The spectrum $m(f)$ of $m(t)$ , is an even triangle pulse , with bandwidth w , find m(t)

Question

So I am trying to find the inverse fourier of m(f) by the definition. $$\begin{aligned}\int_{-\infty}^\infty m(f)e^{j2πft}df&=\int_{-w}^w m(f)e^{j2\pi ft}df\\&=\int_{-w}^wm(0)\left(\frac{-|f|}w+1\right)e^{j2\pi ft}df\\&=m(0)\int_{-w}^0\left(\frac{f}w+1\right)e^{j2πft}df + \int_0^wm(0)\left(\frac{-f}{w}+1\right)e^{j2\pi ft}df\\&=m(0)\frac2{(j2\pi t)^2}(1-\cosh(2\pi wt))\end{aligned}$$ Any feedback?

Answer 1

Your answer is correct. Here's an alternative check.

Since $m(f)$ is an even triangular pulse with support $[-w, w]$, then, like what you said, we have that \begin{align} m(f) = m(0)\Lambda\left(\frac{f}{w}\right) \end{align} where $\Lambda$ is the standard triangular pulse $\Lambda(x) = \max\{1-|x|, 0\}$. In particular, we see that \begin{align} m(f) = \alpha (\Pi\ast\Pi)\left( \frac{f}{w}\right) \end{align} where $\alpha = m(0)$ and $\Pi$ is the rectangular function on the interval $[-\frac{1}{2}, \frac{1}{2}]$ with height $1$.

Finally, we see that \begin{align} \mathcal{F}^{-1}\left(m(f)\right) =& w\alpha\mathcal{F}^{-1}\left(\Pi\ast \Pi \right)\left( wt\right) = w\alpha \mathcal{F}^{-1}(\Pi)^2(wt)\\ =&\ w\alpha \operatorname{sinc}^2(w t) = w\alpha \frac{\sin^2(\pi w t)}{\pi^2 w^2 t^2} \\ =&\ \alpha\frac{1-\cos(2\pi w t)}{2\pi^2 w t^2}. \end{align}