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Use the shift derivative of a step response, writing $y_3(t,b)=H(s)K(s)$ with $s=t-b$. Differentiate with respect to the forcing location:The possible delta term vanishes because $K(0)=0$, while $K^{\prime }(s)=g(s)$. Therefore, as a distribution and also as the ordinary piecewise response,Moving the start of a unit step differentiates its forcing into a negative impulse. The zero initial conditions and linearity ensure that the corresponding responses obey the same identity, including the zero-frequency cases.