10/10,13,16,17 - Fourier Transform Properties: Difference between revisions
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|Superposition, Not X(f_0)H(f_0)?
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===Relation to the Fourier Series===
===Relation to the Fourier Series===
Revision as of 05:46, 24 November 2008
Contents
1
Properties of the Fourier Transform
1.1
Linearity
1.2
Time Invariance (Delay)
1.3
Frequency Shifting
1.4
Double Sideband Modulation
1.5
Differentiation in Time
1.6
The Game (frequency domain)
1.7
The Game (Time Domain??)
1.8
Relation to the Fourier Series
Properties of the Fourier Transform
Linearity
F
[
a
x
(
t
)
+
b
x
(
t
)
]
{\displaystyle F\left[a\,x(t)+b\,x(t)\right]}
=
∫
−
∞
∞
[
a
x
(
t
)
+
b
x
(
t
)
]
e
−
j
2
π
f
t
d
t
{\displaystyle =\int _{-\infty }^{\infty }\left[a\,x(t)+b\,x(t)\right]e^{-j\,2\pi f\,t}\,dt}
=
a
∫
−
∞
∞
x
(
t
)
e
−
j
2
π
f
t
d
t
+
b
∫
−
∞
∞
x
(
t
)
e
−
j
2
π
f
t
d
t
{\displaystyle =a\int _{-\infty }^{\infty }\,x(t)\,e^{-j\,2\pi f\,t}\,dt+b\int _{-\infty }^{\infty }\,x(t)\,e^{-j\,2\pi f\,t}\,dt}
=
a
F
[
x
(
t
)
]
+
b
F
[
x
(
t
)
]
{\displaystyle =a\,F[x(t)]+b\,F[x(t)]}
Time Invariance (Delay)
F
[
x
(
t
−
t
0
)
]
{\displaystyle F[x(t-t_{0})]\,\!}
=
∫
−
∞
∞
x
(
t
−
t
0
)
e
−
j
2
π
f
t
d
t
{\displaystyle =\int _{-\infty }^{\infty }x(t-t_{0})\,e^{-j\,2\pi f\,t}\,dt}
Let
u
=
t
−
t
0
{\displaystyle u=t-t_{0}\,\!}
and
d
u
=
d
t
{\displaystyle du=dt\,\!}
=
∫
−
∞
∞
x
(
u
)
e
−
j
2
π
f
(
u
+
t
0
)
d
u
{\displaystyle =\int _{-\infty }^{\infty }x(u)\,e^{-j\,2\pi f\,(u+t_{0})}\,du}
=
e
−
j
2
π
f
t
0
∫
−
∞
∞
x
(
u
)
e
−
j
2
π
f
u
d
u
{\displaystyle =e^{-j\,2\pi f\,t_{0}}\int _{-\infty }^{\infty }x(u)\,e^{-j\,2\pi f\,u}\,du}
=
e
−
j
2
π
f
t
0
F
[
x
(
t
)
]
{\displaystyle =e^{-j\,2\pi f\,t_{0}}\,F[x(t)]}
Frequency Shifting
F
[
e
j
2
π
f
t
x
(
t
)
]
{\displaystyle F\left[e^{j\,2\pi f\,t}x(t)\right]}
=
∫
−
∞
∞
[
e
j
2
π
f
0
t
x
(
t
)
]
e
−
j
2
π
f
t
d
t
{\displaystyle =\int _{-\infty }^{\infty }\left[e^{j\,2\pi f_{0}\,t}x(t)\right]e^{-j\,2\pi f\,t}\,dt}
=
∫
−
∞
∞
x
(
t
)
e
−
j
2
π
(
f
−
f
0
)
t
d
t
{\displaystyle =\int _{-\infty }^{\infty }x(t)\,e^{-j\,2\pi (f-f_{0})\,t}\,dt}
=
X
(
f
−
f
0
)
{\displaystyle =X(f-f_{0})\,\!}
Double Sideband Modulation
F
[
c
o
s
(
2
π
f
0
t
)
⋅
x
(
t
)
]
{\displaystyle F[cos(2\pi f_{0}\,t)\cdot x(t)]}
=
∫
−
∞
∞
e
j
2
π
f
0
t
+
e
−
j
2
π
f
0
t
2
x
(
t
)
e
−
j
2
π
f
t
d
t
{\displaystyle =\int _{-\infty }^{\infty }{\frac {e^{j\,2\pi f_{0}\,t}+e^{-j\,2\pi f_{0}\,t}}{2}}x(t)\,e^{-j\,2\pi f\,t}\,dt}
=
1
2
∫
−
∞
∞
x
(
t
)
[
e
−
j
2
π
(
f
−
f
0
)
t
+
e
−
j
2
π
(
f
+
f
0
)
t
]
d
t
{\displaystyle ={\frac {1}{2}}\int _{-\infty }^{\infty }x(t)\left[e^{-j\,2\pi (f-f_{0})\,t}+e^{-j\,2\pi (f+f_{0})\,t}\right]\,dt}
=
1
2
X
(
f
−
f
0
)
+
1
2
X
(
f
+
f
0
)
{\displaystyle ={\frac {1}{2}}X(f-f_{0})+{\frac {1}{2}}X(f+f_{0})}
Differentiation in Time
x
(
t
)
{\displaystyle x(t)\,\!}
=
F
−
1
[
X
(
f
)
]
{\displaystyle =F^{-1}\left[X(f)\right]}
F
[
d
x
d
t
]
{\displaystyle F\left[{\frac {dx}{dt}}\right]}
=
F
[
d
d
t
F
−
1
[
X
(
f
)
]
]
{\displaystyle =F\left[{\frac {d}{dt}}F^{-1}\left[X(f)\right]\right]}
=
F
[
d
d
t
∫
−
∞
∞
X
(
f
)
e
j
2
π
f
t
d
f
]
{\displaystyle =F\left[{\frac {d}{dt}}\int _{-\infty }^{\infty }X(f)\,e^{j\,2\pi f\,t}\,df\right]}
=
F
[
∫
−
∞
∞
j
2
π
f
X
(
f
)
e
j
2
π
f
t
d
f
]
{\displaystyle =F\left[\int _{-\infty }^{\infty }j\,2\pi fX(f)e^{j\,2\pi f\,t}\,df\right]}
=
F
[
j
2
π
f
F
−
1
[
X
(
f
)
]
]
{\displaystyle =F\left[j\,2\pi f\,F^{-1}[X(f)]\right]}
=
j
2
π
f
X
(
f
)
{\displaystyle =j\,2\pi f\,X(f)}
Thus
d
x
d
t
{\displaystyle {\frac {dx}{dt}}}
is a linear filter with transfer function
j
2
π
f
{\displaystyle j\,2\pi f}
The Game (frequency domain)
You can play the game in the frequency or time domain, but not both at the same time
Then how can you use the Fourier Transform, but can't build up to it?
Input
LTI System
Output
Reason
δ
(
t
)
{\displaystyle \delta (t)\,\!}
⟹
{\displaystyle \Longrightarrow }
h
(
t
)
{\displaystyle h(t)\,\!}
Given
δ
(
t
)
e
−
j
2
π
f
t
{\displaystyle \delta (t)\,e^{-j\,2\,\pi f\,t}}
⟹
{\displaystyle \Longrightarrow }
h
(
t
)
e
−
j
2
π
f
t
{\displaystyle h(t)\,e^{-j\,2\,\pi f\,t}}
Proportionality
∫
−
∞
∞
δ
(
t
)
e
−
j
2
π
f
t
d
t
=
F
[
δ
(
t
)
]
=
1
{\displaystyle \int _{-\infty }^{\infty }\delta (t)\,e^{-j\,2\,\pi f\,t}\,dt=F[\delta (t)]=1}
⟹
{\displaystyle \Longrightarrow }
∫
−
∞
∞
h
(
t
)
e
−
j
2
π
f
t
d
t
=
F
[
h
(
t
)
]
=
H
(
f
)
{\displaystyle \int _{-\infty }^{\infty }h(t)\,e^{-j\,2\,\pi f\,t}\,dt=F[h(t)]=H(f)}
Superposition
∫
−
∞
∞
δ
(
t
−
λ
)
e
−
j
2
π
f
t
d
t
=
F
[
δ
(
t
−
λ
)
]
=
1
⋅
e
−
j
2
π
f
λ
{\displaystyle \int _{-\infty }^{\infty }\delta (t-\lambda )\,e^{-j\,2\,\pi f\,t}\,dt=F[\delta (t-\lambda )]=1\cdot e^{-j\,2\,\pi f\,\lambda }}
⟹
{\displaystyle \Longrightarrow }
H
(
f
)
⋅
e
−
j
2
π
f
λ
{\displaystyle H(f)\cdot e^{-j\,2\,\pi f\,\lambda }}
Time Invariance
x
(
λ
)
⋅
1
⋅
e
−
j
2
π
f
λ
{\displaystyle x(\lambda )\cdot 1\cdot e^{-j\,2\,\pi f\,\lambda }}
⟹
{\displaystyle \Longrightarrow }
x
(
λ
)
⋅
H
(
f
)
⋅
e
−
j
2
π
f
λ
{\displaystyle x(\lambda )\cdot H(f)\cdot e^{-j\,2\,\pi f\,\lambda }}
Proportionality
∫
−
∞
∞
x
(
λ
)
⋅
1
⋅
e
j
2
π
f
λ
d
λ
=
X
(
F
)
{\displaystyle \int _{-\infty }^{\infty }x(\lambda )\cdot 1\cdot e^{j\,2\,\pi f\,\lambda }\,d\lambda =X(F)}
⟹
{\displaystyle \Longrightarrow }
∫
−
∞
∞
x
(
λ
)
⋅
H
(
f
)
⋅
e
j
2
π
f
λ
d
λ
=
X
(
F
)
H
(
f
)
{\displaystyle \int _{-\infty }^{\infty }x(\lambda )\cdot H(f)\cdot e^{j\,2\,\pi f\,\lambda }\,d\lambda =X(F)\,H(f)}
Superposition
Having trouble seeing
F
[
x
(
t
)
∗
h
(
t
)
]
=
X
(
f
)
⋅
H
(
f
)
{\displaystyle F\left[x(t)*h(t)\right]=X(f)\cdot H(f)}
Since we were dealing in the frequency domain, is that the reason why multiplying one side did not result in a convolution on the other?
The Game (Time Domain??)
Input
LTI System
Output
Reason
1
⋅
e
j
2
π
f
0
t
{\displaystyle 1\cdot e^{j\,2\,\pi f_{0}\,t}}
⟹
{\displaystyle \Longrightarrow }
h
(
t
)
∗
e
j
2
π
f
0
t
=
e
j
2
π
f
0
t
∗
h
(
t
)
{\displaystyle h(t)*e^{j\,2\,\pi f_{0}\,t}=e^{j\,2\,\pi f_{0}\,t}*h(t)}
Proportionality
∫
−
∞
∞
h
(
λ
)
⋅
e
j
2
π
f
0
(
t
−
λ
)
d
λ
{\displaystyle \int _{-\infty }^{\infty }h(\lambda )\cdot e^{j\,2\,\pi f_{0}\,(t-\lambda )}\,d\lambda }
Why d lambda instead of dt?
e
j
2
π
f
0
λ
∫
−
∞
∞
h
(
λ
)
⋅
e
−
j
2
π
f
0
λ
d
λ
{\displaystyle e^{j\,2\,\pi f_{0}\,\lambda }\int _{-\infty }^{\infty }h(\lambda )\cdot e^{-j\,2\,\pi f_{0}\,\lambda }\,d\lambda }
e
j
2
π
f
0
t
H
(
f
0
)
{\displaystyle e^{j\,2\,\pi f_{0}\,t}\,H(f_{0})}
X
(
f
0
)
⋅
e
j
2
π
f
0
t
{\displaystyle X(f_{0})\cdot e^{j\,2\,\pi f_{0}\,t}}
⟹
{\displaystyle \Longrightarrow }
X
(
f
0
)
⋅
e
j
2
π
f
0
t
H
(
f
0
)
{\displaystyle X(f_{0})\cdot e^{j\,2\,\pi f_{0}\,t}\,H(f_{0})}
Proportionality, Why isn't this a convolution?
∫
−
∞
∞
X
(
f
0
)
⋅
e
j
2
π
f
0
t
d
f
0
=
x
(
t
)
{\displaystyle \int _{-\infty }^{\infty }X(f_{0})\cdot e^{j\,2\,\pi f_{0}\,t}\,df_{0}=x(t)}
⟹
{\displaystyle \Longrightarrow }
∫
−
∞
∞
X
(
f
0
)
H
(
f
0
)
⋅
e
j
2
π
f
0
t
d
f
0
=
F
−
1
[
X
(
f
)
H
(
f
)
]
{\displaystyle \int _{-\infty }^{\infty }X(f_{0})H(f_{0})\cdot e^{j\,2\,\pi f_{0}\,t}\,df_{0}=F^{-1}\left[X(f)H(f)\right]}
Superposition, Not X(f_0)H(f_0)?
Relation to the Fourier Series
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