单位冲激函数性质
单位冲激函数筛选性质取样性质尺度性质证明根据冲激函数尺度性质证明cos(w0t)\cos (w_0t)cos(w0t)的傅里叶变换筛选性质设信号s(t)\displaystyle s\left( t \right)s(t)是一个在t=t0t = {t_0}t=t0处连续的函数，则有s(t)δ(t−t0)=s(t0)δ(t−t0)s\displaystyle \left( t \right)\
筛选性质
设信号
s
(
t
)
\displaystyle s\left( t \right)
s(t)是一个在
t
=
t
0
t = {t_0}
t=t0处连续的函数，则有
s
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t
)
δ
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t
−
t
0
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=
s
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0
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δ
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t
−
t
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)
s\displaystyle \left( t \right)\delta (t - {t_0}) = s\left( {{t_0}} \right)\delta (t - {t_0})
s(t)δ(t−t0)=s(t0)δ(t−t0)
取样性质
设信号
s
(
t
)
s\left( t \right)
s(t)是一个在
t
=
t
0
t = {t_0}
t=t0处连续的函数，则有
∫
−
∞
+
∞
s
(
t
)
δ
(
t
−
t
0
)
d
t
=
s
(
t
0
)
\displaystyle \int_{ - \infty }^{ + \infty } {s\left( t \right)} \delta (t - {t_0})dt = s({t_0})
∫−∞+∞s(t)δ(t−t0)dt=s(t0)
特别地，当
t
0
=
0
{t_0} = 0
t0=0时
∫
−
∞
+
∞
s
(
t
)
δ
(
t
)
d
t
=
s
(
0
)
\int_{ - \infty }^{ + \infty } {s\left( t \right)} \delta (t)dt = s(0)
∫−∞+∞s(t)δ(t)dt=s(0)
尺度性质证明
由上图可知矩形的面积如下：
S
[
r
e
c
t
(
t
)
]
=
1
S
[
r
e
c
t
(
a
t
)
]
=
1
∣
a
∣
\begin{aligned} \displaystyle S[rect(t)] &= 1\\ S[rect(at)] &= \frac{1}{{|a|}} \end{aligned}
S[rect(t)]S[rect(at)]=1=∣a∣1
当
τ
→
0
\tau \to {\rm{0}}
τ→0时，有
lim
τ
→
0
r
e
c
t
(
t
)
=
δ
(
t
)
lim
τ
→
0
r
e
c
t
(
a
t
)
=
1
∣
a
∣
δ
(
t
)
\begin{aligned} \mathop {\lim }\limits_{\tau \to {\rm{0}}} rect(t) &= \delta (t)\\ \displaystyle \mathop {\lim }\limits_{\tau \to {\rm{0}}} rect(at) &= \frac{1}{{|a|}}\delta (t) \end{aligned}
τ→0limrect(t)τ→0limrect(at)=δ(t)=∣a∣1δ(t)
证明：
δ
(
a
t
−
b
)
=
1
∣
a
∣
δ
(
t
−
b
a
)
\displaystyle \delta (at - b) = \frac{1}{{|a|}}\delta (t - \frac{b}{a})
δ(at−b)=∣a∣1δ(t−ab)
当
a
>
0
a > 0
a>0时，令
a
t
−
b
=
x
at - b = x
at−b=x
∫
−
∞
+
∞
s
(
t
)
δ
(
a
t
−
b
)
d
t
=
1
a
∫
−
∞
+
∞
s
(
1
a
x
+
b
a
)
δ
(
x
)
d
x
=
1
a
s
(
b
a
)
\begin{aligned} \displaystyle \int_{ - \infty }^{ + \infty } {s\left( t \right)} \delta (at - b)dt &= \frac{1}{a}\int_{ - \infty }^{ + \infty } {s\left( {\frac{1}{a}x + \frac{b}{a}} \right)} \delta (x)dx\\ & = \frac{1}{a}s\left( {\frac{b}{a}} \right) \displaystyle \end{aligned}
∫−∞+∞s(t)δ(at−b)dt=a1∫−∞+∞s(a1x+ab)δ(x)dx=a1s(ab)
根据取样性质
∫
−
∞
+
∞
1
a
s
(
t
)
δ
(
t
−
b
a
)
d
t
=
1
a
s
(
b
a
)
\int_{ - \infty }^{ + \infty } {\frac{1}{a}s\left( t \right)} \delta (t - \frac{b}{a})dt = \frac{1}{a}s\left( {\frac{b}{a}} \right)
∫−∞+∞a1s(t)δ(t−ab)dt=a1s(ab)
当
a
<
0
a < 0
a<0时，令
−
∣
a
∣
t
−
b
=
x
-|a|t - b = x
−∣a∣t−b=x
{
t
:
−
∞
→
+
∞
x
:
+
∞
→
−
∞
\left\{ \begin{array}{l} t: - \infty \to + \infty \\ x: + \infty \to - \infty \end{array} \right.
{t:−∞→+∞x:+∞→−∞
∫ − ∞ + ∞ s ( t ) δ ( a t − b ) d t = − 1 ∣ a ∣ ∫ + ∞ − ∞ s ( − 1 ∣ a ∣ x − b ∣ a ∣ ) δ ( x ) d x = 1 ∣ a ∣ ∫ − ∞ + ∞ s ( − 1 ∣ a ∣ x − b ∣ a ∣ ) δ ( x ) d x = 1 ∣ a ∣ s ( − b ∣ a ∣ ) \begin{aligned} \displaystyle {\int_{ - \infty }^{ + \infty } {s\left( t \right)} \delta (at - b)dt }&={ - \frac{1}{{|a|}}\int_{ + \infty }^{ - \infty } {s\left( { - \frac{1}{{|a|}}x - \frac{b}{{|a|}}} \right)} \delta (x)dx}\\ \displaystyle &= \frac{1}{{|a|}}\int_{ - \infty }^{ + \infty } {s\left( { - \frac{1}{{|a|}}x - \frac{b}{{|a|}}} \right)} \delta (x)dx\\ \displaystyle &= \frac{1}{{|a|}}s\left( { - \frac{b}{{|a|}}} \right) \end{aligned} ∫−∞+∞s(t)δ(at−b)dt=−∣a∣1∫+∞−∞s(−∣a∣1x−∣a∣b)δ(x)dx=∣a∣1∫−∞+∞s(−∣a∣1x−∣a∣b)δ(x)dx=∣a∣1s(−∣a∣b)
同样根据取样性质，且
∣
a
∣
=
−
a
|a| = - a
∣a∣=−a
∫
−
∞
+
∞
1
∣
a
∣
s
(
t
)
δ
(
t
+
b
∣
a
∣
)
d
t
=
1
∣
a
∣
s
(
−
b
∣
a
∣
)
\int_{ - \infty }^{ + \infty } {\frac{1}{{|a|}}s\left( t \right)} \delta (t + \frac{b}{{|a|}})dt = \frac{1}{{|a|}}s\left( { - \frac{b}{{|a|}}} \right)
∫−∞+∞∣a∣1s(t)δ(t+∣a∣b)dt=∣a∣1s(−∣a∣b)
得证。
根据冲激函数尺度性质证明 cos ( w 0 t ) \cos (w_0t) cos(w0t)的傅里叶变换
根据欧拉公式
cos
(
w
t
)
=
1
2
(
e
j
w
t
+
e
−
j
w
t
)
\cos (wt) = \frac{1}{2}({e^{jwt}} + {e^{ - jwt}})
cos(wt)=21(ejwt+e−jwt)
其Fourier变换为
G
(
f
)
=
∫
+
∞
−
∞
cos
(
w
0
t
)
e
−
j
w
t
d
t
=
1
2
∫
+
∞
−
∞
(
e
j
w
0
t
+
e
−
j
w
0
t
)
e
−
j
w
t
d
t
=
1
2
∫
+
∞
−
∞
(
e
−
j
2
π
(
f
−
f
0
)
t
+
e
−
j
2
π
(
f
+
f
0
)
t
)
d
t
=
1
2
[
δ
(
f
−
f
0
)
+
δ
(
f
+
f
0
)
]
\begin{aligned} \displaystyle G(f) &= \int_{ + \infty }^{ - \infty } {\cos ({w_0}t)} {e^{ - jwt}}dt\\ \displaystyle & = \frac{1}{2}\int_{ + \infty }^{ - \infty } {({e^{j{w_0}t}} + {e^{ - j{w_0}t}})} {e^{ - jwt}}dt\\ \displaystyle & = \frac{1}{2}\int_{ + \infty }^{ - \infty } {({e^{ - j2\pi (f - {f_0})t}} + {e^{ - j2\pi (f + {f_0})t}})} dt\\ \displaystyle & = \frac{1}{2}[\delta (f - {f_0}) + \delta (f + {f_0})] \displaystyle \end{aligned}
G(f)=∫+∞−∞cos(w0t)e−jwtdt=21∫+∞−∞(ejw0t+e−jw0t)e−jwtdt=21∫+∞−∞(e−j2π(f−f0)t+e−j2π(f+f0)t)dt=21[δ(f−f0)+δ(f+f0)]
根据冲激函数尺度性质
δ
(
w
−
w
0
)
=
δ
[
2
π
(
f
−
f
0
)
]
=
1
2
π
δ
[
(
f
−
f
0
)
]
\begin{aligned} \displaystyle \delta (w - {w_0}) &= \delta [2\pi (f - {f_0})]\\ & = \frac{1}{{2\pi }}\delta [(f - {f_0})] \end{aligned}
δ(w−w0)=δ[2π(f−f0)]=2π1δ[(f−f0)]
所以
G
(
w
)
=
π
[
δ
(
w
−
w
0
)
+
δ
(
w
+
w
0
)
]
G(w) = \pi [\delta (w - {w_0}) + \delta (w + {w_0})]
G(w)=π[δ(w−w0)+δ(w+w0)]
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