2. b) find the fourier transform of given signal below: $x(t)=\begin{cases}t^{2}&0 < t < 1\\0&otherwiseend{ca…

Explanation:

Step1: Recall Fourier - transform formula

The Fourier - transform of a function $x(t)$ is given by $X(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt$. Since $x(t) = 0$ outside the interval $(0,1)$, we have $X(j\omega)=\int_{0}^{1}t^{2}e^{-j\omega t}dt$.

Step2: Use integration by parts

Let $u = t^{2}$, $dv=e^{-j\omega t}dt$. Then $du = 2tdt$, $v=-\frac{1}{j\omega}e^{-j\omega t}$. By the integration - by - parts formula $\int_{a}^{b}u;dv=uv|{a}^{b}-\int{a}^{b}v;du$, we get: [ \begin{align*} \int_{0}^{1}t^{2}e^{-j\omega t}dt&=-\frac{t^{2}}{j\omega}e^{-j\omega t}\big|{0}^{1}+\frac{2}{j\omega}\int{0}^{1}te^{-j\omega t}dt\ &=-\frac{1}{j\omega}e^{-j\omega}+\frac{2}{j\omega}\int_{0}^{1}te^{-j\omega t}dt \end{align*} ]

Step3: Use integration by parts again

For $\int_{0}^{1}te^{-j\omega t}dt$, let $u = t$, $dv=e^{-j\omega t}dt$. Then $du = dt$, $v = -\frac{1}{j\omega}e^{-j\omega t}$. [ \begin{align*} \int_{0}^{1}te^{-j\omega t}dt&=-\frac{t}{j\omega}e^{-j\omega t}\big|{0}^{1}+\frac{1}{j\omega}\int{0}^{1}e^{-j\omega t}dt\ &=-\frac{1}{j\omega}e^{-j\omega}+\frac{1}{j\omega}\left(-\frac{1}{j\omega}e^{-j\omega t}\big|_{0}^{1}\right)\ &=-\frac{1}{j\omega}e^{-j\omega}-\frac{1}{\omega^{2}}(e^{-j\omega}-1) \end{align*} ]

Step4: Substitute back

[ \begin{align*} X(j\omega)&=-\frac{1}{j\omega}e^{-j\omega}+\frac{2}{j\omega}\left(-\frac{1}{j\omega}e^{-j\omega}-\frac{1}{\omega^{2}}(e^{-j\omega}-1)\right)\ &=-\frac{1}{j\omega}e^{-j\omega}-\frac{2}{\omega^{2}}e^{-j\omega}+\frac{2}{j\omega^{3}}(1 - e^{-j\omega})\ &=\frac{2}{j\omega^{3}}-\left(\frac{1}{j\omega}+\frac{2}{\omega^{2}}+\frac{2}{j\omega^{3}}\right)e^{-j\omega} \end{align*} ]

Answer:

$X(j\omega)=\frac{2}{j\omega^{3}}-\left(\frac{1}{j\omega}+\frac{2}{\omega^{2}}+\frac{2}{j\omega^{3}}\right)e^{-j\omega}$