- - - - - - - - - - - - -~;,-(2(Ja)~ .

Eqs. (20 a) ~nd,. .(20

.,--~~), Sec. 9.1ou. ner cosine~
integral
-oc < t < -p 9. Work Exercise 5 usm : -~Q"ij lO c

~{-l
g the Fo uri er sine integra
--p < 1 < 0 10. Work Ex erc ise 6 using l pair.
the Fo uri er cosine integra
O <t <p 11. Using the Fourier cos l pair.
ine integral pai r, express P(w , t) in
p< l < x ten ns of sine integral fun 0
ctions for
-ao < t < -1 O - oo < t s - I

-{! -I < t < 2

2< t< X
f(t) =
{
I+ t
I-t
-1 :S 't:S O
. O: St :S l
0 l:S t< oo
lrl <
l 12. Verify the assertion ma
={! I< t 2 < 4
lrl > 2
ma xim um
de on p. ~6 0 that the firs
of the function s(z) =
½+ (l/1 r) Si( 1r). Hint: Wh at
t pmitive
½+ (Th r) Si(z) is
is the derivative of the inte
gra l defining Si(z)? -

U IIOPElm OF FOURIER TR
.......
....... ANSFORMS
......
......
......
......
......
......
......
......

So far, ex ce pt for ma kin ......

......
......
......

g use of the altern

ative formulas (27a) and
. . ti, • • •
••• ••••••
••••••
••••••
••••••
•

we ha ve for fin din g (27b), Sec. 9.1, the on ly me

the Fo uri er integral of a thod
na tions of int eg ral for mu function is by dir ect integration using suitab
las as stipulated in Sec. le combi-
tra nsf orm pa ir [E qs . (11 9.1. Ho wever, if the members
) and (12), (20 a) and (20 of a Fourier
primarily as fun cti on als b), or (21 a) and (21b),
, or transformations, the Sec. 9.1] are thought of
en ab le us to fin d the Fo y are found to have ·prop
uri er integral of a functi ert te~ which in many cases·
develop som e of the se on without integration.
pro pe rtie s and illustrate In thi s section we shall
some of their uses.
 . .. ...... .. ~~l -~ ~-~lS Of FotJh.
. ·· d (12) Sec. 9.1], the factor '•,, ··. "q'-
1/271'"·vvasa ··.. -~
, s(ll)an ' .
1 . ral pair fE~ · C(w) because that was where 1t arose th Ssociat
1 h "' ,n,ctz . function b . . ·11 0st nat ect ~i'
, c comT' c
rc co .effi cient .
Founc ,. integrals. However, y tncorporattng th e faCtorUra11 Yi IIL'
;n;cgral {01 ~:uiicr scncs lothe transform pair as
d rcwnung Iii/~
r,11c . ·on fro m
J1Jns1 !I f( I) an 'n1,'I 1

• rcgra Jfur · 1 J oo .
in f(t) =- F(w)e 1<01 dw
( l) 21r - oo

f
00

(2) F(w) = f(t)e - icm dt

- oo

a "mp
)
1
. er coefficient function F(w) ~ 21rC(w) is introduced Which ' since ,r

F("' ,
. easier to write and tabulate than C(w). Thus in dealing With
IS .
. for convenience, we shall henceforth use Formulas (1) and (2) E
112 1s not af
cornp/ex Fourier
. tra r ~~
function
pam. F(w) IS . generally complex. • Ven Whenf (1J IS~ . ~lrq0
f 1
The process o mu tip · lymg
· /( t) by e-iw1 an d then tntegrating
. from "" _ a,if(
single function F to each function f that satisfies the conditions of Th 1 "' to s 3ssig ..
1

process, together wit. h 1ts

. doma1n. o.f .f uncttons,
. ts eorern
. itself a function, Called 1 Sec.1· 09·I· Thi.~
the F~ ill i
lion, whose value at f ts the functton F. Denoting this transformation b :J. ner lransron.
(3)
Y 'We Wnte
F(w) :::: ~[f(t)]

which emphasizes the dependence of F(w) onJ(t). When interpreted as av l f crt

functton . called the Fourier
. F(w) ts • transfonn of f(t). The functionJ(t) isia·nuetuo ~.fthe c~ffic.ii.
mverse Founer transform of F(w) and is denoted by
• •
•
m re erred ~.,,
(4)

Together,f and F, aS given by (1) and (2), are commonly referred to as the com¢exFm
transform pair.

The Fourier cosine and sine integral pairs [Eqs. (20) and (21), Sec. 9.1] can also~"""'.
as pairs of mutually inverse transforms. For reasons of symmetry. this is usually done bt ~
the numerical factor 2/ 1r in A(w) and B(w) into two.equal factors and ass,gnmg ~_,,,
• · '".factor to~
integral
pair in the pair. The cosine integral pair ·then becomes
· the ·so-called Founer
· cosme u=

(5a)
/(t) == {i;_ f 00 Fc(w) cos wt dw

Fe(w) =~
7T

r o
f(t) cos wt dt
J even

U\legral pair is converted ,mto the Fourier sine transform Pair

f(I) = ~ 7T
r 0
F,(w) sin wt dw J odd

F,(w) /2 Joo
= \J-:;;, f(t) sin wt dt
0

··1.ak. .llli.c
2 . _....,4:58 ~ = ~A(w)
,. ,, _- lflTEG'~)IID~rvvth '."··-= e:
sin:
co· ·: tra:ns~
form~pa ~ - (5 ) an d Eq . (7a) th
ir ~~~~(?.~=- =-- - - - - - -
inverse F ~u ; function
F (w) . 56 3
th :s fo rm of J( t), an d f(t ) is the
18
er cosine t;a knoWn
/' In bo
tra ns fo rm at io n CZFc, l'lsforrn of as !he F .
tra h Fo ur ie r co sin e F (cv) ou oe r
cosine th at is, in
(111S of t e c ;
) gf - , = ~( w ) (b
te (a ) ;ffec ff( I) ] r [f'.:.(<.v)] === /(t )
(8) (l b) . the F
F.( {JJ) in th e sin e tra ns fo rm paIiir (6 ) an d Eq . JS
ouner s·
. •
or m of F( w) 1 in_e transtor
fun cti on s
ur
•
ie r sm e tr an s ter ms of the
T he . h in ve rs e Fo s • n
Fo er sin e Iran
un ni ofJen
t) is t e sfo a-.
and J( Tin
. n <!fs•
tJO f(t )
(a ) ~A f( t) ] = F's(w) (b) CZF; '[fr(w)J ===

(9) is co mp. le te ly symm e1nc m that . . .

fo rm pa irs mterchange of the
of the Jast tw o tra ns I . b .
ei th er .eq ua tio n . an
_
Each e se a ar va na Jes m resu 1ts m the O ther equati on of th
. al variables an d of th er ch an ge d . (6
{unction
e, w he n/ an d F's, as we ll as t an d w, ar e mt th
· m a), e result is (6b · 1s
) Th ·e
. for exampl irs ea sy to re m em be r.
pair. try makes bo th pa . . d
en co un t ere and the · .
symme rm s of th e Fo un er m te gr al. that we ha ve fu . mmor differences
e various fo. . tio n m ay so m eti m es see m con smg H
Th
, co ef s, an d no ta
fic ie nt . . ence, before ar. ·
"'"t ween their . limits
. .
op
. .
en 1e s of th e Fo un er tra ns fo nn ati on s ar.
.:r, ..r and ~ ·
may be
l}\;
g m this secti on on pr c, s, It
tratin di sc us se d.
conCen th e fo rm s we ha ve
well to bring together all
al • Its Variants
Wlf 9. r TIie FOllrier l■t,gr
ur ier integral, Eq . ( 10
), Se c. 9.1:
exponential Fo

1: [ 1:
The complex

= ~ f(t )e -iw r dt]eiwr dw

/(t ) 2

2), Se c. 9.1:
Fo uri er integral pa ir, Eq s. (11) an d (1
The complex

f(t ) = J 00

-00
C(w)eiwr dw
[C om pa re Eq s. (7 )
an d (8), Sec. 8.5.]
,
= - 1 Jae f(t )e -iw r dt
C(w)
21r -ac ,

(1 6) , Se c. 9.1:
Fo uri er integral, Eq s. (15a), (1 5b ], an d
The Standard
00

[A (w ) cos w t+ B(w)
sin wt] dw
/(t ) =[
0
). and
Jae, /(t ) sin wt dt
1 [C om pa re Eqs. (1). ( 10 2
A(tu) : ..!. J"" f(t) cos wt dt B(w) = - (11), with d = -p . Se c. 8- -1
.,,.
- . an d
11' -ac ,

'lnepoUrier ·
Sec. 9.1:
COsme integraJ, Eq. (18),
/ even
00

2 [" "[ /(-r) co s w- rc os wt dr dw

/(t ) •- :;;
.,,__ .
0 0
•1ic Ji',,.._.
• "lll1Cr eoa. ·
JJJe integral pair, ), Se c. 9.1:
Eqs. (20a) an d (20b

/(f ) -[CID A(c,,) cos wt dw 8

1, Sec. .4.)
(C om pa re Th eo rem
A(eu)- 2 r• /e ve n
.1«, /(t ) cos wt dt
'•11;1 m ) ~
;
 S6t.
t we obtai n
I
·
ma ,.
11,· hy in
· the ,ym l101'- wand •
· tcn:hangmg.
-
f 'X, •

F ( l )e ,w 1
d
l
2 7T(( - w) - x.

th t 2 7T/( - u.•) .JS the Fourier trans

. form of F(t) .
which is Just the statement a . te Scale) If a .ts a positive real con stant,
THIOllM3 (Change of Tm

at)) = (' oc. f(at)e - iw, dt

~
..
t
Under
PROOF By definition, ~ (f(
. . at - z we have dt - zI a,
the subst1tut1on
d and the integral becomes

as asserted. t
_l_
a
r
- co
f(z)e - ;w,la dz= >[f(z)]_wla = >( :)
The change of_ scale property is a1so
established in the same way. . valid for both the cosine and sine transformations and •
1HEOIEM 4 (Time Shifting)

~[f(t - to)) = e-iwto~[f(t)] = e-iwto F(w)

t PROOF By definition, ~[f(t - to)] = f~ooJ(t - to)e-iwt dt
If we put t - t 0 = z and dt = dz, we have

:,i[J(I - lo)] = r
-oo
f(z)e-;w<<+•~ dz = ,-;,.,,, J~
-oo
f(z),-;w , dz

= e-iwt ~[f(zJ] = e-iwt F(w) t

0
0

In words, the time-shifting theorem says that translating a function an amount lo in ' :
domain results in the transfonn of the function being multiplied by e-iwr • Conversely, as athIOO"'
0
~
recovering_a functi_on from its transfonn, Theorem 4 asserts that if a transform contains in""''
the form e ~". us mverse can be found by first deleting the factor e - ,w,,, then finding ·tbe a factc, l
the remaining factor, and finally translating that inverse an amount equal to lo,
Since the transform F(w) can be Written in the complex fonn

F(w) ::::o r(w)ei9(w)

where r(w)
0 is the magniludc and O(w) is the polar angle of F(w), it f0Uows <ha< trans . g/lfl5ll
amount t in the lime domain adds -w,
0 to the argument of F(w); that is,
laun

(10) 3'[/(t - lo)] ~ F(w), - iw,, = lr(w)e;•<w>]e - ;w,,

== r(w)eil9(w> - wraJ
•
11,e cosine and sine transfOl'lllafions
· have no comparably simple time-sh•·fting prope rtY.

2024.03.\1 8 il~:59.
 ,-lfJ ra.-Sf~MS
-dlP ,,_ ,cncY Shifting ) 1f wo is a real constant, then
J.-, ,,.- (frcQL 567
,fll'"'r fO-E~ 5
/ Ttf ~ lei'<>o'.f(t)l = g; lf(t)J,.,__,. ,.,_,.,0 = F(w - Wo)

By definition.
« p~OOf

h frequency-shiftin. g theorem says that multiplying a f . .

In ~,ords. t. e tl·e transform being s
h'f d
1 te an amount w 0 in the freq
unction by e'Wrl .
· esults in •
domain r d tern1 ining a function when its .
transfor uency dom · Cin the time
ol for e m is known Th ., ain . onversely
as a 10 . . ' eoiem 5 assert h . '
f function is shifted an amount w 0 and the inverse of the res . s_t at 1F the
O
transform ~;%1 times that inverse is the inverse of the original transf 1
, nd then e u tibng function or w is
iou · . 0 rm efore tra I ·
•ne and sine transform ations have no compara bly simple frequ h. . ns ation.
The cos1 ency-s 1fting propeny.
0 t f t f t t t f I' t t t • t t t t t t t t t t ■ t • t t t ■ t t t t t t t t t -~ e t t t t t t t t ~ ~ t t t t t t t t t t t t t t t t I t t t t t t t t t t I t I t t I t t t t t t t t t t t t t t t I

· th Fourier transform of the product f(t) cos at?

What ,s e .
Using Theorem 5 and the complex exponent ial form of the cosine, we have

~[f(t) COS at] = <!f [ f(t)

eiat +
2
e- iat]
· = <!f
[J(t)
+ iat]
+~
[ ·
f(t);-rat]

= ½[F(w - a) + F(w + a)]

If we think of f(t) cos at as a carrier signal, cos at, modulated by f(t), this transform allows
us to express the
transform of the modulated signal in terms of the transform F of the modulating function!,
which determines
the amplitude envelope of the signal.

t ~ t It I It t

EDI
I t I t t I I I t I I I I I I I I I t I I I I I I t t t • t t t t t I t I I I I I t I t t t I I I I I I I I I I I I I I I I t 1 1 1 1 1 1 1 1
I I I I I I t t I I I I I It I t I t' t

Find the inverse of the Fourier transform F(w) 1

, a > 0. =
4a + w 4 4
We first express the denomin ator of F(w) as the product of two quadratic factors with real coefficients
b~
using the four fourth roots of -4a4, namely,

Z1 = a(l + i) Z2 = a(-1 + i) .: 3 = a(-1 - · i) = z2

to write

Z2)l = l(w - a)"+ a21[(w + a) + a~l

404 2
+ w4 = [(w - z 1)(w - ·z 1)][(w - z2)(w -
1

The method f ·
o Partial fractions then gives
- a + ((11 + al__l
F(w ) =~ I 2a - w '2a + w +((II + 0 ) ! ] _ _ l_ [ ~ 'tJ - a) 2 +
8a 1 a2 + (w - a)2 I + a2 + (w + al -
I
,
21(111 I- ,1
l 8aJ a'2 + (w - o) a
1
l
·- I l l ,.,a I 2a I - li(1u - aL, . . - ! rl1,; 1-,1l '
- -8 3 - ~ + - - - - - - - : : - 7·•- - -
a 2 a 2 + (w - a) 2 I- ((tJ - (I ) l 1~ 1 ti
u 2 + (o> + a) 2
•
2 21 tr .
'
·•~t, 1'
Ac · ,t t,1;1,"
cording to b T· . . . hi· 1.,~l pull \
. .. . . o ll't u1s 111~1dl l
' able 9.2, and Theorem 5, the inverse ol the 111st iw

1 flt 1,11 l
{ - ()
2e (' _1 11/e /111 ::: ('Ill COS (I{
, - zl'
g (t) ={ l - ti( ill( L 1 11/l' 1111 ::- {' 111 ' l)S (I{
() - 1
2e e · ir ~t' (.; •
 _u ".'°".HfflS OF FOU111 ~

. . transformation of each side of th·is en

the cosine
510 ["(I) = a 2'j(t). Taking ·1Ua.
. f twice we get have
Differentiat'.ng Corollar)' 3, we
. an d then using
uon
~A f "U)I = a2~Af(t)J
/2 f'(O) = a2F, (w)
- ui2F,(w) - \/-;

we obtain
, = -a
Noting that/ (O)
and solving for Fc(w),
f; a

F.C(W ) = -'TT Q2 + w2

. Ta ble 9.2. Similarly, equa fng

I the sine transforms of f"(t) and a2J(t), Inen
d iven tn
hich is the transfonn g J(O) = I, we have .
w 4 and setltng
using Corollary ~ _,[f"(t)] = 29;_,[f(t).]
0

-w 2F,(
., w) + /2 wf(O) = a2Fs(w)
Y-;;
/2 w
F_,(w) = Y-;; a2 + w2
. .th he transfonn e, Table 9.2.
........... ........................................... . .. .................. ......... . .................. , • , .... ., ........
which agrees w1 I

As we should now expect, the time

. differentiation theorem has a counterpart in the freque110
domain which is

1HEOUM 7 (Frequency Differentiation) If F(w) is the Fourier transfonn of j(t), then

IO I I IO I I I IO I I I I I

F'(w) = :?F[-iif(t)]
f PROOF By definition,

F(w) = f-oc
00 J(t)e-iwr dt
and F' (w) = _:!_ [J oe f(t)e-iwt dt]
dw -oo

Since /(t) is known to be absolute! y integrable on (-co, co), the indicated

.
. ...,.t
. we ~"-
. 1f~' rentiauon
d.
carried out inside the integral sign (see Theorems 1 and 4, Sec. 10.3). Domg th15 "

F'(w) = f-co
00 [-iif(t)]e-iwr dt = :?F[ -itj(t)] t
~-

' 11111 1( ' ,.

In words, this theorem says that differentiating a transform in the frequency donHIII
d'n£ j111't:~.'.,. :,
multi·p1YlllB
. lls. inverse
. b . . h .
11 I for fin , ~ . ,111,, It
Y - In t e time domain. Conversely, as a too heJI _,, .
.,..... 11ys1hat i( an antiderivative of a transfonn has a known inverse/(/). 1 .,." t-
invene is lbe Inverse' of the original transfonn. b/ i. f'
H we choose, we can multiply both sides of the conclusion in Theorem 1 to l
equivalent formula

~[if(t)] == iF'(w)

2024.03.18 1.'4:59. f ,_
 11• r•ANSfOIIIII"
,.r4o FoU• are b
f(t) and successive integral powers of
1
S71
_..-~-'s s can be continued. leading to the form la solute\y inte
h products of
ua grabJe
I'
fll"".. as t e d·ng proces n
J\S 1ong the prece • 0
c-oe, c:,:;) , ~[t''.f(t)] = i"F 1"l(w)

( J 2) •••••••••••••••••••••••••••••••••••••••••••••••••••••• ••••

•••• = + b.,.
.••••.. •• •• • •• ••• • •
I - a a
nd k ~ O an integer. find the Fourier transfor •• · ··· ··············
rn of the funcr, on
a.::. 0. b rea. ~
for
g(I) -
_ {o' -,
,< O
f~e- 0 < I

Sform of the function

for a < 0. the tran

.f<r) = {o-,
e·
I< 0
O<r
is given by
I"' = - - -
F(w) = fo
x 1
e~ e-iw, dr = -----
ela+(/,-.w)i11

Z. - IW O iw - z

Clear IY, g -- 1k'f· Hence ' using (12) and the easily derived relation

we have

_,,.,, ............................................................................................................... .

In Sec. 2.9 we considered a particular combination of two functions called their convolution.
This concept, when extended to functions with more general domains, has many practical applica-
tions and is of fundamental importance to our further study of the Fourier transformation .

.DEflllmoN 1 The function h defined by the integral

h(t) = Jr-a f(A)g(t - A.) dA

a

is caUed the convolution

• . .
of the functions f and g over the mterva
1 from a to t - 0 ·

It ·is customa ~ kin the change o ,

f v·triabk
t - ,\::::: ry to denote h by f * g, !ind it is easily proved, by ma g
.
1 .
µ, 1n the c . f . . . 1 , frl)lll Ll
onvo ut1on integral, that f * g = g * . nvolution ot .I nnu -~
V

In rnan
tot given iyprobiems, a = 0, in which case we have the unilateral co

03)

(J * g )(/) = I ()
I .f(,.\)g(t - ,.\) dA
 9.3 PROPERTIES Of FOURIQ
"-Al!Sf~
. ·r (-oc, oc) given by
uuoo ove
1:11 coovo l
577
If a -
__
00
, we obtain
. the bi11:1tcr

(f*g)(t) -
_ I ,;:,
. (P·
f(A)g(/ - A) dA

(14) . tant role ,in t hc theory ,an d app 1·icattons

. of L
ution ( 11) plays an 1,np_~:tcral convolution ( I 3) and the bi Iatcral con 'Plac,
Toe uni I,icral conv 01 IO 11). 50th the un' •. . Voluti~
transforn1ations (see sec. .' ith Fourier tran sforrnauon s. . ., . , .
. portant in working w h Cauchy product of two series is a su m of convoJ . 1
( 14 ) are 1111, dy know (Sec. • ) that t e
9 • ·1ar resu 1t. In the case f FUt1on
. regrcils yields a s1rn1
2 1
C a n:a
I
' roducl of tWO //1 . • 0 oune r

~
'
Wt is natural to ask 1f the P tly concerned with , we find that 1f two functi on J
t u~ t k d we arc presen and
h
integrals. which arc the ,n '
are repre~ented by I J C/., G(w)ei<,,, dw
g(l) == -
and 2'1T
f( t) === !_ r )C f(w)e;w, dw -½

21T -"'
.
th dummy variable w in the first integral by A and in the second inteoraJ-
then, upon rep Iactng e . o
µ,, the product off and g has the representation

f(t)g(t) == ~ f"'
4~ -00
F(A)e;>.t dA f- oo
00 G(µ,)e;,_,.r dµ,

Because the dummy variables are now different, we can write the product of the integrals ai ;1
repeated integral
j
f(t)g(t) == 4~2 Joo Joo
-oo -oo F(A)G(µ,)ei(Hµ.)1 d.A dµ,

The change of variable µ., == w - A gives dµ, = dw, and

f(t)g(t) = j
~2
Joo Joo F(.A)G(w - .A)eiwr dA dw
4 -oo -oo

_ } Joo [
- 21r -oc
j
1r
Joo
-oc F(.A)G(w - .A) dA ]e;w, dw
2
1
The sLructure of the last expre .
bilaleral convolution of the ssiofn shows that the quantity in brackets, narnd),11
1rans orms F( ) d f l't ui, ►
?..
on the left; that is, · w an G(w), is the Fourier trans on
Ji( I)g(t)

1
?Jilf(t)g(t) I = -27T J"" F(A)G(w - .A) d.A = --
I fl£tl) * GlLtl)
~ ')
_ 'TT

We havei dlu1 established tht i•11ponan1 result

fiIIIFI
, • I 'I (Jc'reque n,y
. ('onvolutlon . . . . ·< ..,,.,,,,,,,,11• •·
lftll.e i.nt«Val and are 11b1mlutt1y · ) 11 J a11d
tnll'grahll· '
~ satisf)' till' \)1ndtl,
· ·
llll ( nt., 0.' ). lht'll

'.IJ. lf<01-1(0I ') I F(,u) >1- (;(,u)

... Tr

With dall dllanlm 11tablished, we are prepured to prove

2024.03.18 1~:59.
~ -~ -
f()C.llllfll fllJNSfOllllilS

:; rseval's Modulus Theorem) If a real function f .

..-,,_-
,.
OREM 9 (Pa I -e polar fonn is F(w)
THE . r transform w ,os
= r(<JJ)c'o<,.n , then . ( 1) With domain (

a fouric J"" .f2(1) dt = 2

I J"' r\cv) dw
00 7T .,,.,
(I 5)

W first note from the definitive integral, written w· h

f p,rOOF c .t , It w replaced h
y 't,

F(T) = J''° f(t)e

00
iTt dt

f (t) 1•s real, both i and the real variable Tenter only through th .
,h 1 when . . b . d e product l't H
u,a
13 )..'.e the conJ
•ugate of F(T) by replacing L Y - L, an subsequently replace b enCJ!. 1f 111~
7 Y - -r w h
nchanged. In other words, whatever the argument may bf' F( ) _ -F ' e ave lef· the
ansfom1 u - . ~, ,. - (- -r) I
tr ' F ( _ w) = F(w - T). The convolution of F(T) with itself can the f be n Paricu ..:r.
then. r re ore wntten

F(T) * F(T) = rx,

- ~
F(w)F(T - w) dw = J -oo
00

F(w)F(w - -r) dw

Now from this result and the frequency convolution theorem, we have

,,------ 27T
1
9f[/ (t)] = -F(T) * F(T) = - I
21r
J 00

-oo
F(w)F(w - T) dw

From this, by putting r = 0, we obtain

Joo-oo
2
f (t) dt =-
27T
1 Joo
-oo
F(w)F(w) dw

th
But e product of any number, real or complex, and its conjugate is the square of the ~c~,cli:t.:
value of the number. Hence the last equation yields

Joo
-oo
f 2 (t) dt =-
27T
1 Joo
-00
r 2 (w) dw

as asserted by the theorem. t

In
the the case of a signal f(t), the square of the modulus of F(w), namt') t>nt'r~~ in :.1.11:, 1-•
rgy SJ>ectr
I
f,:(wl ,~ 1.:ilkd the
~ urn o f the signal
.
because r 2 (w) dw represents ct1c nmounl
O -· ,n n
ihc wed 1:uc:r.~
gment of th f 2 ) ( - ~ cc ) g1H' " t t111, a.,
the f)i , e requency spectrum, and the integral of r (w over , ·. kJ\t' ihc prlx11
iUJ ex.gn~J. Parseval's theorem is also true if J(t) is complex. but we s hu 11
erc1&e.
...... .
··········..
Venr ......... . . ........... ti . . . . . . . . . . . . . . . . . . . .
. ............. .. ........ ... .. ·· ····
'i Par~vaJ' , h
s L eore,n for the fun ction

~
I • ()

f(t) { ' I , ()
In thi
s Particular case

r •''Ill, - 1.
2
 - ,autrlElt Tl~1n: · · .. · · · .. .. .. .. . - - - - - - - --
-:--- -
,_.. ~ ·· wo factors of the last product are, respective) · · ·
S7S
..- '
,. BY •·
fable 9.2, the t

g(t) = e-21
t< O
0< t
{o and
y' the Fourier tr

h(t)::: {O t <0
ansfonns f
o the functions
e- 3,
0< t
. are identically zero for t < 0. Hence, by Corollary 1
h of which , Theorem 10
bOl
J(t) = L
e-2Ae-J<r-AJ dA = e-3,
o
eA dA = {O
, the r .
J'
t :s O equ,red inverse is
o e- 2, - e- 3,
0 :s t
·····························································
,..~•~: inversef(t) of the Fourier transfonn F(w) = [4a sin kw)Jw(a'
Fin .d d by Theorem 10, we express the given transform as the product ), a> 0.
+·~:· ·················•. , , , ,
Gui e

2a ) (eikw_e-ikw)(
~--(2sinkw)(
w a 2
+ w2
=--- 1w
~)
a2 + w2
• --

.th == k and a = b = 1, we see from f, Table 9.2, that the inverse of th e tilfSt factor .in F(w) is
W1 1

g(t) = {o 1
ltl > k
ltl < k
From b, Table 9.2, the inverse of the second factor is

eat ts; 0
h(t) ={ e-a,
t~0

r
Hence, by Theorem 10,

J(t) = r -oo
g(A)h(t - A) dA =
-k
h(t - A) d!.

To evaluate the last integral, we observe that

ea(t-A) ts; A
h(t - A) = { e-a(t-A) As; t
. three cases to consider
Thus, depending on the relation of t to the limits on A, we have the following
1
·ts -k: f(t) =
fk
ea(t->.) dA
eat
= -[e-a,\tk = - - - e
2 sinh ak at

-k -a a at

2. ltl s k: f(t) =
f
I

-k
e-a(t->..) dA + I
t
k
ea(t-A)
e-at
dA == -;-[eaAJ'-1r - -;[e
e -aAJk
,

= 2(1 - e-ak cosh at)

3
' t ?!:. k: f(t) =
f k
a

e-a(1->..) dA
-at
= _e_ [ea>. t-k == ~ e
2 sinh ak -at

-k a
Finally . . e transform
'combming cases, we have for the required mvers

2 sinh ak a1
---e
a

f(t) = ~ (J
,,,.
....
,,,,
2 sinh ak -a1 ,,,,
~ ----e ,,,,
•~ a ···········••'
I I I I I I I I I
....................... • I f I I I I I I f I I
 9.3 PROPIRTilS Of FOURIER llAN
. SF<llll$

of its modulus, we first e,,_press f (

f o find the square co) in
f able 9.2.
574 - (I)) off is given by a,
Th 1ransfor!11 fl
edard con1pleX fon11
stan I ::= - - - : -
~ .
_ ;w
===
I + w
I
- - i + t I + w2
. - w
1
f((t)) === ~ \ + ,w
2 I
I 2 ( - (J) )
__-;; -
- ___.-;;-
2
Froni this we gel _ (- - ;) + I + w2 I + w
r2(w) - I + w2
ee dw 1 - i \CG 1
f "" 1 = -7T Tan O
= -2
and
~f
""
r i(w) d(I) === ~
I
dw
_-,===-
_,,. I + wz 7T o 1
___.-:::-
+ 2
w
J w

27T -a<:
h Ids for the given function f
Hence Parse val' s theorem o •••••••••• • ••••••••••••••••••••••••••••.•• , •, ,
········· ' '"
· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·............ ved the frequency convolution theorem, we can multi
In essenua . 11 Yth e same way that we pro tablish the following companion. theorem. .
ply the integrals o! two transfonns to es .
. C t·on) The inverse of the product of two transforms 1s the convolu-
TMEOREM 10 (Time onvoIu 1
tion of their inverses; that is,
~r'[F(w)G(w)] = f "" f(A)g(t - A) dA = f(t) * g(t)
- OC>

This theorem is often useful for finding the inverses of transforms which can be resolvedinto
factors whose inverses we know. However, when the inverses of the individual factors arefunc-
tions which are identically zero for t < 0, the bilateral convolution appearing in the statement ci
Theorem 10 must be replaced by the unilateral convolution.
To see why this is so, we first note that if f and g are functions which are trivial fort< 0, ilicl
their bilateral convolution reduces to

f(t) * g(t) = f'

0
f(A)g(t - A) dA

Moreover, when t - A< 0 in the cour • f h . . . th inteif;;

becomes zero and the integration effecs~ o t e i~tegratlon, the second factor in thee bil;:e:l<
convolution becomes the - tlvely terrnmates at A = t. In other words,
um1atera1 convolution

f(t) * g(t) = r 0
f(A)g(t - A) dA

1biJ olNervation we formalize in the followmg

. . ,
- (Unllat IT -ro~f coro11 ary to Theorem l O.
era lme Conv l . . r tf!ll1s
two &edor, /(t) and g(t) wh1'ch . o utton) If F(w) and G(w) are the fourie
• are 1dent'ically zero for t < O then

r
'

~ - I [F(w)G(w)) = f(A)g(t - A) dA ,/
,,, ...........
................. 0
........ .
,,,• '

f XAMP!f I 'er transform F(w) == l; ••••••••••••••••••••••••••.•.

6 2
( + 5i -
, the denominato r becom w ).
es easy to ·factor, thus
F("') = l
6
5;g •A APPII' AIIO"\ hf lbllltlJ '"'If W

un be dctermmed b) sc.annmg ( h) tJsr. thr. rcq1111,.,I r,.h,111111 ,,t 1•,,

• Al\ •1111 ' •it
...., OJr?: ~ co;:ba..~ filter 1111
~•4 "'
ol jl1rnmpl1• K, 1,, 11111 I 1111• 111vr, ,. , • M11l11
. II' Pi,
Cl!: p:! rnot:pi e nam,,,. r......, of the function /(r) of
the foaner tr£1StUuu 11
lA \mf) ~ (' ,. ..., I llu I (MJ '~ f
form / (w)
c • ......1-& ~ 1 r!t'rr1
<:.J ••...,- 4a ~m ler, lrr (11 1 \ 1,/
26 \how 1
that if 0111· or tlu· c,tlu r <,I 11 • I i
F.., = "' al ... w2) • 11 11,,,
t\ ah~olutely 1ntq1rnh lt• Mui th,· , , 11 ' 111/r,J,
tll1i1 has domam (0. o:.) and the f-ouner ( ~ • .t:i), then the ir b1h,1 ria l c c1nv,,1ii1i,
I1ll~r • •~11\d,,4if,
lS •l ~"!ll!'.g f
c:ie uzns'orm FJ,,,A c:r.press
lhc tran,fonn Flw) of
Ill , ' l'4l. ~
,...., of rm lenru of F:/,.w).
the odd C:UCOS"'"

94 Als AND TRANSFORMS

_ APPLICATIONS _Of FOU~IER _I_NT_E_G~ .. .... ...... ........ · · · · · · · · · · · · · · · · · · · · · ·: · · · ·· .................
•• • ••••••• • ••• •· ••• • ll ns of Founer integrals and Pou ri er tr,1n, fc m1, arc ..
The mo,1 ,mporumt app 1,ca o . . . pn;,,~
h 11
'ound m fie\~ ,;uc as sign al analysis, communi catiOonsh engtneenn
h g, O!)l t(.1, , ancl ~rii-.Q.r,l ~"
'• here • ,- e phennmena are conspicuous ly involved
f I
. . . n t e 0 1 er hand , there are«\ • ~--l '"~I
to problems onnu a,e , d m tenns of differential . equations,
eall Pla) n
. . inary and pan,,, ,1,o,.
both ord ~
phenomena,
d th h f ) · . 1
o part Many such applications are extens ions of the u<,e (Jf f.,.,,.,.,•
15
an e c ,e difference simpl) the difference
. . between. summa11on .and integratic,n 'r'L
""' ,
dlustraled m the so\uuon of o,dinar, dtfferentta\ equattons w here disturh,n g >nfluen.,
repne,ented m some easil) handle<! Wa) . Thus. m both Exampt_e s I and 2 of Ste ; no,
periodic fos:mg funcuon which was ,ep laced by . a Fonner sen es . Then each term .,6 Ilk 1
senes was treated as though it we,e the only tern, mvolved. and fina ll y al\ these Paniat ""1 "'
added to fonn the actual solution. Had the fomng functton been nonperiodic. it could no~ tie
represented b) a Founer series, but under suitable conditio ns a Fourie, integral reprc1ttt"'1o:
ha, e been found fo, tL In such a case the,e are no discrete terms to be handled indmdua\ , "'~
summe<J. lnstea~ general infinitesimal segments of the continuous-frequency repre<r.t
1'eated as though they were individual terms. and then the Iota\ effect of their infim• '"'tz1
bLuons ts found
mathematical b) integration. Our first example illustrates this process in detail •,,.
exercise

·················· ······ ······· ····· .... , .. ' ... .... .... ····· ... ··························""
'
Find a particular integral of the equation

th
1
y' + 3y' + 2y == f(t)
if
f(•) = l~ \t\ < l
\t\ > \
. In as isgiven
tat1on, problem, f <)(23),
by Eq. tS theSec.
pulse9.lfunction
, is we studie<i in detai \ in Sec 9. I. and ,is Founon•.<P?

f(t) == -
2 f"" -sin w cos wt dw
7T O w
Hence die given "luation can be wnnen tn the fonn
(J)

y" + 3y' + 2y == -2 l· _~in <v w, wt d1u

ne -
u,
·mi•nit-•--•
I
r---~
--......, r-uuu
i~!fy .
-
~ d
d/ bldividuat
of /(I)
111
~
f ,uent:1~s
,n "'
7T O

~ in the Founer integral rcprese ntall\ ,· . ', .in a ~~' ""

,_ _.
(t)

'/ ~ll 10\t~JJ ~,I;

th,, of die ~_;:-~•~pon ing to or ansmg from the ret,

scgJDeot · ·· . ~ y SCale:
 FOURIHI TRANSFOIIMS

"",,_..
-ALS AIID

-oo < r :s - 7r

t, _,l == {~,n r - 1r :s r :s
1r s r <
7T (c) Prove Caroll
(d) Prove Corollary '\, Theorem 6
/l 0
-oo < rs
00

- 1r
17. If F(w) i~ lhc
ff lJ'
;y ~•
oun er
·1hcorem 6
.,. f(s) dt ) - r tran~forrn

i..l
JIil == {~qn
0
rl
- 7r S I S

7r S f <oo
7T 18. Show that P; - ·.(c,J)/iw
?Iex, .i.e., Whenrsevat ~ thcor .
(1f /rri, ih
r,,,, '~ .,
f(I) ::: em ,~al~o l
If f,(t) + if',(I) rue\1/hcn/ ,
f clions an d Theorem 6 to find the inverse of ' ff 1111 F,r ' c '11-
;ir11al ra
, l ti: r r transfom1
F(cu) of Ex ample 2. ,t ,h,,,.., Ih11
I ff [f(t)] ::: F(w
founc .
11-t • a and w given by == ) then ?, -
I L<t .,u,e ,denlllY
10
4a 4 + w4 Then apply Theo [fr,ll"' Fr
19 L
· ct P be a unit rern 8 to h
t e Proctuc · ~,o,
~ and Theorem 10 lo find
-.-- - a)"][a" + (w + ai · I
pulse function be t/(rJ/ flJ"' 11
twee 11 _ 1 . rr

lo' ,. (w f the Fourier transfonn F(w) of Example 2.

die ,nl'crse o
P(I) ::: u(r + I and I ;n lhat f '
) - u(, - 1),,,lo 1,1:.1
I .
S,'iO" !hat
dz 21r_
___ Show by repeated differ . . I Iii,;: I
to w that entiat,on of ?t[p(t)] ••
111 1
-•(a+ iz)(a - ;, + iw) - 2a + iw W1u, re,pcq
• dz =O ~[l"p(t)] " ' ~ ~
(bl f (a_ iz)(a - iz + iw) .
1T dw"
Hml -1•nterprel each integral as a convolut10n and then where S(w) === (sin w)/w. fa lain
use Theorem 8. used to obtain the Fou . p how this result CJn t~
!G. liflw) and G(w) are, respectively, the Fourier transfonns pulse J(t)p(t) defined ~:r transform of a single ar::1tr:iri
of f(t) and g{I), show that power series. t Ween -1 and I bYa ccn·,e;;!:lt

II) r f(t)g(t) dt = 27r f_

00

00
F(r)G(-r) dr. Hint: Suit-
20. Using the definitions of E .
l\erc1se 19, show that
a;y specialize Theorem 8.

(b) [f(r)g(-1) dt
ably specialize Theorem 8.
= 21r t: F(r)G(r) dr. Hint: Suit-
~[ nhr, ( )]
e PI
l
"' - S(w - n.r)
7i

Explain how this result can be used to obtam tl:e FD!.r:c

I"'
transform of a single pulse defined betv.een - l a:.~ 1by
sin z sin (w - z) 1r sin w a Fourier series in either complel\ exponenual er ral L- i·
ILShow tha1 -----dz== - - - onometric form.
-« z w- z w
11 UsmgTheorem 10, find the inverse of each of the follow- 21. If J(t) is a pulse defined between - I and I b) c1L".:r ot
~ Fourier transforms. the equivalent series
(a) - . I _
(a +iw)2 (b) - - ------;;- 2, (an cos n1rt + bn sin n1rt) or
OU11ng Theorem (a+ iw) 3 n=O
e·3.., 10, find the inverse o~ F(w) =
use the results of faercise 20 to sho11 thJt {J l "•:
~ -Check your result by using a different 1rF(n1r) + F(-n1r), (b) bn = i7if(n rrl - f t-n~,• ~~
kltlonution of F(w). (c) Cn = 1rF(n1r), where f(w) is the Founer tr.1.,,,,rn

FIQd the
inverse of each of the following transforms. the pulse. f - , ~ ,~u b• b
. f f the unctmn 11 1. 1
la)~2iw 8 + 4· 22. Is the Founer trans onn o . ·o rne trJn,1,1n11 ~, LI!<
(I+ , ~ (b) !W Table 9.2, the same us the Founer ~ '
2
l1a1 If 'W)_( + iw)2 (1 + iw)2(3 + iw)2 following function? Why?
/(I) IS tde t' 11
lo n •ca) Y
•~)and iff'( zero for 1 < O and continuous on ll
I · t' "'
(1) = I .....
lhJ, '6how th l tsThat least piecewise continuous on
g
or lhl' t1111l~ II JI(
tr.Ill

a eorem 6 becomes . d the Fourier trun~forrn

23. Fm ,, .... ,
1
I) ~lf'(t)] == iw~[J(t)] - f(O)

tt ~ [f"
f(l) = f~lll (lll1f 1,,/ ,
lilt 0~1;P'°Priate conditions, what is (I) ]1 l , 1.1lu~ iii/,
l~c' ll"
~•
II ~t ,,.. itry I , Theorem 6. nnd show that
l l11rg1'1 lIl l ,J J
'-Ul"Ol111n, 2 , l ,~ I . l'rl'l}llt'lld(' ii.\:
-, , fhcorern 6. COllCC llll'lltl'\
llfOlltld t \(

l l '\\
II u1.il J
---- J), 11(,111(,'
+1.
tin . .- - - - - - - )/11 11111/ 111 ft11/ t-/c•(l(II
C P~tcuJar
n problem~. 'f'a/1kv oj r/r1• 1"111wr/011 (.11,1 11
llnb ctge, Ma&s. , will be of co11 \idl·nthll' lu:lp .
 576 I Hint: What is the antideriv .
~ atrveor lhe
of 1hiS (c ) (iw + I)
,11corcrn~
EXfltCISf5 I 1 11sc 1hc . fu11clions rransform?
if Ex11111p c . , is of 1hc
I G,.,c11 rhc rc~11lr < Fouiicr 1r1111sfon1 e '"'Hint: Factor the denom 1 ·nator
• 10 find rhc (d) ~ . . . and
,cc11nn . : 9. 10. . find rhc
,h()\\ 11 111 J ,g. I , ~cell Oil, F( w) == F,(w) + tF;(w) 1s the Fourieq .
of 111s • 5 (a) I f . h . . 'd . ransfo
id 1hcorc111.~ sfonns. · f ction/(t) wh1c 1s I ent1cal ly zero bc,orc r rrn 0r
ll"np rhc 1ran~fpr111~ n,, lio" ing Fouiicr 1ra11. a un . , use pa
f cnd1 of rhc o ,,,,, show that F,(w) rs the trans:orm o_f onc-ha1r,/ '0, 7
• . •
!:_.- • [he fo
11 11r1,cn ,sion of /(I) and that 1F;(w) 1s the t Ceve
1, :"" (b) -;;T2 + itu) ex tcl . ransfo 0
rm of
2. (M) .1( I ➔ 1/u) _,,., one-half the odd extension of /(1). s. use th
('
,,,.. (b) Whal are the functions represented, rcspc .
I - t' (d) 27r( I + iiu)( 2 + itv) . . ctrvcJy b
the real and the imaginary terms in th c lransi, y
(C) 2. + ll•'
- \u,J
J /(a + iw)? 0 rin [a2 +(
e
r f:'-tw1l
of the f . the jnve
(e) ~
(f) J+Ziw
6. (a) Show that the Fourier transform unc1, 00
- 2fo>
9, Show ~h
---
(•IL I
e
w i ~in (h) - wi + 4iw + 3 -ex;,< I :S 0
(g)~
iw

1:-~
Os1 s; 77
3· (a) (iw
I
(b) (iw + I )(iw + 2) 7TSt< oo
+ I )(iw + 2) (b)
e-iw Hint: Inte
(c) iw +
I

i
(d) iw+2 I + e -i1rw use Theor
I -
is F(w) = 10. If F(w) an
e-2,w 21r I - w2
of f(t) andl
(e) (iw

sin (w -
4. ( a ) - - - -
+ 3)2

w-2
2)
(b)--
sin (w/2)
w
(b) Using the result of Part a, find the Fourier lransfoa
of each of the following functions
(a) 1: f(tl)
ably sp

FIGUIE 9.10
(b) 1:
ably spe
J(t)

f(t) f(I) f(t) 11. Show that ~

2 2 2
12. _Using Theor;
Ing FoLJrier
(a)
1J U . (a + iw)2
-2 · sing "rL
2 I -2 :...1 0 •oeor
2 I -2 -I 0 2 I
-:--:----!__- 3 ,6J I
_,
\l+ ieu~
-I
-I faq0 ri ,l
14 ,-.. Zation f
(a) • l'IJJd the . o
(b) inverse
(c) (a) :-:-----2 + 2 ,-J
l~ Cl+.~
'• (a) lt'.11 ~ '_w)(2 i
/(I)
J(t)
.~)an idenj
J(t) [O () IS
2-
2- [O, ~) d if

ll U.1 lI . •ho_,~ j
(11) llder th .
-2 )
Cb) ~'ve Ce app
-I ·1 ·rove OrolJ
2 I 0 C:orolJ