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Inverse Laplace returns hyperbolic trig functions
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| Matthew 2008-01-30, 8:20 pm |
| I am taking the inverse Laplace as shown below. Why does
it return the answer using hyperbolic sine & cosine
functions? Is there a way I can get Matlab to return the
answer using the exp function?
syms X s
X=3/((s^2)-(4*s)+4)/(s+2)
ilaplace(X)
Matt Ridzon
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| vijaykumar.krithivasan@gmail.com 2008-01-30, 8:20 pm |
| On Jan 30, 11:50 am, "Matthew " <m_rid...@hotmail.com> wrote:
> I am taking the inverse Laplace as shown below. Why does
> it return the answer using hyperbolic sine & cosine
> functions? Is there a way I can get Matlab to return the
> answer using the exp function?
>
> syms X s
> X=3/((s^2)-(4*s)+4)/(s+2)
> ilaplace(X)
>
> Matt Ridzon
Matt:
This is what I tired: and the results where in exponential form:
[color=darkred]
X=3/((s^2)-(4*s)+4)/(s+2)
ilaplace(X)
X =
3/(s^2-4*s+4)/(s+2)
ans =
3/16*exp(-2*t)+(3/4*t-3/16)*exp(2*t)
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| Matthew 2008-01-31, 8:28 pm |
| vijaykumar.krithivasan,
It doesn't work for me and I'm not sure why. My answer
returns in hyperbolic trig functions. Do you know of a
setting that needs changed in Matlab to make the answers
return in exp form?
Matt Ridzon
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