- #1

Reshma

- 749

- 6

1] Find the inverse Laplace transform of the given function, i. e.

[tex]L^{-1}\left[1\over {s^3 + 1}}\right][/tex]

And this is how I proceeded about:

[tex]L^{-1}\left[1\over {s^3 + 1}}\right][/tex]

[tex]= L^{-1}\left[\frac{1}{(s+1)(s^2 - s +1)}\right][/tex] (By factorising the denominator)

I am stuck at this step and can't proceed furthur. Someone help.

2]Find the inverse Laplace Transform of the given function, i. e.

[tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]

I did solve this one, just want to know if its correct.

[tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]

[tex]= 3L^{-1}\left[3\over {s(s^2 + 9)}}\right][/tex]

We know:

[tex]L^{-1}\left[3\over {s^2 + 9}}\right] = \sin 3t = F(t)[/tex]

[tex]L^{-1}\left[1\over s}\right] = 1 = G(t)[/tex]

By the convolution theorem;

[tex]L^{-1}{f(s)} = F(t) * G(t) = \int_0^t G(\tau)F(t - \tau)d\tau[/tex]

So,

[tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]

[tex]=3\int_0^t sin3(t - \tau)d\tau = (\cos 3\tau - 1)[/tex]

Please care to see if this is right. Thanks in advance.

[tex]L^{-1}\left[1\over {s^3 + 1}}\right][/tex]

And this is how I proceeded about:

[tex]L^{-1}\left[1\over {s^3 + 1}}\right][/tex]

[tex]= L^{-1}\left[\frac{1}{(s+1)(s^2 - s +1)}\right][/tex] (By factorising the denominator)

I am stuck at this step and can't proceed furthur. Someone help.

2]Find the inverse Laplace Transform of the given function, i. e.

[tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]

I did solve this one, just want to know if its correct.

[tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]

[tex]= 3L^{-1}\left[3\over {s(s^2 + 9)}}\right][/tex]

We know:

[tex]L^{-1}\left[3\over {s^2 + 9}}\right] = \sin 3t = F(t)[/tex]

[tex]L^{-1}\left[1\over s}\right] = 1 = G(t)[/tex]

By the convolution theorem;

[tex]L^{-1}{f(s)} = F(t) * G(t) = \int_0^t G(\tau)F(t - \tau)d\tau[/tex]

So,

[tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]

[tex]=3\int_0^t sin3(t - \tau)d\tau = (\cos 3\tau - 1)[/tex]

Please care to see if this is right. Thanks in advance.

Last edited: