- #106
TFM
- 1,026
- 0
Okay so:
[tex] Y = \frac{\frac{1}{s^2 + 1} + s - 1/2}{s^2 + 1} [/tex]
[tex] Y = \frac{1}{(s^2 + 1)^2} + \frac{s}{s^2 + 1} - \frac{1}{2s^2 + 2} [/tex]
So, the Laplace tranformation of:
[tex] \frac{1}{(s^2 + 1)^2} = = \frac{1}{2} (-sin(t)- tcos(t)) [/tex]
and the Laplace transformation of:
[tex] \frac{s}{s^2+1} = cos(t) [/tex]
And finally, the Laplace Transformation of:
[tex] \frac{1}{2s^2 + 2} [/tex]
would this be similar to the first one, but require first shift theorem:
[tex] \frac{1}{(s^2 + 1)^2} = = \frac{1}{2} (-sin(t)- tcos(t)) [/tex]
[tex] \frac{1}{(2s^2 + 2)^2} = = \left(\frac{1}{2} (-sin(t)- tcos(t))\right) e^2t [/tex]
Is this okay?
TFM
[tex] Y = \frac{\frac{1}{s^2 + 1} + s - 1/2}{s^2 + 1} [/tex]
[tex] Y = \frac{1}{(s^2 + 1)^2} + \frac{s}{s^2 + 1} - \frac{1}{2s^2 + 2} [/tex]
So, the Laplace tranformation of:
[tex] \frac{1}{(s^2 + 1)^2} = = \frac{1}{2} (-sin(t)- tcos(t)) [/tex]
and the Laplace transformation of:
[tex] \frac{s}{s^2+1} = cos(t) [/tex]
And finally, the Laplace Transformation of:
[tex] \frac{1}{2s^2 + 2} [/tex]
would this be similar to the first one, but require first shift theorem:
[tex] \frac{1}{(s^2 + 1)^2} = = \frac{1}{2} (-sin(t)- tcos(t)) [/tex]
[tex] \frac{1}{(2s^2 + 2)^2} = = \left(\frac{1}{2} (-sin(t)- tcos(t))\right) e^2t [/tex]
Is this okay?
TFM