10/10,13,16,17 - Fourier Transform Properties: Difference between revisions
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*Having trouble seeing <math>F\left[x(t)*h(t)\right]=X(f)\cdot H(f)</math>
Revision as of 02:49, 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)
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)}
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