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In the case of continuous-time systems, if the system is causal, its Laplace transfer function is strictly proper (the degree of the numerator is less than the degree of the denominator).

Is this true also in case of discrete-time systems? That is, if the system is causal, is its Z transfer function strictly proper?

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Note that a stable and causal continuous-time transfer function does not need to be strictly proper but only proper, i.e. the degree of the numerator does not exceed the degree of the denominator, but numerator and denominator degree can be equal. E.g.

$$H(s)=\frac{as^2+bs+c}{s^2+ds+e}$$

can represent a causal and stable system, as long as its poles are in the left half of the complex $s$-plane.

For discrete-time systems the same is true. A transfer function for which the degree of the numerator is greater than the degree of the denominator has at least one pole at infinity. Since causal and stable discrete-time systems must have all their poles inside the unit circle, such a system can't be causal and stable. However, as is the case with continuous-time systems, equal degrees of numerator and denominator are possible.

Assume a causal and stable discrete-time system is described by the following linear difference equation with constant coefficients:

$$y[n]+a_1y[n-1]+\ldots+a_Ny[n-N]=b_0x[n]+b_1x[n-1]+\ldots+b_Mx[n-M]\tag{1}$$

The corresponding transfer function is

$$H(z)=\frac{\sum_{m=0}^{M}b_mz^{-m}}{1+\sum_{n=1}^Na_nz^{-n}}\tag{2}$$

Note that there are of course no restrictions on the choices of $M$ and $N$. Nevertheless, the transfer function $H(z)$ is always proper. Assuming $M>N$ gives

$$H(z)=\frac{b_0z^M+b_1z^{M-1}+\ldots+b_M}{z^M+a_1z^{M-1}+\ldots+a_Nz^{M-N}}\tag{3}$$

Obviously, the transfer function given by (3) is proper. $M>N$ just means that the last few coefficients of the denominator polynomial are zero. A very similar argument holds for $M<N$.

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    $\begingroup$ Given the linear system $y(t)+a_{1}y(t-1)+...+a_{na}y(t-na)=b_{0}u(t)+b_{1}u(t-1)+...+b_{nb}u(t-nb)$ the output $y(t)$ depends only on the past values of $u$ and $y$, that means it's always causal, independently from the fact that na is bigger or smaller than nb (I suppose always na and nb positive). na and nb are related to the degree of numerator and denominator of its Z transfer function. So, why do I need proper transfer functions to have a discrete-time causal system? $\endgroup$ – Simone Giorgi Dec 18 '14 at 14:21
  • $\begingroup$ @SimoneGiorgi: I've updated my answer to address that question. $\endgroup$ – Matt L. Dec 18 '14 at 16:18

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