BIBO安定条件 | x ( t ) | < M ⟹ | y ( t ) | < N {\displaystyle |x(t)|<M\Longrightarrow |y(t)|<N} 因果性システムの線形時不変の場合、 ∫ 0 ∞ | g ( t ) | d t < ∞ {\displaystyle \int _{0}^{\infty }|g(t)|dt<\infty } で安定する。
下図に示す系があるとする。
入力関数 : f ( t ) = 1 ⋯ t ≥ 0 {\displaystyle f(t)=1\quad \cdots \quad t\geq 0} 応答関数 : g ( t ) = 2 ( 1 − t 5 ) {\displaystyle g(t)=2\left(1-{\frac {t}{5}}\right)} y ( t ) = ∫ 0 t f ( τ ) g ( t − τ ) d τ {\displaystyle y(t)=\int _{0}^{t}{f(\tau )g(t-\tau )}d\tau } より、 t < 0 {\displaystyle t<0} の時 f ( τ ) {\displaystyle f(\tau )} と g ( t − τ ) {\displaystyle g(t-\tau )} の重なる部分が無いため、 y ( t ) = 0 {\displaystyle y(t)=0} t = 1 {\displaystyle t=1} の時 y ( t ) = ∫ 0 1 f ( τ ) g ( t − τ ) d τ = ∫ 0 1 f ( τ ) g ( 1 − τ ) d τ = ∫ 0 1 2 ( 1 − 1 5 + τ 5 ) d τ = 2 [ 4 5 τ + τ 2 10 ] 0 1 = 2 ( 4 5 + 1 10 ) = 9 5 {\displaystyle {\begin{aligned}y(t)&=\int _{0}^{1}{f(\tau )g(t-\tau )}d\tau \\&=\int _{0}^{1}{f(\tau )g(1-\tau )}d\tau \\&=\int _{0}^{1}{2(1-{\frac {1}{5}}+{\frac {\tau }{5}})}d\tau \\&=2\left[{\frac {4}{5}}\tau +{\frac {\tau ^{2}}{10}}\right]_{0}^{1}\\&=2\left({\frac {4}{5}}+{\frac {1}{10}}\right)\\&={\frac {9}{5}}\end{aligned}}} t = 5 {\displaystyle t=5} の時 y ( t ) = ∫ 0 5 f ( τ ) g ( t − τ ) d τ = ∫ 0 5 f ( τ ) g ( 5 − τ ) d τ = ∫ 0 5 2 ( 1 − 1 + τ 5 ) d τ = 2 [ τ 2 10 ] 0 5 = 2 5 2 = 5 {\displaystyle {\begin{aligned}y(t)&=\int _{0}^{5}{f(\tau )g(t-\tau )}d\tau \\&=\int _{0}^{5}{f(\tau )g(5-\tau )}d\tau \\&=\int _{0}^{5}{2(1-1+{\frac {\tau }{5}})}d\tau \\&=2\left[{\frac {\tau ^{2}}{10}}\right]_{0}^{5}\\&=2{\frac {5}{2}}\\&=5\end{aligned}}} 0 ≤ t ≤ 5 {\displaystyle 0\leq t\leq 5} の時 y ( t ) = ∫ 0 t 2 ( 1 − t 5 + τ 5 ) d τ = 2 [ τ − t 5 τ + τ 2 10 ] 0 t = 2 ( t − t 2 5 + t 2 10 ) = 2 t − t 2 5 {\displaystyle {\begin{aligned}y(t)&=\int _{0}^{t}{2(1-{\frac {t}{5}}+{\frac {\tau }{5}})}d\tau \\&=2\left[\tau -{\frac {t}{5}}\tau +{\frac {\tau ^{2}}{10}}\right]_{0}^{t}\\&=2\left(t-{\frac {t^{2}}{5}}+{\frac {t^{2}}{10}}\right)\\&=2t-{\frac {t^{2}}{5}}\end{aligned}}} 下図に、畳み込み積分の出力波形を示す。
{ y ( t ) = 0 ⋯ t < 0 y ( t ) = 2 t − t 2 5 ⋯ 0 ≤ t ≤ 5 y ( t ) = 5 ⋯ t > 0 {\displaystyle {\begin{cases}y(t)&=0\quad \cdots \quad t<0\\y(t)&=2t-{\frac {t^{2}}{5}}\quad \cdots \quad 0\leq t\leq 5\\y(t)&=5\quad \cdots \quad t>0\end{cases}}}
![{\displaystyle {\begin{aligned}y(t)&=\int _{0}^{1}{f(\tau )g(t-\tau )}d\tau \\&=\int _{0}^{1}{f(\tau )g(1-\tau )}d\tau \\&=\int _{0}^{1}{2(1-{\frac {1}{5}}+{\frac {\tau }{5}})}d\tau \\&=2\left[{\frac {4}{5}}\tau +{\frac {\tau ^{2}}{10}}\right]_{0}^{1}\\&=2\left({\frac {4}{5}}+{\frac {1}{10}}\right)\\&={\frac {9}{5}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9ffd88d284a8b6953b35e86a3b588bf6aec75a4)
![{\displaystyle {\begin{aligned}y(t)&=\int _{0}^{5}{f(\tau )g(t-\tau )}d\tau \\&=\int _{0}^{5}{f(\tau )g(5-\tau )}d\tau \\&=\int _{0}^{5}{2(1-1+{\frac {\tau }{5}})}d\tau \\&=2\left[{\frac {\tau ^{2}}{10}}\right]_{0}^{5}\\&=2{\frac {5}{2}}\\&=5\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee10e860cde010036a1f5fe88dbf037e908e2c10)
![{\displaystyle {\begin{aligned}y(t)&=\int _{0}^{t}{2(1-{\frac {t}{5}}+{\frac {\tau }{5}})}d\tau \\&=2\left[\tau -{\frac {t}{5}}\tau +{\frac {\tau ^{2}}{10}}\right]_{0}^{t}\\&=2\left(t-{\frac {t^{2}}{5}}+{\frac {t^{2}}{10}}\right)\\&=2t-{\frac {t^{2}}{5}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d57bd98e200a231499c580c3a100c1502c34ece0)
