Start
Zufällige Seite
Anmelden
Einstellungen
Spenden
Über Wikiversity
Haftungsausschluss
Suchen
Fourier-Transformation/Translationseigenschaften/Fakt/Beweis
Sprache
Beobachten
Bearbeiten
<
Fourier-Transformation/Translationseigenschaften/Fakt
Beweis
Es ist
1
(
2
π
)
n
/
2
∫
R
n
e
−
i
⟨
u
,
t
⟩
g
(
t
)
d
t
=
1
(
2
π
)
n
/
2
∫
R
n
e
−
i
⟨
u
,
t
⟩
f
(
t
)
e
i
⟨
v
,
t
⟩
d
t
=
1
(
2
π
)
n
/
2
∫
R
n
e
−
i
⟨
u
−
v
,
t
⟩
f
(
t
)
d
t
=
f
^
(
u
−
v
)
.
{\displaystyle {}{\begin{aligned}{\frac {1}{(2\pi )^{n/2}}}\int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},{\mathfrak {t}}\right\rangle }g({\mathfrak {t}})d{\mathfrak {t}}&={\frac {1}{(2\pi )^{n/2}}}\int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},{\mathfrak {t}}\right\rangle }f({\mathfrak {t}})e^{{\mathrm {i} }\left\langle v,{\mathfrak {t}}\right\rangle }d{\mathfrak {t}}\\&={\frac {1}{(2\pi )^{n/2}}}\int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}}-v,{\mathfrak {t}}\right\rangle }f({\mathfrak {t}})d{\mathfrak {t}}\\&={\hat {f}}({\mathfrak {u}}-v).\end{aligned}}}
Es ist wegen der Translationsinvarianz
1
(
2
π
)
n
/
2
∫
R
n
e
−
i
⟨
u
,
t
⟩
g
(
t
)
d
t
=
1
(
2
π
)
n
/
2
∫
R
n
e
−
i
⟨
u
,
t
⟩
f
(
t
−
v
)
d
t
=
1
(
2
π
)
n
/
2
∫
R
n
e
−
i
⟨
u
,
s
+
v
⟩
f
(
s
)
d
s
=
1
(
2
π
)
n
/
2
∫
R
n
e
−
i
⟨
u
,
t
+
v
⟩
f
(
t
)
d
t
=
1
(
2
π
)
n
/
2
⋅
e
−
i
⟨
u
,
v
⟩
∫
R
n
e
−
i
⟨
u
,
t
⟩
f
(
t
)
d
t
=
f
^
(
u
)
⋅
e
−
i
⟨
u
,
v
⟩
.
{\displaystyle {}{\begin{aligned}{\frac {1}{(2\pi )^{n/2}}}\int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},{\mathfrak {t}}\right\rangle }g({\mathfrak {t}})d{\mathfrak {t}}&={\frac {1}{(2\pi )^{n/2}}}\int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},{\mathfrak {t}}\right\rangle }f({\mathfrak {t}}-v)d{\mathfrak {t}}\\&={\frac {1}{(2\pi )^{n/2}}}\int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},s+v\right\rangle }f(s)ds\\&={\frac {1}{(2\pi )^{n/2}}}\int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},{\mathfrak {t}}+v\right\rangle }f({\mathfrak {t}})d{\mathfrak {t}}\\&={\frac {1}{(2\pi )^{n/2}}}\cdot e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},v\right\rangle }\int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},{\mathfrak {t}}\right\rangle }f({\mathfrak {t}})d{\mathfrak {t}}\\&={\hat {f}}({\mathfrak {u}})\cdot e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},v\right\rangle }.\end{aligned}}}
Mit
s
=
c
t
{\displaystyle {}s=c{\mathfrak {t}}}
ist
g
^
(
u
)
=
1
(
2
π
)
n
/
2
⋅
∫
R
n
e
−
i
⟨
u
,
t
⟩
g
(
t
)
d
t
=
1
(
2
π
)
n
/
2
⋅
∫
R
n
e
−
i
⟨
u
,
t
⟩
f
(
c
t
)
d
t
=
1
(
2
π
)
n
/
2
⋅
∫
R
n
e
−
i
⟨
u
,
s
c
⟩
f
(
s
)
d
s
c
=
1
(
2
π
)
n
/
2
⋅
1
|
c
|
n
⋅
∫
R
n
e
−
i
⟨
u
c
,
s
⟩
f
(
s
)
d
s
=
1
|
c
|
n
f
^
(
u
c
)
.
{\displaystyle {}{\begin{aligned}{\hat {g}}({\mathfrak {u}})&={\frac {1}{(2\pi )^{n/2}}}\cdot \int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},{\mathfrak {t}}\right\rangle }g({\mathfrak {t}})d{\mathfrak {t}}\\&={\frac {1}{(2\pi )^{n/2}}}\cdot \int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},{\mathfrak {t}}\right\rangle }f(c{\mathfrak {t}})d{\mathfrak {t}}\\&={\frac {1}{(2\pi )^{n/2}}}\cdot \int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},{\frac {s}{c}}\right\rangle }f(s)d{\frac {s}{c}}\\&={\frac {1}{(2\pi )^{n/2}}}\cdot {\frac {1}{\vert {c}\vert ^{n}}}\cdot \int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\frac {\mathfrak {u}}{c}},s\right\rangle }f(s)ds\\&={\frac {1}{\vert {c}\vert ^{n}}}{\hat {f}}\left({\frac {\mathfrak {u}}{c}}\right).\end{aligned}}}
Zur bewiesenen Aussage
ist
