I suspect this is rather obvious once explained; I must be misunderstanding something.

From reading various answers here (such as this one), I understand that an LTI system must have zero output for zero input, which I think is equivalent to zero initial conditions or initial rest. To quote @MattL from the answer I referenced,

"This explains why a system with non-zero initial conditions can neither be linear nor time-invariant. A linear system must have a zero output for zero input"

In the textbook I am working through, one of the problems states "Find the response for the system described by the following difference equation with the initial conditions given", and the difference equation is as follows:

$$y[n]-0.7y[n-1]=u[n], y[-1]=-3$$

Here the input starts at $n=0$ yet $y[-1]=-3$. The textbook is asking the question in the context of LTI systems.

At first I thought there was a mistake in the question, but I have seen similar setups in answers here on stackexchange as well.

How is the above scenario compatible with zero initial conditions?


The system is described by a linear difference equation with constant coefficients and as such, it is described in the same way as a linear time-invariant system. It is just the non-zero initial condition that makes the system non-linear, at least according to the common definition of linearity in system theory (homogeneity and additivity).

Such a system with non-zero initial conditions is also referred to as incrementally linear. An incrementally linear system responds linearly to changes in the input. For this reason it is very common, and it also makes a lot of sense, to discuss it in the same chapter as general discrete-time LTI systems.

As a final note, in practice we're mainly interested in stable systems. For such systems, the influence of non-zero initial conditions becomes negligible after some time, and, consequently, the system will practically behave as an LTI system after the transient caused by the non-zero initial conditions has died out.

  • 1
    $\begingroup$ Thank you for clarifying! I just wish the textbook I'm using (and other 'classic' ones I have access to) would make clear that the definition of linearity they are using is not all there is, and that more broad definitions that accommodate initial conditions or (as subsequent searching brought up) affine transformations are also possible.🙄 $\endgroup$ – Westerley Aug 10 '19 at 17:08

Many text books, particularly those that develop linear systems from a state variable perspective will define two characteristics.

  1. zero input linearity
  2. zero state linearity

and will require that both properties must hold for a system to be linear, if the system obeys the initial rest condition.

Zero input linearity is concerned with a situation where you look at a system at some time $t_0$ that you consider to be the start of your analysis and there was some input prior to $t_0$ (that was initially at rest). The initial conditions correspond to the memory of the system prior to $t_0$ at $t_0$.

There are other text books that only define zero state linearity but the time limits are $-\infty$ to $\infty$, the complete history (and future)

As an aside, the overlap add algorithm for fir filtering uses the residual output of the previous block which is an initial condition. Any DSP library with a filter() function will have an auxiliary input for initial state and another for final state.

You can effectively split your time invariant convolution in time with initial conditions.

  • $\begingroup$ Thank you for responding. I don't have any familiarity with the state-variable modelling, but now imagine it provides a more full perspective. I was limited in my understanding by thinking that whenever LTI systems are discussed, superposition and only superposition is assumed. $\endgroup$ – Westerley Aug 10 '19 at 17:14
  • $\begingroup$ @Westerley if you want to learn Kalman Filtering, state variables are a prerequisite. The First edition of Oppenheim and Schaefer covered the topic but they dropped it $\endgroup$ – user28715 Aug 10 '19 at 17:18
  • $\begingroup$ zero input states do satisfy superposition and homogeneity provided the initial test condition was satisfied $\endgroup$ – user28715 Aug 10 '19 at 19:14
  • $\begingroup$ I should post a question asking the best introductory textbooks for signal processing; or perhaps, what books did you read to get to your understanding of where you are (though that invariably involves more than reading textbooks) $\endgroup$ – Westerley Aug 11 '19 at 3:24

There's nothing wrong in asking an LCCDE (Linear Constant Coefficient Difference Equation) with nonzero initial conditions.

Books on Signals and Systems shall not only deal with LTI systems but also the other systems to some degree.

Also the concept of LCCDE is not limited to LTI systems, and methods of time-domain solutions, thefore, shall be given in a general perspective first and then in particular for those that correspond to LTI systems.

Therefore the student should definetely learn how to solve LCCDE with nonzero initial conditions as well.

Nevertheless, it's true that the most useful type of LCCDE will correspond to that for LTI systems; those with initial rest (zero initial conditions) and whose outputs can be computed via convolutions.

Considering the extreme importance of convolution in the mathematical development of signal processing, it's no wonder that initial rest based LCCDE, tha correspond to LTI systems, will also be the most encountered.

  • $\begingroup$ Thank you for responding; I understood that LCCDEs could be use in contexts other than LTI systems, and clearly they were being used for LTI systems with initial conditions. I was just missing the step that the definition of LTI systems that I was working with was overly narrow. $\endgroup$ – Westerley Aug 10 '19 at 17:16

It is in the context of LTI systems because despite being non linear as system due of the non zero initial conditions, it is linear as equation, which represents a specific situation of a LTI system (The response of the system to the initial condition as input).

  • $\begingroup$ Thank you for responding! If I've understood what you are saying, I think its along the lines of what @MattL is also saying. $\endgroup$ – Westerley Aug 10 '19 at 17:21

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