Q6Advanced Engineering Mathematics I

We can express $f(t)$ using the Heaviside Unit Step function $u(t)$: $f(t)=\sin t\cdot [u(t)-u(t-\pi )]=\sin t\cdot u(t)-\sin t\cdot u(t-\pi )$

We know that $\sin t=\sin ((t-\pi )+\pi )=-\sin (t-\pi )$. Substituting this back: $f(t)=\sin t\cdot u(t)+\sin (t-\pi )\cdot u(t-\pi )$

Using the second shifting property $L\{f(t-a)u(t-a)\}=e^{-as}F(s)$: $L\{f(t)\}=L\{\sin t\}+e^{-\pi s}L\{\sin t\}$

Since $L\{\sin t\}=\frac{1}{s^2+1}$: $L\{f(t)\}=\frac{1}{s^2+1}+e^{-\pi s}\frac{1}{s^2+1}=\frac{1+e^{-\pi s}}{s^2+1}$