Diracs impulsfunksjon er uendelig når og null ellers:
δ ( x ) = { 0 n a ˚ r x < 0 ∞ n a ˚ r x = 0 0 n a ˚ r x > 0 \delta(x) = \left\{ \begin{array}{ll}
0 & \textnormal{når } x < 0 \\
\infty & \textnormal{når } x = 0 \\
0 & \textnormal{når } x > 0
\end{array} \right. δ ( x ) = ⎩ ⎨ ⎧ 0 ∞ 0 n a ˚ r x < 0 n a ˚ r x = 0 n a ˚ r x > 0
Kalles også enhetspulsen eller impulsfunksjonen
Innført av en engelsk fysiker, Paul Dirac (1902–1984)
Brukes mye i signalbehandling, kvantemekanikk, alle steder der vi har en impuls og Laplace transformer
+ Definisjon og viktig egenskap
Diracs deltafunksjon, , er definert slik at:
∫ a b f ( x ) δ ( x ) d x = f ( 0 ) n a ˚ r a ≤ 0 < b \int_a^b \textcolor{blue}{f(x)} \textcolor{red}{\delta(x)} dx = \textcolor{blue}{f(0)}
\quad \textnormal{ når } a \le 0 < b ∫ a b f ( x ) δ ( x ) d x = f ( 0 ) n a ˚ r a ≤ 0 < b Og, vips, blir noen integraler mye enklere.
Noen eksempler:
∫ − ∞ ∞ e 5 x δ ( x ) d x = e 5 ⋅ 0 = 1 ∫ − 10 10 7 sin ( x ) e 10 + x δ ( x ) d x = 7 sin ( 0 ) e 10 + 0 = 0 ∫ − 7 5 ( x 4 + 5 x 2 − 7 x + 18 ) δ ( x ) d x = 0 4 + 5 ⋅ 0 2 − 7 ⋅ 0 + 18 = 18 \int_{-\infty}^{\infty} \textcolor{blue}{e^{5x}} \textcolor{red}{\delta(x)} dx = \textcolor{blue}{e^{5 \cdot 0}} = 1 \\
\int_{-10}^{10} \textcolor{blue}{7\sin(x) e^{10 + x}} \textcolor{red}{\delta(x)} dx = \textcolor{blue}{7 \sin(0)e^{10 + 0}} = 0 \\
\int_{-7}^5 \textcolor{blue}{(x^4 + 5x^2 - 7x +18)} \textcolor{red}{\delta(x)} dx = \textcolor{blue}{0^4 + 5 \cdot 0^2 - 7 \cdot 0 + 18} = 18 ∫ − ∞ ∞ e 5 x δ ( x ) d x = e 5 ⋅ 0 = 1 ∫ − 10 10 7 s i n ( x ) e 10 + x δ ( x ) d x = 7 s i n ( 0 ) e 10 + 0 = 0 ∫ − 7 5 ( x 4 + 5 x 2 − 7 x + 18 ) δ ( x ) d x = 0 4 + 5 ⋅ 0 2 − 7 ⋅ 0 + 18 = 18 Utfordring
+ Eksempel 1:
∫ 0 5 e 2 x δ ( x ) d x = e 2 ⋅ 0 = 1 \int_0^5 \textcolor{blue}{e^{2x}} \textcolor{red}{\delta(x)} dx = \textcolor{blue}{e^{2 \cdot 0}} = 1 ∫ 0 5 e 2 x δ ( x ) d x = e 2 ⋅ 0 = 1 fordi:
∫ a b f ( x ) δ ( x ) d x = f ( 0 ) n a ˚ r a ≤ 0 < b \int_a^b \textcolor{blue}{f(x)} \textcolor{red}{\delta(x)} dx = \textcolor{blue}{f(0)}
\quad \textnormal{ når } a \le 0 < b ∫ a b f ( x ) δ ( x ) d x = f ( 0 ) n a ˚ r a ≤ 0 < b
+ Eksempel 2:
∫ 0 5 e 2 x δ ( x − 1 ) d x \int_0^5 \textcolor{blue}{e^{2x}} \textcolor{red}{\delta(x-1)} dx ∫ 0 5 e 2 x δ ( x − 1 ) d x Her må vi først bruke substitusjon :
u = x − 1 ⇒ d u = d x Nedre grense: x = 0 gir u = 0 − 1 = − 1 Øvre grense: x = 5 gir u = 5 − 1 = 4 \begin{array}{llcl}
& u = x - 1 & \Rightarrow & du = dx \\
\textnormal{Nedre grense: } & x = 0 & \textnormal{gir} & u = 0 - 1 = -1 \\
\textnormal{Øvre grense: } & x = 5 & \textnormal{gir} &u = 5 - 1 = 4
\end{array} Nedre grense: Øvre grense: u = x − 1 x = 0 x = 5 ⇒ gir gir d u = d x u = 0 − 1 = − 1 u = 5 − 1 = 4 Husk å bytte grensene slik at de gjelder :
∫ 0 5 e 2 x δ ( x − 1 ) d x = ∫ − 1 4 e 2 ( u + 1 ) δ ( u ) d u = e 2 \int_0^5 \textcolor{blue}{e^{2x}} \textcolor{red}{\delta(x-1)} dx
= \int_{-1}^4 \textcolor{blue}{e^{2(u + 1)}} \textcolor{red}{\delta(u)} du
= \textcolor{blue}{e^{2}} ∫ 0 5 e 2 x δ ( x − 1 ) d x = ∫ − 1 4 e 2 ( u + 1 ) δ ( u ) d u = e 2 fordi:
∫ a b f ( x ) δ ( x ) d x = f ( 0 ) n a ˚ r a ≤ 0 < b \int_a^b \textcolor{blue}{f(x)} \textcolor{red}{\delta(x)} dx = \textcolor{blue}{f(0)}
\quad \textnormal{ når } a \le 0 < b ∫ a b f ( x ) δ ( x ) d x = f ( 0 ) n a ˚ r a ≤ 0 < b
+ Eksempel 3:
∫ − ∞ ∞ f ( x ) δ ( x − c ) d x \int_{-\infty}^{\infty} \textcolor{blue}{f(x)} \textcolor{red}{\delta(x-c)} dx ∫ − ∞ ∞ f ( x ) δ ( x − c ) d x Her må vi først bruke substitusjon :
u = x − c ⇒ d u = d x Nedre grense: x = − ∞ gir u = − ∞ Øvre grense: x = ∞ gir u = ∞ \begin{array}{llcl}
& u = x - c & \Rightarrow & du = dx \\
\textnormal{Nedre grense: } & x = -\infty & \textnormal{gir} & u = - \infty \\
\textnormal{Øvre grense: } & x = \infty & \textnormal{gir} &u = \infty
\end{array} Nedre grense: Øvre grense: u = x − c x = − ∞ x = ∞ ⇒ gir gir d u = d x u = − ∞ u = ∞ Husk å bytte grensene slik at de gjelder :
∫ − ∞ ∞ f ( x ) δ ( x − c ) d x = ∫ − ∞ ∞ f ( u + c ) δ ( u ) d u = f ( c ) \int_{-\infty}^{\infty} \textcolor{blue}{f(x)} \textcolor{red}{\delta(x-c)} dx
= \int_{-\infty}^{\infty} \textcolor{blue}{f(u+c)} \textcolor{red}{\delta(u)} du
= f(c) ∫ − ∞ ∞ f ( x ) δ ( x − c ) d x = ∫ − ∞ ∞ f ( u + c ) δ ( u ) d u = f ( c ) fordi:
∫ a b f ( x ) δ ( x ) d x = f ( 0 ) n a ˚ r a ≤ 0 < b \int_a^b \textcolor{blue}{f(x)} \textcolor{red}{\delta(x)} dx = \textcolor{blue}{f(0)}
\quad \textnormal{ når } a \le 0 < b ∫ a b f ( x ) δ ( x ) d x = f ( 0 ) n a ˚ r a ≤ 0 < b
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