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I was always wondering about the acausal nature of solutions obtained by Fourier transforms in the case of inhomogeneous equations. The solution usually revolves around the integration of the transform of the inhomegeneous term - and that transform necessarily depends on all future values of that term - so is it really breaking causality?

Example: Analysis of an integrator circuit

Consider a resistor $R$ and capacitor $C$ connected in series to each other, and an external voltage $V(t)$ is applied to the circuit. To find the voltage drop across the capacitor at any moment, we must solve the equation $V(t) = \dot{Q}R + \frac{1}{C}Q$. We may transform the equation to the frequency domain and obtain that $Q_\omega = \frac{CV_\omega}{1+i\omega RC}$ so the final solution for the voltage across the capacitor would be (using the unitary FT convention):

$V_C=\frac{Q(t)}{C}=\frac{1}{\sqrt{2\pi}}\int{\frac{V_\omega d\omega}{1+i\omega RC}}e^{i\omega t}$

But expanding the term $V_\omega$ clearly shows it involves the integration of $V(t)$ from the dawn till the end of time. This would imply that the solution depends on future values of the input function. Is this really acausal?

Note: Of course, one may take the limit, either $\omega << RC$ or $\omega >> RC $, of the solution and execute the inverse transform analytically and obtain a solution in terms of either $V(t)$ or it's time integral until time $t$, thus removing the problem of causality. But I'm talking about this as a general difficulty, and it's implications on other problems as well.

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Causality in the Fourier domain is manifest in the behavior of the transfer function as a complex function of frequency, i.e. the location of poles etc. Your example has a pole at +i/RC, so one may deform the integration countour into the lower complex plane for positive tau and show that it vanishes.

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  • $\begingroup$ Do you mean, show that the entire integral vanishes? Wouldn't it be equal to the residue at $i/RC$? $\endgroup$
    – Deep Blue
    Commented Oct 5, 2012 at 6:17
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    $\begingroup$ @BenjiRemez: Only when the appropriate t term in the exponent has the right sign to close the contour to encircle the pole. This enforces causality. $\endgroup$
    – Ron Maimon
    Commented Oct 5, 2012 at 7:19
  • $\begingroup$ So, if I get this right, for $t - t'>0$, I can only integrate the contour on the upper half of the plane, where the pole contributes a non-trivial residue, but for $t - t' < 0$ the exponent only vanishes in the lower half, so I must take the contour there - but no poles means no residue, so the expression is guaranteed to vanish, regardless of the choice of input? Neat. Very Neat. $\endgroup$
    – Deep Blue
    Commented Oct 5, 2012 at 7:25
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I) It is not surprising that a solution that uses time-frequency Fourier transformation can superficially look acausal (without actually being acausal), because the Fourier transform $$V_{\omega}~ :=~ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\! dt~V(t)e^{-i\omega t} $$ depends by definition on all times $t$ from the far past to the far future.

II) However, the differential equation $$V(t) ~=~ R\dot{Q}(t) + \frac{Q(t)}{C}$$ can also be solved in a manifestly causal form

$$ Q(t)~=~Q(t_i)\exp\left( \frac{t_i-t}{RC}\right) + \int_{t_i}^{t} \! dt^{\prime}~ \frac{V(t^{\prime})}{R}\exp\left( \frac{t^{\prime}-t}{RC}\right), $$

where we integrate time $t^{\prime}$ from the initial time $t_i$ to now $t$, i.e. only in the past.

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  • $\begingroup$ I agree that the problem may be solved like you have wrote, but what step goes wrong in the FT approach that seems to break causality? If the solution I outlined above it valid, what would happen if I try to evaluate it for some input function? I suspect that if I switch the order of integration, the first (homogeneous) solution term you wrote may pop out from the transform solution, but I've yet to check it out. Also, as I said in, I'm interested in this difficulty in a more general situation, and only used the RC circuit to illustrate the problem. $\endgroup$
    – Deep Blue
    Commented Oct 4, 2012 at 23:01
  • $\begingroup$ The very first step: to use an object (the FT) that depends on the far future, cf. section I. $\endgroup$
    – Qmechanic ♦
    Commented Oct 4, 2012 at 23:08
  • $\begingroup$ I agree, but does that solution not solve the original equation? Since the equation is causal, I expect the solution to maintain that property. I remember from school that when solving wave equations in classical EM, a delta function pops out during integration and enforces this causality (this raises an interesting question of what would happen when it is solved on a finite interval where the wavenumbers are discrete, but that's for another time) - but I can't see a way that such a causality-ex-machina would show up in any general problem. $\endgroup$
    – Deep Blue
    Commented Oct 4, 2012 at 23:20
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    $\begingroup$ Right, after you impose the appropriate initial condition, then your FT formula for $Q(t)$ must be equivalent to the manifestly causal formula in my answer. $\endgroup$
    – Qmechanic ♦
    Commented Oct 4, 2012 at 23:39
  • $\begingroup$ I'll see if I can get that result this evening. $\endgroup$
    – Deep Blue
    Commented Oct 5, 2012 at 0:09
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This would imply that the solution depends on future values of the input function.

This isn't true, your reasoning here is faulty.

First, an LTI system is causal if:

$h(t) = 0, t < 0$

where $h(t)$ is the impulse response.

Consider the RC network given. The impulse response is:

$h(t) = \frac{e^{\frac{-t}{RC}}}{RC}u(t)$

where $u(t)$ is the unit step so this system is causal.

The fact that the transform involves integration over all time does not imply acausality. For example, consider the identity system where $v_o(t) = v_i(t)$ which is clearly causal.

Now,

$v_o(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\! ~V_i(\omega)e^{j\omega t}d\omega = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\! ~V_o(\omega)e^{j\omega t}d\omega$

Since $V_o$ involves "integration of $v_o(t)$ from the dawn till the end of time", by your reasoning, the value of $v_o(t)$ depends on future values of itself!

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  • $\begingroup$ Yes, but I can always expand this expression as $v_{o}(t)=\frac{1}{2\pi}\int\int d\omega dt'v_{i}(t')e^{j\omega (t-t')}$ Integrating first over $\omega$, this clearly yields a delta function $\delta (t-t') $ and causality is restored analytically for every input $v_i$. With the RC circuit, the extra $\omega$ dependent terms in the integrand don't allow this triviality. And again, I'm concerned about this is a general issue, not necessarily in circuit analysis. $\endgroup$
    – Deep Blue
    Commented Oct 4, 2012 at 23:14
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    $\begingroup$ Since $V_o(\omega) = V_i(\omega)H(\omega)$, it is $H(\omega)$ that is of interest, the transform of the impulse response. It is its nature that determines causality: en.wikipedia.org/wiki/Causal_filter $\endgroup$
    – Alfred Centauri
    Commented Oct 5, 2012 at 0:14
  • $\begingroup$ So, in principle, could I construct a differential equation which would produce such an $H(\omega )$ that violates causality? $\endgroup$
    – Deep Blue
    Commented Oct 5, 2012 at 0:27
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    $\begingroup$ @BenjiRemez, yes. For example, let $H(\omega) = \dfrac{e^{i\omega \tau}}{i \omega}$. This is non-causal since the corresponding differential equation is $\dot y(t) = x(t + \tau)$ $\endgroup$
    – Alfred Centauri
    Commented Oct 5, 2012 at 2:17

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