Advancing a causal IIR filter's impulse response

Given an infinite impulse response (IIR) filter as coefficients in a topology of your choice, what is a straightforward way to modify the filter so that the impulse response is shifted in time (advanced) by one sample to the left, discarding the sample that would make the filter non-causal?

Or is it necessary to modify the impulse response and solve the new coefficients from scratch?

(I asked this question because I wanted to know if it's possible to snip away, with a time shift, an arbitrary number of the initial samples of an IIR filter's impulse response without increasing the filter's complexity.)

• Please check out my edited answer. Commented Aug 22, 2018 at 9:48
• The method is exact for any shift of the impulse response, and for all possible relations between $M$ and $N$ ($M=N$, $M>N$, $M<N$). Commented Aug 22, 2018 at 10:29
• Hey Olli, I would be curious to know the application of such a method. I edited the answer to add a what I think better explanation, including a very simple implementation. Commented Aug 26, 2018 at 19:56
• @MattL. No application other than filter design considerations. In your answer to dsp.stackexchange.com/questions/51122/… all your impulse responses start "abruptly". Whereas in my answer the Gaussian impulse response starts smoothly. I had written that it may be useful to delay the prototype impulse response to accommodate the rising left-hand tail of for example an approximate Gaussian function. But the tail would add to the total error and can be snipped away by your method, so what I wrote was wrong. Commented Aug 27, 2018 at 5:31

Let's consider a causal IIR filter with the following transfer function:

$$H(z)=\frac{B(z)}{A(z)}=\frac{\displaystyle\sum_{n=0}^{M}b[n]z^{-n}}{1+\displaystyle\sum_{n=1}^{N}a[n]z^{-n}}\tag{1}$$

Note that $M$ need not be equal to $N$. Let $K$ be the number of samples that must be removed from the impulse response corresponding to the system $(1)$, so the new desired impulse response is

$$\tilde{h}[n]=h[n+K],\qquad n=0,1,\ldots\tag{2}$$

Let $H_K(z)$ denote the FIR transfer function corresponding to the first $K$ samples of $h[n]$:

$$H_K(z)=h[0]+h[1]z^{-1}+\ldots +h[K-1]z^{-(K-1)}\tag{3}$$

The transfer function of the new filter is then given by

$$\tilde{H}(z)=z^K\left[\frac{B(z)}{A(z)}-H_K(z)\right]=\frac{z^K\big[B(z)-H_K(z)A(z)\big]}{A(z)}\tag{4}$$

As expected, the denominator polynomial remains unchanged, we just get a new numerator polynomial

$$\tilde{B}(z)=z^K\big[B(z)-H_K(z)A(z)\big]\tag{5}$$

Let's first consider the case $K=1$. In this case, the FIR transfer function $H_1(z)$ defined in $(3)$ is trivially given by

$$H_1(z)=h[0]=b[0]\tag{6}$$

Consequently, the new numerator polynomial becomes

$$\tilde{B}(z)=\big(b[1]-b[0]a[1]\big)+\big(b[2]-b[0]a[2]\big)z^{-1}+\ldots +\\+\big(b[N]-b[0]a[N]\big)z^{-(N-1)}\tag{7}$$

if $M=N$. If $M>N$, there are additional terms $b[k]z^{-(k-1)}$, $k=N+1,\ldots,M$, and if $M<N$, there are additional terms $-b[0]a[k]z^{-(k-1)}$, $k=M+1,\ldots,N$.

The computation of the new numerator coefficients according to $(7)$ is very simple and efficient. For $K>1$ this procedure can be applied iteratively. In Matlab/Octave this is (almost) a one-liner:

for k = 1:K
b = [b(2:M+1);zeros(N-M,1)] - b(1)*[a(2:N+1);zeros(M-N,1)];
M = length(b) - 1;
end


Note that for any $K$, the order of the new numerator polynomial $\tilde{B}(z)$ equals $\max(M-K,N-1)$.

Original Answer (correct, but the presentation and solution given above is probably better; the examples below are still valid and hopefully useful)

A solution can be obtained by recomputing the numerator coefficients as follows:

$$\tilde{b}[n]=(\tilde{h}\star a)[n],\qquad n=0,1,\ldots,L-1\tag{3}$$

where $\star$ denotes convolution, $L=\max(M,N)$, and $\tilde{h}[n]$ is related to the impulse response $h[n]$ of the original IIR filter by

$$\tilde{h}[n]=h[n+1],\qquad n=0,1,\ldots,L-1\tag{4}$$

The method can be summarized as follows:

1. Compute $L=\max(M,N)$ coefficients of the original impulse response $h[n]$ for indices $n=1,2,\ldots,L$
2. Compute the new numerator coefficients from the impulse response and from the denominator coefficients using Eq. $(4)$

I chose a random filter to show the result of the suggested procedure in the figure below (blue: original impulse repsonse: red: new impulse response shifted to the right by one sample for comparison):

The left column below shows the original numerator coefficients alongside the modified numerator coefficients, aligned in such a way that it becomes obvious that for higher indices the new coefficients equal the original coefficients. (This is the case because in this example $M>N$; c.f. Eq. $(7)$ of the new answer above, including the text below Eq. $(7)$).


-0.67010
0.65673   0.92477
-0.61219  -0.74621
-0.60645  -0.53944
-0.50316  -0.50316
0.82855   0.82855
0.20396   0.20396
1.15680   1.15680
-0.29830  -0.29830
0.10605   0.10605



The denominator coefficients of both filters are

a = [1, 0.4, -0.2, 0.1]


EDIT:

This method can be generalized to remove $K$ initial values of the original impulse response, even if $K>M$:

1. Compute $L=\max(M,N)$ coefficients of the original impulse response $h[n]$ for indices $n=K,K+1,\ldots,K+L-1$
2. Compute the new numerator coefficients according to Eq. $(4)$ with $\tilde{h}[n]=h[n+K],\qquad n=0,1,\ldots,L-1$

The resulting numerator coefficients will generally have several trailing zeros (or values sufficiently close to zero) which can be removed.

The following example shows the result of removing $20$ initial values of the impulse response of the filter with coefficients

[b,a] =

1.000000   1.000000
2.000000  -0.187178
3.000000  -0.075960
-1.000000   0.812250


The numerator coefficients of the new filter are

   0.95236
0.90498
0.42252


Of course, these coefficients can no longer be interpreted as (an approximation of) a shifted version of the original numerator coefficients.

• Nicely done, Matt!
– Peter K.
Commented Aug 21, 2018 at 12:45
• Great addition/edit! This settles the matter. It also means that if one tries to approximate a given prototype impulse response using an IIR filter, then it doesn't make sense to delay the prototype impulse response (by an integer number of samples) by adding leading zeros, as that doesn't decrease the IIR complexity. Commented Aug 22, 2018 at 13:08
• @OlliNiemitalo: I'm not sure about the delay, especially if the desired impulse response is unrealistic for an IIR filter of a given order. Commented Aug 22, 2018 at 14:26
• With regard to applications, your answer indicates that the late tail of the impulse response of an IIR filter can be predicted with zero residual error by a linear predictor given $N$ warm-up samples. Commented Nov 6, 2018 at 12:21
• @OlliNiemitalo: Yes sure, for $n>M$, we have $h[n]=-a_1h[n-1]-\ldots - a_Nh[n-N]$ (with $M$ and $N$ defined as in Eq. (1) of my answer). I suppose that is what you mean. Commented Nov 7, 2018 at 10:29