Circular convolution: Difference between revisions

Revision as of 19:04, 3 January 2022

Revision as of 22:26, 11 February 2022

Line 15:

\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}}}}

{{Collapsemath topproof|title=Derivation of Eq.1}}|proof=

<math display="block">\begin{align}

\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}\\

Line 23:

&=\int_{t_o}^{t_o+T} \underbrace{\left[\sum_{k=-\infty}^\infty h(u + kT)\right]}_{\triangleq \ h_{_T}(u)}\cdot x_{_T}(t - u)\ du\\

&=\int_{t_o}^{t_o+T} h_{_T}(\tau)\cdot x_{_T}(t - \tau)\ d\tau \quad \text{substituting } \tau \triangleq u

\end{align}</math>}}

{{Collapse bottom}}

Both forms can be called ''periodic convolution''.{{efn-la