I have scoured the internet for a fully-explained solution to this problem but have found none:

The problem asks to solve this differential equation for $y(t)$ using Fourier Transforms, and then consider cases where $b > w_0$, $b < w_0$, and $b = w_0$

$$ \frac{d^2y(t)}{dt^2} + 2b\frac{dy(t)}{dt} + w_0^2y(t) = \delta(t) $$

So far I have, by differential properties of Fourier Transforms, converting the function of $t$ to a function of $w$,

$$ -w^2\hat{y}(w) + 2biw\hat{y}(w) + w_0^2\hat{y}(w) = \frac{1}{\sqrt{2\pi}} $$

So algebraically,

$$ \hat{y}(w) = \frac{\frac{1}{\sqrt{2\pi}}}{-w^2 + 2biw + w_0^2} $$

And the Inverse Fourier Transform of $\hat{y}(w)$ can yield $y(t)$ as

$$ y(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{\frac{e^{-iwt} \frac{1}{\sqrt{2\pi}}}{-w^2 + 2biw + w_0^2}dw} $$

My professor says that this is an "easy integral; just do it". I have found no way of doing it thusfar, even for the simplest(?) case of $b = w_0$. I don't even know if this is of the right form (i.e. using the "easiest" definition of FT to solve the problem). He looked at it for like 2 seconds.

All help is appreciated!

  • $\begingroup$ What is the integration variable? $\endgroup$ Commented Oct 22, 2015 at 17:44
  • $\begingroup$ @AdamHughes dw, sorry, fixed in problem. $\endgroup$
    – khaverim
    Commented Oct 22, 2015 at 18:18
  • 1
    $\begingroup$ Here is something that might help: $$-\omega^2 + 2bi\omega + \omega_0^2 = (i\omega)^2 + 2bi\omega + b^2 -b^2 + \omega_0^2 = (i\omega + b)^2 + \omega_0^2-b^2.$$ With the appearance of $\omega_0^2-b^2$, you can start to see why they suggested that you break down the solution into three cases. $\endgroup$ Commented Oct 22, 2015 at 18:22
  • $\begingroup$ I've noted that -- for $b = w_0$, that terms goes to $(iw + b)^2$, but even that integral is unsolvable, as far as I know. $\endgroup$
    – khaverim
    Commented Oct 22, 2015 at 18:37
  • $\begingroup$ @khaverim I would say the integral is easy if you know a little bit of complex analysis. It will not be so fun to do via common methods $\endgroup$
    – DaveNine
    Commented Oct 23, 2015 at 15:47

1 Answer 1


Such kind of integral usually evaluate via residue. $$ y(t) = -\frac{1}{2\pi}\int_{-\infty}^{\infty}{\frac{e^{-iwt} }{(w-ib)^2 - w_0^2-b^2}dw} $$ The function $\frac{e^{-iwt} }{(w-ib)^2 - w_0^2-b^2}$ is a holomorphic function. Tis function has two residue $w_{\pm}=i b \pm\sqrt{w_0^2+b^2}$

You should consider two different case $t>0$ and $t<0$ In the first case $t>0$ you should close path in the lower half plane and the integral equal zero becouse of all residue is located in the upper half-plane. In the second case $t<0$ one can close path in the upper half plane and using Cauchy's theorem we get $$ y(t) = -\frac{i}{2}{\frac{e^{-iw_+t}-e^{-iw_-t} }{w_+-w_-}} $$ Thus$$ y(t) = -\frac{i}{2}{\frac{e^{-iw_+t}-e^{-iw_-t} }{w_+-w_-}}\theta(-t)$$ where $\theta(x)$ is a Heaviside step function.

  • $\begingroup$ Do you mean meromorphic? The function is indeed holomorphic except at the poles you gave. $\endgroup$
    – DaveNine
    Commented Oct 23, 2015 at 1:12
  • $\begingroup$ It also seems you didn't consider the three cases given by the OP, as those also might change the locations of the poles. OP didn't specify, but it matter if $b$ is positive or not, and will determine which sort of Contour you will take. It would also change your cases for $t$ as well. $\endgroup$
    – DaveNine
    Commented Oct 23, 2015 at 1:21
  • $\begingroup$ Yes I mean that $b>0$. In case of $b<0$ the calculation are the same. It means you should substitute $\theta(-t)$ by $\theta(-t b)$. $\endgroup$
    – Peter
    Commented Oct 23, 2015 at 1:48
  • $\begingroup$ I mean that one can use Residue theorem en.wikipedia.org/wiki/Residue_theorem $\endgroup$
    – Peter
    Commented Oct 23, 2015 at 1:53
  • $\begingroup$ Are you saying that your integral expression of y (t) here is equivalent to mine? $\endgroup$
    – khaverim
    Commented Oct 23, 2015 at 11:43

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