# 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. Commented Feb 4, 2016 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 $\mathcal{Z}\{f[n+1]\}=z(F(z)-f[0])$ Commented Feb 7, 2016 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 $n=0$. Commented Feb 7, 2016 at 16:28