巡回畳み込み
↑ 証明: ∫ − ∞ ∞ h ( τ ) ⋅ x T ( t − τ ) d τ {\displaystyle \int _{-\infty }^{\infty }h(\tau )\cdot x_{T}(t-\tau )\,d\tau } = ∑ k = − ∞ ∞ [ ∫ t o + k T t o + ( k + 1 ) T h ( τ ) ⋅ x T ( t − τ ) d τ ] = τ → τ + k T ∑ k = − ∞ ∞ [ ∫ t o t o + T h ( τ + k T ) ⋅ x T ( t − τ − k T ) d τ ] = ∫ t o t o + T [ ∑ k = − ∞ ∞ h ( τ + k T ) ⋅ x T ( t − τ − k T ) ⏟ x T ( t − τ ) , by periodicity ] d τ = ∫ t o t o + T [ ∑ k = − ∞ ∞ h ( τ + k T ) ] ⏟ = d e f h T ( τ ) ⋅ x T ( t − τ ) d τ ( Q E D ) {\displaystyle {\begin{aligned}&=\sum _{k=-\infty }^{\infty }\left[\int _{t_{o}+kT}^{t_{o}+(k+1)T}h(\tau )\cdot x_{T}(t-\tau )\ d\tau \right]\\&{\stackrel {\tau \rightarrow \tau +kT}{=}}\ \sum _{k=-\infty }^{\infty }\left[\int _{t_{o}}^{t_{o}+T}h(\tau +kT)\cdot x_{T}(t-\tau -kT)\ d\tau \right]\\&=\int _{t_{o}}^{t_{o}+T}\left[\sum _{k=-\infty }^{\infty }h(\tau +kT)\cdot \underbrace {x_{T}(t-\tau -kT)} _{x_{T}(t-\tau ),{\text{by periodicity}}}\right]\ d\tau \\&=\int _{t_{o}}^{t_{o}+T}\underbrace {\left[\sum _{k=-\infty }^{\infty }h(\tau +kT)\right]} _{{\stackrel {\mathrm {def} }{=}}\ h_{T}(\tau )}\cdot x_{T}(t-\tau )\ d\tau \quad \quad \scriptstyle {(QED)}\end{aligned}}}