082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017
476 Chapter 15 Applications of integrationExample15.4 Calculate the r.m.s.value of f (t) =Asin(ωt + φ) across [0,2π/ω].√ ∫ 2π/ωA0Solution 2 sin 2 (ωt + φ)dtr.m.s. =2π/ω√∫A=2 ω 2π/ω1−cos2(ωt +φ)dt4π 0√[]A=2 ω sin2(ωt + φ) 2π/ωt −4π 2ω0=√(A 2 ω 2π4π ω−sin2(2π + φ)2ω+ sin2φ )2ωNow sin2(2π + φ) = sin(4π + 2φ) and since sin(t + φ) has period 2π we see thatsin(4π +2φ) = sin2φ. Hence,√ √A2 ω 2π Ar.m.s. =4π ω = 22 = √ A = 0.707A2Notethatsin(ωt + φ)hasperiod2π/ω.TheresultofExample15.4illustratesageneralresult:The r.m.s. value of any sinusoidal waveform taken across an interval of length oneperiodis0.707 ×amplitudeofthe waveformRoot mean square value is an effective measure of the energy transfer capability of atime-varyingelectricalcurrent.Toseewhythisisso,considerthefollowingEngineeringapplication.Engineeringapplication15.3Average power developed across a resistor by a time-varyingcurrentConsideracurrenti(t)whichdevelopsapowerp(t)inaloadresistorR.Thiscurrentflowsfromtimet =t 1totimet =t 2.LetP avbetheaveragepowerdissipatedbytheresistor during the time interval [ ]t 1,t 2 . We require that total energy transfer,E,bethe same inboth cases. So wehaveE =P av(t 2−t 1)=∫ t2t 1p(t)dt
15.3 Root mean square value of a function 477Nowand sop(t) = (i(t)) 2 RP av(t 2−t 1)=∫ t2t 1i 2 RdtIfwenowconsidertheaveragepowerdissipatedbytheresistortobetheresultofaneffective currentI effthen wehaveI 2 eff R(t 2 −t 1 )= ∫ t2I 2 eff (t 2 −t 1 )= ∫ t2t 1i 2 Rdtt 1i 2 dt∫ t2i 2 dtI 2 eff = t 1t 2−t 1∫ t2√it 2 dtI eff=1t 2−t 1We see that the equivalent direct current is the r.m.s. value of the time-varyingcurrent.Engineeringapplication15.4Averagevalueandr.m.s.valueofaperiodicwaveformConsider the periodicwaveform shown inFigure 15.5.Thecurrenti(t) isi(t) =20e −t 0 t <10, period T =10(a) Calculate the average valueofthe currentover a complete period.(b) Calculate the r.m.s. valueofthe currentover a complete period.i(t)200 10 2030 tFigure15.5Waveform forEngineering application15.4.➔
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15.3 Root mean square value of a function 477
Now
and so
p(t) = (i(t)) 2 R
P av
(t 2
−t 1
)=
∫ t2
t 1
i 2 Rdt
Ifwenowconsidertheaveragepowerdissipatedbytheresistortobetheresultofan
effective currentI eff
then wehave
I 2 eff R(t 2 −t 1 )= ∫ t2
I 2 eff (t 2 −t 1 )= ∫ t2
t 1
i 2 Rdt
t 1
i 2 dt
∫ t2
i 2 dt
I 2 eff = t 1
t 2
−t 1
∫ t2
√
i
t 2 dt
I eff
=
1
t 2
−t 1
We see that the equivalent direct current is the r.m.s. value of the time-varying
current.
Engineeringapplication15.4
Averagevalueandr.m.s.valueofaperiodicwaveform
Consider the periodicwaveform shown inFigure 15.5.
Thecurrenti(t) is
i(t) =20e −t 0 t <10, period T =10
(a) Calculate the average valueofthe currentover a complete period.
(b) Calculate the r.m.s. valueofthe currentover a complete period.
i(t)
20
0 10 20
30 t
Figure15.5
Waveform forEngineering application
15.4.
➔
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