Question

Find all Fourier coefficients of the absolute value function

$\mathit{f}\left(t\right)\mathbf{=}\left|t\right|$

Short answer

The Fourier coefficients are

$$\begin{gathered} a_0=\frac{\tau \tau }{\sqrt{2}} \\ {b}_k=0 \\ {c}_k=\frac{-4}{\tau \tau k^2},0 \end{gathered}$$

k is even.

Step-by-step solution

The Fourier coefficient of the function   f(t)=|t|

Here it needs to find out the function $f\left(t\right)=\left|t\right|$. Recall that the coefficients are given by

The given function is $f\left(t\right)=\left|t\right|$. So,

$$\begin{gathered} a_0=\left|t\right|,\frac{1}{\sqrt{2}} \\ =\frac{1}{\sqrt{2\tau \tau }}{\int }_{-\tau \tau }^{\tau \tau }\left|t\right|dt \\ =\frac{2}{\sqrt{2\tau \tau }}{\int }_0^{\tau \tau }tdt \\ =\frac{\tau \tau }{2} \end{gathered}$$

Similarly, the value of is,

And now for the last coefficient $c_k$consider the solution,

Further solve the above expression,

$$\begin{gathered} c_k=\frac{2}{\tau \tau }{\left[\frac{1}{k}t\sin \left(kt\right)\right]}_0^{\tau \tau }-{\int }_0^{\pi }-\sin \left(kt\right)dt \\ =\frac{2}{\tau \tau }.\frac{1}{k^2}{\left[\cos \left(kt\right)\right]}_o^{\tau \tau } \\ =\frac{2}{\tau \tau k^2}\left(\cos \left(k\tau \tau \right)-1\right) \\ =\left\{\begin{array}{ll}\frac{-4}{\tau \tau k^2}& kiseven\\ 0& kiseven0\end{array}\right. \end{gathered}$$

Hence, the Fourier coefficients are $a_0=\frac{\tau \tau }{\sqrt{2}},b_k=0$and $c_k=\frac{-4}{\tau \tau k^2},0$.