Find the impulse response to a RLC filter I have a serial RLC-filter for which I should first determine the transfer function and then the impulse response.
I figured out that the transfer function is: $H(s)=V(s)/U(s)$ And my circuit has the formula (Ri(t) = v(t) which is the output, and u(t) is input): $$ Li'(t)+\frac{1}{C}\int_0^ti(t)dt+Ri(t)=u(t)$$ then by transforming it with Laplace and rearranging it I get:
$$H(s) = \frac{R}{Ls+\frac{1}{Cs}+R}$$ 
But how do I get the impulse response for that? I think that I should transform H(s) back to t-domain again, but I don't really know how to do that when the expression is as it looks above.
EDIT
Completing the square of the denominator gives me this:
$$H(t) = \frac{Rs}{L(s+\frac{R}{2L})^2+\frac{1}{C}-\frac{R^2}{4L}}$$
But how can I use this? 
 A: So I got the expression:
$$H(t) = \frac{Rs}{L(s+\frac{R}{2L})^2+\frac{1}{C}-\frac{R^2}{4L}}$$
And then I divide everything by L and break things out until I get this:
$$\frac{Rs}{L}\frac{1}{(s+a)^2+\omega^2}$$ where $ a=\frac{R}{2L}$ and $\omega^2=\frac{1}{CL}-(\frac{R}{2L})^2$
I can then rewrite it again as:
$$\frac{a}{2}\frac{s+a}{(s+a)^2+\omega^2}-a\frac{\omega}{(s+a)^2+\omega^2}\frac{1}{\omega}$$
And from here we have two parts depending on s, which fits in to two of the entries in the Laplace transform table at Wikipedia. 
Transforming back to t domain gives this:
$$\frac{a}{2}e^{-at}\cos(\omega t)u(t)-ae^{-at}\sin(\omega t)u(t)\frac{1}{\omega}$$
A: First, let me fill in my comments regarding why the inverse Laplace transform of $$H(s)=\frac{R}{Ls+\frac{1}{Cs}+R}=\frac{Rs/L}{(s+R/2L)^2+1/LC-(R/2L)^2}=\frac{2as}{(s+a)^2+\omega^2}$$ (where I've used the definitions from my comments and theva's answer) can't quite be  
$$H(t)=\frac{a}{2}e^{-at}\cos(\omega t)u(t)-ae^{-at}\sin(\omega t)u(t)\frac{1}{\omega}.$$
The big red flag is seen by inspecting units: The only dimensionful part of the first term is $a$ (units of frequency), whereas the second term has $1/\omega$ as well. So the first term has units of frequency, but the second is dimensionless; these aren't compatible, so at least one of them must be flawed.
To see which is right, note that the $s$-domain transfer function is a ratio of input and output voltages. On the other hand, taking a Laplace transform to get from $t$- to $s$-domain involves integrating $H(t)$ over all times. Thus $H(t)$ must have units of frequency for $H(s)$ to be dimensionless. So there must have been an algebra error for the second coefficient.
To resolve this, note that $s=(s+a)-a$ and therefore
$$H(s)=2a\frac{(s+a)-a}{(s+a)^2+\omega^2}=2a\left[\frac{s+a}{(s+a)^2+\omega^2}-\frac{a}{\omega}\frac{\omega}{(s+a)^2+\omega^2}\right].$$ We can recognize these from Wikipedia's table of Laplace transforms as those of exponentially decaying cosine and sine waves respectively. Consequently one has
$$H(t)=2a e^{-at}\left[\cos \omega t-\frac{a}{\omega}\sin \omega t\right]u(t).$$
Since $a/\omega$ is dimensionless, the units of $H(t)$ are correctly those of $a$ i.e. frequency.
For independent cross-checks, it's helpful to take limiting values. For instance, consider the non-damping limit $\omega\to 0$. Then $$H(s)=2a\dfrac{(s+a)-a}{(s+a)^2}=2a\left[\frac{1}{s+a}-\frac{a}{(s+a)^2}\right].$$ This is the Laplace transform of $H(t)=2ae^{-at}(1-at)u(t)$, which agrees with the limiting form of the $H(t)$ obtained above. One can also check that the initial and final value theorems for $H(t)$ are satisfied:
$$\lim_{t\to0^+}H(t)=\lim_{s\to \infty}s H(s)=2a,\quad \lim_{t\to\infty}H(t)=\lim_{s\to 0}s H(s)=0.$$ The first check is especially helpful since it catches any missing factors in the first term.
