# System Response Terminology

If I have a system specified by

$$P(D)y(t)=Q(D)x(t)$$

and I specify initial conditions $$y(0^-)=a, \ y'(0^-)=b,\ x(0^-)=c$$ does the term $$x(0^-)=c$$ correspond to the zero state response or zero input response?

Assuming that you denote the output by $$y(t)$$ and the input by $$x(t)$$, the initial conditions are given by the values of $$y(t)$$ and its derivatives at a certain time $$t_0$$ (choose $$t_0=0^-$$ if you like). A value $$x(t_0)$$ has nothing to do with an initial condition.

For the zero-input response (ZIR) you just set $$x(t)=0$$ and compute the output caused by possibly non-zero initial conditions. For the zero-state response (ZSR), you set $$y^{(k)}(t_0)=0$$, $$k=0,1,\ldots$$, and compute the output for a given input $$x(t)$$. The total response is then the sum of the ZIR and the ZSR.

• Thank you once again! – Colin Hicks Apr 8 at 20:55

$$x(0^-) \ \triangleq \ \lim_{0<\epsilon \to 0} x(-\epsilon)$$

using the traditional undergraduate one-sided definition of the Laplace Transform,

$$\mathscr{L}\Big\{ x(t) \Big\} \triangleq X(s) = \int_{0^-}^{\infty} x(t) \, e^{-st} \, \mathrm{d}t$$

the purpose of the notation "$$0^-$$" is to make sure you include all of any dirac delta $$\delta(t)$$ that may occur at $$t=0$$.

this one-sided Laplace is useful only if you are omitting all of time $$t<0$$ from consideration in your problem and you want a handy way to include initial conditions (which are not strictly part of the differential equation) of your problem in the solution of the problem.