Fourier series clearly open the frequency domain as an
interesting and useful way of determining how circuits and
systems respond to periodic input signals.
Can we use similar techniques for nonperiodic signals? What is
the response of the filter to a single pulse? Addressing these
issues requires us to find the Fourier spectrum of all signals,
both periodic and nonperiodic ones. We need a definition for
the Fourier spectrum of a signal, periodic
or not. This spectrum is calculated by what is known as the
Fourier transform.
Let
s
T
t
s
T
t
be a periodic signal having period
T
T.
We want to consider what happens to this signal's spectrum as we
let the period become longer and longer. We denote the spectrum
for any assumed value of the period by
c
k
T
c
k
T
.
We calculate the spectrum according to the familiar formula
c
k
T=1T∫−T2T2
s
T
te−i2πktTd
t
c
k
T
1
T
t
T
2
T
2
s
T
t
2
k
t
T
(1)
where we have used a symmetric placement of the integration
interval about the origin for subsequent derivational
convenience. Let
ff be a
fixed frequency equaling
kT
k
T
;
we vary the frequency index
k
k
proportionally as we increase the period. Define
S
T
f≡T
c
k
T=∫−T2T2
s
T
te−(i2πft)d
t
S
T
f
T
c
k
T
t
T
2
T
2
s
T
t
2
f
t
(2)
making the corresponding Fourier series
s
T
t=∑
k
=−∞∞
S
T
fei2πft1T
s
T
t
k
S
T
f
2
f
t
1
T
(3)
As the period increases, the spectral lines become closer
together, becoming a continuum. Therefore,
limit
T
→
∞
s
T
t≡st=∫−∞∞Sfei2πftd
f
T
s
T
t
s
t
f
S
f
2
f
t
(4)
with
Sf=∫−∞∞ste−(i2πft)d
t
S
f
t
s
t
2
f
t
(5)
Sf
S
f
is the Fourier transform of
st
s
t
(the Fourier transform is symbolically denoted by the
uppercase version of the signal's symbol) and is defined
for
any signal for which the integral (
Equation 5) converges.
Let's calculate the Fourier transform of the pulse
signal,
pt
p
t
.
Pf=∫−∞∞pte−(i2πft)d
t
=∫0Δe−(i2πft)d
t
=1−(i2πf)(e−(i2πfΔ)−1)
P
f
t
p
t
2
f
t
t
0
Δ
2
f
t
1
2
f
2
f
Δ
1
Pf=e−(iπfΔ)sinπfΔπf
P
f
f
Δ
f
Δ
f
Note how closely this result resembles the expression for
Fourier
series coefficients of the periodic pulse signal.
Figure 1 shows how increasing the
period does indeed lead to a continuum of coefficients, and that
the Fourier transform does correspond to what the continuum
becomes. The quantity
sintt
t
t
has a special name, the sinc (pronounced "sink")
function, and is denoted by
sinct
sinc
t
.
Thus, the magnitude of the pulse's Fourier transform equals
ΔsincπfΔ
Δ
sinc
f
Δ
.
The Fourier transform relates a signal's time and frequency
domain representations to each other. The direct Fourier
transform (or simply the Fourier transform) calculates a
signal's frequency domain representation from its timedomain
variant (Equation 6). The inverse
Fourier transform (Equation 7)
finds the timedomain representation from the frequency
domain. Rather than explicitly writing the required integral, we
often symbolically express these transform calculations as
ℱs
ℱ
s
and
ℱ1S
ℱ
S
,
respectively.
ℱs=Sf=∫−∞∞ste−(i2πft)d
t
ℱ
s
S
f
t
s
t
2
f
t
(6)
ℱ1S=st=∫−∞∞Sfei2πftdf
ℱ
S
s
t
f
S
f
2
f
t
(7)
We must have
st=ℱ1ℱst
s
t
ℱ
ℱ
s
t
and
Sf=ℱℱ1Sf
S
f
ℱ
ℱ
S
f
,
and these results are indeed valid with minor exceptions.
Recall that the Fourier series for a square wave gives a value
for the signal at the discontinuities equal to the average
value of the jump. This value may differ from how the signal
is defined in the time domain, but being
unequal at a point is indeed minor.
Showing that you "get back to where you started" is difficult
from an analytic viewpoint, and we won't try here. Note that the
direct and inverse transforms differ only in the sign of the
exponent.
The differing exponent signs means that some curious results
occur when we use the wrong sign. What is
ℱSf
ℱ
S
f
?
In other words, use the wrong exponent sign in evaluating
the inverse Fourier transform.
ℱSf=∫−∞∞Sfe−(i2πft)d
f
=∫−∞∞Sfei2πf(−t)d
f
=s−t
ℱ
S
f
f
S
f
2
f
t
f
S
f
2
f
t
s
t
Properties of the Fourier transform and some useful transform
pairs are provided in the accompanying tables (Table 1 and Table 2).
Especially
important among these properties is Parseval's
Theorem, which states that power computed in either
domain equals the power in the other.
∫−∞∞s2td
t
=∫−∞∞Sf2d
f
t
s
t
2
f
S
f
2
(8)
Of practical importance is the conjugate symmetry property: When
st
s
t
is realvalued, the spectrum at negative frequencies equals the
complex conjugate of the spectrum at the corresponding positive
frequencies. Consequently, we need only plot the positive
frequency portion of the spectrum (we can easily determine the
remainder of the spectrum).
How many Fourier transform operations need to be applied to
get the original signal back:
ℱ⋯ℱs=st
ℱ
⋯
ℱ
s
s
t
?
ℱℱℱℱst=st
ℱ
ℱ
ℱ
ℱ
s
t
s
t
.
We know that
ℱSf=∫−∞∞Sfe−(i2πft)d
f
=∫−∞∞Sfei2πf(−t)d
f
=s−t
ℱ
S
f
f
S
f
2
f
t
f
S
f
2
f
t
s
t
. Therefore, two Fourier transforms applied to
st
s
t
yields
s−t
s
t
.
We need two more to get us back where we started.
Note that the mathematical relationships between the time domain
and frequency domain versions of the same signal are termed
transforms. We are transforming (in the
nontechnical meaning of the word) a signal from one
representation to another. We express Fourier transform
pairs as
st↔Sf
↔
s
t
S
f
. A signal's time and frequency domain representations
are uniquely related to each other. A signal thus "exists" in
both the time and frequency domains, with the Fourier transform
bridging between the two. We can define an information carrying
signal in either the time or frequency domains; it behooves the
wise engineer to use the simpler of the two.
A common misunderstanding is that while a signal exists in both
the time and frequency domains, a single formula expressing a
signal must contain only time or frequency:
Both cannot be present simultaneously. This situation mirrors
what happens with complex amplitudes in circuits: As we reveal
how communications systems work and are designed, we will define
signals entirely in the frequency domain without explicitly
finding their time domain variants. This idea is shown in another module where we
define Fourier series coefficients according to letter to be
transmitted. Thus, a signal, though most familiarly defined in
the timedomain, really can be defined equally as well (and
sometimes more easily) in the frequency domain. For example,
impedances depend on frequency and the time variable cannot
appear.
We will learn that
finding a linear, timeinvariant system's output in the time
domain can be most easily calculated by determining the input
signal's spectrum, performing a simple calculation in the
frequency domain, and inverse transforming the result.
Furthermore, understanding communications and information
processing systems requires a thorough understanding of signal
structure and of how systems work in both
the time and frequency domains.
The only difficulty in calculating the Fourier transform of any
signal occurs when we have periodic signals (in either
domain). Realizing that the Fourier series is a special case of
the Fourier transform, we simply calculate the Fourier series
coefficients instead, and plot them along with the spectra of
nonperiodic signals on the same frequency axis.
Table 1: Short Table of Fourier Transform Pairs
st
s
t

Sf
S
f

e−(at)ut
a
t
u
t

1i2πf+a
1
2
f
a

e−(at)
a
t

2a4π2f2+a2
2
a
4
2
f
2
a
2

pt={1 if t<Δ20 if t>Δ2
p
t
1
t
Δ
2
0
t
Δ
2

sinπfΔπf
f
Δ
f

sin2πWtπt
2
W
t
t

Sf={1 if f<W0 if f>W
S
f
1
f
W
0
f
W

Table 2: Fourier Transform Properties

TimeDomain

Frequency Domain

Linearity

a
1
s
1
t+
a
2
s
2
t
a
1
s
1
t
a
2
s
2
t

a
1
S
1
f+
a
2
S
2
f
a
1
S
1
f
a
2
S
2
f

Conjugate Symmetry

st∈R
s
t

Sf=S−f¯
S
f
S
f

Even Symmetry

st=s−t
s
t
s
t

Sf=S−f
S
f
S
f

Odd Symmetry

st=−s−t
s
t
s
t

Sf=−S−f
S
f
S
f

Scale Change

sat
s
a
t

1aSfa
1
a
S
f
a

Time Delay

st−τ
s
t
τ

e−(i2πfτ)Sf
2
f
τ
S
f

Complex Modulation

ei2π
f
0
tst
2
f
0
t
s
t

Sf−
f
0
S
f
f
0

Amplitude Modulation by Cosine

stcos2π
f
0
t
s
t
2
f
0
t

Sf−
f
0
+Sf+
f
0
2
S
f
f
0
S
f
f
0
2

Amplitude Modulation by Sine

stsin2π
f
0
t
s
t
2
f
0
t

Sf−
f
0
−Sf+
f
0
2i
S
f
f
0
S
f
f
0
2

Differentiation

dd
t
st
t
s
t

i2πfSf
2
f
S
f

Integration

∫−∞tsαd
α
α
t
s
α

1i2πfSf
1
2
f
S
f
if
S0=0
S
0
0

Multiplication by tt

tst
t
s
t

1−(i2π)dSfd
f
1
2
f
S
f

Area

∫−∞∞std
t
t
s
t

S0
S
0

Value at Origin

s0
s
0

∫−∞∞Sfd
f
f
S
f

Parseval's Theorem

∫−∞∞st2d
t
t
s
t
2

∫−∞∞Sf2d
f
f
S
f
2

In communications, a very important operation on a signal
st
s
t
is to amplitude modulate it. Using this operation
more as an example rather than elaborating the communications
aspects here, we want to compute the Fourier transform —
the spectrum — of
(1+st)cos2π
f
c
t
1
s
t
2
f
c
t
Thus,
(1+st)cos2π
f
c
t=cos2π
f
c
t+stcos2π
f
c
t
1
s
t
2
f
c
t
2
f
c
t
s
t
2
f
c
t
For the spectrum of
cos2π
f
c
t
2
f
c
t
,
we use the Fourier series. Its period is
1
f
c
1
f
c
,
and its only nonzero Fourier coefficients are
c
±
1
=12
c
±
1
1
2
.
The second term is not periodic unless
st
s
t
has the same period as the sinusoid. Using Euler's relation,
the spectrum of the second term can be derived as
stcos2π
f
c
t=∫−∞∞Sfei2πftd
f
cos2π
f
c
t
s
t
2
f
c
t
f
S
f
2
f
t
2
f
c
t
Using Euler's relation for the cosine,
(stcos2π
f
c
t)=12∫−∞∞Sfei2π(f+
f
c
)td
f
+12∫−∞∞Sfei2π(f−
f
c
)td
f
s
t
2
f
c
t
1
2
f
S
f
2
f
f
c
t
1
2
f
S
f
2
f
f
c
t
(stcos2π
f
c
t)=12∫−∞∞Sf−
f
c
ei2πftd
f
+12∫−∞∞Sf+
f
c
ei2πftd
f
s
t
2
f
c
t
1
2
f
S
f
f
c
2
f
t
1
2
f
S
f
f
c
2
f
t
(stcos2π
f
c
t)=∫−∞∞Sf−
f
c
+Sf+
f
c
2ei2πftd
f
s
t
2
f
c
t
f
S
f
f
c
S
f
f
c
2
2
f
t
Exploiting the uniqueness property of the Fourier transform,
we have
ℱstcos2π
f
c
t=Sf−
f
c
+Sf+
f
c
2
ℱ
s
t
2
f
c
t
S
f
f
c
S
f
f
c
2
(9)
This component of the spectrum consists of the original
signal's spectrum delayed and advanced
in
frequency. The spectrum of the amplitude modulated
signal is shown in
Figure 2.
Note how in this figure the signal
st
s
t
is defined in the frequency domain. To find its time domain
representation, we simply use the inverse Fourier transform.
What is the signal
st
s
t
that corresponds to the spectrum shown in the upper panel of
Figure 2?
The signal is the inverse Fourier transform of the
triangularly shaped spectrum, and equals
st=WsinπWtπWt2
s
t
W
W
t
W
t
2
What is the power in
xt
x
t
,
the amplitudemodulated signal? Try the calculation in
both the time and frequency domains.
The result is most easily found in the spectrum's formula:
the power in the signalrelated part of
xt
x
t
is half the power of the signal
st
s
t
.
In this example, we call the signal
st
s
t
a baseband signal because its power is contained at
low frequencies. Signals such as speech and the Dow Jones
averages are baseband signals. The baseband signal's
bandwidth equals WW,
the highest frequency at which it has power. Since
xt
x
t
's
spectrum is confined to a frequency band not close to the origin
(we assume
f
c
≫W
≫
f
c
W
), we have a bandpass signal. The
bandwidth of a bandpass signal is not its
highest frequency, but the range of positive frequencies where
the signal has power. Thus, in this example, the bandwidth is
2WHz
2
W
Hz
. Why a signal's bandwidth should depend on
its spectral shape will become clear once we develop
communications systems.
"Electrical Engineering Digital Processing Systems in Braille."