Solution Part1

Problem

3.1.11

a) Dete

in

th

func functi ti

f(t)

wh se Fourie urie tran transf sf

is sh

in figu figu

-3.l -3.l.a .a

Dete

in

th

func functio tio

f(t)

wh se Fourie urie tran transf sf

is sh wn in figu figu

-3.l -3.l.b .b

c) Sk tc

(t

and

near

th

waveform? IF(w)1

IG(w)1

-w

-w

e,(w)

eg(w) 11"/2

-11"/2

(b

Solution:

Th

functi functi

f(t)

an

Figu Figu

9:

-3.1 -3.1.1 .1

tain tain

fr

F(w)

doin

an inve inve se Fourie urie

tran transf sf

i.e.,

(t

y-l{F(w)}

F(w)ejwtdt

(75)

a) f(

-w

27

(76)

rr

b) f(t)

-w

re

11

/2

eJwtdw

7re-

11

/2

eJwtdw

1-

W--:----:-(Wt)

iv

ir ti

tran tran fo

Problem

is

3.1.21

im in

an

ti

F(w)

Y{f(t)},

(77)

in

io ti

th fu ctio ctio

then

a) f(t)dt;

F(O)

24

(78)

0.8

0.6

0.4

0.2

-0.2

- --- -

igur igur

-' --

10 Th fu ctio ctio

L _

_ _

f(t)

sin(Wt)/t

for

1, even even symm symmet etry ry

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8 -30

igur igur

-20

11 Th func functi tion on f(t)

-10

[1

cos(Wt)]/t

25

for

1, od

symm symmet etry ry

0.8

0.6

0.4

0.2

-0.2

- --- -

igur igur

-' --

10 Th fu ctio ctio

L _

_ _

f(t)

sin(Wt)/t

for

1, even even symm symmet etry ry

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8 -30

igur igur

-20

11 Th func functi tion on f(t)

-10

[1

cos(Wt)]/t

25

for

1, od

symm symmet etry ry

b) IF(w)1

(79)

If(t)ldt;

(80)

i:i:i:

f(

r)e-

iW

(81)

2rrF(0).

drdwdt

Solution:

a) (82) sing sing th

tria triang ng

IBI,

in qual qualit ity, y,

(83) 1.

le-iwtl

Note tha

(84)

F(w)eiwtdwl

that that

(85) th n,

have have

Problem

3.2.11

a) Fi

th

Fo ie tran transf sf

(t

fo

th

aise aise

cosi

ulse ulse sign signal al

fine fine

co nt

(87)

elsewhere ss se quat quatio io

sw .1

in

Sa

(equat quatio io

se ie co ffic fficie ient nt fo th

fo

y:

).

3.15 3.15 is as fo

wing wing pe iodi iodi

s:

pu se trai trai

F(nwo)/T) fo th

case case

to find find th

xp nent nentia ia

Fourie urie

2.

(X

kT) k=-(X)

(88)

Fourie

se ie co fficient

di ct

usin

ul 's id ntities.

Solution:

a) -jwtdt

F(w)

(W-7r)tdt

2Sa(w)

Sa(w

(w+7r)tdt 7r

te that

e-Jwtdt

Sa(w)

(89) 27r/T

In this case

7r th

ab ve we get, {2Sa[n7r] Sinc

Sa(k7r

Sa[(n fo

0, it fo

(90)

1)7r]}

that

(91)

F-1 Fo

(92)

1,

and,

(93)

1.

(94) 1.

Fl

Problem 3.2.21

tr

sity

th

ti

f(t)

(95)

exp(at) and sp

to th

fo

in

ha t:

f(t)

27

-a

pa ticu ar

it

1.6

1.4 1.2

0.8 0.6 0.4

Figu

Solution:

a)

ie

sf

F(w) Fr

th

12 Rais

ca

ab ve xp ssio

it

tx

be(a-jw)tdt we

cosine pu se

s:

ae-(a+jw)tdt

tain th

fo

(b

a)

(96)

wing cases:

a2

if

(97)

F(w) -j2wa Th se conc usions ar consistent an

to th

Problelll 3.2.31 in th

th

char

iv

ab

it

co

sp ndin

to th

fi st

second

f(t)

pu se signal desc ib ti

ti pa

Fo ie

xp (-altl)rect(t/T)

is

peate

pe iodicall

tain th

Solution:

a) Fo

ie

in in

th th

ie transf

(-aim

ies. your answ itio

to

th

-b

T.

with pe io

us

it

if

with pr ca

.7. atio

.1 (e

ti

.1

fo

s:

~F(nwo))

Fourie se ie co fficients? ne pu se (fr

tab

3.1

(98)

Using

3.15:

quatio

F(w)

2a/T /T2)

(99)

_!_F(w) But, pr

(100)

2.7.1 gives:

le

e-acos7l'n)

a(l-

(101)

a2 (100) is inde

Relationship

th

te

th

ie

ie

tati

th

io ic fu

se tati

th

io ic

while (101)

(-altl) ie

ie

ti

tain

atin

th

f(t)

fu ctio

xp (-altl)rect(t/T).

f(t), pe iodic signal

f(t)

is ze

the fo

L:~-oo

tsid

ar th

ti

igin

ig al

ve th

th

F(w)

In ca ying

urie

su ting pe iodi

ti

inte va

th

fo

fficie ts

th

f(t)

co fficie ts in

ig

L:~-oo

signal na

0' 21

ut your so utio

athe atical

ie

is qual

T.

it

th sa

to th

th

in have th

iv

th

f(t) Fourie transf

~F(nwo)

inte al T.

do no have ve ap an to th

ul

u2

il

he pful to co

J~oo

te

i. ., th

0'

aussia

e-

functi

du is it

in

ponents, co pl te

y'iF. Note tha

f(t)

and

Fourie transf

!So uti n:

(102) Le

(t

j0'2w)/V2;

then,

(103)

!P

3. .2

ho

that

gene al stat

nt

arseva 's th

fo fo

ne gy signal

(3.21) is f(t)g*(t)dt

271'

29

F(w)G*(w)dw.

(104)

Solution:

00f(t)g*(t)dt in

an inte change in th

- 0 0 f (t )g * (t )d t in

th

Problem

b)

dx/(a

),

c)

dx/(a

)2

kn

co

F(w)G*(x)

s:

(106)

e-j(x-w)tdwdxdt.

is

G*

3.4.3! Evaluate th [sin x/x]2dx,

21l'

integration, this

(21l'F

21l'

a)

!S ution:

de

te

f(t)g*(t)dt

F(w)ejwtdw]

21l'

fo

in

21l'

definite integral

usin

F(w)G*(w)dw.

(107)

arseva 's th

that

a) rect(t/2)