Lecture Note
Introduction to Frequency Domain Processing
-
University:
Massachusetts Institute of Technology -
Course:
2.161 | Signal Processing: Continuous and Discrete Academic year:
2019
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Views:
117
Pages:
44
Author:
Annie Lynch
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
DEPARTMENT OF MECHANICAL ENGINEERING
2.161 Signal Processing – Continuous and Discrete
Introduction to Frequency Domain Processing
1
Introduction
In this set of notes we examine an alternative to the time-domain convolution operations describing
the input-output operations of a linear processing system. The methods developed here use Fourier
techniques to transform the temporal representation of a waveform, x(t), to a frequency domain
representation, X(jΩ), where the operation of convolution is replaced by simple multiplication.
In addition, an understanding of Fourier methods gives qualitative insights to signal processing
techniques, such as filtering.
The methods are known collectively as Fourier analysis methods, after Jean Baptiste Joseph
Fourier, who in the early part of the 19th century proposed that an arbitrary repetitive function
could be written as an infinite sum of sine and cosine functions [1]. The Fourier series representation
of periodic functions may be extended through the Fourier transform to represent non-repeating
aperiodic (or transient) functions as a continuous distribution of sinusoidal components. A further
generalization produces the Laplace transform representation of waveforms.
The methods of representing and analyzing waveforms and system responses in terms of the
action of the frequency response function on component sinusoidal or exponential waveforms are
known collectively as the frequency-domain methods. Such methods, developed in this chapter,
have important theoretical and practical applications throughout engineering, especially in signal
processing, system dynamics and control system theory.
2
Fourier Analysis of Periodic Waveforms
Consider the steady-state response of linear time-invariant systems to two periodic waveforms, the
real sinusoid x(t) = sin Ωt, and the complex exponential x(t) = ejΩt = cos Ωt + j sin Ωt. Both
functions are repetitive; that each of them has repeating values at intervals of t = 2π/Ω seconds
apart. In general, a periodic function is a function that satisfies the relationship:
x(t) = x(t + T )
for all t, or x(t) = x(t + nT ) for n = ±1, ±2, ±3, . . ., where T is defined as the period. Figure 1
shows some examples of periodic functions.
The fundamental angular frequency Ω0 (in radians/second) of a periodic waveform is defined
directly from the period
2π
Ω0 =
,
T
and the fundamental frequency F0 (in Hz) is
F0 =
Rowell, September 16, 2008
1
1
T x 1 (t)
x
T
t
2
(t)
x 3 (t)
T
t
0
T
0
Figure 1: Examples of periodic functions of time.
so that Ω0 = 2πF0 .
Any periodic function with period T is also be periodic at intervals of nT for any positive
integer n. Similarly any waveform with a period of T /n is periodic at intervals of T seconds.
Two waveforms whose periods, or frequencies, are related by a simple integer ratio are said to be
harmonically related.
Consider, for example, a pair of periodic functions; the first x1 (t) with a period of T1 = 12
seconds, and the second x2 (t) with a period of T2 = 4 seconds. If the fundamental frequency Ω0
is defined by x1 (t), that is Ωo = 2π/12, then x2 (t) has a frequency of 3Ω0 . The two functions
are harmonically related, and x2 (t) is said to have a frequency which is the third harmonic of the
fundamental Ω0 . If these two functions are summed together to produce a new function g(t) =
x1 (t) + x2 (t), then g(t) will repeat at intervals defined by the longest period of the two, in this
case every 12 seconds. In general, when harmonically related waveforms are added together the
resulting function is also periodic with a repetition period equal to the fundamental period.
Example 1
A family of waveforms gN (t) (N = 1, 2 . . . 5) is formed by adding together the first N
of up to five component functions, that is
gN (t) =
N
�
xn (t)
1 Δ), but at our discretion we can let T become very large. Figure
11 shows the hypothetical periodic extension xp (t) created from the observed x(t). As observers of
the function xp (t) we need not be concerned with its pseudo-periodicity because we will never be
given the opportunity to experience it outside the first period, and furthermore we can assume that
if xp (t) is the input to a linear system, T is so large that the system response decays to zero before
the arrival of the second period. Therefore we assume that the response of the system to x(t) and
xp (t) is identical within our chosen observation interval. The important difference between the two
functions is that xp (t) is periodic, and therefore has a Fourier series description.
The development of Fourier analysis methods for transient phenomena is based on the limiting
behavior of the Fourier series describing xp (t) as the period T approaches infinity. Consider the
behavior of the Fourier series of a simple periodic function as its period T is varied; for example
an even periodic pulse function x(t) of fixed width Δ:
x(t) =
|t| < Δ/2
Δ/2 ≤ |t| ≤ T − Δ/2
1
0
(29)
as shown in Figure 12. Assume that the pulse width Δ remains constant as the period T varies.
The Fourier coefficients in complex form are
Xn =
=
=
=
1
T
Δ/2
−Δ/2
e−jnΩ0 t dt
j −jnΩ0 Δ/2
e
− ejnΩ0 Δ/2
2nπ
1
sin (nπΔ/T )
nπ
Δ sin (nπΔ/T )
n = 0
T (nπΔ/T )
19
(30) x (t)
1 .0
0 .5
-2 T
-T
0
T
2 T
3 T
4 T
t
Figure 12: A periodic rectangular pulse function of fixed duration Δ but varying period T .
and
1
X0 =
T
Δ/2
−Δ/2
1dt =
Δ
T
(31)
and the spectral lines are spaced along the frequency axis at intervals of Ω0 = 2π/T rad/s. To
investigate the behavior of the spectrum as the period T is altered, we define a continuous function
of frequency
sin(ΩΔ/2)
X(Ω) =
(ΩΔ/2)
and note that the Fourier coefficients may be computed directly from X(Ω)
Xn =
=
Δ sin (ΩΔ/2)
T (ΩΔ/2) Ω=2πn/T
(32)
Δ
X(Ω)|Ω=nΩ0
T
(33)
The function X(Ω) depends only on the pulse width Δ, and is independent of the period T . As T
is changed, apart from the amplitude scaling factor Δ/T , the frequency dependence of the Fourier
coefficients is defined by X(Ω); the relative strength of the nth complex harmonic component, at
a frequency nΩ0 , is defined by X(nΩ0 ). The function X(Ω) is therefore an envelope function that
depends only on x(t) and not on the length of the assumed period. Figure 13 shows examples
of the line spectra for the periodic pulse train as the period T is changed. The following general
observations on the behavior of the Fourier coefficients as T varies can be made:
(1) As the repetition period T increases, the fundamental frequency Ω0 decreases, and
the spacing between adjacent lines in the spectrum decreases.
(2) As the repetition period T increases, the scaling factor Δ/T decreases, causing the
magnitude of all of the spectral lines to be diminished. In the limit as T approaches
infinity, the amplitude of the individual lines becomes infinitesimal.
(3) The “shape” of the spectrum is defined by the function X(Ω) and is independent
of T .
Assume that we have an aperiodic function x(t) that is non-zero only for a defined time interval
Δ, and without loss of generality assume that the interval is centered around the time origin (t = 0).
Then assume a periodic extension xp (t) of x(t) with period T that fully encompasses the interval
Δ. The Fourier series description of xp (t) is contained in the analysis and synthesis equations
Xn =
1
T
T /2
−T /2
20
xp (t)e−jnΩ0 t dt
(34) x (t)
T X
T = 4
1 .0
n
M
0
1
x (t)
2
3
t
4
n
T X
n
M
0
x (t)
1
2
3
t
4
T X
1
2
3
t
4
ra d /s e c
n
n
M
0
= p
0
0
T = 1
1 .0
ra d /s e c
0
T = 2
1 .0
= p /2
0
0
= 2 p
ra d /s e c
0
n
Figure 13: Line spectra of periodic extensions of an even rectangular pulse function.
∞
xp (t) =
n=−∞
Xn ejnΩ0 t .
(35)
These two equations may be combined by substituting for Xn in the synthesis equation,
xp (t) =
∞
n=−∞
Ω0
2π
T /2
−T /2
xp (t)e−jnΩ0 t dt ejnΩ0 t
(36)
where the substitution Ω0 /2π = 1/T has also been made.
The period T is now allowed to become arbitrarily large, with the result that the fundamental
frequency Ω0 becomes very small and we write Ω0 = δΩ. We define x(t) as the limiting case of
xp (t) as T approaches infinity, that is
x(t) =
lim xp (t)
T →∞
∞
1
= lim
T →∞
2π
n=−∞
=
T /2
−∞
∞
1
−∞
2π
−∞
−T /2
−jnδΩt
xp (t)e
dt ejnδΩt δΩ
x(t)e−jΩt dt ejΩt dΩ
(37)
where in the limit the summation has been replaced by an integral. If the function inside the
braces is defined to be X(jΩ), Equation 37 may be expanded into a pair of equations, known as
the Fourier transform pair:
X(jΩ) =
x(t) =
∞
−∞
x(t)e−jΩt dt
1 ∞
2π
−∞
21
X(jΩ)ejΩt dΩ
(38)
(39) which are the equations we seek.
Equation 38 is known as the forward Fourier transform, and is analogous to the analysis equation
of the Fourier series representation. It expresses the time-domain function x(t) as a function of
frequency, but unlike the Fourier series representation it is a continuous function of frequency.
Whereas the Fourier series coefficients have units of amplitude, for example volts or Newtons, the
function X(jΩ) has units of amplitude density, that is the total “amplitude” contained within a
small increment of frequency is X(jΩ)δΩ/2π.
Equation 39 defines the inverse Fourier transform. It allows the computation of the time-domain
function from the frequency domain representation X(jΩ), and is therefore analogous to the Fourier
series synthesis equation. Each of the two functions x(t) or X(jΩ) is a complete description of the
function and Equations 38 and 39 allow the transformation between the domains.
We adopt the convention of using lower-case letters to designate time-domain functions, and the
same upper-case letter to designate the frequency-domain function. We also adopt the nomenclature
F
x(t) ⇐⇒ X(jΩ)
as denoting the bidirectional Fourier transform relationship between the time and frequency-domain
representations, and we also frequently write
X(jΩ) = F{x(t)}
x(t) = F −1 {X(jΩ)}
as denoting the operation of taking the forward F{}, and inverse F −1 {} Fourier transforms respec
tively.
4.1
Alternate Definitions
Although the definitions of Eqs. (38) and (39) flow directly from the Fourier series, definitions
for the Fourier transform varies appreciably from text to text and in different disciplines. The
main objection to the convention adopted here is the asymmetry introduced by the factor 1/2π
that appears in the inverse transform. Some authors, usually in physics texts, define the so-called
unitary Fourier transform pair as
X(jΩ) =
x(t) =
∞
1
√
x(t)e−jΩt dt
2π −∞
∞
1
√
X(jΩ)ejΩt dΩ
2π −∞
so as to distribute the constant symmetrically over the forward and inverse transforms.
Many engineering texts address the issue of the asymmetry by defining the transform with
respect to frequency F = 2πΩ in Hz, instead of angular frequency Ω in radians/s. The effect,
through the change in the variable in the inverse transform, is to redefine the transform pair as
X(F ) =
x(t) =
∞
−∞
∞
−∞
x(t)e−j2πF t dt
X(F )ej2πF t dF
Some authors also adopt the notation of dropping the j from the frequency domain representation
and write X(Ω) or X(F ) as above.
22 Even more confusing is the fact that some authors (particularly in physics) adopt a definition
that reverses the sign convention on the exponential terms in the Fourier integral, that is they
define
∞
X(jΩ) =
−∞
1
2π
x(t) =
x(t)ejΩt dt
∞
X(jΩ)e−jΩt dΩ
−∞
These various definitions of the transform pair mean that care must be taken to understand the
particular definition adopted by texts and software packages. Throughout this course we will retain
the definitions in Eqs. (38) and (39).
4.2
Fourier Transform Examples
In this section we present five illustrative examples of Fourier transforms of common time domain
functions.
Example 8
Find the Fourier transform of the pulse function
x(t) =
|t| < T /2
otherwise.
a
0
shown in Figure 14.
X ( j9 )
a
x (t)
a T
F o u r ie r T r a n s fo r m
-T /2
0
T /2
t
-3 0
T
-2 0
T
-1 0
T
0
1 0
T
2 0
T
3 0
T
9
Figure 14: An even aperiodic pulse function and its Fourier transform
Solution: From the definition of the forward Fourier transform
X(jΩ) =
∞
x(t)e−jΩt dt
−∞
T /2
= a
−T /2
e−jΩt dt
(ii)
j −jΩt T /2
= a
e
Ω
−T /2
23
(i)
(iii)
ja −jΩT /2
e
− ejΩT /2
Ω
sin(ΩT /2)
= aT
.
ΩT /2
=
(iv)
(v)
The Fourier transform of the rectangular pulse is a real function, of the form (sin x)/x
centered around the jΩ = 0 axis. Because the function is real, it is sufficient to plot
a single graph showing only |X(jΩ)| as in Figure 14. Notice that while X(jΩ) is a
generally decreasing function of Ω it never becomes identically zero, indicating that the
rectangular pulse function contains frequency components at all frequencies.
The function (sin x)/x = 0 when the argument x = nπ for any integer n (n = 0). The
main peak or “lobe” of the spectrum X(jΩ) is therefore contained within the frequency
band defined by the first two zero-crossings |ΩT /2| < π or |Ω| < 2π/T . Thus as the
pulse duration T is decreased, the spectral bandwidth of the pulse increases as shown in
Figure 15, indicating that short duration pulses have a relatively larger high frequency
content.
a T
x (t)
a
X ( j9 )
F o u r ie r T r a n s fo r m
T
0
-T /2
t
T /2
X ( j9 )
x (t)
a
9
F o u r ie r T r a n s fo r m
T /4
-T /8
0
a T /4
t
T /8
9
Figure 15: Dependence of the bandwidth of a pulse on its duration
Example 9
Find the Fourier transform of the Dirac delta function δ(t).
Solution: The delta impulse function, is an important theoretical function in system
dynamics, defined as
δ(t) =
0
undefined
with the additional defining property that
∞
−∞
δ(t)dt = 1.
24
t = 0
t = 0,
(i) The delta function exhibits a “sifting” property when included in an integrand:
∞
δ(t − T )x(t)dt = x(T ).
−∞
(ii)
When substituted into the forward Fourier transform
∞
Δ(jΩ) =
−∞
δ(t)e−jΩt dt
= 1
(iii)
by the sifting property. The spectrum of the delta function is therefore constant over
all frequencies. It is this property that makes the impulse a very useful test input for
linear systems.
Example 10
Find the Fourier transform of a finite duration sinusoidal “tone-burst”
x(t) =
|t| < T /2
otherwise.
sin Ω0 t
0
shown in Figure 16.
1 .0
x (t)
4 .0
X ( j9 )
F o u r ie r T r a n s fo r m
t
2 .0
0
-1 0
1 0
9
-2 .0
-1 .0
-4 .0
Figure 16: A sinusoidal tone-burst and its Fourier transform
Solution: The forward Fourier transform is
X(jΩ) =
=
∞
x(t)e−jΩt dt
−∞
T /2
−T /2
sin(Ω0 t)e−jΩt dt.
(i)
(ii)
The sinusoid is expanded using the Euler formula:
X(jΩ) =
1
j2
T /2
−T /2
e−j(Ω−Ω0 )t − e−j(Ω+Ω0 )t dt
1
1
T /2
T /2
e−j(Ω−Ω0 )t
−
e−j(Ω+Ω0 )t
−T /2
−T /2
2(Ω − Ω0 )
2(Ω + Ω0 )
T sin ((Ω − Ω0 ) T /2) sin ((Ω + Ω0 ) T /2)
= −j
−
2
(Ω − Ω0 ) T /2
(Ω + Ω0 ) T /2
=
25
(iii)
(iv)
(v) which is a purely imaginary odd function that is the sum of a pair of shifted imaginary
(sin x)/x functions, centered on frequencies ±Ω0 , as shown in Fig. 16. The zero-crossings
of the main lobe of each function are at Ω0 ± 2π/T indicating again that as the duration
of a transient waveform is decreased its spectral width increases. In this case notice
that as the duration T is increased the spectrum becomes narrower, and in the limit as
T → ∞ it achieves zero width and becomes a simple line spectrum.
Example 11
Find the Fourier transform of the one-sided real exponential function
x(t) =
0
e−at
t<0
t ≥ 0.
( for a > 0) as shown in Figure 17.
|X ( jW ) |
1 /a
x (t)
1 .0
F o u r ie r T r a n s fo r m
0 .5
0
e
1
-2 0 a
-a t
0
-1 0 a
|X ( jW ) |
2
1 .5
3
4
1 0 a
2 0 a
1 0 a
2 0 a
W
1 .0
a t
0 .5
-2 0 a
-1 0 a
-0 .5
0
W
-1 .0
-1 .5
Figure 17: The one-sided real exponential function and its spectrum
Solution: From the definition of the forward Fourier transform
X(jΩ) =
=
∞
−∞
∞
0
x(t)e−jΩt dt
(i)
e−at e−jΩt dt
(ii)
−1 −(a+jΩ)t ∞
e
=
a + jΩ
(iii)
0
=
1
a + jΩ
26
(iv) which is complex, and in terms of a magnitude and phase function is
1
+ Ω2
−Ω
X(jΩ) = tan−1
a
√
|X(jΩ)| =
(v)
a2
(vi)
Example 12
Find the Fourier transform of a damped one-sided sinusoidal function
x(t) =
0
e−σt sin Ω0 t
t<0
t ≥ 0.
(for σ > 0) as shown in Figure 18.
|X ( jW ) |
1 .0
x (t)
0 .5
F o u r ie r T r a n s fo r m
0 .5
0
1
2
3
4
-4 0
t
-2 0
0
Ð F ( jW )
2 0
4 0
2 0
4 0
W
3 .0
2 .0
-0 .5
1 .0
-1 .0
-4 0
-2 0
-1 .0
0
W
-2 .0
-3 .0
Figure 18: A damped sinusoidal function and its spectrum
Solution: From the definition of the forward Fourier transform
X(jΩ) =
=
=
∞
−∞
x(t)e−jΩt dt
1 ∞ −σt jΩ0 t
e
e
− e−jΩ0 t e−jΩt dt
j2 0
∞
1
1
−(σ+j(Ω−Ω0 ))t
−
e
−
j2
σ + j(Ω0 − Ω)
0
∞
1
1
−(σ+j(Ω+Ω0 ))t
e
−
j2
σ + j(Ω0 + Ω)
0
=
Ω0
.
(σ + jΩ)2 + Ω20
27
(i)
(ii)
(iii)
(iv)
(v) The magnitude and phase of this complex quantity are
|X(jΩ)| =
4.3
Ω0
(vi)
(σ 2 + Ω20 − Ω2 )2 + (2σΩ)2
X(jΩ) = tan−1
σ2
−2σΩ
+ Ω20 − Ω2
(vii)
Properties of the Fourier Transform
A full description of the properties of the Fourier transform is beyond the scope and intent of this
handout, and the interested reader is referred to the many texts devoted to the Fourier transform
[2,4,5]. The properties listed below are presented because of their importance in the study of system
dynamics. In the following properties it is assumed that
F
x(t) ⇐⇒ X(jΩ)
(1) Existence of the Fourier Transform A modified form of the three Dirichlet
conditions presented for the Fourier series guarantees the existence of the Fourier
transform. These are sufficient conditions, but are not strictly necessary. For the
Fourier transform the conditions are
• The function x(t) must be integrable in the absolute sense over all time, that
is
∞
|x(t)| dt < ∞.
−∞
• There must be at most a finite number of maxima and minima in the function
x(t). Notice that periodic functions are excluded by this and the previous
condition.
• There must be at most a finite number of discontinuities in the function x(t),
and all such discontinuities must be finite in magnitude.
(2) Linearity of the Fourier Transform Like the Fourier series, the Fourier trans
form is a linear operation. If
F
F
g(t) ⇐⇒ G(jΩ) and h(t) ⇐⇒ H(jΩ)
then for arbitrary constants a and b,
F
ag(t) + bh(t) ⇐⇒ aG(jΩ) + bH(jΩ).
(40)
(3) Duality If
1
x(t) =
2π
∞
−∞
jΩt
X(jΩ)e
F
dΩ ⇐⇒ X(jΩ) =
∞
−∞
x(t)e−jΩt dt
then by definition
F {X(jt)} =
∞
−∞
−jΩt
X(jt)e
dt =
28
∞
−∞
X(jt)ej(−Ω)t dt = 2πx(−Ω) or
F
X(jt) ⇐⇒ 2πx(−Ω)
where X(jt) is X(jΩ) where Ω has been replaced by t, and x(−Ω) is x(t) where
t is replaced by −Ω. Therefore if we know the Fourier transform of one function,
we also know it for another.
(4) Even and Odd Functions Let x(t) be a real function, and write it in terms of
an even function xe (t) and an odd function xo (t)
x(t) = xe (t) + xo (t).
Then the Fourier transform is
X(jΩ) =
=
∞
−∞
∞
−∞
(xe (t) + xco (t)) (cos(Ωt) − j sin(Ωt)) dt
xe (t) cos(Ωt)dt − j
∞
−∞
xo (t) sin(Ωt)dt
(41)
since odd terms in the integrand contribute zero to the integral. Equation (41)
allows the following conclusions:
• For real x(t) the Fourier transform shows conjugate symmetry, that is
X(−jΩ) = X(jΩ).
(42)
• The Fourier transform of a real even function of time is a real even function
• The Fourier transform of an real odd function is an imaginary odd function.
Similar arguments may be used to make similar conclusions for the inverse Fourier
transform, for example if X(jΩ) is a real function, then
x(−t) = x(t).
(5) Time Shifting The Fourier transform of x(t + τ ), a time shifted version of x(t),
is
F{x(t + τ )} = ejΩτ X(jΩ).
This result can be shown easily, since by definition
F{x(t + τ )} =
∞
−∞
x(t + τ )e−jΩt dt.
With a change of variable ν = t + τ ,
F{x(t + τ )} =
∞
−∞
jΩτ
= e
x(ν)e−jΩ(ν−τ ) dν
∞
−∞
x(ν)e−jΩν dν
= ejΩτ X(jΩ)
X(jΩ)+Ωτ )
= |X(jΩ)| ej(
.
(43)
which indicates that the Fourier transform of a time shifted waveform has the
same magnitude function as the original waveform, but has an additional phaseshift term that is directly proportional to frequency.
29 (4) Time Scaling
1
X (jΩ/a) ,
|a|
F
x(at) ⇐⇒
a = 0
(44)
(9) Time Reversal If a = −1, the time scaling property gives
F
x(−t) ⇐⇒ X(−jΩ).
(45)
(5) Waveform Energy We have asserted that the time domain representation x(t)
and the frequency domain representation X(jΩ) are both complete descriptions of
the function related through the Fourier transform
F
x(t) ⇐⇒ X(jΩ).
If we consider the function x(t) to be a system variable, for example a voltage or
current in an electrical system, the instantaneous power dissipated in a resistive
element with a value of unity is equal to x2 (t). For example, the power dissipated
when voltage v(t) is applied to an electrical resistance of 1 ohm is v 2 (t). The
power associated with a complex variable v(t) is |v(t)|2 . The “energy” of an ape
riodic function in the time domain is defined as the integral of this hypothetical
instantaneous power over all time
E=
∞
−∞
|x(t)|2 dt
(46)
Parseval’s theorem [3] asserts the equivalence of the total waveform energy in the
time and frequency domains by the relationship
∞
−∞
2
|x(t)| dt =
=
1 ∞
|X(jΩ)|2 dΩ
2π −∞
1 ∞
X(jΩ)X(jΩ)dΩ.
2π −∞
(47)
In other words, the quantity |X(jΩ)|2 is a measure of the energy of the function
per unit bandwidth.
The energy ΔE contained between two angular frequencies Ω1 and Ω2 is
ΔE =
1
2π
Ω2
Ω1
X(jΩ)X(jΩ)dΩ.
(48)
Notice that this is a one-sided energy content and that because the Fourier trans
form is a two-sided spectral representation, the total energy in a real function
includes contributions from both positive and negative frequencies. The function
Φ(jΩ) = |X(jΩ)|2
is a very important quantity in experimental system dynamics, and is known as
the energy density spectrum. It is a real function, with units of energy per unit
bandwidth, and shows how the energy of a waveform x(t) is distributed across the
spectrum.
30 (6) Relationship Between the Fourier Transform and the Fourier
Series of a Periodic Extension of an Aperiodic Function Let x(t) be a
function, with Fourier transform X(jΩ), that exists only in a defined interval
|t| < Δ/2 centered on the time origin. Then if xp (t) is a periodic extension of x(t),
formed by repeating x(t) at intervals T > Δ, each period contains x(t). Then the
Fourier coefficients describing xp (t) are
1 T /2
xp (t)e−jnΩ0 t dt
T −T /2
1 ∞
x(t)e−jnΩ0 t dt
T −∞
1
X(jnΩ0 ).
T
cn =
=
=
(49)
The Fourier coefficients of a periodic function are scaled samples of the Fourier
transform of the function contained within a single period. The transform thus
forms the envelope function for the definition of the Fourier series as discussed in
Section 2.
(7) The Fourier Transform of the Derivative of a Function If a function x(t)
has a Fourier transform X(jΩ) then
F
dx
dt
= jΩX(jΩ),
which is easily shown using integration by parts:
F
dx
dt
∞
dx −jΩt
=
dt
e
−∞
dt
∞
∞
−jΩt
= x(t)e
−
x(t)(−jΩ)e−jΩt dt
−∞
−∞
= 0 + jΩX(jΩ)
since by definition x(t) = 0 at t = ±∞.
This result can be applied repetitively to show that the Fourier transform of the
nth derivative of x(t) is
n
d x
F
= (jΩ)n X(jΩ)
(50)
dtn
(8) The Fourier Transform of the Integral of a Function If a function x(t) has
a Fourier transform X(jΩ) then
F
t
−∞
x(ν)dν
= πX(0)δ(jΩ) +
1
X(jΩ)
jΩ
(51)
The additive term πX(j0)δ(jΩ) accounts for the possibility of a non-zero dc com
ponent in x(t), where
X(j0) =
∞
−∞
x(t)dt
is the average value of x(t).
(9) Time Reversal If a function x(t) has a Fourier transform X(jΩ) then
F {x(−t)} = X(−jΩ).
31
(52) 4.4
Extension of the Fourier Transform to Functions for which the Fourier
Integral does not Converge.
The Dirichelet conditions (described in Sec. 4.3) are sufficient but not necessary conditions for the
existence of the Fourier transform. If the use of the Dirac delta function δ(x) is allowed, the Fourier
transform of many functions with a non-convergent Fourier integral may de defined. In this section
we examine the approach through several examples.
Example 13
Define the Fourier transform of a constant x(t) = a. The Fourier integral
X(jΩ) =
∞
ae−jΩt dt
−∞
does not converge. In Ex. 8 the Fourier transform of a even pulse of duration T
g(t) =
|t| < T /2
otherwise.
a
0
was shown to be
G(jΩ) = aT
sin(ΩT /2)
.
ΩT /2
It is also shown in the example that as the duration T is increased, the peak ampli
tude of G(jΩ) (at jΩ = 0) increases, and its bandwidth decreases. Since x(t) = a =
limT →∞ g(t), we have
X(jΩ) = lim G(jΩ)] = lim
T →∞
T →∞
sin(ΩT /2)
aT
ΩT /2
= 2πaδ(Ω)
where δ() is the Dirac delta function. In other words the Fourier transform of a constant
is an impulse at Ω = 0.
Example 14
Define the Fourier transform of the unit-step (Heaviside) function us (t), where
us (t) =
0
1
t<0
t ≥ 0.
Clearly the Fourier integral
Us (jΩ) =
∞
∞
us (t)ejΩt dt =
∞
0
e−jΩt dt
does not converge, and we seek an alternative approach.
32
(i) Consider the one-sided real exponential function
x(t) =
0
e−at
t<0
t ≥ 0.
as described in Example 12, and note that us (t) = lima→0 x(t) which implies Us (jΩ) =
lima→0 X(jΩ). From Example 12
X(jΩ) =
1
a
Ω
= 2
−j 2
2
a + jΩ
a +Ω
a + Ω2
(ii)
The real part is in the form of a Cauchy distribution, and we note that for a > 0
∞
−∞
a2
a
dΩ = π
+ Ω2
and that as a → 0, a/(a2 + Ω2 ) becomes impulse-like, Therefore, as a → 0
a2
a
→ πδ(Ω)
+ Ω2
and
−j
a2
Ω
1
→ −j
2
+Ω
Ω
so that we may define the Fourier transform of the unit-step function as
Us (jΩ) = πδ(Ω) +
1
jΩ
(iii)
Note that the same result may be derived by considering that the unit-step as the
running integral of the Dirac delta function, and using the integral property (Property
8 in Section 4.2) of the Fourier transform.
Example 15
Find the Fourier transform of the signum (sign) function
⎧
⎪
⎨ −1
sgn(t) =
⎪
⎩
for t < 0
for t = 0 .
for t > 0
0
1
We note that
sgn(x) = 2us (t) − 1
and using the linearity property
F {sgn(t)} = 2F {us (t)} − F {1}
1
=
2πδ(Ω) + 2
− 2πδ(Ω)
jΩ
2
=
jΩ
33 Example 16
Define the Fourier transform of the sinusoid g(t) = sin(Ω0 t).
In Example 11 the Fourier transform of a finite sinusoidal tone-burst of duration T
gT (t) =
was shown to be
T
GT (jΩ) = −j
2
|t| < T /2
otherwise.
sin Ω0 t
0
sin ((Ω − Ω0 ) T /2) sin ((Ω + Ω0 ) T /2)
−
.
(Ω − Ω0 ) T /2
(Ω + Ω0 ) T /2
(i)
Since g(t) = limT →∞ gT (t), we define G(jΩ) = limT →∞ GT (jΩ), and we note that
∞
sin ax
ax
−∞
dx =
π
.
a
As a → ∞, a sin(ax)/ax becomes impulse-like with area π, so that
G(jΩ) = lim GT (jΩ) = −jπ (δ(Ω − Ω0 ) − δ(Ω + Ω0 )) .
T →∞
(ii)
The Fourier transform of a sinusoid is therefore purely imaginary, with a pair of impulses
at plus and minus the frequency Ω0 .
A similar argument shows
F {cos(Ω0 t)} = π (δ(Ω − Ω0 ) + δ(Ω + Ω0 ))
which is a real function.
Example 17
Find the inverse Fourier transform of a one-sided spectrum X(jΩ) = us (Ω), where us (x)
is the unit step function in x. Then X(jΩ) ≡ 0 for Ω < 0.
In Ex. 14 it is shown that
us (t)
F
⇐⇒
πδ(Ω) +
F
1
jΩ
F
Using the duality property, that is if x(t) ⇐⇒ X(jΩ) then X(jt) ⇐⇒ 2πx(−Ω), we
have
1 F
X(jt) = πδ(t) +
⇐⇒ 2πx(−Ω) = 2πus (−Ω).
jt
Therefore
1
j
F
⇐⇒
us (Ω)
(time reversal)
δ(−t) +
2
2πt
1
j
F
⇐⇒
us (Ω)
(δ(t) assumed to be even).
δ(t) +
2
2πt
and
1
j
F −1 {us (Ω)} = δ(t) +
.
2
2πt
34 5
5.1
Fourier Transform Based Properties of Linear Systems
Response of Linear Systems to Aperiodic Inputs
The Fourier transform provides an alternative method for computing the response of a linear system
to a transient input. Assume that a linear system with frequency response H(jΩ) is initially at
rest, and is subsequently subjected to an aperiodic input u(t) having a Fourier Transform U (jΩ).
The task is to compute the response y(t).
Assume that the input function u(t), is a periodic function up (t) with an arbitrarily large period
T , and that since up (t) is periodic it can be described by a set of complex Fourier coefficients {Un }.
The output yp (t) is periodic with period T and is described by its own set of Fourier coefficients
{Yn }. In Section 3.2 it was shown that the output coefficients are
Yn = H(jnΩ0 )Un ,
(53)
and that the Fourier synthesis equation for the output is
yp (t) =
=
=
∞
n=−∞
∞
n=−∞
∞
n=−∞
Yn ejnΩ0 t
(54)
H(jnΩ0 )Un ejnΩ0 t
(55)
H(jnΩ0 )
T /2
Ω0
2π
−T /2
up (t)e−jnΩ0 t dt ejnΩ0 t
(56)
As in the development of the Fourier transform, we let T → ∞ so that up (t) → u(t), and in the
limit write the summation as an integral
y(t) =
=
=
lim yp (t)
T →∞
∞
1 ∞
H(jΩ)
u(t)e−jΩt dt ejΩt dΩ
2π −∞
−∞
1 ∞
H(jΩ)U (jΩ)ejΩt dΩ.
2π −∞
(57)
This equation expresses the system output y(t) in the form of the inverse Fourier transform of the
product H(jΩ)U (jΩ) or
y(t) = F −1 {H(jΩ)U (jΩ)} .
(58)
We can therefore write
Y (jΩ) = H(jΩ)U (jΩ),
(59)
which is the fundamental frequency-domain input/output relationship for a linear system. The
output spectrum is therefore the product of the input spectrum and the system frequency response
function:
Given a relaxed linear system with a frequency response H(jΩ) and an input that
possesses a Fourier transform, the response may be found by the following three steps:
(1) Compute the Fourier transform of the input
U (jΩ) = F {u(t)} .
35 u (t)
U ( j9 )
L in e a r S y s te m
y (t)
H ( j9 )
-1
Y ( j9 ) = U ( j9 ) H ( j9 )
m u ltip lic a tio n
Figure 19: Frequency domain computation of system response.
(2) Form the output spectrum as the product
Y (jΩ) = H(jΩ)U (jΩ),
(3) Compute the inverse Fourier transform
y(t) = F −1 {Y (jΩ)} .
Figure 19 illustrates the steps involved in computing the system response using the Fourier trans
form method.
Example 18
Use the Fourier transform method to find the response of a linear first-order system
with a transfer function
1
H(s) =
τs + 1
to a one-sided decaying exponential input
u(t) =
0
e−at
t<0
t ≥ 0.
where a > 0.
Solution: The frequency response of the system is
H(jΩ) =
1/τ
τ + jΩ
(i)
and from Example 11 the Fourier transform of a decaying exponential input is
U (jΩ) = F {u(t)}
= F e−aτ
1
=
.
a + jΩ
(ii)
The output spectrum is the product of the transfer function and the frequency response
Y (jΩ) = H(jΩ)U (jΩ)
1/τ
1
=
.
.
1/τ + jΩ a + jΩ
36
(iii) In order to compute the time domain response through the inverse transform, it is
convenient to expand Y (jΩ) in terms of its partial fractions
Y (jΩ) =
1
1
1
−
aτ − 1 1/τ + jΩ a + jΩ
(iv)
provided a =
1/τ , and using the linearity property of the inverse transform
y(t) = F −1 {Y (jΩ)}
1
1
1
−1
−1
=
F
−F
aτ − 1
1/τ + jΩ
a + jΩ
(v)
Using the results of Example 11 once more
F
e−aτ ⇐⇒
1
,
a + jΩ
(vi)
the desired solution is
1 −t/τ
e
− e−at
aτ − 1
These input/output relationships are summarized in Figure 20.
y(t) =
(vii)
�
������
��������
� ����
������
� ���
������
����
�
���
�
�
�
�
�
�
���������������
Figure 20: First-order system response to an exponential input
Before the 1960’s frequency domain analysis methods were of theoretical interest, but the dif
ficulty of numerically computing Fourier transforms limited their applicability to experimental
studies. The problem lay in the fact that numerical computation of the transform of n samples
of data required n2 complex multiplications, which took an inordinate amount of time on existing
digital computers. In the 1960’s a set of computational algorithms, known as the Fast Fourier
transform (FFT) methods, that required only n log2 n multiplications for computing the Fourier
transform of experimental data were developed. The computational savings are very great, for
example in order to compute the transform of 1024 data points, the FFT algorithm is faster by
a factor of more than 500. These computational procedures revolutionized spectral analysis and
37 frequency domain analysis of system behavior, and opened up many new analysis methods that
had previously been impractical. FFT based system analysis is now routinely done in both software
and in dedicated digital signal-processing (DSP) electronic hardware. These techniques are based
on a “discrete-time” version of the continuous Fourier transforms described above, and have some
minor differences in definition and interpretation.
5.2
The Frequency Response Defined Directly from the Fourier Transform
The system frequency response function H(jΩ) may be defined directly using the transform prop
erty of derivatives. Consider a linear system described by the single input/output differential
equation
an
dn y
dn−1 y
dy
+
a
+ . . . + a1
+ a0 y =
n−1
n
n−1
dt
dt
dt
dm u
dm−1 u
du
bm m + bm−1 m−1 + . . . + b1
+ b0 u
dt
dt
dt
(60)
and assume that the Fourier transforms of both the input u(t) and the output y(t) exist. Then the
Fourier transform of both sides of the differential equation may be found by using the derivative
property (Property (7) of Section 4.2):
F
n
d f
dtn
= (jΩ)n F (jΩ)
to give
an (jΩ)n + an−1 (jΩ)n−1 + . . . + a1 (jΩ) + a0 Y (jΩ) =
bm (jΩ)m + bm−1 (jΩ)m−1 + . . . + b1 (jΩ) + b0 U (jΩ),
(61)
which has reduced the original differential equation to an algebraic equation in jΩ. This equation
may be rewritten explicitly in terms of Y (jΩ) in terms of the frequency response H(jΩ)
bm (jΩ)m + bm−1 (jΩ)m−1 + . . . + b1 (jΩ) + b0
U (jΩ)
an (jΩ)n + an−1 (jΩ)n−1 + . . . + a1 (jΩ) + a0
= H(jΩ)U (jΩ),
Y (jΩ) =
(62)
(63)
showing again the generalized multiplicative frequency domain relationship between input and
output.
5.3
Relationship between the Frequency Response and the Impulse Response
In Example 9 it is shown that the Dirac delta function δ(t) has a unique property; its Fourier
transform is unity for all frequencies
F {δ(t)} = 1,
The impulse response of a system h(t) is defined to be the response to an input u(t) = δ(t), the
output spectrum is then Yδ (jΩ) = F {h(t)},
Y (jΩ) = F {δ(t)} H(jΩ)
= H(jΩ).
38
(64) or
h(t) = F −1 {H(jΩ)} .
(65)
In other words, the system impulse response h(t) and its frequency response H(jΩ) are a Fourier
transform pair:
F
h(t) ⇐⇒ H(jΩ).
(66)
In the same sense that H(jΩ) completely characterizes a linear system in the frequency response,
the impulse response provides a complete system characterization in the time domain.
Example 19
An unknown electrical circuit is driven by a pulse generator, and its output is connected
to a recorder for subsequent analysis, as shown in Figure 21.
������������
����������
�����
���������
����
����
�
�
�������
�������
Figure 21: Setup for estimating the frequency response of an electrical circuit
The pulse generator produces pulses of 1 msec. duration and an amplitude of 10 volts.
When the circuit is excited by a single pulse the output is found to be very closely
approximated by a damped sinusoidal oscillation of the form
y(t) = 0.02e−t sin(10t).
Estimate the frequency response of the system.
Solution: The input u(t) is a short rectangular pulse, much shorter in duration than the
observed duration of the system response. The impulse function δ(t) is the limiting case
of a unit area rectangular pulse, as its duration approached zero. For this example we
assume that the duration of the pulse is short enough to approximate a delta function,
and because this pulse has an area of 10 × 0.001 = 0.01 v-s, we assume
u(t) = 0.01δ(t)
(i)
and therefore assume that the observed response is a scaled version of the system impulse
response,
y(t) = 0.01h(t),
(ii)
39 or
h(t) = 100y(t)
= 2e−5t sin(12t).
The frequency response is
H(jΩ) = F {h(t)}
�
�
= 2F e−5t sin(12t) .
(iii)
In Example 12 it is shown that
�
�
F e−σt sin Ω0 t =
Ω0
,
(σ + jΩ)2 + Ω20
and substituting Ω0 = 12, σ = 5 gives
H(jΩ) =
24
(jΩ)2 + j20Ω + 169
(iv)
We therefore make the substitution s = jΩ and conclude that our unknown electrical
network is a second-order system with a transfer function
H(s) =
s2
24
,
+ 20s + 169
(v)
which has an undamped natural frequency Ωn = 13 radians/sec. and a damping ratio
ζ = 10/13. The input/output differential equation is
dy
d2 y
+ 20 + 169y = 24u(t).
2
dt
dt
(vi)
Example 20
Find the impulse response of an all-pass filter that provides a phase shift of −π/2 rad.
(−90◦ deg.) at all frequencies. The required frequency response function is
⎧
−jπ/2
⎪
⎨ −j = e
H(jΩ) = −jsgn(Ω) =
0
⎪
⎩ j = ejπ/2
Ω>0
Ω=0
Ω<0
For example, if a signal f (t) = cos(Ω0 (t) is applied to such a filter, the output y(t) may
be found
F
f (t) = cos(Ωo t) ⇐⇒ π (δ(Ω − Ω0 ) + δ(Ω + Ω0 ))
and
Y (jΩ) = F (jΩ)H(jΩ) = −jπ (δ(Ω − Ω0 ) − δ(Ω + Ω0 )) = F {sin(Ω0 t)}
and y(t) = sin(Ω0 t) = cos(Ω0 t − π/2). Similarly if f (t) = cos(Ω0 (t), then y(t) =
sin(Ω0 t − π/2).
40 Solution: The impulse response is h(t) = F −1 {H(jΩ)} = F −1 {−jsgn(Ω)}. In Ex. 15
it was shown
2
F {sgn(t)} =
jΩ
F
and using the duality property of the Fourier transform, that is if f (t) ⇐⇒ F (jΩ), then
F
F (jt) ⇐⇒ 2πf (−Ω), we have
1
πt
h(t) = F {−jsgn(Ω)} =
which is clearly acausal, and has a singularity at time t = 0. Although not physically
realizable, this filter, known as a Hilbert transformer, plays an important role in com
munication theory. (Such filters may be approximated as digital filters using discrete
time methods.)
5.4
The Convolution Property
A system with an impulse response h(t), driven by an input u(t), responds with an output y(t)
given by the convolution integral
y(t) = h(t) ⊗ u(t)
=
� ∞
−∞
u(τ )h(t − τ )dτ
(67)
or alternatively by changing the variable of integration
y(t) =
� ∞
−∞
u(t − τ )h(τ )dτ.
(68)
In the frequency domain the input/output relationship for a linear system is multiplicative, that is
Y (jΩ) = U (jΩ)H(jΩ). Because by definition
F
y(t) ⇐⇒ Y (jΩ),
we are lead to the conclusion that
F
h(t) ⊗ u(t) ⇐⇒ H(jΩ)U (jΩ).
(69)
The computationally intensive operation of computing the convolution integral has been replaced
by the operation of multiplication. This result, known as the convolution property of the Fourier
transform, can be shown to be true for the product of any two spectra, for example F (jΩ) and
G(jΩ)
F (jΩ)G(jΩ) =
=
� ∞
−∞
−jΩν
f (ν)e
� ∞ � ∞
−∞ −∞
41
dν.
� ∞
−∞
g(τ )e−jΩτ dτ
−jΩ(ν+τ )
f (ν)g(τ )e
dτ dν, and with the substitution t = ν + τ
F (jΩ)G(jΩ) =
=
� ∞ �� ∞
−∞
� ∞
−∞
−∞
�
f (t − τ )g(τ )dτ e−jΩt dt
(f (t) ⊗ g(t)) e−jΩt dt
= F {f (t) ⊗ g(t)} .
(70)
A dual property holds: if any two functions, f (t) and g(t), are multiplied together in the time
domain, then the Fourier transform of their product is a convolution of their spectra. The dual
convolution/multiplication properties are
F
f (t) ⊗ g(t) ⇐⇒ F (jΩ)G(jΩ)
1
F
f (t)g(t) ⇐⇒
F (jΩ) ⊗ G(jΩ).
2π
Example 21
A real signal f (t) has a two-sided spectrum F (jΩ). Find the signal fˆ(t) that has a
right-sided spectrum F̂ (jΩ), that is
�
F̂ (jΩ) =
F (jΩ),
0,
Ω≥0
Ω < 0,
or F̂ (jΩ) = us (Ω)F (jΩ), where u(s (Ω) is the unit step function in Ω.
In Ex. 17 it is shown that
1
j
F −1 {us (Ω)} = δ(t) +
.
2
2πt
If F̂ (jΩ) is formed as the product F̂ (jΩ) = us (Ω)F (jΩ), then
�
fˆ(t) =
=
=
�
1
j
δ(t) +
⊗ f (t)
2
2πt
� ∞
1
j
f (τ )
f (t) +
dτ
2
2π −∞ t − τ
1
(f (t) + jH {f (t)})
2
where
�
1 ∞ f (τ )
dτ
π −∞ t − τ
is known as the Hilbert transform, and and is used frequently in communication systems.
The impulse response of the Hilbert transformer is
H {f (t)} =
1
πt
and because of the singularity at t = 0, the convolution should be written as a Cauchy
principal value:
� ∞
1
f (τ )
H {f (t)} = P V
dτ
π
t
−∞ − τ
h(t) =
42
(71)
(72) u (t)
u (t)
H 1 ( jW )
H
1
( jW )
H 2 ( jW )
H 2 ( jW )
+
+
y (t)
u (t)
y (t)
u (t)
H 1 ( jW ) H 2 ( jW )
y (t)
H 1 ( jW ) + H 2 ( jW )
y (t)
Figure 22: Frequency Response of Cascaded and Parallel Linear Systems
5.5
The Frequency Response of Interconnected Systems
If two linear systems H1 (jΩ) and H2 (jΩ) are connected in cascade, or series, as shown if Fig. 22 so
that the output variable of the first is the input to the second, then provided the interconnection
does not affect y1 (t), overall frequency response is
Y2 (jΩ) = H2 (jΩ)Y1 (jΩ)
= H2 (jΩ) {H1 (jΩ)U (jΩ)}
= {H2 (jΩ)H1 (jΩ)} U (jΩ)
(73)
The overall frequency response is therefore the product of the two cascaded frequency responses
H(jΩ) = H1 (jΩ)H2 (jΩ).
(74)
Similarly, if two linear systems are connected in parallel so that their outputs are summed
together, then
Y (jΩ) = H1 (jΩ)U (jΩ) + H2 (jΩ)U (jΩ)
= {H1 (jΩ) + H2 (jΩ)} U (jΩ),
(75)
so that the overall frequency response is the sum of the two component frequency responses
H(jΩ) = H1 (jΩ) + H2 (jΩ).
(76)
References:
1. Gratton-Guiness, I., Joseph Fourier 1768–1830, The MIT Press, Cambridge MA, 1972
2. Bracewell, R. N., The Fourier Transform and Its Applications, (2nd Edition), McGraw-Hill,
New York NY, 1978
3. Papoulis, A., The Fourier Integral and Its Applications, McGraw-Hill, New York NY, 1962
43 4. Dym, H. and McKean, H. P., Fourier Series and Integrals, Academic Press, New York NY,
1972
5. Oppenheim, A. V., Willsky, A. S., Young, I. T., Signals and Systems, Prentice Hall Inc.
Englewood Cliffs NJ, 1983
6. Brigham, E. O., The Fast Fourier Transform, Prentice Hall Inc. Englewood Cliffs NJ, 1974
44
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