Question

Find the Fourier coefficients of the piecewise continuous function

$\mathit{f}\left(t\right)\mathbf{=}\{\begin{array}{l}\begin{array}{ll}0& ift\le 0\end{array}\\ \begin{array}{ll}1& ift>0\end{array}\end{array}$

Sketch the graphs of the first few Fourier polynomials.

Short answer

The coefficients are

$$\begin{gathered} a_0=0 \\ {b}_k=\left\{\begin{array}{l}0\\ \frac{4}{k\tau \tau }\end{array}\right. \\ {c}_k=0 \end{gathered}$$

The Fourier polynomials are

$$\begin{gathered} f_1\left(t\right)=\frac{4}{\mathrm{\pi }}\sin t \\ {f}_2\left(t\right)=\frac{4}{\mathrm{\pi }}\sin t \\ {f}_3\left(t\right)=\frac{4}{\mathrm{\pi }}\sin t+\frac{4}{\mathrm{\pi }}\sin \left(3t\right) \\ {f}_4\left(t\right)=\frac{4}{\mathrm{\pi }}\sin t+\frac{4}{\mathrm{\pi }}\sin \left(3t\right) \\ {f}_5\left(t\right)=\frac{4}{\mathrm{\pi }}\sin t+\frac{4}{\mathrm{\pi }}\sin \left(3t\right)+\frac{4}{5\mathrm{\pi }} \\ {f}_6\left(t\right)=\frac{4}{\mathrm{\pi }}\sin t+\frac{4}{\mathrm{\pi }}\sin \left(3t\right)+\frac{4}{5\mathrm{\pi }} \end{gathered}$$

Step-by-step solution

Fourier coefficients and the Fourier polynomials.

It wants to find the Fourier coefficients of the above function. Observe that the function f is an odd function and therefore $a_0=0$and $c_k=0$ for all k=1,2,3,…..

$$\begin{gathered} b_k=\frac{1}{\tau \tau }{\int }_{-\tau \tau }^{\tau \tau }f\left(t\right)\sin \left(kt\right)dt \\ =\frac{1}{\tau \tau }{\left[\frac{\cos \left(kt\right)}{k}\right]}_{-\tau \tau }^0-{\left[\frac{\cos \left(kt\right)}{k}\right]}_{\tau \tau }^0 \\ =\frac{1}{\tau \tau }\times \frac{1}{k}-\frac{\cos \left(k\tau \tau \right)}{k} \\ =\left\{\begin{array}{ccc}0& k& iceven\\ \frac{4}{k\tau \tau }& k& isodd\end{array}\right. \end{gathered}$$

Consider the diagrams below.

For$k_1$

For$f_3$

For$f_5$

For$f_7$

Hence, the coefficients will be

$$\begin{gathered} a_0=0 \\ {b}_k=\left\{\begin{array}{l}0\\ \frac{4}{k\tau \tau }\end{array}\right. \\ {c}_k=0 \end{gathered}$$

And The Fourier polynomials will be

$$\begin{gathered} f_1\left(t\right)=\frac{4}{\tau \tau }\sin t \\ {f}_2\left(t\right)=\frac{4}{\mathrm{\pi }}\sin t \\ {f}_3\left(t\right)=\frac{4}{\mathrm{\pi }}\sin t+\frac{4}{\mathrm{\pi }}\sin \left(3t\right) \\ {f}_4\left(t\right)=\frac{4}{\mathrm{\pi }}\sin t+\frac{4}{\mathrm{\pi }}\sin \left(3t\right) \\ {f}_5\left(t\right)=\frac{4}{\mathrm{\pi }}\sin t+\frac{4}{\mathrm{\pi }}\sin \left(3t\right)+\frac{4}{5\mathrm{\pi }} \\ {f}_6\left(t\right)=\frac{4}{\mathrm{\pi }}\sin t+\frac{4}{\mathrm{\pi }}\sin \left(3t\right)+\frac{4}{5\mathrm{\pi }} \end{gathered}$$