Series Combination is Commutative

Since multiplication of complex numbers is commutative, we have

$$H_1(z)H_2(z)=H_2(z)H_1(z),$$

which implies that any ordering of filters in series results in the same overall transfer function. Note, however, that the numerical performance of the overall filter is usually affected by the ordering of filter stages in a series combination [102].

By the convolution theorem for z transforms, commutativity of a product of transfer functions implies that convolution is commutative:

$$h_1 \ast h_2 \leftrightarrow H_1\cdot H_2 = H_2\cdot H_1 \leftrightarrow h_2 \ast h_1$$