# Are discrete systems defined by LCCDE always LTI?

Suppose a discrete-time system is defined by linear constant-coefficient difference equation

$$\sum_{k=0}^{N} a_k y[n-k] = \sum_{k=0}^{M} b_k x[n-k]$$

where at least two different coefficients $a_i,a_j$ are nonzero (so the equation is recursive). Isn't it true that such system is LTI and causal regardless of any auxiliary condition? After all, it is causal because it does not depend on any future inputs (RHS of the equation has only present or past inputs). And it is LTI because: if $x_1[n] \to y_1 [n]$ and $x_2[n] \to y_2 [n]$ the above difference equation is valid for $x_1,y_1$ and $x_2,y_2$ and by adding them together we can show that

$$\sum_{k=0}^{N} a_k (y_1[n-k]+y_2[n-k]) = \sum_{k=0}^{M} b_k (x_1[n-k]+x_2[n-k])$$

which means $x_1[n]+x_2[n] \to y_1[n]+y_2[n]$.

(similarly we can prove the homogeneity and the time-invariance property).

However it doesn't seem to be true according to Signals & Systems by Oppenheim A.:

As in the continuous-time case, such equation does not completely specify the output in terms of the input. To do this, we must also specify some auxiliary conditions.

Up to this point I agree. But then they write:

[..] we will focus for the most part of the condition of initial rest - i.e., if $x[n] = 0$ for $n < n_0$, then $y[n]=0$ for $n<n_0$ as well. With initial rest, the system described by the [first] equation is LTI and causal

In other words, the author seems to imply that without the initial rest condition, there's no guarantee that the system is LTI and causal (why contradicts my previous statements). Where's my mistake?

What you have is a so-called behavior which truncates the signal space $\mathcal{X}\oplus\mathcal{Y}$ of all possible signal pairs $(x, y)$ to the ones that satisfies this equality.

This does not specify an input-output mapping and hence a dependence and hence causality does not apply here. They simultaneously satisfy this. However regardless of causality, it is LTI since you can take arbitrarily and superposition applies.

Just like you can integrate any differential equation sufficiently many times and end up with a integral equation, you can also forward time shift these equations then if one of them depends on future inputs and that is your choice of dependent variable then you can speculate about causality.

This is actually the tool to allow for nondifferentiable input signals and other discontinuities and the results are investigated under the name of weak solutions

• What conditions are sufficient/necessary for a discrete system described by a linear, constant coefficient, difference equation to be LTI and causal? – user241601 Jun 9 '18 at 10:25
• Also, I'm not so sure about linearity - what if we set $y[-1]=1$ as the auxiliary condition? Then homogeneity is violated because if $x[n] \to y[n]$ then $2 x[n] \not \to 2 y[n]$. – user241601 Jun 9 '18 at 16:21
• @user241601 you can't constrain the output, linearity should satisfy the superposition. – percusse Jun 9 '18 at 16:34
• but why it's forbidden to constrain the output? After all, defining initial conditions is necessary for obtaining a unique solution. There's a problem in Oppenheim's book that starts with the following: "The initial rest assumption corresponds to a zero-valued auxiliary condition being imposed at a time determined in accordance with the input signals. In this problem we show that if the auxiliary condition used is nonzero or if it is always applied at a fixed time (regardless of the input signal), the corresponding system cannot be LTI". – user241601 Jun 9 '18 at 16:52
• @user241601 I don't know how it is related in this context. It says if there is a hard coded $u=4$. Then it is time dependent and hence not time invariant if you shift everything else in time. – percusse Jun 9 '18 at 17:42

Your initial set of difference equations has a unique Z transform but the Z transform can have more than one region of convergence (ROC) that depends on the direction of time. The ROC correspond to a casual region, an anti-causal region (reverse in time). I think there will be a section in your book with all the figures detailing the various ROC, There is another ROC that I forget the name of. You can have up to 3 ROC. The ROC that contains the unit circle is the one that is stable.

Causal and stable, are separate conditions. An LTI can be anti-causal and unstable.

The initial rest condition is interesting because $n_0$ isn't constrained. $n_0$ can be negative. One is free to choose where $n=0$ is as well. Time invariance is after all invariant. So there is an ambiguous situation, where at $n=0$ an initial condition may exist where you may not know if the initial condition was initially at rest at some $n_0 <0$ or the initial condition was due to some sort of mechanism independent of prior input.

So for a system to be Linear $$\alpha x_1[n] + \beta x_2[n] \Rightarrow \alpha y_1[n] + \beta y_2[n]$$ without the initial rest condition, you can have $\alpha \; 0 \Rightarrow \alpha y[n]$ doesn't hold for any $\alpha$. Or using different words, any $y[n] \ne 0$ in a causal system where $x[k]=0$ $k \le n$ is not linear. Or to start beating the dog, a spontaneous $y[n]$ unrelated to any $x[k]$, causal (or anti causal) is not linear.

There is probably an eventual rest condition for anti causal systems.

Many people will argue that any initial condition at $n=0$ for a causal system will render a system to be nonlinear.

If on the other hand, you look at a book that develops the linear systems theory from a state variable perspective. A system will be defined as being linear if it is both zero state linear and zero input linear. The initial rest condition is still required but $n_0 < 0$ is more obviously recognized.

The concept of a spontaneous $y[n] \ne 0$ might seem peculiar but no system is perfectly linear and electronic components like capacitors can have residual voltages that may not dissipate. This is one physical justification for requiring the initial rest condition.

• Thank you. But then where is the mistake in my proof? – user241601 Jun 2 '18 at 16:51
• Mistake? I would say jumping to causality bit prematurely, and specifically not accounting for the intimal rest condition . Without the initial rest condition, you can have a $y[n]$ that is non zero when all $x[k]$ for $k \le n$ are zero. – Stanley Pawlukiewicz Jun 2 '18 at 18:06
• Oppenheim's definition of causality: "A system is causal if the output at any time depends only on values of the input at present time and in the past". Since there's no $x[n+k]$ (where $k>0$ integer) on the RHS of the difference equation I thought we can deduce the system is causal. Thank you, I have to process your answer. – user241601 Jun 2 '18 at 21:08