What is the Inverse Laplace Transform for V(s)=\frac{2s}{(s^{2}+4)^{2}}?

photonsquared
Messages
15
Reaction score
0
1. Find v(t) if V(s)=\frac{2s}{(s^{2}+4)^{2}}

Ans: v(t)=\frac{1}{2}tsin2tu(t)

2. Homework Equations :

V(s)=\frac{a_{n}}{(s-p)^{n}}+\frac{a_{n-1}}{(s-p)^{n-1}}+\cdots+\frac{a_{1}}{(s-p)}
a_{n-k}=\frac{1}{k!}\frac{d^{k}}{ds^{k}}[(s-p)^{n}V(s)]_{s=p}

3. Attempt at a solution:

V(s)=\frac{2s}{(s^{2}+4)^{2}}

V(s)=\frac{2s}{(s^{2}+4)^{2}}=\frac{A}{(s^{2}+4)^{2}}+\frac{B}{(s^{2}+4)}

A=\left[2s-B(s^{2}+4)\right]_{s=2i}

A=4i

B=\frac{d}{ds}\left[2s-B(s^{2}+4)\right]_{s=2i}

B=2

V(s)=\frac{4i}{(s^{2}+4)^{2}}+\frac{2}{(s^{2}+4)}

I am not sure what to do with the imaginary term, but it does not translate to 1/2t, which is what is required for the answer.

?+sin2tu(t)






 
Physics news on Phys.org
photonsquared said:
1. Find v(t) if V(s)=\frac{2s}{(s^{2}+4)^{2}}

Ans: v(t)=\frac{1}{2}tsin2tu(t)

2. Homework Equations :

V(s)=\frac{a_{n}}{(s-p)^{n}}+\frac{a_{n-1}}{(s-p)^{n-1}}+\cdots+\frac{a_{1}}{(s-p)}
a_{n-k}=\frac{1}{k!}\frac{d^{k}}{ds^{k}}[(s-p)^{n}V(s)]_{s=p}

3. Attempt at a solution:

V(s)=\frac{2s}{(s^{2}+4)^{2}}

V(s)=\frac{2s}{(s^{2}+4)^{2}}=\frac{A}{(s^{2}+4)^{2}}+\frac{B}{(s^{2}+4)}
No. Since the denominator, s^2+ 4 is quadratic you need
\frac{2s}{(s^2+4)^2}= \frac{Ax+ B}{(x^2+4)^2}+ \frac{Cx+ D}{x^2+4}
 
Thanks, I'll attempt again.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top