HW 05: Difference between revisions
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Line 65:
|<math>=\alpha_{fT}\,\!</math>
|<math>=\alpha_{fT}\,\!</math>
|}
|}
Is the last problem done correctly?
Revision as of 02:15, 22 November 2008
Find the following Fourier Transforms
F
[
e
j
ω
0
t
]
{\displaystyle F[e^{j\omega _{0}t}]}
F
[
cos
ω
0
t
]
{\displaystyle F[\cos {\omega _{0}t}]\,\!}
F
[
∑
−
∞
∞
α
n
e
j
2
π
n
t
/
T
]
{\displaystyle F[\sum _{-\infty }^{\infty }\alpha _{n}e^{j2\pi nt/T}]}
F
[
sin
ω
0
t
]
{\displaystyle F[\sin {\omega _{0}t}]\,\!}
Solutions
F
[
e
j
ω
0
t
]
{\displaystyle F[e^{j\omega _{0}t}]}
=
∫
−
∞
∞
e
j
ω
0
t
e
−
j
ω
t
d
t
{\displaystyle =\int _{-\infty }^{\infty }e^{j\omega _{0}t}e^{-j\omega t}dt}
=
∫
−
∞
∞
e
j
(
ω
0
−
ω
)
t
d
t
{\displaystyle =\int _{-\infty }^{\infty }e^{j(\omega _{0}-\omega )t}dt}
=
2
π
[
1
2
π
∫
−
∞
∞
e
j
(
ω
0
−
ω
)
t
d
t
]
{\displaystyle =2\pi \left[{\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{j(\omega _{0}-\omega )t}dt\right]}
=
2
π
δ
(
ω
0
−
ω
)
{\displaystyle =2\pi \delta (\omega _{0}-\omega )\,\!}
F
[
cos
ω
0
t
]
{\displaystyle F[\cos {\omega _{0}t}]\,\!}
=
∫
−
∞
∞
e
j
ω
0
t
+
e
−
j
ω
0
t
2
e
−
j
ω
t
d
t
{\displaystyle =\int _{-\infty }^{\infty }{\frac {e^{j\omega _{0}t}+e^{-j\omega _{0}t}}{2}}e^{-j\omega t}dt}
=
1
2
∫
−
∞
∞
(
e
j
ω
0
t
+
e
−
j
ω
0
t
)
2
e
−
j
ω
t
d
t
{\displaystyle ={\frac {1}{2}}\int _{-\infty }^{\infty }\left(e^{j\omega _{0}t}+e^{-j\omega _{0}t}\right)2e^{-j\omega t}dt}
=
1
2
∫
−
∞
∞
[
2
e
j
(
ω
0
−
ω
)
t
+
2
e
−
j
(
ω
0
+
ω
)
t
]
d
t
{\displaystyle ={\frac {1}{2}}\int _{-\infty }^{\infty }\left[2e^{j(\omega _{0}-\omega )t}+2e^{-j(\omega _{0}+\omega )t}\right]dt}
=
2
π
[
1
2
π
∫
−
∞
∞
(
e
j
(
ω
0
−
ω
)
t
+
e
−
j
(
ω
0
+
ω
)
t
)
d
t
]
{\displaystyle =2\pi \left[{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\left(e^{j(\omega _{0}-\omega )t}+e^{-j(\omega _{0}+\omega )t}\right)\,dt\right]}
=
2
π
δ
(
ω
0
−
ω
)
+
2
π
δ
(
ω
0
+
ω
)
{\displaystyle =2\pi \delta (\omega _{0}-\omega )+2\pi \delta (\omega _{0}+\omega )\,\!}
F
[
sin
ω
0
t
]
{\displaystyle F[\sin {\omega _{0}t}]\,\!}
=
∫
−
∞
∞
e
j
ω
0
t
−
e
−
j
ω
0
t
2
j
e
−
j
ω
t
d
t
{\displaystyle =\int _{-\infty }^{\infty }{\frac {e^{j\omega _{0}t}-e^{-j\omega _{0}t}}{2j}}e^{-j\omega t}dt}
=
1
2
j
∫
−
∞
∞
(
e
j
ω
0
t
−
e
−
j
ω
0
t
)
2
j
e
−
j
ω
t
d
t
{\displaystyle ={\frac {1}{2j}}\int _{-\infty }^{\infty }\left(e^{j\omega _{0}t}-e^{-j\omega _{0}t}\right)2je^{-j\omega t}dt}
=
∫
−
∞
∞
(
e
j
(
ω
0
−
ω
)
t
−
e
−
j
(
ω
0
+
ω
)
t
)
d
t
{\displaystyle =\int _{-\infty }^{\infty }\left(e^{j(\omega _{0}-\omega )t}-e^{-j(\omega _{0}+\omega )t}\right)dt}
=
2
π
[
1
2
π
∫
−
∞
∞
(
e
j
(
ω
0
−
ω
)
t
−
e
−
j
(
ω
0
+
ω
)
t
)
d
t
]
{\displaystyle =2\pi \left[{\frac {1}{2\pi }}\int _{-\infty }^{\infty }\left(e^{j(\omega _{0}-\omega )t}-e^{-j(\omega _{0}+\omega )t}\right)\,dt\right]}
=
2
π
δ
(
ω
0
−
ω
)
−
2
π
δ
(
ω
0
+
ω
)
{\displaystyle =2\pi \delta (\omega _{0}-\omega )-2\pi \delta (\omega _{0}+\omega )\,\!}
F
[
∑
−
∞
∞
α
n
e
j
2
π
n
t
/
T
]
{\displaystyle F[\sum _{-\infty }^{\infty }\alpha _{n}e^{j2\pi nt/T}]}
=
∫
−
∞
∞
(
∑
−
∞
∞
α
n
e
j
2
π
n
t
/
T
)
e
−
j
ω
t
d
t
{\displaystyle =\int _{-\infty }^{\infty }\left(\sum _{-\infty }^{\infty }\alpha _{n}e^{j2\pi nt/T}\right)e^{-j\omega t}dt}
=
∑
−
∞
∞
α
n
(
∫
−
∞
∞
e
j
2
π
n
t
/
T
e
−
j
2
π
f
t
d
t
)
{\displaystyle =\sum _{-\infty }^{\infty }\alpha _{n}\left(\int _{-\infty }^{\infty }e^{j2\pi nt/T}e^{-j2\pi ft}dt\right)}
=
∑
−
∞
∞
α
n
(
∫
−
∞
∞
e
j
2
π
t
(
n
T
−
f
)
d
t
)
{\displaystyle =\sum _{-\infty }^{\infty }\alpha _{n}\left(\int _{-\infty }^{\infty }e^{j2\pi t({\frac {n}{T}}-f)}dt\right)}
=
∑
−
∞
∞
α
n
δ
n
T
−
f
{\displaystyle =\sum _{-\infty }^{\infty }\alpha _{n}\delta _{{\frac {n}{T}}-f}}
=
α
f
T
{\displaystyle =\alpha _{fT}\,\!}
Is the last problem done correctly?
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