Notes: a) showing details of the work, write an expansion in Fourier series of the signal f(x) which is assumed to have the period 2`pi`f(x) = "Please...

Saturday, 21 February 2015

a) showing details of the work, write an expansion in Fourier series of the signal f(x) which is assumed to have the period 2 $\displaystyle\pi$ f(x) = "Please...

This function is odd, thus its Fourier expansions contains only $\displaystyle{\sin{{\left({n}{x}\right)}}}$ terms, i.e. $\displaystyle{a}_{{n}}={0},{n}\geq{0}.$ This expansion has the form $\displaystyle{\sum_{{{n}={1}}}^{{\infty}}}{b}_{{n}}{\sin{{\left({n}{x}\right)}}}$ where $\displaystyle{b}_{{n}}=\frac{{1}}{\pi}{\int_{{-\pi}}^{{\pi}}}{f{{\left({x}\right)}}}{\sin{{\left({n}{x}\right)}}}{\left.{d}{x}\right.}.$

Find these coefficients:

$\displaystyle{b}_{{n}}=\frac{{1}}{\pi}{\int_{{-\pi}}^{{\pi}}}{f{{\left({x}\right)}}}{\sin{{\left({n}{x}\right)}}}{\left.{d}{x}\right.}=\frac{{2}}{\pi}{\int_{{0}}^{{\pi}}}{f{{\left({x}\right)}}}{\sin{{\left({n}{x}\right)}}}{\left.{d}{x}\right.}=$

$\displaystyle=\frac{{2}}{\pi}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{f{{\left({x}\right)}}}{\sin{{\left({n}{x}\right)}}}{\left.{d}{x}\right.}+\frac{{2}}{\pi}{\int_{{\frac{\pi}{{2}}}}^{{\pi}}}{f{{\left({x}\right)}}}{\sin{{\left({n}{x}\right)}}}{\left.{d}{x}\right.}{\)}=$

$\displaystyle=\frac{{2}}{\pi}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{x}{\sin{{\left({n}{x}\right)}}}{\left.{d}{x}\right.}+\frac{{2}}{\pi}{\int_{{\frac{\pi}{{2}}}}^{{\pi}}}\frac{\pi}{{2}}{\sin{{\left({n}{x}\right)}}}{\left.{d}{x}\right.}.$

The second integral...

Find these coefficients:

The second integral is obviously

$\displaystyle-{\left({\cos{{\left({n}{x}\right)}}}\right)}{{\mid}_{{{x}=\frac{\pi}{{2}}}}^{\pi}}=-\frac{{1}}{{n}}{\left({\cos{{\left({n}\pi\right)}}}-{\cos{{\left(\frac{{{n}\pi}}{{2}}\right)}}}\right)}.$

To find the second, use integration by parts:

$\displaystyle\frac{{2}}{\pi}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{x}{\sin{{\left({n}{x}\right)}}}{\left.{d}{x}\right.}=$

$\displaystyle={\left|{u}={x},{d}{u}={\left.{d}{x}\right.},{d}{v}={\sin{{\left({n}{x}\right)}}}{\left.{d}{x}\right.},{v}=-\frac{{1}}{{n}}{\cos{{\left({n}{x}\right)}}}\right|}=$

$\displaystyle=-\frac{{2}}{{{n}\pi}}{\left({x}{\cos{{\left({n}{x}\right)}}}\right)}{{\mid}_{{{x}={0}}}^{{\frac{\pi}{{2}}}}}+\frac{{2}}{{{n}\pi}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\cos{{\left({n}{x}\right)}}}{\left.{d}{x}\right.}=$

$\displaystyle=-\frac{{1}}{{n}}{\cos{{\left(\frac{{{n}\pi}}{{2}}\right)}}}+\frac{{2}}{{{{n}}^{{2}}\pi}}{\left({\sin{{\left({n}{x}\right)}}}\right)}{{\mid}_{{{x}={0}}}^{{\frac{\pi}{{2}}}}}=$

$\displaystyle=\frac{{2}}{{{{n}}^{{2}}\pi}}{\sin{{\left(\frac{{{n}\pi}}{{2}}\right)}}}-\frac{{1}}{{n}}{\cos{{\left(\frac{{{n}\pi}}{{2}}\right)}}}.$

This way $\displaystyle{b}_{{n}}=\frac{{2}}{{{{n}}^{{2}}\pi}}{\sin{{\left(\frac{{{n}\pi}}{{2}}\right)}}}-\frac{{1}}{{n}}{\cos{{\left({n}\pi\right)}}}.$ Therefore $\displaystyle{b}_{{1}}={1}+\frac{{2}}{\pi},{b}_{{3}}=\frac{{1}}{{3}}-\frac{{2}}{{{9}\pi}},{b}_{{5}}=\frac{{1}}{{5}}+\frac{{2}}{{{25}\pi}}.$

The graphs are attached (the function is in blue, the approximation is in green). They are not so close but are somewhat similar. Note that on $\displaystyle{\left(-{4}\pi,{4}\pi\right)}$ the function is $\displaystyle{2}\pi$ -periodic.