Laplace Transform Help Watch

tgodkin
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#1
Report Thread starter 8 years ago
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Hey, I have to perform this Laplace transform and I just wanted to check if my solution is correct:

F(s) = \int^{\infty}_{0}{sinh(at)e^{-st}}\ dt

Consider: G(s) = \int{sinh(at)e^{-st}}\ dt

Then, G(s)=e^{-st}\int{sinh(at)}\ dt-\int{\frac{de^{-st}}{dt}}\int{sinh(at)}\ dt \ dt

G(s)=\frac{cosh(at)e^{-st}}{a}+\frac{s}{a}\int{e^{-st}}cosh(at)\ dt

Now, let K(s)=\int{cosh(at){e^{-st}}}\ dt

Then K(s)=\frac{sinh(at)e^{-st}}{a}+\frac{s}{a}\int{sinh(at)  e^{-st}}\ dt

Or: K(s)=\frac{sinh(at)e^{-st}}{a}+\frac{s}{a}I

It then follows that:
G(s)=\frac{cosh(at)e^{-st}}{a}+\frac{sinh(at)e^{-st}s}{a^2}+\frac{s^2}{a^2}G(s)

G(s)(1-{\frac{s^2}{a^2})}=\frac{cosh(at  )e^{-st}}{a}+\frac{sinh(at)e^{-st}s}{a^2}

G(s)({\frac{a^{2}-s^2}{a^2})}=\frac{cosh(at)e^{-st}}{a}+\frac{sinh(at)e^{-st}s}{a^2}

G(s)({a^{2}-s^2})={cosh(at)e^{-st}a}+{sinh(at)e^{-st}s}

G(s)=\frac{{cosh(at)a}+{sinh(at)  s}}{(a^{2}-s^2)e^{st}}


At t=0---->F(s)=\frac{a}{a^{2}-s^2}

As t-->\infty, sinh(at)--->cosh(at)

So, G(s)=\frac{(a+s){cosh(at)}}{(a^{  2}-s^2)e^{st}}

G(s)=\frac{{cosh(at)e^{-st}}}{(a-s)}

As t-->\infty, e^{-st}-->

So, F(s)=\frac{a}{(s-a)(s+a)}

Is this correct? Many thanks.
Last edited by tgodkin; 8 years ago
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ghostwalker
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(Original post by tgodkin)

As t-->\infty, sinh(at)--->cosh(at)

So, G(s)=\frac{(a+s){cosh(at)}}{(a^{  2}-s^2)e^{st}}

G(s)=\frac{{cosh(at)e^{-st}}}{(a-s)}

As t-->\infty, e^{-st}-->

So, F(s)=\frac{a}{(s-a)(s+a)}

Is this correct? Many thanks.
Nice LaTex!

Your final answer is correct, and working looks OK all the way up to your processing for t goes to infinity.

Can't recall much on limits, but that methodology doesn't feel right; but really needs someone with more expertise to give you a definitive answer. Your working on that part seems to lose the necessary restriction between a and s. Personally, (not an expert), I'd break up the cosh and sinh into exponentials, rearrange, and then take the limit.

Someone more qualified to respond please!

Edit: Spellinge etc.
Last edited by ghostwalker; 8 years ago
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little_wizard123
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Yeah, use the exponential definitions for cosh and sinh, like ghostwalker says .
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ghostwalker
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(Original post by tgodkin)
So, G(s)=\frac{(a+s){cosh(at)}}{(a^{  2}-s^2)e^{st}}

G(s)=\frac{{cosh(at)e^{-st}}}{(a-s)}

As t-->\infty, e^{-st}-->

So, F(s)=\frac{a}{(s-a)(s+a)}

Is this correct? Many thanks.
Just realised (being dim!), although e^{-st}\to 0, cosh(at)\to \infty depending on the values of a and s of course.
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tgodkin
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Report Thread starter 8 years ago
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Thanks guys, just brought the exponentials together, was 0 anyway .
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ghostwalker
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(Original post by tgodkin)
Thanks guys, just brought the exponentials together, was 0 anyway .
The outcome wasn't in doubt, just the methodology, and it's only 0 subject to certain restrictions on a and s.
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