עריכה
ניקח גל מישורי סטנדרטי:
y
(
x
,
t
)
=
A
⋅
cos
(
ω
t
−
k
x
)
=
A
⋅
R
e
(
e
−
i
(
k
x
−
ω
t
)
)
{\displaystyle \ y\left(x,t\right)=A\cdot \cos \left(\omega t-kx\right)=A\cdot Re\left(e^{-i\left(kx-\omega t\right)}\right)}
נגדיר :
y
¯
~
(
x
,
t
)
=
A
⋅
e
−
i
(
k
x
−
ω
t
)
{\displaystyle \ {\tilde {\bar {y}}}(x,t)=A\cdot e^{-i\left(kx-\omega t\right)}}
k
{\displaystyle \ k}
k
=
2
π
λ
{\displaystyle \ k={\frac {2\pi }{\lambda }}}
k
∈
[
k
1
,
k
2
]
{\displaystyle \ k\in [k_{1},k_{2}]}
y
¯
(
x
,
t
)
=
A
k
2
−
k
1
∫
k
2
k
1
d
k
⏟
∗
⋅
e
i
(
k
x
−
ω
t
)
⏟
∗
∗
{\displaystyle \ {\bar {y}}(x,t)={\frac {A}{k_{2}-k_{1}}}\underbrace {\int \limits _{k_{2}}^{k_{1}}dk} _{*}\cdot \underbrace {e^{i\left(kx-\omega t\right)}} _{**}}
∗
{\displaystyle \ *}
∗
∗
{\displaystyle \ **}
⇒
y
¯
(
x
,
t
)
=
A
k
2
−
k
1
⋅
e
−
i
ω
t
∫
k
2
k
1
d
k
⋅
e
i
k
x
=
{\displaystyle \ \Rightarrow {\bar {y}}(x,t)={\frac {A}{k_{2}-k_{1}}}\cdot e^{-i\omega t}\int \limits _{k_{2}}^{k_{1}}dk\cdot e^{ikx}=}
=
A
k
2
−
k
1
⋅
e
−
i
ω
t
⋅
1
i
x
(
e
i
k
2
x
−
e
i
k
1
x
)
=
{\displaystyle \ ={\frac {A}{k_{2}-k_{1}}}\cdot e^{-i\omega t}\cdot {\frac {1}{ix}}\left(e^{ik_{2}x}-e^{ik_{1}x}\right)=}
=
A
⋅
e
i
ω
t
k
2
−
k
1
1
i
x
e
i
(
k
1
+
k
2
2
)
x
(
e
i
k
2
−
k
1
2
x
−
e
−
i
k
2
−
k
1
2
x
)
⏟
∗
{\displaystyle \ ={\frac {A\cdot e^{i\omega t}}{k_{2}-k_{1}}}{\frac {1}{ix}}e^{i\left({\frac {k_{1}+k_{2}}{2}}\right)x}\underbrace {\left(e^{i{\frac {k_{2}-k_{1}}{2}}x}-e^{-i{\frac {k_{2}-k_{1}}{2}}x}\right)} _{*}}
∗
=
(
e
i
k
2
−
k
1
2
x
−
e
−
i
k
2
−
k
1
2
x
)
=
2
i
sin
(
k
2
−
k
1
2
)
{\displaystyle \ *=\left(e^{i{\frac {k_{2}-k_{1}}{2}}x}-e^{-i{\frac {k_{2}-k_{1}}{2}}x}\right)=2i\sin \left({\frac {k_{2}-k_{1}}{2}}\right)}
Δ
K
=
k
2
−
k
1
{\displaystyle \ \Delta K=k_{2}-k_{1}}
y
¯
(
x
,
t
)
=
A
⋅
e
−
i
ω
t
e
i
(
k
1
+
k
2
)
2
x
sin
(
1
2
Δ
k
x
)
1
2
Δ
k
x
{\displaystyle \ {\bar {y}}(x,t)=A\cdot e^{-i\omega t}e^{{\frac {i\left(k_{1}+k_{2}\right)}{2}}x}{\frac {\sin \left({\frac {1}{2}}\Delta kx\right)}{{\frac {1}{2}}\Delta kx}}}
⇒
y
(
x
,
t
)
=
R
e
(
y
^
(
x
,
t
)
)
=
sin
(
1
2
Δ
k
x
)
1
2
Δ
k
x
⋅
cos
(
k
1
+
k
2
2
x
−
ω
t
)
{\displaystyle \ \Rightarrow y(x,t)=Re\left({\hat {y}}(x,t)\right)={\frac {\sin \left({\frac {1}{2}}\Delta kx\right)}{{\frac {1}{2}}\Delta kx}}\cdot \cos \left({\frac {k_{1}+k_{2}}{2}}x-\omega t\right)}
x
2
−
x
1
=
Δ
x
{\displaystyle \ x_{2}-x_{1}=\Delta x}
Δ
k
{\displaystyle \ \Delta k}
sin
(
1
2
Δ
k
x
)
=
0
{\displaystyle \ \sin \left({\frac {1}{2}}\Delta kx\right)=0}
1
2
Δ
k
x
=
n
π
⇐
{\displaystyle \ {\frac {1}{2}}\Delta kx=n\pi \Leftarrow }
n
∈
Z
{\displaystyle \ n\in \mathbb {Z} }
x
1
{\displaystyle \ x_{1}}
1
2
Δ
k
x
1
=
−
π
{\displaystyle \ {\frac {1}{2}}\Delta kx_{1}=-\pi }
x
2
{\displaystyle \ x_{2}}
1
2
Δ
k
x
2
=
+
π
{\displaystyle \ {\frac {1}{2}}\Delta kx_{2}=+\pi }
1
2
Δ
k
Δ
x
=
2
π
⇐
1
2
Δ
k
(
x
2
−
x
1
)
=
2
π
{\displaystyle \ {\frac {1}{2}}\Delta k\Delta x=2\pi \Leftarrow {\frac {1}{2}}\Delta k\left(x_{2}-x_{1}\right)=2\pi }
וקיבלנו:
Δ
x
Δ
k
=
4
π
{\displaystyle \ \Delta x\Delta k=4\pi }
כלומר: אם נרצה לקבל
Δ
x
{\displaystyle \ \Delta x}
Δ
k
{\displaystyle \ \Delta k}
עריכה
כאמור, אסכמת (אינטגרל) של גלים מישוריים בתחום התדרים
[
k
1
,
k
2
]
{\displaystyle \ \left[k_{1},k_{2}\right]}
ψ
(
x
,
t
)
=
A
k
1
−
k
2
∫
k
1
k
2
d
k
⋅
e
i
(
k
x
−
ω
t
)
{\displaystyle \ \psi (x,t)={\frac {A}{k_{1}-k_{2}}}\int \limits _{k_{1}}^{k_{2}}dk\cdot e^{i\left(kx-\omega t\right)}}
k
∈
[
k
1
,
k
2
]
{\displaystyle \ k\in \left[k_{1},k_{2}\right]}
A
k
2
−
k
1
{\displaystyle \ {\frac {A}{k_{2}-k_{1}}}}
k
∈
(
−
∞
,
∞
)
{\displaystyle \ k\in \left(-\infty ,\infty \right)}
g
(
k
)
{\displaystyle \ g\left(k\right)}
g
(
k
)
{\displaystyle \ g\left(k\right)}
k
{\displaystyle \ k}
k
{\displaystyle \ k}
k
{\displaystyle \ k}
g
(
k
)
{\displaystyle \ g\left(k\right)}
k
{\displaystyle \ k}
בתחילת ההרצאה הקודמת .
1
2
π
{\displaystyle \ {\frac {1}{\sqrt {2\pi }}}}
ψ
(
x
,
t
)
=
1
2
π
∫
−
∞
∞
d
k
⋅
g
(
k
)
⋅
e
i
(
k
x
−
ω
t
)
{\displaystyle \ \psi (x,t)={\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\cdot g(k)\cdot e^{i\left(kx-\omega t\right)}}
t
=
o
{\displaystyle \ t=o}
ψ
(
x
,
t
=
0
)
=
ψ
(
x
,
0
)
−
1
2
π
∫
−
∞
∞
d
k
⋅
g
(
k
)
⋅
e
i
k
x
{\displaystyle \ \psi (x,t=0)=\psi (x,0)-{\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\cdot g(k)\cdot e^{ikx}}
g
(
k
)
=
1
2
π
∫
−
∞
∞
d
x
⋅
ψ
(
x
,
0
)
e
−
i
k
x
{\displaystyle \ g\left(k\right)={\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dx\cdot \psi (x,0)e^{-ikx}}
הפונקציה
ψ
(
x
,
0
)
{\displaystyle \ \psi (x,0)}
g
(
k
)
{\displaystyle \ g(k)}
g
(
k
)
=
|
g
(
k
)
|
⋅
e
i
α
(
k
)
{\displaystyle \ g(k)=\left|g(k)\right|\cdot e^{i\alpha (k)}}
Δ
k
=
∞
{\displaystyle \ \Delta k=\infty }
g
(
k
)
{\displaystyle \ g(k)}
α
(
k
)
≃
α
(
k
0
)
+
(
k
−
k
0
)
⏟
=
Δ
k
(
∂
α
∂
k
)
|
k
=
k
0
+
.
.
.
(
∗
)
{\displaystyle \ \alpha (k)\simeq \alpha \left(k_{0}\right)+\underbrace {\left(k-k_{0}\right)} _{=\Delta k}\left.\left({\frac {\partial \alpha }{\partial k}}\right)\right|_{k=k_{0}}+...(*)}
⇒
ψ
(
x
,
0
)
≃
1
2
π
∫
−
∞
∞
d
k
|
g
(
k
)
|
e
i
(
α
(
k
0
)
+
(
k
−
k
o
)
(
∂
α
∂
k
)
|
k
=
k
0
)
e
i
k
x
{\displaystyle \ \Rightarrow \psi \left(x,0\right)\simeq {\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(\alpha \left(k_{0}\right)+\left(k-k_{o}\right)\left.\left({\frac {\partial \alpha }{\partial k}}\right)\right|_{k=k_{0}}\right)}e^{ikx}}
נסמן:
x
0
≡
−
(
∂
α
∂
k
)
|
k
=
k
0
{\displaystyle \ x_{0}\equiv -\left.\left({\frac {\partial \alpha }{\partial k}}\right)\right|_{k=k_{0}}}
ψ
(
x
,
0
)
≃
1
2
π
∫
−
∞
∞
d
k
|
g
(
k
)
|
e
i
(
α
(
k
0
)
+
(
k
−
k
0
)
x
0
+
k
x
)
=
{\displaystyle \ \psi \left(x,0\right)\simeq {\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(\alpha \left(k_{0}\right)+\left(k-k_{0}\right)x_{0}+kx\right)}=}
=
1
2
π
∫
−
∞
∞
d
k
|
g
(
k
)
|
e
i
(
α
(
k
0
)
+
k
x
0
−
k
0
x
0
+
k
x
)
=
{\displaystyle \ ={\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(\alpha \left(k_{0}\right)+kx_{0}-k_{0}x_{0}+kx\right)}=}
1
2
π
∫
−
∞
∞
d
k
|
g
(
k
)
|
e
i
(
α
(
k
0
)
+
k
x
0
−
k
0
x
0
+
k
x
+
k
0
x
−
k
0
x
)
=
{\displaystyle \ {\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(\alpha \left(k_{0}\right)+kx_{0}-k_{0}x_{0}+kx+k_{0}x-k_{0}x\right)}=}
1
2
π
∫
−
∞
∞
d
k
|
g
(
k
)
|
e
i
(
α
(
k
0
)
+
k
0
x
)
+
i
(
k
x
0
−
k
0
x
0
+
k
x
−
k
0
x
)
{\displaystyle \ {\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(\alpha \left(k_{0}\right)+k_{0}x\right)+i\left(kx_{0}-k_{0}x_{0}+kx-k_{0}x\right)}}
α
(
k
0
)
{\displaystyle \ \alpha \left(k_{0}\right)}
k
0
x
{\displaystyle \ k_{0}x}
ψ
(
x
,
0
)
=≃
1
2
π
e
i
α
(
k
0
)
+
k
0
x
∫
−
∞
∞
d
k
|
g
(
k
)
|
e
i
(
k
−
k
0
)
(
x
−
x
0
)
{\displaystyle \ \psi \left(x,0\right)=\simeq {\frac {1}{\sqrt {2\pi }}}e^{i\alpha \left(k_{0}\right)+k_{0}x}\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(k-k_{0}\right)\left(x-x_{0}\right)}}
A
=
∫
−
∞
∞
d
k
|
g
(
k
)
|
e
i
(
k
−
k
0
)
(
x
−
x
0
)
{\displaystyle \ A=\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(k-k_{0}\right)\left(x-x_{0}\right)}}
נסמן: ., ונרצה ללמוד משהו על הקשר בינו לבין (שהוא, כזכור, טווח בתדירויות שלנו). לשם כך, נבדוק מתי הסינוס מתאפס: 
![{\displaystyle \ k\in [k_{1},k_{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0575acb821bbf4cf6d1a484113352c46e3e090ff)
![{\displaystyle \ \left[k_{1},k_{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c594d1cc1c823d4d2ab131aaa400b280304c3a5a)
![{\displaystyle \ k\in \left[k_{1},k_{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0213db827413a1027cfa403cb9c0b50bfa06eb51)
