Step response of a damped oscillator ($H(t)K(t),K(t)={\int }_0^{t}g(s)ds$)

Integrating the smooth positive-time impulse kernel gives the zero-data response to the Heaviside step function. For nonzero $\omega$,At zero frequency and nonzero $k$ the limit is $[1-e^{-kt}(1+kt)]/k^2$; when both parameters vanish it is $t^2/2$. The integral representation avoids ambiguous divisions in these limiting cases.

Adding a load $M$ to a spring-supported table shifts its static downward position by $\Delta =Mg/k$. If the load is deposited with negligible impulse and moves with the table, the moving mass is $m+M$. Starting at the former equilibrium with zero velocity, an underdamped response isIts transient envelope decays over $1/\beta$. Critical damping $b=2\sqrt{k(m+M)}$ gives $x=\Delta [1-(1+{\omega }_{0}t)e^{-{\omega }_{0}t}]$. A nonzero impact impulse changes the initial velocity and therefore the transient.

For a step-response kernel with $K(0)=0$, differentiating the start time produces no delta term from the response itself, because its coefficient is zero. The negative shift derivative is the impulse response. More generally an additional $\delta (t-b)K(0)$ term appears if the kernel has a nonzero initial value. The identity holds distributionally as well as for the ordinary smooth pieces.