Why do we need the initial conditions to be zero for the LTI systems described as a difference equations?

  1. First question is why do we need it for linearity? I can't think of any example of the non linear system described as a difference equation with constant coefficients.

  2. The definition of TI system is that it does not depend on particular time the input is applied. I can't understand how does this definition relate to the initial conditions? Those initial conditions will be the same whenever we apply the input so why they have to be zero?


  • $\begingroup$ Given different initial conditions, the output can be different (in systems with memory) even given the same identical input, but at different times after those initial conditions. $\endgroup$
    – hotpaw2
    Commented Feb 4, 2016 at 19:37

1 Answer 1


The problem is that non-zero initial conditions cause a term in the output signal that does not depend on the input signal. 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. With non-zero initial conditions the output will generally be non-zero, even for a zero input signal. Alternatively, think of scaling a given input signal. A linear system will have a response that is scaled in the same way. However, the part of the output signal caused by non-zero initial conditions will not scale accordingly, because it's independent of the input signal.

The same is true concerning time-invariance. For a time-invariant system, a shifted version of the input signal must result in an output signal with the same shift. However, the output term caused by non-zero initial conditions will not shift accordingly, as it is independent of the input signal.

Consequently, a system described by a linear difference equation with constant coefficients plus initial conditions is only linear and time-invariant if the initial conditions are zero.

  • $\begingroup$ if the time that the initial conditions varies along with the arbitrary time-offset of the signal, might then we call the system "time-invariant"? i guess that is asking for too much. as much as asking if both the time and amplitudes of all the initial conditions vary with the time-delay and the scale of the input and calling it LTI. but there it still doesn't pass superposition, so it's not linear. i guess initial conditions of the LTI have to all be zero, in order to use Fourier or Laplace or Z transforms on it. $\endgroup$ Commented Feb 6, 2016 at 1:14
  • $\begingroup$ @robertbristow-johnson: You can use the unilateral Laplace or Z-transform with non-zero initial conditions if the system is otherwise LTI, due to the following properties: $\mathcal{L}\{f'(t)\}=sF(0)-f(0)$ and $\mathcal{Z}\{f[n+1]\}=z(F(z)-f[0])$ $\endgroup$
    – Matt L.
    Commented Feb 7, 2016 at 8:52
  • $\begingroup$ yeah, we know that. it's the original way we are taught it in our EE curriculum. i consider the unilateral $\mathcal{L}$ or $\mathcal{Z}$ (with initial conditions) to be simply a special case of the bilateral with a specific set of Dirac or Kroneckers preceding $t=0$ or $n=0$. $\endgroup$ Commented Feb 7, 2016 at 16:28

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