Understanding Engineering Mathematics

So we get

πbm=

∫π

−π

f(t)sinmtdt

=

∫ 0

−π

(−A)sinmtdt+

∫π

0

Asinmtdt

=A

[

cosmt

m

] 0

−π

−A

[

cosmt

m

]π

0

=

A

m

( 1 −(− 1 )m)−

A

m

((− 1 )m− 1 )

(cosmπ=(− 1 )m)

=

2 A

m

( 1 −(− 1 )m)

Note that sincef(t)is odd we could have used (284
➤
)
∫π

−π

f(t)sinmtdt= 2

∫π

0

f(t)sinmtdt

So finally

bm=

2 A

mπ

( 1 −(− 1 )m)

=0ifmeven

=

4 A

mπ

ifmodd

or, reverting ton

bn=

2 A

nπ

( 1 −(− 1 )n)

Theseriesisthus

f(t)=

∑∞

n= 1

bnsinnt=

∑∞

n= 1

2 A

nπ

( 1 −(− 1 )n)sinnt

=

4 A

π

sint+

4 A

3 π

sin 3t+

4 A

5 π

sin 5t+···

=

4 A

π

(

sint+

1

3

sin 3t+

1

5

sin 5t+···

)

Exercise on 17.10

Verify the series given in Exercise on 17.9 for the triangular wave

f(t)=t 0 <t<π

−t −π<t< 0

f(t)=f(t+ 2 π)