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{{Equation box 1
\int_{t_o}^{t_o+T} h_{_T}(\tau)\cdot x_{_T}(t - \tau)\,d\tau = \int_{-\infty}^\infty h(\tau)\cdot x_{_T}(t - \tau)\,d\tau\ \triangleq\ (h *x_{_T})(t) = (x * h_{_T})(t).</math>|{{EquationRef|Eq.1}}}}▼
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<math>
▲\int_{t_o}^{t_o+T} h_{_T}(\tau)\cdot x_{_T}(t - \tau)\,d\tau = \int_{-\infty}^\infty h(\tau)\cdot x_{_T}(t - \tau)\,d\tau\ \triangleq\ (h *x_{_T})(t) = (x * h_{_T})(t).</math>
|{{EquationRef|Eq.1}} }} }}
{{math proof|title=Derivation of Eq.1|proof=
:<math
\int_{-\infty}^\infty h(\tau)\cdot x_{_T}(t - \tau)\,d\tau
&=\sum_{k=-\infty}^\infty \left[\int_{t_o+kT}^{t_o+(k+1)T} h(\tau)\cdot x_{_T}(t - \tau)\ d\tau\right] \quad t_0 \text{ is an arbitrary parameter}\\
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