Question

T 14 3 5 13. Find the half period series for f(x) given in the range (o,) by the graph OPQ as shown in figure. (U.P.T.U. 2009) P(a, d) Q(1,) x- a o<x<a Hint. f(x) = a Hint. f(x) = 4(1-x) a<x<1 11-a. f(x) = Ź bn sin ma

Answer 1

The coefficient bn of the Fourier Series is:

$b_{n} = \frac{2}{T}\int_{0}^{a}f(x)\cdot \sin \frac{n\pi x}{l}+\frac{2}{T}\int_{a}^{l}f(x)\cdot \sin \frac{n\pi x}{l}$

Substituting f(x):

$b_{n} = \frac{2}{T}\int_{0}^{a}\left [\left (\frac{xd}{a} \right )\cdot \sin \frac{n\pi x}{l} \: dx \right ]+\frac{2}{T}\int_{a}^{l}\left [\left (\frac{d(l-x)}{l-a} \right )\cdot \sin \frac{n\pi x}{l} \: dx \right ]$

Separating the integrals, and getting out the constant terms:

2dl пп dr sin T-sin ) d аT

$-\: \frac{2d\cdot x}{T\left (l-a \right )}\int_{a}^{l}x\cdot \sin \left (\frac{n\pi x}{l} \right ) \: dx$

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The first integral is calculated integrating by parts:

$\int_{0}^{a}x\cdot \sin \left (\frac{n\pi x}{l} \right ) \: dx \rightarrow u = x;\: dv = \sin \left (\frac{n\pi x}{l} \right )$

$du = dx;\: v = -\cos \left (\frac{n\pi x}{l} \right )$;

Presenting the general equation of integral by parts, and substituting:

$\int u\: dv=uv-\int v\: du \rightarrow -x\cdot \cos \left (\frac{n\pi x}{l} \right )+\int \cos \left (\frac{n\pi x}{l} \right )\: dx$

Solving the integral and evaluating:

$=\left [-x\cdot \cos \left (\frac{n\pi x}{l} \right )+ \sin \left (\frac{n\pi x}{l} \right ) \right ]_{0}^{a} = -a\cdot \cos \left (\frac{n\pi a}{l} \right )+ \sin \left (\frac{n\pi a}{l} \right )$

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The second integral is as just the integral of sin(x)

$\int_{a}^{l}\: \sin\left ( \frac{n\pi x}{l} \right )\: dx =\left [ \cos\left ( \frac{n\pi x}{l} \right ) \right ]_{a}^{l} = \cos\left ( n\pi \right )-\cos\left ( \frac{n\pi a}{l} \right )$

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The third integral is the result of the integral by parts of the first integral, but evaluated between a and l:

$=\left [-x\cdot \cos \left (\frac{n\pi x}{l} \right )+ \sin \left (\frac{n\pi x}{l} \right ) \right ]_{a}^{l} =\left ( -l\cdot \cos \left (n\pi \right )+ \sin \left (n\pi \right ) \right )-\left ( -a\cdot \cos \left (\frac{n\pi a}{l} \right )+ \sin \left (\frac{n\pi a}{l} \right ) \right )$

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Substituting the result of the integrals, bn is:

$b_{n} =\frac{2d}{al} \left [-a\cdot \cos \left (\frac{n\pi a}{l} \right )+ \sin \left (\frac{n\pi a}{l} \right ) \right ]+ \frac{2dl}{l(l-a)}\left [ \cos\left ( n\pi \right )-\cos\left ( \frac{n\pi a}{l} \right ) \right ]$

$-\frac{2d}{l(l-a)}\left [\left ( -l\cdot \cos \left (n\pi \right )+ \sin \left (n\pi \right ) \right )-\left ( -a\cdot \cos \left (\frac{n\pi a}{l} \right )+ \sin \left (\frac{n\pi a}{l} \right ) \right ) \right ]$