一、向量空间背景
(1) 具有如下点内积或标量内积的实数域
R
R
R上的欧式空间
R
N
R^N
RN:
⟨
u
,
v
⟩
=
u
T
v
=
u
0
v
0
+
u
1
v
1
+
⋯
+
u
N
−
1
v
N
−
1
=
∑
i
=
0
N
−
1
u
i
v
i
\langle\boldsymbol{u}, \boldsymbol{v}\rangle=\boldsymbol{u}^{\mathrm{T}} \boldsymbol{v}=u_{0} v_{0}+u_{1} v_{1}+\cdots+u_{N-1} v_{N-1}=\sum_{i=0}^{N-1} u_{i} v_{i}
⟨u,v⟩=uTv=u0v0+u1v1+⋯+uN−1vN−1=i=0∑N−1uivi
其中, u \boldsymbol{u} u和 v \boldsymbol{v} v是 N × 1 N \times 1 N×1列向量。
(2) 具有如下内积函数的复数域 C C C上的酉空间(unitary linear space) C N C^N CN:
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v
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=
u
∗
T
v
=
∑
i
=
0
N
−
1
u
i
∗
v
i
=
<
v
,
u
>
∗
\langle\boldsymbol{u}, \boldsymbol{v}\rangle=\boldsymbol{u}^{* \mathrm{~T}} \boldsymbol{v}=\sum_{i=0}^{N-1} u_{i}^{*} v_{i}=<\boldsymbol{v}, \boldsymbol{u}>^{*}
⟨u,v⟩=u∗ Tv=i=0∑N−1ui∗vi=<v,u>∗
(3) 内积空间
C
(
[
a
,
b
]
)
C([a, b])
C([a,b]), 其中向量是区间
a
≤
x
≤
b
a≤x≤b
a≤x≤b上的连续函数, 内积函数是积分内积:
< f ( x ) , g ( x ) > = ∫ a b f ∗ ( x ) g ( x ) d x <f(x), g(x)>=\int_{a}^{b} f^{*}(x) g(x) \mathrm{d} x <f(x),g(x)>=∫abf∗(x)g(x)dx
向量
z
z
z的范数或长度为:
∥
z
∥
=
⟨
z
,
z
⟩
\|\boldsymbol{z}\|=\sqrt{\langle\boldsymbol{z}, \boldsymbol{z}\rangle}
∥z∥=⟨z,z⟩
若 z z z的范数是1, 则称 z z z是归一化的。
两个非零向量z和w之间的夹角为:
θ
=
arccos
⟨
z
,
w
⟩
∥
z
∥
∥
w
∥
\theta=\arccos \frac{\left \langle\boldsymbol{z}, \boldsymbol{w} \right\rangle }{\|\boldsymbol{z}\|\|\boldsymbol{w}\|}
θ=arccos∥z∥∥w∥⟨z,w⟩
若
⟨
z
,
w
⟩
=
0
\left \langle\boldsymbol{z}, \boldsymbol{w}\right\rangle=0
⟨z,w⟩=0,则称
z
z
z和
w
w
w是正交的(orthogonal)。即当且仅当
<
w
k
,
w
l
>
=
0
,
k
≠
l
<\boldsymbol{w}_{k}, \boldsymbol{w}_{l}>=0, \quad k \neq l
<wk,wl>=0,k=l
时, 非零向量
w
0
,
w
1
,
w
2
,
⋯
w_{0}, w_{1}, w_{2}, \cdots
w0,w1,w2,⋯是所张成内积空间的正交基。若基向量是归一化的,则它们是一个正交基,并且有
⟨
w
k
,
w
l
⟩
=
δ
k
l
=
{
0
,
k
≠
l
1
,
k
=
l
\left\langle\boldsymbol{w}_{k}, \boldsymbol{w}_{l}\right\rangle=\delta_{k l}=\left\{\begin{array}{l} 0, k \neq l \\ 1, k=l \end{array}\right.
⟨wk,wl⟩=δkl={0,k=l1,k=l
类似地, 若
<
w
~
k
,
w
l
>
=
0
,
k
≠
l
<\tilde{\boldsymbol{w}}_{k}, \boldsymbol{w}_{l}>=0, k \neq l
<w~k,wl>=0,k=l
则称向量集合
w
0
,
w
1
,
w
2
,
⋯
\boldsymbol{w}_{0}, \boldsymbol{w}_{1}, \boldsymbol{w}_{2}, \cdots
w0,w1,w2,⋯和对偶向量补集
w
~
0
,
w
~
1
,
w
~
2
,
⋯
\tilde{\boldsymbol{w}}_{0},\tilde{\boldsymbol{w}}_{1}, \tilde{\boldsymbol{w}}_{2}, \cdots
w~0,w~1,w~2,⋯是双正交的,并且是所张成向量空间的一个双正交基。 当且仅当
<
w
~
k
,
w
l
>
=
δ
k
l
=
{
0
,
k
≠
l
1
,
k
=
l
<\tilde{\boldsymbol{w}}_{k}, \boldsymbol{w}_{l}>=\delta_{k l}=\left\{\begin{array}{l} 0, k \neq l \\ 1, k=l \end{array}\right.
<w~k,wl>=δkl={0,k=l1,k=l
时, 它们才是双规范正交基。
令
W
=
{
w
0
,
w
1
,
w
2
,
⋯
}
W=\left\{\boldsymbol{w}_{0}, \boldsymbol{w}_{1}, \boldsymbol{w}_{2}, \cdots\right\}
W={w0,w1,w2,⋯}是内积空间
V
V
V的一个正交基,并且令
z
∈
V
z\in V
z∈V,则向量
z
z
z可 表示为基向量的线性组合:
z
=
α
0
w
0
+
α
1
w
1
+
α
2
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2
+
⋯
\boldsymbol{z}=\alpha_{0} \boldsymbol{w}_{0}+\alpha_{1} \boldsymbol{w}_{1}+\alpha_{2} \boldsymbol{w}_{2}+\cdots
z=α0w0+α1w1+α2w2+⋯
计算得:
α
i
=
⟨
w
i
,
z
⟩
⟨
w
i
,
w
i
⟩
\alpha_{i}=\frac{\left\langle\boldsymbol{w}_{i}, \boldsymbol{z}\right\rangle}{\left\langle\boldsymbol{w}_{i}, \boldsymbol{w}_{i}\right\rangle}
αi=⟨wi,wi⟩⟨wi,z⟩
若基向量的范数为1, 则简化为:
α
i
=
<
w
i
,
z
>
\alpha_{i}=<\boldsymbol{w}_{i}, \boldsymbol{z}>
αi=<wi,z>
二、基于矩阵的变换
一维离散傅里叶变换是一类重要的变换, 这类变换可用如下通式表示:
T
(
u
)
=
∑
x
=
0
N
−
1
f
(
x
)
r
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x
,
u
)
(16)
T(u)=\sum_{x=0}^{N-1} f(x) r(x, u)\tag{16}
T(u)=x=0∑N−1f(x)r(x,u)(16)
其中,
x
x
x是空间变量;
T
(
u
)
T(u)
T(u)是
f
(
x
)
f(x)
f(x)的变换;
r
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x
,
u
)
r(x,u)
r(x,u)是正变换核;整数
u
u
u是变换变量, 其值域为
0
,
1
,
2
,
⋯
,
N
−
1
0,1,2,\cdots,N-1
0,1,2,⋯,N−1。
T
(
u
)
T(u)
T(u)的反变换:
f
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x
)
=
∑
u
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N
−
1
T
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s
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x
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(17)
f(x)=\sum_{u=0}^{N-1} T(u)s(x,u)\tag{17}
f(x)=u=0∑N−1T(u)s(x,u)(17)
其中,
s
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x
,
u
)
s(x,u)
s(x,u)是反变换核, x的值域为
0
,
1
,
2
,
⋯
,
N
−
1
0,1,2,\cdots,N-1
0,1,2,⋯,N−1。
f
(
x
)
f(x)
f(x)是N个反变换核的加权和,
T
(
u
)
T(u)
T(u)是权重。
展开式
(
17
)
(17)
(17)右侧得到:
f
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=
T
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0
)
s
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x
,
0
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+
T
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1
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s
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1
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+
⋯
+
T
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N
−
1
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s
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x
,
N
−
1
)
(18)
f(x)=T(0) s(x, 0)+T(1) s(x, 1)+\cdots+T(N-1) s(x, N-1)\tag{18}
f(x)=T(0)s(x,0)+T(1)s(x,1)+⋯+T(N−1)s(x,N−1)(18)
假设式
(
17
)
(17)
(17)中的s(x, u)是内积空间的正交基向量,且基向量的范数为1,则:
T
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u
)
=
⟨
s
(
x
,
u
)
,
f
(
x
)
⟩
(19)
T(u)=\left \langle s(x,u),f(x) \right \rangle \tag{19}
T(u)=⟨s(x,u),f(x)⟩(19)
即变换的每个元素 T ( u ) T(u) T(u), 可通过内积来计算。
现在准备利用矩阵来表达式
(
16
)
(16)
(16)和式
(
17
)
(17)
(17)。首先将函数
f
(
x
)
f(x)
f(x),
T
(
u
)
T(u)
T(u)和
s
(
x
,
u
)
s(x, u)
s(x,u)定义为列向量:
f
=
[
f
(
0
)
f
(
1
)
⋯
f
(
N
−
1
)
]
T
=
[
f
0
f
1
⋯
f
N
−
1
]
T
(20)
\boldsymbol{f}=\left[\begin{array}{lll} f(0) & f(1) \cdots f(N-1) \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{lll} f_{0} & f_{1} \cdots & f_{N-1} \end{array}\right]^{\mathrm{T}}\tag{20}
f=[f(0)f(1)⋯f(N−1)]T=[f0f1⋯fN−1]T(20)
t
=
[
T
(
0
)
T
(
1
)
⋯
T
(
N
−
1
)
]
T
=
[
t
0
t
1
⋯
t
N
−
1
]
T
(21)
\boldsymbol{t}=\left[\begin{array}{lll} T(0) & T(1) \cdots T(N-1) \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{lll} t_{0} & t_{1} \cdots & t_{N-1} \end{array}\right]^{\mathrm{T}}\tag{21}
t=[T(0)T(1)⋯T(N−1)]T=[t0t1⋯tN−1]T(21)
s u = [ s ( 0 , u ) s ( 1 , u ) ⋯ s ( N − 1 , u ) ] T = [ s u , 0 s u , 1 ⋯ s u , N − 1 ] T s_{u}=[s(0, u)\quad s(1, u) \cdots s(N-1, u)]^{\mathrm{T}}=\left[s_{u, 0}\quad s_{u, 1} \cdots s_{u, N-1}\right]^{\mathrm{T}}\ su=[s(0,u)s(1,u)⋯s(N−1,u)]T=[su,0su,1⋯su,N−1]T
其中,
u
=
0
,
1
,
⋯
,
N
−
1
u=0,1, \cdots, N-1
u=0,1,⋯,N−1
利用这些列向量, 重写式
(
19
)
(19)
(19), 得:
T
(
u
)
=
<
s
u
,
f
>
,
u
=
0
,
1
,
⋯
,
N
−
1
(23)
T(u)=<s_{u}, f>, \quad u=0,1, \cdots, N-1\tag{23}
T(u)=<su,f>,u=0,1,⋯,N−1(23)
A = [ s 0 T s 1 T ⋮ s N − 1 T ] = [ s 0 s 1 … s N − 1 ] T (24) \mathbf{A}=\left[\begin{array}{c} \mathbf{s}_{0}^{T} \\ \mathbf{s}_{1}^{T} \\ \vdots \\ \mathbf{s}_{N-1}^{T} \end{array}\right]=\left[\begin{array}{llll} \mathbf{s}_{0} & \mathbf{s}_{1} & \ldots & \mathbf{s}_{N-1} \end{array}\right]^{T}\tag{24} A= s0Ts1T⋮sN−1T =[s0s1…sN−1]T(24)
然后将式
(
23
)
(23)
(23)代入式
(
21
)
(21)
(21), 并利用式
(
1
)
(1)
(1), 得
A
A
T
=
[
s
0
T
s
1
T
⋮
s
N
−
1
T
]
[
s
0
s
1
…
s
N
−
1
]
=
[
s
0
T
s
0
s
0
T
s
1
…
s
0
T
s
N
−
1
s
1
T
s
0
s
1
T
s
1
⋮
⋮
⋱
s
N
−
1
T
s
0
…
s
N
−
1
T
s
N
−
1
]
=
[
⟨
s
0
,
s
0
⟩
⟨
s
0
,
s
1
⟩
…
⟨
s
0
,
s
N
−
1
⟩
⟨
s
1
,
s
0
⟩
⟨
s
1
,
s
1
⟩
⋮
⋮
⋱
⟨
s
N
−
1
,
s
0
⟩
…
⟨
s
N
−
1
,
s
N
−
1
⟩
]
=
[
1
0
…
0
0
1
⋮
⋮
⋱
0
…
1
]
=
I
(27)
\begin{aligned} \mathbf{A A}^{T} & =\left[\begin{array}{c} \mathbf{s}_{0}^{T} \\ \mathbf{s}_{1}^{T} \\ \vdots \\ \mathbf{s}_{N-1}^{T} \end{array}\right]\left[\begin{array}{llll} \mathbf{s}_{0} & \mathbf{s}_{1} & \ldots & \mathbf{s}_{N-1} \end{array}\right] \\ & =\left[\begin{array}{cccc} \mathbf{s}_{0}^{T} \mathbf{s}_{0} & \mathbf{s}_{0}^{T} \mathbf{s}_{1} & \ldots & \mathbf{s}_{0}^{T} \mathbf{s}_{N-1} \\ \mathbf{s}_{1}^{T} \mathbf{s}_{0} & \mathbf{s}_{1}^{T} \mathbf{s}_{1} & & \vdots \\ \vdots & & \ddots & \\ \mathbf{s}_{N-1}^{T} \mathbf{s}_{0} & \ldots & & \mathbf{s}_{N-1}^{T} \mathbf{s}_{N-1} \end{array}\right] \\ & =\left[\begin{array}{cccc} \left\langle\mathbf{s}_{0}, \mathbf{s}_{0}\right\rangle & \left\langle\mathbf{s}_{0}, \mathbf{s}_{1}\right\rangle & \ldots & \left\langle\mathbf{s}_{0}, \mathbf{s}_{N-1}\right\rangle \\ \left\langle\mathbf{s}_{1}, \mathbf{s}_{0}\right\rangle & \left\langle\mathbf{s}_{1}, \mathbf{s}_{1}\right\rangle & & \vdots \\ \vdots & & \ddots & \\ \left\langle\mathbf{s}_{N-1}, \mathbf{s}_{0}\right\rangle & \ldots & & \left\langle\mathbf{s}_{N-1}, \mathbf{s}_{N-1}\right\rangle \end{array}\right] \\ & =\left[\begin{array}{cccc} 1 & 0 & \ldots & 0 \\ 0 & 1 & & \vdots \\ \vdots & & \ddots & \\ 0 & \ldots & & 1 \end{array}\right]=\mathbf{I} \end{aligned}\tag{27}
AAT=
s0Ts1T⋮sN−1T
[s0s1…sN−1]=
s0Ts0s1Ts0⋮sN−1Ts0s0Ts1s1Ts1……⋱s0TsN−1⋮sN−1TsN−1
=
⟨s0,s0⟩⟨s1,s0⟩⋮⟨sN−1,s0⟩⟨s0,s1⟩⟨s1,s1⟩……⋱⟨s0,sN−1⟩⋮⟨sN−1,sN−1⟩
=
10⋮001……⋱0⋮1
=I(27)
或
t
=
A
f
(26)
t=Af\tag{26}
t=Af(26)
对于一维信号, 利用矩阵表达式
(
16
)
(16)
(16)和式
(
17
)
(17)
(17)为
t
=
A
f
(28)
t=Af \tag{28}
t=Af(28)
f
=
A
T
t
(29)
f=A^Tt\tag{29}
f=ATt(29)
式
(
28
)
(28)
(28)和式
(
29
)
(29)
(29)是可逆变换对。
本节中给出的大部分概念, 可推广到形如下式的连续展开:
f
(
x
)
=
∑
u
=
−
∞
+
∞
α
u
s
u
(
x
)
(51)
f(x)=\sum_{u=-\infty}^{+\infty} \alpha_{u} s_{u}(x)\tag{51}
f(x)=u=−∞∑+∞αusu(x)(51)
其中,
α
u
\alpha_{u}
αu和
s
u
s_{u}
su,
u
=
0
,
±
1
,
±
2
,
±
3
,
⋯
u=0, \pm 1, \pm 2, \pm 3, \cdots
u=0,±1,±2,±3,⋯分别表示内积空间
C
(
[
a
,
b
]
)
C([a, b])
C([a,b])的展开系数和基向量。
若
s
u
s_{u}
su,
u
=
0
,
±
1
,
±
2
,
±
3
,
⋯
u=0, \pm 1, \pm 2, \pm 3, \cdots
u=0,±1,±2,±3,⋯是
C
(
[
a
,
b
]
)
C([a, b])
C([a,b])的正交基向量, 则展开系数:
α
u
=
⟨
s
u
(
x
)
,
f
(
x
)
⟩
(52)
\alpha_{u}=\left \langle s_{u}(x), f(x) \right \rangle\tag{52}
αu=⟨su(x),f(x)⟩(52)
三、相关
本节介绍这些系数与相关之间的关系。