082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017
744 Chapter 23 Fourier seriesNotethatthevalueofa nistheamplitudeofthenthharmonic.Further,thefrequencyof the nth harmonic is nf, that is integer multiples of the frequency of the PWMsignal.Agraphof |a n|against fnisoftenreferredtoasthefrequencyspectrum.Thisconcept isdiscussed further inChapter 24.Consider the specific case of a 12 V signal at a frequency of f = 300 Hz.Figures 23.19(a)--(d) show the frequency spectra when d = 0.01,0.15,0.30 and0.50 respectively.It can be seen that the frequency spectrum becomes narrower for higher valuesof the duty cycle. The sinc function (see Section 3.5) forms the envelope for thespectrum, which is most clearly evident for d = 0.15. For low values of the dutycycletheharmonicsmayextenduptoradiofrequenciesandhencethereisapotentialforinterference tooccur.Note thatwhend = 0.5a n= 2A ( nπ)nπ sin 2( nπ)When n is even the value of sin is zero and so all the even harmonics are2absent. In other words, when d = 0.5 the PWM signal possesses only oddharmonics.a n (V)——0.240.220.200.180.160.14a n (V) 3.5——3.02.52.01.51.00.5a n (V)——6(a)2 4 6 8f (kHz)a n (V)——(b)2 4 6 8f (kHz)56434212(c)2 4 6 8f (kHz)Figure23.19Spectrum for(a)d = 0.01, (b)d = 0.15,(c)d = 0.30,(d)d = 0.50.(d)2 4 6 8f (kHz)
23.6 Half-range series 745EXERCISES23.51 Find the Fourier seriesrepresentation ofthe function{0 −5 <t < 0 period10f(t)=1 0<t<52 Find the Fourier seriesrepresentation ofthe function{−t −π <t < 0 period2πf(t)=0 0<t<π3 Find the Fourier seriesrepresentation ofthe functionf(t)=t 2 +πt −π<t <π period2π4 Find the Fourier seriesrepresentation ofthe function{−4 −π <t 0 period 2πf(t)=4 0<t<π5 Find the Fourier seriesrepresentation ofthe function{2(1+t) −1<t0 period2f(t)=0 0<t<16 Find the Fourier seriesrepresentation ofthe functionwith period2π given by{t 2 0t<πf(t)=0 πt<2π7 Find the Fourier seriesrepresentation ofthe functionf(t)=2sint 0<t <2πperiod2π8 Forthe signal in Engineering application 23.1, showthata n = 2Anπ sin(nπd)Solutions12312 + 2 πtsinπ 5 + 2 3πtsin3π 5 + 2 5πtsin5π 5+ 247πtsin7π 5 · · ·π4 − 2 π cost−sint+1 2 sin2t− 259π cos3t−1 3 sin3t...π 23 +2πsint−4cost−πsin2t+cos2t+ 62π3 sin3t − 4 9 cos3t···8(2sint sin5t+···)+ 3 2 sin3t + 5 2π1{2 +2 (2/π)cosπt−sinπt− 1 2 sin2πt+(2/9π)cos3πt − 1 3 sin3πt+···}/ππ 26 + π2 −4sint−2cost− π π 2 sin2t + 1 2 cos2t+ 9π2 −427πsin3t − 2 9 cos3t+···7 2sint23.6 HALF-RANGESERIESSometimes an engineering function is not periodic but is only defined over a finiteinterval, 0 < t < T say, as shown in Figure 23.20. In cases like this Fourier analysiscan still be useful. Because the region of interest is only that between t = 02andt = T we may choose to define the function arbitrarily outside the interval. In particular,we can make our choice so that the resulting function is periodic, with periodT.2Thereismorethanonewaytoproceed.Forexample,wecanreflecttheabovefunctionin the vertical axis and then repeat it periodically so that the result is the periodic evenfunction shown in Figure 23.21. We have performed what is called a periodic extensionof the given function. Note that within the interval of interest nothing has altered
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23.6 Half-range series 745
EXERCISES23.5
1 Find the Fourier seriesrepresentation ofthe function
{
0 −5 <t < 0 period10
f(t)=
1 0<t<5
2 Find the Fourier seriesrepresentation ofthe function
{
−t −π <t < 0 period2π
f(t)=
0 0<t<π
3 Find the Fourier seriesrepresentation ofthe function
f(t)=t 2 +πt −π<t <π period2π
4 Find the Fourier seriesrepresentation ofthe function
{
−4 −π <t 0 period 2π
f(t)=
4 0<t<π
5 Find the Fourier seriesrepresentation ofthe function
{
2(1+t) −1<t0 period2
f(t)=
0 0<t<1
6 Find the Fourier seriesrepresentation ofthe function
with period2π given by
{
t 2 0t<π
f(t)=
0 πt<2π
7 Find the Fourier seriesrepresentation ofthe function
f(t)=2sint 0<t <2π
period2π
8 Forthe signal in Engineering application 23.1, show
that
a n = 2A
nπ sin(nπd)
Solutions
1
2
3
1
2 + 2 πt
sin
π 5 + 2 3πt
sin
3π 5 + 2 5πt
sin
5π 5
+ 2
4
7πt
sin
7π 5 · · ·
π
4 − 2 π cost−sint+1 2 sin2t− 2
5
9π cos3t−1 3 sin3t...
π 2
3 +2πsint−4cost−πsin2t+cos2t+ 6
2π
3 sin3t − 4 9 cos3t···
8
(2sint sin5t+···)
+ 3 2 sin3t + 5 2
π
1
{
2 +2 (2/π)cosπt−sinπt− 1 2 sin2πt
+(2/9π)cos3πt − 1 3 sin3πt+···}/π
π 2
6 + π2 −4
sint−2cost− π π 2 sin2t + 1 2 cos2t
+ 9π2 −4
27π
sin3t − 2 9 cos3t+···
7 2sint
23.6 HALF-RANGESERIES
Sometimes an engineering function is not periodic but is only defined over a finite
interval, 0 < t < T say, as shown in Figure 23.20. In cases like this Fourier analysis
can still be useful. Because the region of interest is only that between t = 0
2
and
t = T we may choose to define the function arbitrarily outside the interval. In particular,
we can make our choice so that the resulting function is periodic, with periodT.
2
Thereismorethanonewaytoproceed.Forexample,wecanreflecttheabovefunction
in the vertical axis and then repeat it periodically so that the result is the periodic even
function shown in Figure 23.21. We have performed what is called a periodic extension
of the given function. Note that within the interval of interest nothing has altered