Does "improper" imply that a system cannot be stable and causal? This answer and the comments in it made me wonder whether the following statement is true:


  If a transfer function is improper, then that system cannot be causal and stable at the same time.


I had thought that this was true for a while. But the other day I wondered why. For example, the transfer function

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

is improper, but the ROC $\{s:\mathrm{Re}(s)&gt;-1\}$ would make it stable (it contains the imaginary axis) and causal (it consists of a left-sided plane).

So how are causality and stability related in improper systems?
An improper system cannot be causal and stable. If the order of the numerator is greater than the order of the denominator, you'll always have at least one pole at infinity. Consequently, not all poles are in the left half-plane (or inside the unit circle in the case of discrete-time systems).

The system in your example is clearly unstable:

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

Part of it is an ideal differentiator ($s$), which is unstable.

Also take a look at this related answer.