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

23.5 Fourier series 735
point out that the integrals in Equations (23.3), (23.4) and (23.5) can be evaluated over
anycompleteperiod,forexamplefromt = − T 2 tot =T 2 .Prudentchoiceoftheinterval
ofintegrationcanoftensaveeffort.Theexpressionappearinginther.h.s.oftheFourier
representation, Equation (23.2), is an infinite series. We list conditions, often called the
Dirichlet conditions, sufficient for the series to converge to the value of the function
f (t). The integral ∫ |f (t)|dt over a complete period must be finite, and f (t) may have
no more than a finite number of discontinuities in any finite interval. Fortunately, most
signals of interest to engineers satisfy these conditions. At a point of discontinuity the
Fourier series converges to the average of the two function values at either side of the
discontinuity.
Example23.13 Findthe Fourier series representation ofthe function with periodT = 1 given by
50
{
1 0 t<0.01
f(t)=
0 0.01 t < 0.02
Solution Thefunction f (t)isshowninFigure23.14.Notethat f (t)isdefinedtobezerobetween
t = 0.01 and t = 0.02. This means we need only consider 0 t < 0.01. Using
Equations (23.3)--(23.5) wefind
∫ 0.02
a 0
= 100 f(t)dt =100
0
a n
= 100
b n
= 100
∫ 0.01
0
∫ 0.01
0
Noting thatcosnπ = (−1) n wefind
b n
= 1
nπ (1 − (−1)n )
∫ 0.01
0
= 100[t] 0.01
0
= 1
1dt+100
∫ 0.02
[ sin100nπt
cos100nπtdt = 100
100nπ
0.01
] 0.01
= 0
[ −cos100nπt
sin100nπtdt = 100
100nπ
0
0dt
] 0.01
= − 1 (cosnπ −cos0)
nπ
0
f(t)
1
1 –––
100
T =
1 ––
50
t
Figure23.14
Graph forExample 23.13.