples. It isn't long before you come up with the solution shown in (b),
that is, and . f ' 3 2 ' &B/4
If you stopped your analysis at this point, you only get 1/3 credit for the
problem. This is because there are two other solutions that you have
missed. As shown in (c), the second solution is and . Even f ' &3 2 ' B/4
if the idea of a negative frequency offends your sensibilities, it doesn't
Chapter 10- Fourier Transform Properties 199
Sample number
0 8 16 24 32
-2
-1
0
1
2
a. Samples
Sample number
0 8 16 24 32
-2
-1
0
1
2
b. Solution #1
f = 3, 2 = -B/4
FIGURE 10-11
The meaning of negative frequencies. The problem is to find the frequency spectrum of the discrete signal shown
in (a). That is, we want to find the frequency and phase of the sinusoid that passed through all of the samples.
Figure (b) is a solution using a positive frequency, while (c) is a solution using a negative frequency. Figure (d)
represents a family of solutions to the problem.
Sample number
0 8 16 24 32
-2
-1
0
1
2
c. Solution #2
f = -3, 2 = B/4
Sample number
0 8 16 24 32
-2
-1
0
1
2
d. Solution #3
f = 35, 2 = -B/4
Amplitude Amplitude
Amplitude Amplitude
change the fact that it is a mathematically valid solution to the defined
problem. Every positive frequency sinusoid can alternately be expressed
as a negative frequency sinusoid. This applies to continuous as well as
discrete signals
The third solution is not a single answer, but an infinite family of solutions.
As shown in (d), the sinusoid with and passes through all of f ' 35 2' &B/4
the discrete points, and is therefore a correct solution. The fact that it shows
oscillation between the samples may be confusing, but it doesn't disqualify it
from being an authentic answer. Likewise, , , , and f ' ±29 f ' ±35 f ' ±61
are all solutions with multiple oscillations between the points. f ' ±67
Each of these three solutions corresponds to a different section of the
frequency spectrum. For discrete signals, the first solution corresponds to
frequencies between 0 and 0.5 of the sampling rate. The second solution
The Scientist and Engineer's Guide to Digital Signal Processing 200
results in frequencies between 0 and -0.5. Lastly, the third solution makes up
the infinite number of duplicated frequencies below -0.5 and above 0.5. If the
signal we are analyzing is continuous, the first solution results in frequencies
from zero to positive infinity, while the second solution results in frequencies
from zero to negative infinity. The third group of solutions does not exist for
continuous signals.
Many DSP techniques do not require the use of negative frequencies, or an
understanding of the DFT's periodicity. For example, two common ones were
described in the last chapter, spectral analysis, and the frequency response of
systems. For these applications, it is completely sufficient to view the time
domain as extending from sample 0 to N-1, and the frequency domain from
zero to one-half of the sampling frequency. These techniques can use a
simpler view of the world because they never result in portions of one period
moving into another period. In these cases, looking at a single period is just
as good as looking at the entire periodic signal.
However, certain procedures can only be analyzed by considering how signals
overflow between periods. Two examples of this have already been presented,
circular convolution and analog-to-digital conversion. In circular
convolution, multiplication of the frequency spectra results in the time domain
signals being convolved. If the resulting time domain signal is too long to fit
inside a single period, it overflows into the adjacent periods, resulting in time
domain aliasing. In contrast, analog-to-digital conversion is an example of
frequency domain aliasing. A nonlinear action is taken in the time domain,
that is, changing a continuous signal into a discrete signal by sampling. The
problem is, the spectrum of the original analog signal may be too long to fit
inside the discrete signal's spectrum. When we force the situation, the ends of
the spectrum protrude into adjacent periods. Let's look at two more examples
where the periodic nature of the DFT is important, compression & expansion
of signals, and amplitude modulation.
Compression and Expansion, Multirate methods
As shown in Fig. 10-12, a compression of the signal in one domain results in
an expansion in the other, and vice versa. For continuous signals, if is the X(f )
Fourier Transform of , then is the Fourier Transform of , x( t ) 1/k ×X( f / k) x(k t)
where k is the parameter controlling the expansion or contraction. If an event
happens faster (it is compressed in time), it must be composed of higher
frequencies. If an event happens slower (it is expanded in time), it must be
composed of lower frequencies. This pattern holds if taken to either of the two
extremes. That is, if the time domain signal is compressed so far that it
becomes an impulse, the corresponding frequency spectrum is expanded so far
that it becomes a constant value. Likewise, if the time domain is expanded
until it becomes a constant value, the frequency domain becomes an impulse.
Discrete signals behave in a similar fashion, but there are a few more details.
The first issue with discrete signals is aliasing. Imagine that the
Chapter 10- Fourier Transform Properties 201
Sample number
0 16 32 48 64 80 96 112 128
-1
0
1
2
3
a. Signal compressed
127
Frequency
0 0.1 0.2 0.3 0.4 0.5
0
4
8
12
16
20
b. Expanded frequency spectrum
Frequency
0 0.1 0.2 0.3 0.4 0.5
0
6
12
18
24
30
d. Frequency spectrum
Frequency Domain Time Domain
Sample number
0 16 32 48 64 80 96 112 128
-1
0
1
2
3
c. Signal
127
samples
underlying
waveform
continious
Sample number
0 16 32 48 64 80 96 112 128
-1
0
1
2
3
e. Signal expanded
127
Frequency
0 0.1 0.2 0.3 0.4 0.5
0
10
20
30
40
50
f. Compressed frequency spectrum
FIGURE 10-12
Compression and expansion. Compressing a signal in one domain results in the signal being expanded in the
other domain, and vice versa. Figures (c) and (d) show a discrete signal and its spectrum, respectively. In (a) and
(b), the time domain signal has been compressed, resulting in the frequency spectrum being expanded. Figures
(e) and (f) show the opposite process. As shown in these figures, discrete signals are expanded or contracted by
expanding or contracting the underlying continuous waveform. This underlying waveform is then resampled
to find the new discrete signal.
Amplitude
Amplitude
Amplitude Amplitude
Amplitude Amplitude
pulse in (a) is compressed several times more than is shown. The frequency
spectrum is expanded by an equal factor, and several of the humps in (b) are
pushed to frequencies beyond 0.5. The resulting aliasing breaks the simple
expansion/contraction relationship. This type of aliasing can also happen in the
The Scientist and Engineer's Guide to Digital Signal Processing 202
time domain. Imagine that the frequency spectrum in (f) is compressed much
harder, resulting in the time domain signal in (e) expanding into neighboring
periods.
A second issue is to define exactly what it means to compress or expand a
discrete signal. As shown in Fig. 10-12a, a discrete signal is compressed by
compressing the underlying continuous curve that the samples lie on, and then
resampling the new continuous curve to find the new discrete signal. Likewise,
this same process for the expansion of discrete signals is shown in (e). When
a discrete signal is compressed, events in the signal (such as the width of the
pulse) happen over a fewer number of samples. Likewise, events in an
expanded signal happen over a greater number of samples.
An equivalent way of looking at this procedure is to keep the underlying
continuous waveform the same, but resample it at a different sampling rate.
For instance, look at Fig. 10-13a, a discrete Gaussian waveform composed of
50 samples. In (b), the same underlying curve is represented by 400 samples.
The change between (a) and (b) can be viewed in two ways: (1) the sampling
rate has been kept constant, but the underlying waveform has been expanded
to be eight times wider, or (2) the underlying waveform has been kept constant,
but the sampling rate has increased by a factor of eight. Methods for changing
the sampling rate in this way are called multirate techniques. If more samples
are added, it is called interpolation. If fewer samples are used to represent
the signal, it is called decimation. Chapter 3 describes how multirate
techniques are used in ADC and DAC.
Here is the problem: if we are given an arbitrary discrete signal, how do
we know what the underlying continuous curve is? It depends on if the
signal's information is encoded in the time domain or in the frequency
domain. For time domain encoded signals, we want the underlying
continuous waveform to be a smooth curve that passes through all the
samples. In the simplest case, we might draw straight lines between the
points and then round the rough corners. The next level of sophistication
is to use a curve fitting algorithm, such as a spline function or polynomial
fit. There is not a single "correct" answer to this problem. This approach
is based on minimizing irregularities in the time domain waveform, and
completely ignores the frequency domain.
When a signal has information encoded in the frequency domain, we ignore
the time domain waveform and concentrate on the frequency spectrum. As
discussed in the last chapter, a finer sampling of a frequency spectrum (more
samples between frequency 0 and 0.5) can be obtained by padding the time
domain signal with zeros before taking the DFT. Duality allows this to work
in the opposite direction. If we want a finer sampling in the time domain
(interpolation), pad the frequency spectrum with zeros before taking the
Inverse DFT. Say we want to interpolate a 50 sample signal into a 400
sample signal. It's done like this: (1) Take the 50 samples and add zeros to
make the signal 64 samples long. (2) Use a 64 point DFT to find the
frequency spectrum, which will consist of a 33 point real part and a 33 point
imaginary part. (3) Pad the right side of the frequency spectrum
Chapter 10- Fourier Transform Properties 203
Sample number
0 10 20 30 40 50
-1
0
1
2
a. Smooth waveform
Sample number
0 80 160 240 320 400
-1
0
1
2
b. Fig. (a) interpolated
Sample number
0 10 20 30 40 50
-1
0
1
2
c. Sharp edges
Sample number
0 80 160 240 320 400
-1
0
1
2
d. Fig. (c) interpolated
Amplitude Amplitude
Amplitude Amplitude
FIGURE 10-13
Interpolation by padding the frequency domain. Figures (a) and (c) each consist of 50 samples. These are interpolated
to 400 samples by padding the frequency domain with zeros, resulting in (b) and (d), respectively. (Figures
(b) and (d) are discrete signals, but are drawn as continuous lines because of the large number of samples).
(both the real and imaginary parts) with 224 zeros to make the frequency
spectrum 257 points long. (4) Use a 512 point Inverse DFT to transform the
data back into the time domain. This will result in a 512 sample signal that is
a high resolution version of the 64 sample signal. The first 400 samples of this
signal are an interpolated version of the original 50 samples.
The key feature of this technique is that the interpolated signal is composed of
exactly the same frequencies as the original signal. This may or may not
provide a well-behaved fit in the time domain. For example, Figs. 10-13 (a)
and (b) show a 50 sample signal being interpolated into a 400 sample signal
by this method. The interpolation is a smooth fit between the original points,
much as if a curve fitting routine had been used. In comparison, (c) and (d)
show another example where the time domain is a mess! The oscillatory
behavior shown in (d) arises at edges or other discontinuities in the signal.
This also includes any discontinuity between sample zero and N-1, since the
time domain is viewed as being circular. This overshoot at discontinuities is
called the Gibbs effect, and is discussed in Chapter 11. Another frequency
domain interpolation technique is presented in Chapter 3, adding zeros between
the time domain samples and low-pass filtering.
The Scientist and Engineer's Guide to Digital Signal Processing 204
Multiplying Signals (Amplitude Modulation)
An important Fourier transform property is that convolution in one domain
corresponds to multiplication in the other domain. One side of this was
discussed in the last chapter: time domain signals can be convolved by
multiplying their frequency spectra. Amplitude modulation is an example of the
reverse situation, multiplication in the time domain corresponds to convolution
in the frequency domain. In addition, amplitude modulation provides an
excellent example of how the elusive negative frequencies enter into everyday
science and engineering problems.
Audio signals are great for short distance communication; when you speak,
someone across the room hears you. On the other hand, radio frequencies are
very good at propagating long distances. For instance, if a 100 volt, 1 MHz
sine wave is fed into an antenna, the resulting radio wave can be detected in
the next room, the next country, and even on the next planet. Modulation is
the process of merging two signals to form a third signal with desirable
characteristics of both. This always involves nonlinear processes such as
multiplication; you can't just add the two signals together. In radio
communication, modulation results in radio signals that can propagate long
distances and carry along audio or other information.
Radio communication is an extremely well developed discipline, and many
modulation schemes have been developed. One of the simplest is called
amplitude modulation. Figure 10-14 shows an example of how amplitude
modulation appears in both the time and frequency domains. Continuous
signals will be used in this example, since modulation is usually carried out in
analog electronics. However, the whole procedure could be carried out in
discrete form if needed (the shape of the future!).
Figure (a) shows an audio signal with a DC bias such that the signal always
has a positive value. Figure (b) shows that its frequency spectrum is composed
of frequencies from 300 Hz to 3 kHz, the range needed for voice
communication, plus a spike for the DC component. All other frequencies
have been removed by analog filtering. Figures (c) and (d) show the carrier
wave, a pure sinusoid of much higher frequency than the audio signal. In the
time domain, amplitude modulation consists of multiplying the audio signal by
the carrier wave. As shown in (e), this results in an oscillatory waveform that
has an instantaneous amplitude proportional to the original audio signal. In the
jargon of the field, the envelope of the carrier wave is equal to the modulating
signal. This signal can be routed to an antenna, converted into a radio wave,
and then detected by a receiving antenna. This results in a signal identical to
(e) being generated in the radio receiver's electronics. A detector or
demodulator circuit is then used to convert the waveform in (e) back into the
waveform in (a).
Since the time domain signals are multiplied, the corresponding frequency
spectra are convolved. That is, (f) is found by convolving (b) & (d). Since the
spectrum of the carrier is a shifted delta function, the spectrum of the
Chapter 10- Fourier Transform Properties 205
Time (milliseconds)
0.0 0.2 0.4 0.6 0.8 1.0
-2
-1
0
1
2
a. Audio signal
Frequency (kHz)
0 5 10 15 20 25 30 35 40
0
1
2
3
4
b. Audio signal
Frequency (kHz)
0 5 10 15 20 25 30 35 40
0
1
2
3
4
d. Carrier signal
Frequency Domain Time Domain
Time (milliseconds)
0.0 0.2 0.4 0.6 0.8 1.0
-2
-1
0
1
2
c. Carrier signal
Frequency (kHz)
0 5 10 15 20 25 30 35 40
0
1
2
3
4
f. Modulated signal
lower
sideband
upper
sideband carrier
Time (milliseconds)
0.0 0.2 0.4 0.6 0.8 1.0
-2
-1
0
1
2
e. Modulated signal
FIGURE 10-14
Amplitude modulation. In the time domain, amplitude modulation is achieved by multiplying the audio signal, (a),
by the carrier signal, (c), to produce the modulated signal, (e). Since multiplication in the time domain corresponds
to convolution in the frequency domain, the spectrum of the modulated signal is the spectrum of the audio signal
shifted to the frequency of the carrier.
Amplitude
Voltage
Amplitude Amplitude
Voltage Voltage
modulated signal is equal to the audio spectrum shifted to the frequency of the
carrier. This results in a modulated spectrum composed of three components:
a carrier wave, an upper sideband, and a lower sideband.
These correspond to the three parts of the original audio signal: the DC
component, the positive frequencies between 0.3 and 3 kHz, and the negative
The Scientist and Engineer's Guide to Digital Signal Processing 206
EQUATION 10-1
The DTFT analysis equation. In this
relation, is the time domain signal x[n]
with n running from 0 to . The N&1
frequency spectrum is held in: (T) Re X
and (T), with T between 0 and pi. Im X
ReX(T) ' j%4
n ' &4
x [n] cos(Tn)
ImX(T) ' & j%4
n ' &4
x [n] sin(Tn)
x[n] ' 1
B m
B
0
ReX(T) cos (Tn) & ImX(T) sin(Tn) dT EQUATION 10-2
The DTFT synthesis
equation.
frequencies between -0.3 and -3 kHz, respectively. Even though the negative
frequencies in the original audio signal are somewhat elusive and abstract, the
resulting frequencies in the lower sideband are as real as you could want them
to be. The ghosts have taken human form!
Communication engineers live and die by this type of frequency domain
analysis. For example, consider the frequency spectrum for television
transmission. A standard TV signal has a frequency spectrum from DC to 6
MHz. By using these frequency shifting techniques, 82 of these 6 MHz wide
channels are stacked end-to-end. For instance, channel 3 is from 60 to 66
MHz, channel 4 is from 66 to 72 MHz, channel 83 is from 884 to 890 MHz,
etc. The television receiver moves the desired channel back to the DC to 6
MHz band for display on the screen. This scheme is called frequency
domain multiplexing.
The Discrete Time Fourier Transform
The Discrete Time Fourier Transform (DTFT) is the member of the Fourier
transform family that operates on aperiodic, discrete signals. The best way
to understand the DTFT is how it relates to the DFT. To start, imagine
that you acquire an N sample signal, and want to find its frequency
spectrum. By using the DFT, the signal can be decomposed into N/2% 1
sine and cosine waves, with frequencies equally spaced between zero and
one-half of the sampling rate. As discussed in the last chapter, padding
the time domain signal with zeros makes the period of the time domain
longer, as well as making the spacing between samples in the frequency
domain narrower. As N approaches infinity, the time domain becomes
aperiodic, and the frequency domain becomes a continuous signal. This is
the DTFT, the Fourier transform that relates an aperiodic, discrete signal,
with a periodic, continuous frequency spectrum.
The mathematics of the DTFT can be understood by starting with the
synthesis and analysis equations for the DFT (Eqs. 8-2, 8-3 and 8-4), and
taking N to infinity:
Chapter 10- Fourier Transform Properties 207
There are many subtle details in these relations. First, the time domain signal,
, is still discrete, and therefore is represented by brackets. In comparison, x[n]
the frequency domain signals, & , are continuous, and are thus ReX(T) Im X(T)
written with parentheses. Since the frequency domain is continuous, the
synthesis equation must be written as an integral, rather than a summation.
As discussed in Chapter 8, frequency is represented in the DFT's frequency
domain by one of three variables: k, an index that runs from 0 to N/2; f, the
fraction of the sampling rate, running from 0 to 0.5; or T, the fraction of the
sampling rate expressed as a natural frequency, running from 0 to B. The
spectrum of the DTFT is continuous, so either or T can be used. The f
common choice is T, because it makes the equations shorter by eliminating the
always present factor of . Remember, when T is used, the frequency 2B
spectrum extends from 0 to B, which corresponds to DC to one-half of the
sampling rate. To make things even more complicated, many authors use S (an
upper case omega) to represent this frequency in the DTFT, rather than T (a
lower case omega).
When calculating the inverse DFT, samples 0 and N/2 must be divided by
two (Eq. 8-3) before the synthesis can be carried out (Eq. 8-2). This is not
necessary with the DTFT. As you recall, this action in the DFT is related
to the frequency spectrum being defined as a spectral density, i.e.,
amplitude per unit of bandwidth. When the spectrum becomes continuous,
the special treatment of the end points disappear. However, there is still a
normalization factor that must be included, the 2/N in the DFT (Eq. 8-3)
becomes 1/B in the DTFT (Eq. 10-2). Some authors place these terms in
front of the synthesis equation, while others place them in front of the
analysis equation. Suppose you start with some time domain signal. After
taking the Fourier transform, and then the Inverse Fourier transform, you
want to end up with what you started. That is, the 1/B term (or the 2/N
term) must be encountered somewhere along the way, either in the forward
or in the inverse transform. Some authors even split the term between the
two transforms by placing in front of both. 1/ B
Since the DTFT involves infinite summations and integrals, it cannot be
calculated with a digital computer. Its main use is in theoretical problems as
an alternative to the DFT. For instance, suppose you want to find the
frequency response of a system from its impulse response. If the impulse
response is known as an array of numbers, such as might be obtained from an
experimental measurement or computer simulation, a DFT program is run on
a computer. This provides the frequency spectrum as another array of
numbers, equally spaced between 0 and 0.5 of the sampling rate.
In other cases, the impulse response might be know as an equation, such as
a sinc function (described in the next chapter) or an exponentially decaying
sinusoid. The DTFT is used here to mathematically calculate the frequency
domain as another equation, specifying the entire continuous curve between
0 and 0.5. While the DFT could also be used for this calculation, it would
only provide an equation for samples of the frequency response, not the
entire curve.
The Scientist and Engineer's Guide to Digital Signal Processing 208
EQUATION 10-3
Parseval's relation. In this equation, is x[i]
a time domain signal with i running from 0
to N-1, and is its modified frequency X[k]
spectrum, with k running from 0 to N/2.
The modified frequency spectrum is found
by taking the DFT of the signal, and
dividing the first and last frequencies
(sample 0 and N/2) by the square-root of
two.
j N&1
i' 0
x [i ]2 ' 2
N
j N /2
k' 0
Mag X [k ]2
Parseval's Relation
Since the time and frequency domains are equivalent representations of the
same signal, they must have the same energy. This is called Parseval's
relation, and holds for all members of the Fourier transform family. For the
DFT, Parseval's relation is expressed:
The left side of this equation is the total energy contained in the time domain
signal, found by summing the energies of the N individual samples. Likewise,
the right side is the energy contained in the frequency domain, found by
summing the energies of the sinusoids. Remember from physics that N/2% 1
energy is proportional to the amplitude squared. For example, the energy in
a spring is proportional to the displacement squared, and the energy stored in
a capacitor is proportional to the voltage squared. In Eq. 10-3, is the X[f ]
frequency spectrum of , with one slight modification: the first and last x[n]
frequency components, & , have been divided by . This X[0 ] X[N/2] 2
modification, along with the 2/N factor on the right side of the equation,
accounts for several subtle details of calculating and summing energies.
To understand these corrections, start by finding the frequency domain
representation of the signal by using the DFT. Next, convert the frequency
domain into the amplitudes of the sinusoids needed to reconstruct the signal, as
previously defined in Eq. 8-3. This is done by dividing the first and last points
(sample 0 and N/2) by 2, and then dividing all of the points by N/2. While this
provides the amplitudes of the sinusoids, they are expressed as a peak
amplitude, not the root-mean-square (rms) amplitude needed for energy
calculations. In a sinusoid, the peak amplitude is converted to rms by dividing
by . This correction must be made to all of the frequency domain values, 2
except sample 0 and N/2. This is because these two sinusoids are unique; one
is a constant value, while the other alternates between two constant values. For
these two special cases, the peak amplitude is already equal to the rms value.
All of the values in the frequency domain are squared and then summed. The
last step is to divide the summed value by N, to account for each sample in the
frequency domain being converted into a sinusoid that covers N values in the
time domain. Working through all of these details produces Eq. 10-3.
While Parseval's relation is interesting from the physics it describes
(conservation of energy), it has few practical uses in DSP.