CrazyEngineers Archive
Old, but evergreen and popular discussions on CrazyEngineers, presented to you in read-only mode.
@Brad Grillakis • 26 Oct, 2011
Hello! I am working on my Laplace Transform homework and I am trying to work out the steps on how my professor got this answer but I am not understanding it. The question is:

Q: What is the inverse Laplace transform of {32 / (s(s[SUP]2[/SUP] + 64))} ?
A: sin[SUP]2[/SUP](4t)

The Laplace Look Up Table formula is :

```Signal                                Transform
sin[SUP]2[/SUP](w[SUB]0[/SUB]t)                          2w[SUB]0[/SUB][SUP]2[/SUP] / (s(s[SUP]2[/SUP] + 4w[SUB]0[/SUB][SUP]2[/SUP]))```
When I am breaking down this function and I am using the transform equation I can only break down the 64 (since 8[SUP]2[/SUP] = 64) but that 4 right next to it and that 32 is throwing me for a loop.

I really appreciate it if someone can help me understand how the answer is sin[SUP]2[/SUP](4t)
@Sagar07 • 27 Oct, 2011 Didn't get your question exactly.......
@Brad Grillakis • 27 Oct, 2011
Sagar07
My question is, from this function 32/(s(s[SUP]2[/SUP]+64)), how is the answer sin[SUP]2[/SUP](4t)? (I have been trying to work out how my professor got this answer but I can not seem to figure this out. I tried doing partial fractions, but u cant because of the s[SUP]2[/SUP].)

The 2 formula I gave is from the Laplace Transform Table.
@ritesh05 • 28 Oct, 2011 Here as we can see that in the denominator there is 's' term so we can directly use the division by s formula that is so after using this formula you will get -.5(cos8t-1)=-.5(1-2sin^(4t)-1)=sin^2(4t)...............{cos2t=1-2sin^2(t)}
@rashidsohail • 01 Nov, 2011 i dont understand that
Here as we can see that in the denominator there is 's' term so we can directly use the division by s formula that is
@simplycoder • 01 Jan, 2012
My question is, from this function 32/(s(s[SUP]2[/SUP]+64)), how is the answer sin[SUP]2[/SUP](4t)? (I have been trying to work out how my professor got this answer but I can not seem to figure this out. I tried doing partial fractions, but u cant because of the s[SUP]2[/SUP].)

The 2 formula I gave is from the Laplace Transform Table.
This can be solved by partial fractions. All you need is proper substitution.

Substitute it as (A/s)+(Bs+C)/(s^2+64).

A=0.5,B=-0.5,C=0.

so you get 0.5/s- 0.5/(s^2+64)

Laplace inverse of the first term is 1, and that of second term is cos(8t).
So you get 0.5*(1-cos8t).By trignometry, cos8t=1-2sin^2 (4t).
So you get 0.5*(1-(1-2sin^2(4t))).
=0.5*2sin^2(4t).
=sin^2(4t).//
@silverscorpion • 02 Jan, 2012 Write the fraction 32/(s(s2+64) as follows..

32/(s(s2+64) = (2*4[sup]2[/sup])/(s*(s[sup]2[/sup] + 4*4[sup]2[/sup]))

So, you get w[sub]0[/sub] to be 4. Hence, the inverse Laplace transform is, from the formula you have given, sin[sup]2[/sup] 4t

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