# How to modify an LTI differential equation to be acausal or anti-causal?

I'm trying to wrap my head around causality in LTI-systems. Considering continuous time only, I'm happy with the fact that the system is causal iff the impulse response function $h(t)=0$ for $t<0$.

What I don't understand is how this is embodied in an LTI differential equation. Specifically, what makes the positive and negative time directions fundamentally different? (a point which is especially clear if considering e.g. length as the independent variable, rather than time)

I'm hoping to shed some light on this by considering the LTI equation

$\frac{dy}{dt} = u(t)$

For this, or a more complex LTI system, I could set the RHS to $\delta(t)$ and then use Laplace transforms to solve for $y(t)=h(t)$. In this specific case $h(t)$ is the Heavyside step-function, i.e. $h(t)=0$ for $t<0$ and $h(t)=1$ for $t>0$. Thus this is a causal system.

If I wanted to modify the above differential equation to make the system either acausal, or anti-causal, what would I have to do?

It's important to realize that generally the differential equation (DE) alone doesn't tell us anything about causality. You claim that the system given in your question is causal. However, an anti-causal system with impulse response $h(t)=-u(-t)$ (where $u(t)$ is the step function) also satisfies the very same DE! So the given DE actually describes two systems, a causal one and an anti-causal one.

What you need to enforce causality (and linearity and time-invariance) is the auxiliary condition that the system is initially at rest. "Initially" means the moment in time when the input signal becomes non-zero. If we denote this moment by $t_0$, these conditions are

$$y(t_0)=y'(t_0)=y''(t_0)=\ldots =0\tag{1}$$

where $y'(t)$ denotes the first derivative of the output signal, etc. Solving the DE with the initial rest conditions $(1)$ for $t>t_0$ gives you the output signal of the causal system described by the DE.

A similar thing is true when using the Laplace transform to compute the transfer function of an LTI system. The expression for the Laplace transform of the transfer function does not uniquely identify the corresponding system. You also have to specify the region of convergence (ROC). In your example, the system's transfer function is

$$H(s)=\frac{1}{s}\tag{2}$$

Depending on the ROC, this function corresponds to two different time domain functions:

$$h(t)=u(t),\quad \text{Re}(s)>0$$ and $$h(t)=-u(-t),\quad \text{Re}(s)<0$$

You need to choose which solution makes sense in your case.

• What about this LCCDE: $y(t)'=x(t+2)$ ? Do we need aux conditions to tell whether this is causal or not ? I'm in a "temporary" confusion... – Fat32 Jun 14 '16 at 15:06
• @Fat32: No, this is not causal, but the general question if its impulse response is right-sided or left-sided remains the same. You basically just shifted $t=0$ to $t=2$. – Matt L. Jun 14 '16 at 16:03
• as expected! But then the first sentence of yours (and mine as well) becomes open to skepticism ? as it is then possible to tell , but not always, when a system cannot be causal form its LCCDE alone ? – Fat32 Jun 14 '16 at 16:20
• If I understand this correctly, in the above case you would actually have the choice between a non-causal solution (corresponds to right-sided impulse response...?), or an anti-causal solution with a "delay" (in quotes since the delay is now from point of view of the anti-causal direction) – funklute Jun 14 '16 at 16:24
• @Fat32: I agree that our first sentences are not absolutely water-proof, but I wonder if it makes things clearer if we add all possible disclaimers right away. "Usually" the DE is given as $$\sum_ka_ky^{(k)}(t)=\sum_kb_kx^{(k)}(t)$$ and in that case the assertion is correct. – Matt L. Jun 14 '16 at 18:41

A differential equation alone cannot tell whether the system being modeled is causal or anti-causal. Causality is an external constraint put on the system, which will guide you through the solution process. i.e. if you somehow (by other means) know that a system is causal, then you constraint your solution to be zero for all $t<0$.

Otherwise, (if you don't know whether your system is causal or not), the LCCDE (Linear Constant Coefficient Differential Equation) itself cannot tell you this and both solutions which extend from say $t=0$ to $t=\infty$ and the other from $t=0$ to $t=-\infty$ will be a valid solution of the given LCCDE.

However note that the following system: $y(t)' = x(t+2)$ is immediately known to be non-causal as the output depends on a value of input at a future time.