# Is the filter described by specific difference equation time independent

Is the filter described by $$y[n] + 2 y[n-1] + y[n-2] = x[n] - 2 x[n-1]$$ time invariant? ($x[n]$ - input, $y[n]$ - output)

Theorem of Foures-Segal says no (?)

When I asked about Foures-Segal I meant the following. F-S theorem says that if transfer function $T$:

$$y=Tu$$

is time invariant then there exists a function $G(z)$ which is analytic and bounded in exterior of unit disk on complex plane, such that

$$T = Z^{-1} G(z) Z$$,

where $Z$ is the Z-transform.

In my case the function $G(z) = \frac{z(z-2)}{(z+1)^2}$ having pole at $z=-1$, is not bounded. I would say it means the system is not time invariant?

• Fourés-Segal were pretty hardcore functional analysts; I think it would be very worth putting down the theorem you're referring in exactly the wording you're planning to use it; I think this might give the discussion a "kick" in the right direction. – Marcus Müller Jul 3 '16 at 16:29
• Foures-Segal say that the Direct Form I is not time invariant? i wonder if this is related to this other question? @MattL. might you weigh in? – robert bristow-johnson Jul 3 '16 at 16:34
• @robertbristow-johnson: thanks for pointing this out; see below for my attempt. BTW: I must admit that I've never heard of Foures-Segal (and I strongly believe that we engineers don't even need them to answer that question). – Matt L. Jul 3 '16 at 19:20

From the given difference equation, nothing can be said about linearity or time-invariance. Note that the difference equation (DE) doesn't even uniquely specify the output for a given input $x[n]$. To see this, assume that $y_1[n]$ is a solution to the DE. Let $y_h[n]$ be a solution to the corresponding homogeneous equation

$$y[n]+2y[n-1]+y[n-2]=0\tag{1}$$

Then any sequence $y_2[n]=y_1[n]+y_h[n]$ must also be a solution to the original DE. Note that if $y_h[n]\neq 0$, the corresponding system is neither linear nor time-invariant because that part of the output signal does not depend on the input $x[n]$. So no matter if you shift the input signal, or if you multiply it with a constant, that term remains unchanged.

In order to uniquely specify $y[n]$ for a given input $x[n]$, we need auxiliary conditions. There are several options, but the auxiliary conditions that make sure that the system described by the DE is linear and time-invariant (and causal) are initial rest conditions. If we assume that $x[n]=0$ for $n<0$, then the initial rest conditions are $y[-1]=y[-2]=\ldots=0$. This makes sure that $y_h[n]=0$ for $n\ge 0$, and consequently, the output $y[n]$ contains no term that is independent of the input $x[n]$.

A system described by a linear DE with constant coefficients (such as the one in your question) describes a linear and time-invariant (and causal) system only in combination with the aforementioned initial rest conditions.

• i continue to feel that the initial conditions can be defined in terms of $x[n]$ and $y[n]$ for $-\infty<n<0$. i continue to think that if the difference equation is presented as the complete text of this story, then it's both linear and time-invariant. – robert bristow-johnson Jul 4 '16 at 6:15
• @robertbristow-johnson: Maybe you could provide an answer going into a little bit more detail; generally you can split up the output into a zero-state response and a zero-input response. If the zero-input response is not zero (this is determined by the initial conditions), then the system cannot be LTI, because that part of the ouput doesn't depend on $x[n]$. Initial rest makes the zero-input response vanish, in which case the system is LTI. – Matt L. Jul 4 '16 at 13:56
• my answer would be no different than this previous answer. i will concede that i am requiring the system at the beginning of time (when $n=-\infty$) to be a completely "relaxed system" (that's a term from the control theory folks to indicate all states are = 0). whatever the states are at $n=0$ can be reconstructed from whatever input we had from $n=-\infty$ until $n=-1$. – robert bristow-johnson Jul 4 '16 at 16:45
• @robertbristow-johnson: Yes, I would say that that's just a special case (starting point at $-\infty$ instead of $n=0$ or any other finite $n=n_0$). Whenever you start, the system has to be relaxed. If you choose $n=-\infty$ that's fine, but you can choose any other $n$ as well, and the requirement of "initial rest" for the system to be LTI remains the same. – Matt L. Jul 4 '16 at 19:18
• i don't disagree with what you just wrote, @MattL. what my main point is, is that whatever non-relaxed state you get at $n=0$ or any other finite $n$ (which we may as well just call it $0$), is obtainable with whatever $x[n]$ exists for $n<0$. and then the definitions of linearity and time-invariancy apply because i am including all of $n<0$ in my representation of $x[n]$ and $y[n]$. – robert bristow-johnson Jul 4 '16 at 20:04

Time variance is defined by a change of filter coefficients over time. Since your coefficients are not changing a lot (1, 2, 1 and 1, -2), the filter should be time invariant ;)

• careful Jan. let's hear what @MattL. might have to say. (question doesn't say anything about initial conditions.) also, time-invariance has a more fundamental definition that says nothing about filter coefficients. – robert bristow-johnson Jul 3 '16 at 16:40

To check your assumption that it isn't LTI is straight forward as the definition of an LTI system is easy to look up. Logically it only has to fail one necessary condition of LTI to be discounted as LTI such as:

• BIBO stability
• Impulse response must have finite energy
• Poles inside unit disk

Given there is a pole on the unit disk it fails one of the necessary conditions, so at least you know you're correct.

The conditions I've listed here are pretty much restatements of the same thing and your Foures-Segal theorem also looks like a mathematical restatement of stability. Applied to this LTI, it doesn't look like Foures-Segal is revealing anything new. It may be a useful tool for more abstract systems though so an understanding of it may prove useful in other settings.