# Applying Kramers-Kronig relation to a simple damped oscillator

I just discovered the Kramers-Kronig relation and am trying to apply it to a simple damped oscillator of the form subjected to an impulse at $$t=0$$, which is a causal system:

$$m\ddot x + c\dot x + k x = \delta(t).$$

In the time-domain, the response can be decomposed in odd/even parts as $$x(t) = \operatorname{sign}(t)h_0(t) + h_0(t)$$. The Kramers-Kronig relation implies that because the signal is causal, the real part of the Fourier transform of a solution $$x(t)$$ is equal to the Hilbert transform of the imaginary part of the Fourier transform of $$x(t)$$. That is what I am trying to illustrate here (and I am afraid that it would lead to a constraint between $$m$$, $$c$$ and $$k$$, which would probably suggest a misunderstanding on my side...).

The Fourier transform of the ODE gives $$(-m \omega^2 +k + ic\omega )\hat x = 1$$

so

$$\hat x = \dfrac{1}{-m \omega^2 +k + ic\omega } = \dfrac{k-m\omega^2}{(k-m\omega^2)^2 + (c\omega )^2} + i \dfrac{-c\omega}{(k-m\omega^2)^2 + (c\omega )^2}$$

If I am not mistaken, the KK relation would imply that the Hilbert transform of $$\dfrac{-c\omega}{(k-m\omega^2)^2 + (c\omega )^2}$$ is $$\dfrac{k-m\omega^2}{(k-m\omega^2)^2 + (c\omega )^2}$$, ie

$$\dfrac{1}{\pi}\int_{-\infty}^\infty -\dfrac{1}{u-\omega}\dfrac{c\omega}{(k-m\omega^2)^2 + (c\omega )^2} d\omega \stackrel{?}{=} \dfrac{k-m u^2}{(k-m u^2)^2 + (cu )^2}$$ which does not hold (take for instance $$u=m=c=k=1$$, the integral does not converge). Edit Actually, it does hold for $$m=c=k=1$$, in the sense of the principal value (see hint and answer) ! I was not able to verify for arbitrary $$m,c,k$$.

Mathematica code for whoever is interested:

m = k = c = 1;
LHS  = Assuming[Element[u, Reals], 1/Pi*Integrate[-1/(u - w)*c*
w/((k - m*w^2)^2 + (c^2*w^2)), {w, -Infinity, Infinity},
PrincipalValue -> True]];
RHS = (k - m*u^2)/((k - m*u^2)^2 + c^2*u^2);
LHS == RHS // Simplify

• Quick hint : Hilbert transform of a function $f(x)$ is $\frac{1}{\pi}P.V.\int_{-\infty}^{+\infty}dx'\frac{f(x')}{x-x'}$. Your integral certainly converges in the sense of a principal value. – Sunyam Mar 1 at 21:36
• @Sunyam Out of curiosity, how do you know it converges in the PV sense? – anderstood Mar 7 at 19:13
• Only singularity your integral (without interpreting it in the sense of a principal value) along real axis is at $\omega=u$ which disappears when interpreted as a principal value (calculated by using semi circular contour in upper/lower plane + real axis with a dent bypassing the singular point). – Sunyam Mar 7 at 20:32

This work!

$$x=\frac{1}{-m{\omega}^{2}+k+ic\omega}=\Re{(x(\omega))}+i\,\Im{(x(\omega))}$$ with: $$\Re{(x(\omega))}={\frac {-m{\omega}^{2}+k}{ \left( -m{\omega}^{2}+k \right) ^{2}+{c}^{2 }{\omega}^{2}}}\tag 1$$ and $$\Im{(x(\omega))}=-{\frac {c\omega}{ \left( -m{\omega}^{2}+k \right) ^{2}+{c}^{2}{\omega }^{2}}} \tag 2$$

Kramers-Kronig relation :

$$I_R=-\frac{2}{\pi}\int_0^\inf \frac{\omega\,\Im{(x(\omega))}}{\omega^2-s^2}\,d\omega$$

for $$m=1\,,k=1\,,c=1$$

we get:

$$I_R=\frac{1-s^2}{s^4-s^2+1}$$

this is the value of equation (1) for $$\Re{(x(s))|_{m=1\,,k=1\,,c=1}}$$

$$I_I=\frac{2}{\pi}\int_0^\inf \frac{s\,\Re{(x(\omega))}}{\omega^2-s^2}\,d\omega$$

we get:

$$I_I=-\frac{s}{s^4-s^2+1}$$

this is the value of equation (2) for $$\Im{(x(s))|_{m=1\,,k=1\,,c=1}}$$

to get the right answer you must take the Cauchy-Principal -Value of the integral result :

$$I= \text{Cauchy-Principal -Value}\,\left[\int_0^\inf (...)\,d\omega\right]$$

I did the calculation with Maple symbolic program