10/09 - Fourier Transform
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⟨
e
j
2
π
n
t
/
T
∣
e
j
2
π
m
t
/
T
⟩
{\displaystyle \left\langle e^{j2\pi nt/T}\mid e^{j2\pi mt/T}\right\rangle }
=
∫
−
∞
∞
e
j
2
π
n
t
/
T
e
−
j
2
π
m
t
/
T
d
t
{\displaystyle =\int _{-\infty }^{\infty }e^{j2\pi nt/T}e^{-j2\pi mt/T}\,dt}
=
∫
−
∞
∞
e
j
2
π
(
n
−
m
)
t
/
T
d
t
{\displaystyle =\int _{-\infty }^{\infty }e^{j2\pi (n-m)t/T}\,dt}
=
∫
−
T
/
2
T
/
2
e
j
2
π
(
n
−
m
)
t
/
T
d
t
{\displaystyle =\int _{-T/2}^{T/2}e^{j2\pi (n-m)t/T}\,dt}
Assuming the function is perodic with the period T
=
T
δ
m
,
n
{\displaystyle =T\delta _{m,n}\,\!}
Contents
1
Fourier Transform
2
Definitions
3
Examples
3.1
Sifting property of the delta function
Fourier Transform
Remember from
10/02 - Fourier Series
α
m
=
1
T
∫
−
T
/
2
T
/
2
x
(
t
)
e
−
j
2
π
m
t
/
T
d
t
{\displaystyle \alpha _{m}={\frac {1}{T}}\int _{-T/2}^{T/2}x(t)e^{-j2\pi mt/T}\,dt}
x
(
t
)
=
x
(
t
+
T
)
=
∑
n
=
−
∞
∞
α
m
e
j
2
π
m
/
T
{\displaystyle x(t)=x(t+T)=\sum _{n=-\infty }^{\infty }\alpha _{m}e^{j2\pi m/T}}
If we let
T
→
∞
{\displaystyle T\rightarrow \infty }
1
T
{\displaystyle {\frac {1}{T}}}
→
d
f
{\displaystyle \rightarrow df}
n
T
{\displaystyle {\frac {n}{T}}}
→
f
{\displaystyle \rightarrow f}
Remember
f
=
2
π
n
T
{\displaystyle f={\frac {2\pi n}{T}}\,\!}
T
{\displaystyle T\,\!}
→
∞
{\displaystyle \rightarrow \infty }
∑
n
=
−
∞
∞
1
T
{\displaystyle \sum _{n=-\infty }^{\infty }{\frac {1}{T}}}
→
∫
−
∞
∞
(
)
d
f
{\displaystyle \rightarrow \int _{-\infty }^{\infty }()df}
Definitions
F
[
x
(
t
)
]
{\displaystyle F[x(t)]\,\!}
=
X
(
f
)
{\displaystyle =X(f)\,\!}
=
∫
−
∞
∞
x
(
t
)
e
−
j
2
π
f
t
d
t
{\displaystyle =\int _{-\infty }^{\infty }x(t)e^{-j2\pi ft}dt}
=
⟨
x
(
t
)
∣
e
j
2
π
f
t
⟩
t
{\displaystyle =\left\langle x(t)\mid e^{j2\pi ft}\right\rangle _{t}}
F
−
1
[
x
(
t
)
]
{\displaystyle F^{-1}[x(t)]\,\!}
=
x
(
t
)
{\displaystyle =x(t)\,\!}
=
∫
−
∞
∞
X
(
f
)
e
j
2
π
f
t
d
f
{\displaystyle =\int _{-\infty }^{\infty }X(f)e^{j2\pi ft}df}
=
⟨
X
(
f
)
∣
e
−
j
2
π
f
t
⟩
f
{\displaystyle =\left\langle X(f)\mid e^{-j2\pi ft}\right\rangle _{f}}
Examples
∫
−
∞
∞
e
j
2
π
f
t
e
−
j
2
π
f
λ
d
f
{\displaystyle \int _{-\infty }^{\infty }e^{j2\pi ft}e^{-j2\pi f\lambda }df}
=
⟨
e
j
2
π
f
t
∣
e
j
2
π
f
t
⟩
f
{\displaystyle =\left\langle e^{j2\pi ft}\mid e^{j2\pi ft}\right\rangle _{f}}
=
δ
(
t
−
λ
)
{\displaystyle =\delta (t-\lambda )\,\!}
∫
−
∞
∞
e
j
2
π
t
f
e
−
j
2
π
t
f
0
d
t
{\displaystyle \int _{-\infty }^{\infty }e^{j2\pi tf}e^{-j2\pi tf_{0}}dt}
=
⟨
e
j
2
π
t
f
∣
e
j
2
π
t
f
0
⟩
t
{\displaystyle =\left\langle e^{j2\pi tf}\mid e^{j2\pi tf_{0}}\right\rangle _{t}}
=
δ
(
f
−
f
0
)
{\displaystyle =\delta (f-f_{0})\,\!}
F
−
1
[
F
[
x
(
t
)
]
]
{\displaystyle F^{-1}[F[x(t)]]\,\!}
=
∫
−
∞
∞
[
∫
−
∞
∞
x
(
λ
)
e
−
j
2
π
f
λ
d
λ
]
e
j
2
π
f
t
d
f
{\displaystyle =\int _{-\infty }^{\infty }\left[\int _{-\infty }^{\infty }x(\lambda )e^{-j2\pi f\lambda }d\lambda \right]e^{j2\pi ft}df}
=
∫
−
∞
∞
X
(
f
)
e
j
2
π
f
t
d
f
{\displaystyle =\int _{-\infty }^{\infty }X(f)e^{j2\pi ft}df}
=
x
(
t
)
{\displaystyle =x(t)\,\!}
=
∫
−
∞
∞
x
(
λ
)
∫
−
∞
∞
e
j
2
π
f
(
t
−
λ
)
d
f
d
λ
{\displaystyle =\int _{-\infty }^{\infty }x(\lambda )\int _{-\infty }^{\infty }e^{j2\pi f(t-\lambda )}dfd\lambda }
=
∫
−
∞
∞
x
(
λ
)
δ
(
t
−
λ
)
d
λ
{\displaystyle =\int _{-\infty }^{\infty }x(\lambda )\delta (t-\lambda )d\lambda }
=
x
(
t
)
{\displaystyle =x(t)\,\!}
=
∫
−
∞
∞
[
∫
−
∞
∞
x
(
λ
)
e
−
j
ω
λ
d
λ
]
e
j
ω
t
1
2
π
d
ω
{\displaystyle =\int _{-\infty }^{\infty }\left[\int _{-\infty }^{\infty }x(\lambda )e^{-j\omega \lambda }d\lambda \right]e^{j\omega t}{\frac {1}{2\pi }}d\omega }
=
∫
−
∞
∞
x
(
λ
)
[
1
2
π
∫
−
∞
∞
e
j
(
t
−
ω
)
λ
d
ω
]
d
λ
{\displaystyle =\int _{-\infty }^{\infty }x(\lambda )\left[{\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{j(t-\omega )\lambda }d\omega \right]d\lambda }
=
∫
−
∞
∞
x
(
λ
)
δ
(
t
−
ω
)
d
λ
{\displaystyle =\int _{-\infty }^{\infty }x(\lambda )\delta (t-\omega )d\lambda }
=
x
(
t
)
{\displaystyle =x(t)\,\!}
Sifting property of the delta function
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