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

3.7 Modelling waves using sint and cost 135
Solution
(a) v 1
(t) has an amplitude of 3 volts and an angular frequency ω = 1 rad s −1 . v 2
(t)
has an amplitude of 2 volts and an angular frequency ω = 1 rad s −1 . Note that
both of these signals have the same angular frequency.
(b) v 3
(t)=v 1
(t)+2v 2
(t)
=3sint+2(2cost)
=3sint+4cost
(c) We wish to write v 3
(t) in the formRsin(t + φ). The choice of sine is arbitrary.
We could have chosen cosine instead.Ris the amplitude of the single sinusoid
and φ isits phase angle.
Using the trigonometric identity sin(t + φ) = sintcos φ +sinφcost found
inTable 3.1 wecan write
Rsin(t +φ) =R(sintcosφ +sinφcost)
= (Rcosφ)sint +(Rsinφ)cost
Comparing thisexpression with thatfor v 3
(t) wenotethat, inorder tomake the
expressions identical,
Rcosφ =3 (3.1)
Rsinφ =4 (3.2)
Weneedtosolve(3.1)and(3.2)toobtainRand φ.Squaringeachequationgives
R 2 cos 2 φ=9
R 2 sin 2 φ = 16
Adding these equations together weobtain
R 2 cos 2 φ+R 2 sin 2 φ=9+16
R 2 (cos 2 φ +sin 2 φ) = 25
Usingthe identitycos 2 φ +sin 2 φ = 1 thissimplifies to
R 2 =25
R = 5
Next wedetermine φ. Dividing (3.2) by (3.1) wefind
Rsinφ
Rcosφ = 4 3
tanφ = 4 3
)
φ = tan −1 ( 4
3
From (3.1) and (3.2) we can see that both sinφ and cos φ are positive, and so
φ must lie in the first quadrant. Calculating tan −1( 4
3)
using a calculator gives
φ = 0.927 radians. So we can express v 3
(t) as
v 3
(t) =3sint +4cost =5sin(t +0.927)
Finally v 3
(t) has amplitude 5 volts and phase 0.927 radians.