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

724 Chapter 23 Fourier series
all angular frequencies are integer multiples of the fundamental angular frequency ω 1
,
which is missing. Nevertheless, f (t) has the same angular frequency as the fundamental.
A common value for ω 1
is 100π as this corresponds to a frequency of 50 Hz, the
frequency of the UK mains supply.
Example23.1 Describe the frequency and amplitude characteristics of the different harmonic components
of the function
f (t) = cos20πt +0.6cos60πt −0.2sin140πt
Solution Thefundamentalangularfrequencyis20πarisingthroughthetermcos20πt.Thiscorresponds
to a frequency of 10 Hz. This term has amplitude 1. The second, fourth, fifth
andsixthharmonicsaremissing,whilethethirdandseventhhaveamplitudes0.6and0.2,
respectively.
Example23.2 If f (t) = 2sint + 3cost, express f (t) as a single sinusoid and hence determine its
amplitude and phase.
Solution Both terms have angular frequency ω = 1. Recalling the trigonometric identity (Section
3.7)
Rcos(ωt−θ)=acosωt+bsinωt
whereR = √ a 2 +b 2 ,tanθ = b a , we see that in this caseR = √ 3 2 +2 2 = √ 13 and
tanθ = 2 , that is θ = 0.59 radians. Therefore we can express f (t)inthe form
3
f(t)= √ 13cos(t −0.59)
We see immediately that this is a sinusoid of amplitude √ 13 and phase angle −0.59
radians.
Example23.3 Findthe amplitudeand phase ofthe fundamentalcomponent ofthe function
f(t) =0.5sinω 1
t +1.5cosω 1
t +3.5sin2ω 1
t −3cos3ω 1
t
Solution Contributions to the fundamentalcomponent -- that is, that with the lowest frequency --
come from the terms 0.5sin ω 1
t and 1.5cos ω 1
t only. To find the amplitude and phase
wemustexpress theseasasinglecomponent.Usingthe trigonometricidentity
Rcos(ωt−θ)=acosωt+bsinωt
whereR = √ a 2 +b 2 ,tanθ = b a , wefind
R = √ 1.5 2 +0.5 2 = 1.58
tanθ = 0.5
1.5 = 1 thatis, θ = 0.32 radians
3
Therefore the fundamental can bewritten1.58cos(ω 1
t −0.32) and has amplitude1.58
and phase angle −0.32 radians.