17.2 The Convergence of the Fourier Integral A function of time / 0 has a Fourier transform if the integral in Eq.. A single-valued function that is nonzero over an infinite interval
Trang 1646 The Fourier Transform
17.1 The Derivation of the
Fourier Transform
We begin the derivation of the Fourier transform, viewed as a limiting case
of a Fourier series, with the exponential form of the series:
00
n oa
where
1 f r/2
* J-T/2
In Eq 17.2, we elected to start the integration at t 0 = - 7 / 2
Allowing the fundamental period T to increase without limit
accom-plishes the transition from a periodic to an aperiodic function In other
words, if T becomes infinite, the function never repeats itself and hence is aperiodic As T increases, the separation between adjacent harmonic
fre-quencies becomes smaller and smaller In particular,
Aw = {n + 1)0)() - HWQ = o>() = — , (17.3)
and as 7 gets larger and larger, the incremental separation &co approaches
a differential separation dxo From Eq 17.3,
1 doj „
T ^ Z T T " ^7 - ^0 0 , ( 1 7'4)
As the period increases, the frequency moves from being a discrete vari-able to becoming a continuous varivari-able, or
In terms of Eq 17.2, as the period increases, the Fourier coefficients C„
get smaller In the limit, C„—»0 as T—> oo This result makes sense,
because we expect the Fourier coefficients to vanish as the function loses
its periodicity Note, however, the limiting value of the product C„T; that is,
In writing Eq 17.6 we took advantage of the relationship in Eq 17.5
The integral in Eq 17.6 is the Fourier transform of /(f) and is denoted
/
00
00
Trang 217.1 The Derivation of the Fourier Transform 647
We obtain an explicit expression for the inverse Fourier transform by
investigating the limiting form of Eq 17.1 as T—» DO We begin by
multi-plying and dividing by T:
/w= f^ c « r )^(f) (17.8)
As 7*—»oo, the summation approaches integration, C n T —* F(co),
n<x) {) —> OJ, and 1/7 —* dco/2TT Thus in the limit, Eq 17.8 becomes
m 2ir F{co)e io)t dco (17.9) -4 Inverse Fourier transform
Equations 17.7 and 17.9 define the Fourier transform Equation 17.7
trans-forms the time-domain expression f(t) into its corresponding
frequency-domain expression F(co) Equation 17.9 defines the inverse operation of
transforming F(co) into/(f)
Let's now derive the Fourier transform of the pulse shown in Fig 17.1
Note that this pulse corresponds to the periodic voltage in Example 16.6 if
we let T —* oo.The Fourier transform of v(t) comes directly from Eq 17.7:
V(co)
r/2
V „£->*" lit
-r/2
(,->">'
= v, (-jio)
r/2
- r / 2
o(t)
-T/2 0 r/2
Figure 17.1 • A voltage pulse
V m ( n • 0)T
—— -2} sin —
which can be put in the form of (sin x)/x by multiplying the numerator
and denominator by T.Then,
V(co) = V m r sin COT/2
COT/2 (17.11) For the periodic train of voltage pulses in Example 16.6, the expression for
the Fourier coefficients is
C V„,T sin nco () T/2
Comparing Eqs 17.11 and 17.12 clearly shows that, as the time-domain
function goes from periodic to aperiodic, the amplitude spectrum goes
from a discrete line spectrum to a continuous spectrum Furthermore, the
envelope of the line spectrum has the same shape as the continuous
spec-trum Thus, as T increases, the spectrum of lines gets denser and the
ampli-tudes get smaller, but the envelope doesn't change shape The physical
interpretation of the Fourier transform V{co) is therefore a measure of the
frequency content of v(t) Figure 17.2 illustrates these observations The
amplitude spectrum plot is based on the assumption that r is constant and
T is increasing
Trang 3C„
o.i v„
-4TT/T —2TT/T
Ji u ^ _
2TT/T Tvn^p
(c) Figure 17.2 • Transition of the amplitude spectrum a s / ( / )
goes from periodic to aperiodic, (a) C„ versus /zw () , JT/T = 5;
(b) C„ versus nw(), T/T = 10; (c) V(a>) versus to
17.2 The Convergence of the
Fourier Integral
A function of time / ( 0 has a Fourier transform if the integral in Eq 17.7
converges If f(t) is a well-behaved function that differs from zero over a finite interval of time, convergence is no problem Well-behaved implies
that / ( 0 is single valued and encloses a finite area over the range of inte-gration In practical terms, all the pulses of finite duration that interest us are well-behaved functions The evaluation of the Fourier transform of the rectangular pulse discussed in Section 17.1 illustrates this point
If /(f) is different from zero over an infinite interval, the convergence
of the Fourier integral depends on the behavior of / ( 0 as f—>oo A single-valued function that is nonzero over an infinite interval has a Fourier transform if the integral
[/(01 dt
Trang 417.2 The Convergence of the Fourier Integral 649
exists and if any discontinuities in / ( 0 are finite An example is the
decaying exponential function illustrated in Fig 17.3 The Fourier
trans-form of / ( 0 is
F(co) = J f{t)e~ Jto ' dt = Ke- a 'e- ]a * dt
fc e -(a+jco)t
(a + jco)
K
o -(a + jw)
K
(0-1)
Figure 17.3 • The decaying exponential function
Ke- ( "u(t)
A third important group of functions have great practical interest but
do not in a strict sense have a Fourier transform For example, the integral
in Eq 17.7 doesn't converge i f / ( 0 is a constant The same can be said if
/ ( 0 is a sinusoidal function, cos oj Q t, or a step function, Ku{t) These
func-tions are of great interest in circuit analysis, but, to include them in Fourier
analysis, we must resort to some mathematical subterfuge First, we create
a function in the time domain that has a Fourier transform and at the same
time can be made arbitrarily close to the function of interest Next, we find
the Fourier transform of the approximating function and then evaluate the
limiting value of F(w) as this function approaches f{t) Last, we define
the limiting value of F(co) as the Fourier transform of f(t)
Let's demonstrate this technique by finding the Fourier transform of a
constant We can approximate a constant with the exponential function
/ ( 0 = Ae- £{ ' 1 , e > 0 (17.14)
As e —* 0, / ( 0 —* A Figure 17.4 shows the approximation graphically The
Fourier transform of / ( 0 is
Carrying out the integration called for in Eq 17.15 yields
„ , , A A 2eA F(to) = — + — = — - (17.16)
e - j(o e + a) e 2 + a) 2
f(t)
A£^^^
Ae^*^^-0
e 2 < e l
A
-^^Ae^l
^^_A£2*
Figure 17.4 A The approximation of a constant with an
exponential function
The function given by Eq 17.16 generates an impulse function at w = 0 as
e —>0 You can verify this result by showing that (1) F(co) approaches
infinity at co = 0 as € —» 0; (2) the width of F(eo) approaches zero as e —>• 0;
and (3) the area under F(w) is independent of e The area under F((o) is
the strength of the impulse and is
2eA
e L + (o l
o e2 + a 2
= 277-/1 (17.17)
Trang 5650 The Fourier Transform
In the limit, f(t) approaches a constant A, and F(co) approaches an
impulse function 2TTA8((O). Therefore, the Fourier transform of a constant
A is defined as 2TTA8(co), or
&{A) = 2TTA8((O) (17.18)
In Section 17.4, we say more about Fourier transforms defined through a limit process Before doing so, in Section 17.3 we show how to take advantage of the Laplace transform to find the Fourier transform of functions for which the Fourier integral converges
/ A S S E S S M E N T PROBLEMS
Objective 1—Be able to calculate the Fourier transform of a function
17.1 Use the defining integral to find the Fourier 17.2 The Fourier transform of f(t) is given by
transform of the following functions: _,,
a) f(t) = -A, -r/2 < t < 0; F((o) = ^ _3 < <, < _2 ;
f{t) « A 0 < f < r / 2 ; F ( C ) = 1, - 2 < O , < 2 ;
fit) = 0 elsewhere F ( f t ) ) m 4> 2 < co < 3;
Answer: (a) - / I — II 1 - c o s — I;
1 1
(b) — « Answer: f(?) = — ( 4 sin 3t - 3 sin 2f)
(rt + JO))" TTt
NOTE: Also try Chapter Problems 17.1 and 17.2
17.3 Using Laplace Transforms
to Find Fourier Transforms
We can use a table of unilateral, or one-sided, Laplace transform pairs to find the Fourier transform of functions for which the Fourier integral
con-verges The Fourier integral converges when all the poles of F(s) lie in the left half of the s plane Note that if F(s) has poles in the right half of the
s plane or along the imaginary axis,/(/) does not satisfy the constraint that
Xoo 1 / ( 0 1 * exists
The following rules apply to the use of Laplace transforms to find the Fourier transforms of such functions
1 If / ( / ) is zero for t ^ 0~, we obtain the Fourier transform of f(t) from the Laplace transform of /(f) simply by replacing 5 by jco Thus
*{/(0) «2{/(0W (1 7-1 9)
Trang 617.3 Using Laplace Transforms to Find Fourier Transforms 651
For example, say that
fit) = e "fcosw(/, 0 + Then
(s + a) 2 + (nl x =j lo (jco + a) 2 + (x)f)
2 Because the range of integration on the Fourier integral goes from
— oo to +oo, the Fourier transform of a negative-time function
exists A negative-time function is nonzero for negative values of
time and zero for positive values of time To find the Fourier
trans-form of such a function, we proceed as follows First, we reflect the
negative-time function over to the positive-time domain and then
find its one-sided Laplace transform We obtain the Fourier
trans-form of the original time function by replacing s with -jco
Therefore, when f(t) = 0 for t > 0+,
For example, if
then
/ ( / ) = A o s f t t f , (for / < 0")
/ ( - 0 - o (for t < 0");
/ ( - 0 = <T'"cosw(/, (for t > 0+)
Both /(/') and its mirror image are plotted in Fig 17.5
The Fourier transform of / ( / ) is
-jco + a
s + a (s + a) + COQ s=- /w
(-jo + a) 2 + &Ȥ
Figure 17.5 • The reflection of a negative-time
function over to the positive-time domain
3 Functions that are nonzero over all time can be resolved into
positive-and negative-time functions We use Eqs 17.19 positive-and 17.20 to find the
Trang 7Fourier transform of the positive- and negative-time functions, respec-tively The Fourier transform of the original function is the sum of the two transforms Thus if we let
then
and
A / ) = / ( 0 (for r > 0 ) ,
/ - ( / ) = / ( 0 ( f o r / < 0 ) ,
/(o = n o + /"(o
= ^{/+(0},=ya> + ie{/-(-0}.v=-; w (17.21)
An example of using Eq 17.21 involves finding the Fourier trans-form of e_a''L For the original function, the positive- and negative-time functions are
/+( 0 = e~ ul and / " ( / ) = e'"
Then
2{/+(0} i
S + a
1
s + a
Therefore, from Eq 17.21,
&{ e -<to\} 1
s + a
1
+
S= I (it
+
s + a
1
s=—ju)
jo) + a —jo) + a
2a
2 '< 2 "
o) + a
I f / ( 0 is even, Eq 17.21 reduces to
9{f(*)) = %{f(t)}.s= jt o +
^{/(01,=-/0,-I f / ( / ) is odd, then Eq 17.21 becomes
9{f{t)} = %{f(t)} s=ia - 2 { / ( / ) } , ~ /
(17.22)
(17.23)
Trang 817.4 Fourier Transforms in the Limit 653
/ A S S E S S M E N T PROBLEM
Objective 1—Be able to calculate the Fourier transform of a function
17.3 Find the Fourier transform of each function In
each case, a is a positive real constant
a) f{t) = 0,
fit) = e "'smcoyt,
b) f(t) = 0,
/ ( 0 = -te at ,
c) f(t) = te- (,t ,
/(0 = to",
t < 0,
t > 0
t > 0,
t < 0
t > 0,
t s 0
Answer: ( a )
(b)
OiQ
(c)
(a + jco) 2 + col'
1
(a - jco) 2,
-j4aa)
NOTE: Also try Chapter Problem 17.5
17A Fourier Transforms in the Limit
As we pointed out in Section 17.2, the Fourier transforms of several
prac-tical functions must be defined by a limit process We now return to these
types of functions and develop their transforms
{a 1 + <«/) 2 \ 2 '
The Fourier Transform of a Signum Function
We showed that the Fourier transform of a constant A is 2TTA8((O) in
Eq 17.18 The next function of interest is the signum function, defined as
+ 1 for / > 0 and - 1 for t < O.The signum function is denoted sgn(0 and
can be expressed in terms of unit-step functions, or
sgn(0 - u(t) - u(-t) (17.24)
Figure 17.6 shows the function graphically
To find the Fourier transform of the signum function, we first create a
function that approaches the signum function in the limit:
sgn(r) = lim[e~ €t u(t) - e et u(-t)], e > 0
The function inside the brackets, plotted in Fig 17.7, has a Fourier
trans-form because the Fourier integral converges Because / ( 0 is an odd
func-tion, we use Eq 17.23 to find its Fourier transform:
S + €
1
1
s=)to S + €
1
s=—]u>
joo + e —jco + e
_ -2jco a) 2 + e2'
As e -> 0, / ( 0 - » sgn(0, and ^ { / ( 0 } ~* 2 / > Therefore,
^{sgn(0} = —•
(17.26)
sgn(/)
1.0
-1.0
Figure 17.6 • The signum function
(17 27) Rw* 17.7 A A function that approaches sgn(r) as
e approaches zero
Trang 9The Fourier Transform of a Unit Step Function
To find the Fourier transform of a unit step function, we use Eqs 17.18 and
17.27 We do so by recognizing that the unit-step function can be
expressed as
"(') = - + -sgn(f) (17.28)
Thus,
= TT8(O)) + — (17.29)
7*>
The Fourier Transform of a Cosine Function
To find the Fourier transform of cos o) () t, we return to the inverse-transform
integral of Eq 17.9 and observe that if
F(<o) = 2TT8(O) - a>o), (17.30)
then
l r
/(?) = — / \2TTS{O) - o) {) )]e"°' do) (17.31)
Using the sifting property of the impulse function, we reduce Eq (17.31) to
Then, from Eqs 17.30 and 17.32,
We now use Eq 17.33 to find the Fourier transform of cos w()r, because
e hvf + e ~m\t
cos a y = (17.34)
Thus,
^{cosa>0/} = - ( 3 = ( ^ } + ^{e' j0 ^})
— [2TT8((O — wo) + 2TT8((O + £%)]
Trang 1017.5 Some Mathematical Properties 655
The Fourier transform of sin co 0 t involves similar manipulation, which
we leave for Problem 17.4 Table 17.1 presents a summary of the transform
pairs of the important elementary functions
We now turn to the properties of the Fourier transform that enhance our
ability to describe aperiodic time-domain behavior in terms of
frequency-domain behavior
TABLE 17.1 Fourier Transforms of Elementary Functions
Type
impulse
constant
signum
step
positive-time exponential
negative-time exponential
positive- and negative-time exponential
complex exponential
cosine
sine
/(')
8(t)
A
sgn(/)
u(t) e- in u{t)
e m u(-t)
e -«\'\
e/w
COS 0)()1
sin corf
F(co)
I
2TTA8(CO)
2//oi
7T8((X)) + 1//(1)
l / ( n + /ftj), a > 0
l/(« - / « ) , a > 0
2a/(a 2 + or), a > 0
2TT8{(0 — too)
7r[5(o> + (0()) + (5((0
-/V[5(o) + (o () ) — 5(a; •
mi)]
~ «,,)]
17.5 Some Mathematical Properties
The first mathematical property we call to your attention is that F(co) is a
complex quantity and can be expressed in either rectangular or polar
form Thus from the defining integral,
/
00
f{t)e-' Mt dt
CXJ
•J
/(/)(cos cot - j sin cot) dt
L f(t) cos cot cit - j f (t) sin cot dt (17.36)
Now we let
M&) = / f(0 COS cot cit (17.37)
-X
fi(o)) = - / / ( / ) sin mtdt (17.38)
J—no
Thus, using the definitions given by Eqs 17.37 and 17.38 in Eq 17.36, we get
F(co) = A(a>) + /£(<«>) = |F(ft>)|e^ (17.39)