Question

In Exercises 5-14, the space is \(C\left[ {0,2\pi } \right]\) with inner product (6).

10. Find the third-order Fourier approximation to square wave function \(f\left( t \right) = 1{\rm{ for }}0 \le t < \pi \) and \(f\left( t \right) = - 1{\rm{ for }}\pi \le t < 2\pi \).

Short answer

The required Approximation is\(f\left( t \right) = \frac{4}{\pi }\sin t + \frac{4}{{3\pi }}\sin 3t\).

Step-by-step solution

Inner Product 

The Inner Productfor any two arbitrary functions is given by:

\(\left\langle {f,g} \right\rangle = \int_0^{2\pi } {f\left( t \right)g\left( t \right)dt} \)

Find the third-order Approximation 

As per the question, we have:

\(f\left( t \right) = \left\{ \begin{array}{l}1\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ for }}0 \le t < \pi \\ - 1\,\,\,\,\,\,\,\,{\rm{ for }}\pi \le t < 2\pi \end{array} \right.\)

Using theinner productrule, we have the Fourier Transformation as:

\(\begin{array}{c}\frac{{{a_0}}}{2} = \frac{1}{2}\left\{ {\frac{1}{\pi }\int_0^{2\pi } {\left( 1 \right)dt} } \right\}\\ = \frac{1}{{2\pi }}\int_0^\pi {\left( 1 \right)dt} - \frac{1}{{2\pi }}\int_\pi ^{2\pi } {\left( 1 \right)dt} \\ = 0\end{array}\)

Similarly,

\[\begin{array}{c}{a_k} = \frac{1}{\pi }\int_0^{2\pi } {\cos ktdt} \\ = \frac{1}{\pi }\int_0^\pi {\cos ktdt} - \frac{1}{\pi }\int_\pi ^{2\pi } {\cos ktdt} \\ = 0\end{array}\]

And,

\(\begin{array}{c}{b_k} = \frac{1}{\pi }\int_0^{2\pi } {\sin ktdt} \\ = \frac{1}{\pi }\int_0^\pi {\sin ktdt} - \frac{1}{\pi }\int_\pi ^{2\pi } {\sin ktdt} \\ = \left\{ \begin{array}{l}\frac{4}{{k\pi }},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{k is odd}}\\0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{k is even}}\end{array} \right.\end{array}\)

Hence, the required Approximation at k=3 is \(f\left( t \right) = \frac{4}{\pi }\sin t + \frac{4}{{3\pi }}\sin 3t\).