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?


3 Answers 3


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

  • 1
    $\begingroup$ What conditions are sufficient/necessary for a discrete system described by a linear, constant coefficient, difference equation to be LTI and causal? $\endgroup$
    – user241601
    Commented Jun 9, 2018 at 10:25
  • $\begingroup$ 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 $2 x[n] \not \to 2 y[n]$. $\endgroup$
    – user241601
    Commented Jun 9, 2018 at 16:21
  • $\begingroup$ @user241601 you can't constrain the output, linearity should satisfy the superposition. $\endgroup$
    – percusse
    Commented Jun 9, 2018 at 16:34
  • $\begingroup$ 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". $\endgroup$
    – user241601
    Commented Jun 9, 2018 at 16:52
  • $\begingroup$ @user241601 I don't know how it is related in this context. It says if there is a hard coded $u[6]=4$. Then it is time dependent and hence not time invariant if you shift everything else in time. $\endgroup$
    – percusse
    Commented Jun 9, 2018 at 17:42

There is a distinction between a linearity in the context of a difference or differential equation and linearity in the context of a linear system. You are assuming that because a difference/differential equation involves a linear operator, the associated system is linear. This is not so. Linearity in terms of the associated system depends on a unique output for a given input and this depends on the initial/auxiliary conditions. In particular, for a differential equation, the initial conditions depend on the all derivative up to the highest derivative in the system. Looking at the homogeneous solution, you will find that if any of these derivatives is non-zero at a given initial time $t_0$, it means the system will generate a non-zero output for zero input. This is clearly a non-linear system because zero input should always generate zero output (by linearity). The same considerations apply to a difference equations whereby the initial conditions are not derivatives up to $n^{th}$ order but rather current and past values of the output (i.e., $y[n]$, $y[n-1]$, $y[n-2]$, etc.).

Thus, linearity with respect to a linear system requires consideration not only of the underlying linearity/non-linearity of the operator describing the differential/difference equation but also the initial conditions.

With respect to time invariance, then, it can also be seen that if the system has non-zero initial conditions, it is not necessarily time-invariant as the output of a shifted input is not necessarily the shifted output of the non-shifted input (as it depends on the specific values of the initial conditions (which are a function of the independent variable)). In particular, the general solution is a linear combination of solutions to the homogeneous equation and a particular solution. However, the homogeneous solution will depend upon initial conditions and therefore the general solution cannot be inferred to be time invariant if these initial conditions are non-zero.

With respect to causality, this cannot be determined by a simple examination of the differential or difference equation. This is so because a difference equation can be recast in an form whereby a change of variables is performed to shift the independent variable forward in time. In this context, what appeared as a causal difference equation may be made to appear non-causal. Similarly, for differential equations, an equation may be recast as an integral equation rather than a differential equation.

In general, a unique solution depends upon the underlying differential/difference equation and the initial conditions.
For a system to be both LTI and causal requires even more. In particular, in requires initial rest conditions meaning that if the input $x(t)$ is $0$ for all $t<t_0$, the output $y(t)$ is also $0$ for all $t<t_0$.


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.

  • $\begingroup$ Thank you. But then where is the mistake in my proof? $\endgroup$
    – user241601
    Commented Jun 2, 2018 at 16:51
  • $\begingroup$ 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. $\endgroup$
    – user28715
    Commented Jun 2, 2018 at 18:06
  • 1
    $\begingroup$ 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. $\endgroup$
    – user241601
    Commented Jun 2, 2018 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.