# Initial conditions for the LTI systems described as a difference equations

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?

Thanx

• 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. – hotpaw2 Feb 4 '16 at 19:37

• @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 $\mathcal{Z}\{f[n+1]\}=z(F(z)-f)$ – Matt L. Feb 7 '16 at 8:52
• 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 $n=0$. – robert bristow-johnson Feb 7 '16 at 16:28