The question is laplace transform of 4sin(at+b) where a and b are constants. The answer is is ((a)cos(b)+(s)sin(b))*4/(s^2+a^2)-
i think table of integrals is used but how do they get to the answer.....
Help would be greatly appreciated
GAB
The question is laplace transform of 4sin(at+b) where a and b are constants. The answer is is ((a)cos(b)+(s)sin(b))*4/(s^2+a^2)-
i think table of integrals is used but how do they get to the answer.....
Help would be greatly appreciated
GAB
$\displaystyle \mathcal{L} [4 \sin(at+b)] = 4 \int_{0}^{\infty} e^{-st}\sin(at+b) \ dt $
$\displaystyle = 4 \int_{0}^{\infty} e^{-st} \Big(\sin(b) \cos(at) + \cos(b) \sin(at)\Big) \ dt $
$\displaystyle = 4 \sin(b) \int_{0}^{\infty} e^{-st} \cos(at) \ dt + 4 \cos(b) \int_{0}^{\infty} e^{-st} \sin(at) \ dt$
$\displaystyle 4 \sin(b) \frac {s}{s^{2}+a^{2}} + 4 \cos(b) \frac {a}{s^{2}+a^{2}} $
$\displaystyle = \frac {4 \Big(a \cos(b) + s \sin(b) \Big) } {s^{2}+a^{2}} $
The Laplace transform:
$\displaystyle [\mathcal{L}f](s)=\int_{-\infty}^{\infty} f(x) e^{-st}\ dt$
In this case:
$\displaystyle \mathcal{L}\left\{4 \sin(at+b)\right\}=4 \int_{-\infty}^{\infty} \sin(at+b) e^{-st}\ dt=$ $\displaystyle
=4\ \text{Im} \left[ \int_{-\infty}^{\infty} e^{-st}e^{i(at+b)} \ dt \right] =$ $\displaystyle
4\ \text{Im} \left[e^{ib} \int_{-\infty}^{\infty} e^{(-s+ia)t} \ dt\right]$
and the rest is routine.
CB