How/why is the relative degree of a transfer function related to the causality of the system it represents?

Question

A transfer function can be classified as strictly proper, proper or improper depending on its relative degree, i.e. the difference between the degree of the polynomial in the denominator and the degree of the polynomial in the numerator. If a transfer is improper, it is said that the system it represents isn't causal. What exactly does this mean?

There are a few arguments I can think of, but none of them seem to make complete sense:

Argument 1) An improper transfer function involves differentiation of the input, which would imply knowledge of the future.

Problem I see: Any transfer function with a zero involves differentiation of the input, no matter how many poles it has. If differentiation of the input is the issue, any system with a zero would be termed non-causal.

Argument 2) The issue is having the output directly reflect a derivative of the input, which would imply knowledge of the future.

Problem I see: In this case, any transfer function with more than a pole would be classified as causal, since, no matter how many input derivatives are involved, a derivative term in the output would always "delay" the information coming from the input reaching the output. For example, given the system:

$$\dot{y}+y=\ddot{u}+\dot{u}+u$$

I can freely fix $y$ at a point in time. The instantaneous derivatives of $u$ would only directly affect $\dot{y}$. Another problem with this argument is that it sounds somewhat arbitrary. Isn't knowledge of the future reflected in a derivative of the output still knowledge of the future, as pointed out in the first argument?

Argument 3) These conventions were made so that state-feedback control would make sense.

Problem I see: It doesn't seem to really address the causality aspect of the transfer function itself, but rather of the closed-loop system when feedback is used, which also seems like an arbitrary definition, since it is specific to a particular form of control.

Can anyone point me in the right direction, please?

Answer 1

Any transfer function with a zero involves differentiation of the input

Let $$H(s) = \frac{Y(s)}{U(s)} = \frac{b_1 s + b_0}{s^2 + a_1 s + a_0} \tag 1$$.

You could express this as a differential equation:

$$\ddot y + a_1 \dot y + y = b_1 \dot x + b_0 x \tag 2$$

But you don't have to. You can express it as an integral equation, with no derivatives:

$$\begin{align} \begin{bmatrix}x_1 \\ x_2\end{bmatrix} &= \int {\begin{bmatrix}0 & 1 \\ -a_0 & -a_1\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix} dt} + \begin{bmatrix}0 \\ 1\end{bmatrix}u(t) \\ y(t) &= \begin{bmatrix}b_0 & b_1\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix} \end{align}. \tag 3$$

So by (3), (1) has a zero, but doesn't involve differentiation of the input -- so your assertion is disproved.

If (1) had a $b_2 s^2$ term, then in (3), $y(t)$ would be a function of $x(t)$ and $u(t)$, which is considered non-causal* because of that direct dependence.

If (1) had a $b_3 s^3$ term, then in (3), $y(t)$ be a function of $x(t)$, $u(t)$, and $\dot u(t)$. So there is no way you can represent the transfer function without using a derivative -- and since the derivative operation is non-causal in anyone's book, this would make the system non-causal.

* Personally, I think it's teetering on the edge of causality -- meta-causal? But the community wants it to be called non-causal, and the community has mathematicians who are way better than me, so I'll roll with that.