Finite impulse response FIR filters

Question

• If H(Z) is linear phase FIR filter, then what can 1/H(Z) represent?
• Can it be causal and stable?
• Can it be stable if it is not required to be causal?

I think it represents non linear phase FIR filter since linear phase FIR filter have poles at Z = 0 , but 1/H(z) has poles different from zero and that what would introduce non linear phase.
Concerning the stability, It is causal and it is stable if all poles lie inside the unit circle, it can be stable also and anti causal if the unit circle lies is inside the region of convergence .

Is it correct?

Answer 1

A real coefficient, linear phase, FIR filter will have the following property :

$$ H(z) = H^*(z^*) = H(\frac{1}{z}) = H^*(\frac{1}{z^*}) $$

which implies that for every zero $z_0$ of $H(z)$ there will be three more zeros at $z_0^*$, $1/z_0$, and $1/z_0^*$; at conjugate, reciprocal, and conjugate-reciprocal locations respectively. Note that the reciprocal zeros are out of the unit circle if the corresponding zero was inside it, and vice-versa.

The all-pole inverse filter $1/H(z)$ will have all those zeros of $H(z)$ as poles. And there will be poles out of unit circle making a causal-stable inverse not possible.

Furtermore, an anti-causal and stable inverse is also not possible, as the ROC (region of convergence) smaller than the minimum pole will not include the unit circle.

But if you select a region of convergence that's between the maximum pole inside the unit circle and its reciprocal pole outside of the unit circle, then you will have a stable two sided IIR impulse response.

Answer 2

In order for H(z) to be a linear phase filter, it must zeros both on the inside of the unit circle and at the complementary locations (1/z) which are outside the unit circle. Therefore a linear phase circle has no stable causal inverse (since this would necessitate poles outside of the unit circle.)

Note that a linear phase filter can be decomposed into the cascade of a minimum phase filter (all zeros inside the unit circle) and a maximum phase filter with the same magnitude response (all zeros outside the unit circle). For this reason, only causal minimum phase FIR filters have a stable causal inverse.

Further interesting detail, the reverse filter (all the coefficients in reverse order) of a minimum phase filter IS the maximum phase filter. This is all now intuitive: the coefficients for a minimum phase filter will be dominant toward the start of the filter (resulting in the minimum delay as the signal will emerge from the filter sooner). Reversing this results in the same magnitude response but the filter will have the maximum delay as the signal will emerge from the filter later. When you cascade two filters, you convolve their coefficients, so we can also see how the cascade of a linear phase filter with a maximum phase filter results in symmetric coefficients. And it is well understood that any FIR filter with symmetric coefficients is a linear phase filter!

Example [2 1 1] is a minimum phase filter, and [1 1 2] is the reverse maximum phase filter. The cascade of the two results in the convolution of [2 1 1] and [1 1 2] which is [2 3 6 3 2] which is a linear phase filter.

If the filter has real coefficients, the all zeros must also have a complex conjugate. (But whether the filter has real or complex coefficients does not change the answer--- both real and complex linear phase filters will not have a stable causal inverse, nor will they have a stable anti-causal inverse).

Regarding causality and the ROC, for causal FIR filter the ROC is $|z|>0$ (all the poles are at $z = 0$), and for anti-causal filters it is $|z| < \infty$ (and all the poles are at $z = \infty$). In either case the requirement for zeros to be at reciprocal locations holds in order for the result to have linear phase.