082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

15.3 Root mean square value of a function 475
Solutions
1 (a) 2 (b) −1 (c)
1
3
(d)
2 (a) 10 (b) 0.6931 (c) 0.9428
1
(d) −1 (e)
6
2 2
3 (a) (b)
π π
4
3
(e)
19
3
1
(c)
πω [1 −cos(πω)] (d) 2
π
sin(πω)
(e) 0 (f)
πω
1+sinω−cosω
(g)
ω
14
4 (a) (b) 1.1752 (c) 2.1752
9
15.3 ROOTMEANSQUAREVALUEOFAFUNCTION
If f (t)isdefinedon [a,b], theroot mean square (r.m.s.)value is
r.m.s. =
√ ∫b
a (f(t))2 dt
b −a
Example15.2 Findthe r.m.s. valueof f (t) =t 2 across[1, 3].
Solution r.m.s. =
√ ∫ 3
1 (t2 ) 2 dt
3 −1
=
√ ∫ 3
1 t4 dt
2
=
√ √
[t 5 /5]
1
3 242
=
2 10 = 4.92
Example15.3 Calculate the r.m.s. valueof f (t) =Asint across [0,2π].
√ ∫ 2π
A
0 2 sin 2 tdt
Solution r.m.s. =
2π
√
A ∫ 2 2π
(1 −cos2t)/2dt
0
=
2π
√
[
A
=
2
t − sin2t ] 2π
4π 2
0
√
A2 2π
=
4π = √ A = 0.707A
2
Thus the r.m.s.value is0.707 ×the amplitude.