# Impulse response rest conditions of LTI systems

Why the impulse response (initial) rest conditions are taken like $h(t)=0$ and $h'(t) =1$, if the order of differential equation is $2$.

Consider the second order LCCDE related to an LTI system with initial rest is given as:

$$a_0 y^{''}(t) + a_1 y^{'}(t) + a_2 y(t) = x(t)$$

The solution of the given equation for the input $x(t) = \delta(t)$ is defined as the impulse response of the LTI system and denoted as $h(t)$. The formal solution of the associated second order differential equation in time domain requires that a homogeneous and a particular solution be found and are added to yield the total solution; $$h(t) = y_h(t) + y_p(t)$$

Since the particular input is an impulse, then the particular solution is zero and therefore $$h(t) = y_h(t)$$ which is computed based on the initial conditions. For a second order equation you need two initial conditions and these are derived based on initial rest assumption so that: $$y_h(0) = 0$$ and $$y_h^{'}(0) = 1/a_0$$.

Since most typically $a_0 = 1$ then $y_h^{'}(0) = 1$. This result comes from integrating the differential equation from $t=0^{-}$ to $t=0^{+}$ while recognising the fact that the output, $y_h(t)$, cannot involve any impulses in its defintion. In fact neither $y_h(t)$ nor any derivate of it upto (but not including) second order can include any impulses in their definition, because otherwise, then on the right side of the equation derivatives of impulse should have appeared, which is not the case, by definition, however. Hence

$$\int_{0^{-}}^{0^{+}} \left( a_0 y_h^{''}(t) + a_1 y_h^{'}(t) + a_2 y_h(t) \right)dt = \int_{0^{-}}^{0^{+}} \delta(t)$$

$$a_0 \int_{0^{-}}^{0^{+}} y_h^{''}(t) dt + a_1 \int_{0^{-}}^{0^{+}} y_h^{'}(t)dt + a_2 \int_{0^{-}}^{0^{+}} y_h(t)dt = 1$$

Now because of the intuition based on the previous paragraph, the second and the third integrals turn out to be zero (since they don't include impulses, their integrals at a point from $0-$ to $0+$ iz zero) but the first integral includes an impulse and it will be nonzero as:

$$a_0 \int_{0^{-}}^{0^{+}} y^{''}(t) dt = a_0 ( y_h^{'}(0^{+}) - y_h^{'}(0^{-}) ) = a_0 y_h^{'}(0^{+}) = 1$$ and it follows that

$$y_h^{'}(0^{+}) = \frac{1}{a_0}$$

since $y_h(t) = h(t)$ and since we can replace $0+$ with $0$, and taking $a_0 = 1$ as usual, then it follows that: $$\boxed{ h'(0) = \frac{1}{a_0} = 1 }$$

As the required initial condition on th impulse response $h(t)$. Similar argumentation follows for systems of higher orders.