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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?

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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.

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  • $\begingroup$ Thank you once again! $\endgroup$ – Colin Hicks Apr 8 at 20:55
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$$ x(0^-) \ \triangleq \ \lim_{0<\epsilon \to 0} x(-\epsilon) $$

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.

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