"Order" of a digital filter (FIR and IIR)
In general, the order refers to the underlying degree of the polynomial. Just like a polynomial, there are the powers of \$x\$ and the coefficients. If the order is 2, then the polynomial has 3 powers of \$x\$ and 3 coefficients, but since \$x^0=1\$, we say it has 2 powers and 3 coefficients.



For filters without feedback (FIR), the powers of \$x^n\$ are the delays, \$z^{-n}\$, and the transfer function has only zeroes (the denominator is 1, all the poles are at 0), so the numerator's coefficients form the polynomial:

$$h_nz^{-n}+h_{n-1}z^{-(n-1)}+...+h_1z^{-1}+h_0$$

The samples do not require to have a memory (e.g. zeroth order sample&amp;hold). But if you need to work with samples from memory, then you'd need one sample for the current value, and \$n\$ samples for the ones in memory. This means that the number of samples is given by the length of the filter, \$n+1\$.

As a side comment, you'll see many notations for length and order, such as the length, \$L=N+1\$, being the order plus one, and \$M=\frac N2\$ being the midpoint, or half the order.



For filters with feedback (IIR), the transfer function should be proper, so then the degree represents the order of the denominator:

$$\prod_{k=0}^N\frac{z-z_k}{z-p_k}$$

which expands into (see this for the formula):

$$\sum_{k=0}^N(-1)^{N-k}\left(\sum_{0\leq i_1&lt;i_2&lt;\dots&lt;i_{N-k}\leq N-1}\prod_{j=1}^{N-k}p_{i_j}\right)z^{-k}$$

So, yes, since the control theory makes use of filters, it obeys their definitions, in general.