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Prove It
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Question: L^-1 [ 2(s^2 - 32) / s^4 + 1024]
My Initial thoughts:
- formula 37/38 from table because s^4 denominator
- partial fractions?
= 2(s^2 - 32) / s^4 + 32^2]= 2(s^2 - 32) / (s^2 + 32)^2]
How would you go about cancelling the (s^2 -32) and (s^2 + 32)^2 ?
Im sure this the right approach...
Hi Douglas, I agree, partial fractions would be the best approach. Notice that
$\displaystyle \begin{align*} s^4 + 1024 &= \left( s^2 \right) ^2 + 2\cdot s^2 \cdot 32 + \left( 32 \right) ^2 - 2\cdot s^2 \cdot 32 ^2 \\ &= \left( s^2 + 32 \right) ^2 - 64s^2 \\ &= \left( s^2 + 32 \right) ^2 - \left( 8s \right) ^2 \\ &= \left( s^2 - 8s + 32 \right) \left( s^2 + 8s + 32 \right) \end{align*}$
The partial fraction decomposition of $\displaystyle \begin{align*} \frac{2 \left( s^2 - 32 \right) }{ s^4 + 1024 } \end{align*}$ is
$\displaystyle \begin{align*} \frac{2 \left( s^2 - 32 \right) }{ s^4 + 1024} &= -\frac{1}{4} \left( \frac{s + 4}{s^2 + 8s + 32} \right) + \frac{1}{4} \left( \frac{s - 4}{s^2 - 8s + 32} \right) \\ &= -\frac{1}{4} \left[ \frac{s + 4}{ \left( s + 4 \right) ^2 + 16} \right] + \frac{1}{4} \left[ \frac{s - 4}{ \left( s - 4 \right) ^2 + 16 } \right] \end{align*}$
and so now
$\displaystyle \begin{align*} \mathcal{L} \left\{ -\frac{1}{4} \left[ \frac{s + 4}{ \left( s + 4 \right) ^2 + 16 } \right] + \frac{1}{4} \left[ \frac{s - 4}{ \left( s - 4 \right) ^2 + 16} \right] \right\} &= -\frac{1}{4} e^{-4t} \mathcal{L} \left\{ \frac{s}{s^2 + 4^2} \right\} + \frac{1}{4} e^{4t} \mathcal{L} \left\{ \frac{s}{s^2 + 4^2} \right\} \textrm{ (First Shift Theorem)} \\ &= -\frac{1}{4}e^{-4t} \cos{(4t)} + \frac{1}{4}e^{4t} \cos{(4t)} \end{align*}$