Laplace transfrom
\[L(e^{2t}*\sinh(t))\]
I tried integration by parts, and I ended up with \[y=e^{t*(2-s)}*\cosh(t)-(\cosh(t)*e^{t*(2-s)}-\int\limits_{0}^{\infty}\sinh(t)*e^{t(2-s)})\] where the above integral becomes y, but then the y's cancel out and become 0 :(
When a function is multiplied by an exponential in time domain, in frequency somain (Laplace tranformed signal) is shifted in frequency \[\mathcal{L}[e^{at}f(t)]=F(s-a)\]
so, we first find \[\mathcal{L}[\sinh(t)]=\frac{1}{s^2-1}\\ \mathcal{L}[e^{2t}\sinh(t)]=\frac{1}{(s-2)^2-1} \] you can simplify from there
domain... \(Re(s)>1\)
Ah, thanks this makes sense now, I forgot about that theorem. Is it possible to obtain the laplace transform of sinh(t) manually by taking the infinite integral, or do you just have to know it from the laplace table?
@electrokid
sure.. it'd be easier if you express the hyperbolic t-ratios as exponentials..
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