整波滤波器是一类能够整合具有任意频谱密度的静定随机信号的滤波器。其输入信号往往是白噪声。
1. 整波滤波器推导
由统计动力学笔记(二)频谱密度与线性随机系统的动态准确性(自留用)一文可以知道系统输出
x
x
x和输入
u
u
u之间的互频谱密度:
S
x
(
ω
)
=
W
(
j
ω
)
W
(
−
j
ω
)
S
u
(
ω
)
(1)
S_x (\omega) = W(j \omega) W(-j \omega) S_u (\omega) \tag{1}
Sx(ω)=W(jω)W(−jω)Su(ω)(1)当输入为白噪声时,
S
u
(
ω
)
=
S
n
(
ω
)
=
1
S_u (\omega) = S_n (\omega) = 1
Su(ω)=Sn(ω)=1,则
S
x
(
ω
)
=
W
(
j
ω
)
W
(
−
j
ω
)
(2)
S_x (\omega) = W(j \omega) W(-j \omega) \tag{2}
Sx(ω)=W(jω)W(−jω)(2)这样一来,只要将输出端
x
x
x的频谱密度分解为2个共轭的部分,就可以得到系统的传递函数。这一步也称为频谱密度的分解。
例:输出端的频谱密度为
S
x
(
ω
)
=
4
4
ω
2
+
1
=
2
2
j
ω
+
1
⋅
2
2
(
−
j
ω
)
+
1
S_x (\omega) = \frac{4}{4\omega^2 + 1} = \frac{2}{2 j \omega +1} \cdot \frac{2}{2 (- j\omega) + 1 }
Sx(ω)=4ω2+14=2jω+12⋅2(−jω)+12则系统的传函为
W
(
j
ω
)
=
2
2
j
ω
+
1
W(j \omega) = \frac{2}{2 j \omega +1}
W(jω)=2jω+12即
W
(
s
)
=
2
2
s
+
1
W({\rm s}) = \frac{2}{2 {\rm s} +1}
W(s)=2s+12
2. 线性动态系统输出端的随机信号的方差
方差的定义式在统计动力学笔记(二)频谱密度与线性随机系统的动态准确性(自留用)一文的式(5)已给出:
D
x
=
R
x
(
0
)
=
1
2
π
∫
−
∞
∞
S
x
(
ω
)
d
ω
D_x = R_x (0) = \frac{1}{2\pi} \int_{-\infty} ^\infty S_x (\omega) {\rm d} \omega
Dx=Rx(0)=2π1∫−∞∞Sx(ω)dω代入式(1)
D
x
=
1
2
π
∫
−
∞
∞
W
(
j
ω
)
W
(
−
j
ω
)
S
u
(
ω
)
d
ω
=
1
2
π
∫
−
∞
∞
∣
W
(
j
ω
)
∣
2
S
u
(
ω
)
d
ω
(3)
D_x = \frac{1}{2\pi} \int_{-\infty} ^\infty W(j \omega) W(-j \omega) S_u (\omega) {\rm d} \omega = \frac{1}{2\pi} \int_{-\infty} ^\infty \big\lvert W(j \omega) \big\rvert^2 S_u (\omega) {\rm d} \omega \tag{3}
Dx=2π1∫−∞∞W(jω)W(−jω)Su(ω)dω=2π1∫−∞∞
W(jω)
2Su(ω)dω(3)式(3)的计算方式,有如下一套固定的方法,称为“
I
n
I_n
In – 积分法”:
I
n
=
1
2
π
∫
−
∞
∞
∣
G
(
j
ω
)
∣
2
∣
H
n
(
j
ω
)
∣
2
d
ω
=
1
2
π
∫
−
∞
∞
G
n
(
j
ω
)
H
n
(
j
ω
)
H
n
(
−
j
ω
)
d
ω
(4)
I_n = \frac{1}{2\pi} \int_{-\infty} ^\infty \frac{ \big\lvert G(j \omega) \big\rvert^2 }{ \big\lvert H_n(j \omega) \big\rvert^2 } {\rm d} \omega = \frac{1}{2\pi} \int_{-\infty} ^\infty \frac{ G_n (j \omega) }{ H_n(j \omega) H_n(-j \omega) } {\rm d} \omega \tag{4}
In=2π1∫−∞∞
Hn(jω)
2
G(jω)
2dω=2π1∫−∞∞Hn(jω)Hn(−jω)Gn(jω)dω(4)其中
G
n
(
j
ω
)
=
b
0
(
j
ω
)
2
n
−
2
+
b
1
(
j
ω
)
2
n
−
4
+
⋯
+
b
n
−
1
,
H
n
(
j
ω
)
=
a
0
(
j
ω
)
n
+
a
1
(
j
ω
)
n
−
1
+
⋯
+
a
n
(5)
G_n (j \omega) = b_0 (j \omega)^{2n-2} + b_1 (j \omega)^{2n-4} + \cdots + b_{n-1}, \\ H_n (j \omega) = a_0 (j \omega)^{n} + a_1 (j \omega)^{n-1} + \cdots + a_n \tag{5}
Gn(jω)=b0(jω)2n−2+b1(jω)2n−4+⋯+bn−1,Hn(jω)=a0(jω)n+a1(jω)n−1+⋯+an(5)关于式(4)(5)有如下几点:
(1)若积分式的分母阶数为
n
n
n,则实际系统中,分子的阶数不会超过
2
n
−
2
2n-2
2n−2。
(2)积分式分母
H
n
(
j
ω
)
H
n
(
−
j
ω
)
H_n(j \omega) H_n(-j \omega)
Hn(jω)Hn(−jω)为
ω
\omega
ω的偶函数。
(3)积分式分子
G
n
(
j
ω
)
G_n(j \omega)
Gn(jω)只含有
j
ω
j\omega
jω的偶次幂。若出现了奇次幂,则可以直接忽视掉,因为积分后奇次幂将等于零。
(4)积分式分母中的
H
n
(
j
ω
)
H_n(j \omega)
Hn(jω)应当是稳定的。
则对于
I
n
I_n
In – 积分,其计算方法如下:
I
n
=
(
−
1
)
n
+
1
N
n
2
a
0
D
n
(6)
I_n = (-1) ^{n+1} \frac{N_n}{2a_0 D_n} \tag{6}
In=(−1)n+12a0DnNn(6)其中
D
n
=
∣
a
1
a
0
0
⋯
0
a
3
a
2
a
1
⋯
0
a
5
a
4
a
3
⋯
0
⋮
⋮
⋮
⋱
⋮
0
0
0
⋯
a
n
∣
,
(7)
D_n = \begin{vmatrix} a_1 & a_0 & 0 & \cdots & 0 \\ a_3 & a_2 & a_1 & \cdots & 0 \\ a_5 & a_4 & a_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & a_n \end{vmatrix} \tag{7},
Dn=
a1a3a5⋮0a0a2a4⋮00a1a3⋮0⋯⋯⋯⋱⋯000⋮an
,(7)
N
n
=
∣
b
0
a
0
0
⋯
0
b
1
a
2
a
1
⋯
0
b
2
a
4
a
3
⋯
0
⋮
⋮
⋮
⋱
⋮
b
n
−
1
0
0
⋯
a
n
∣
(8)
N_n = \begin{vmatrix} b_0 & a_0 & 0 & \cdots & 0 \\ b_1 & a_2 & a_1 & \cdots & 0 \\ b_2 & a_4 & a_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ b_{n-1} & 0 & 0 & \cdots & a_n \end{vmatrix} \tag{8}
Nn=
b0b1b2⋮bn−1a0a2a4⋮00a1a3⋮0⋯⋯⋯⋱⋯000⋮an
(8)
N
n
N_n
Nn只是把
D
n
D_n
Dn中的第一列替换成了
b
i
b_i
bi。文章来源:https://www.toymoban.com/news/detail-675134.html
例:设系统的传递函数为
W
(
s
)
=
K
T
s
+
1
W({\rm s}) = \frac{K}{T {\rm s} +1}
W(s)=Ts+1K输入信号的频谱密度为
S
u
(
ω
)
=
D
u
α
2
+
ω
2
S_u (\omega) = \frac{D_u}{\alpha^2 + \omega^2}
Su(ω)=α2+ω2Du计算该系统的均方差。
首先得到系统误差的传函:
Φ
e
(
s
)
=
1
1
+
W
(
s
)
=
T
s
+
1
T
s
+
1
+
K
\Phi_e ({\rm s}) = \frac{1}{1 + W( {\rm s})} = \frac{T{\rm s} +1}{T{\rm s} + 1 + K}
Φe(s)=1+W(s)1=Ts+1+KTs+1代入式(1)计算误差的频谱密度
S
e
(
ω
)
=
∣
Φ
e
(
j
ω
)
∣
2
S
u
(
ω
)
=
∣
T
(
j
ω
)
+
1
T
(
j
ω
)
+
1
+
K
∣
2
D
u
α
2
+
ω
2
=
D
u
(
T
2
ω
2
+
1
)
∣
(
T
(
j
ω
)
+
1
+
K
)
(
j
ω
+
α
)
∣
2
=
D
u
(
T
2
ω
2
+
1
)
∣
T
(
j
ω
)
2
+
(
α
T
+
1
+
K
)
j
ω
+
(
1
+
K
)
α
∣
2
\begin{aligned} S_e (\omega) &= \left| \Phi_e (j \omega) \right|^2 S_u (\omega) = \left| \frac{T(j \omega) +1}{T(j \omega) + 1 + K} \right|^2 \frac{D_u}{\alpha^2 + \omega^2} \\ &= \frac{D_u \left( T^2 \omega^2 + 1\right) }{ \left| \left( T( j\omega) + 1 + K \right) \left( j\omega + \alpha \right) \right|^2 } \\ &= \frac{D_u \left( T^2 \omega^2 + 1 \right) }{ \left| T(j\omega)^2 + \left( \alpha T + 1 +K \right) j\omega + (1 + K) \alpha \right|^2 } \end{aligned}
Se(ω)=∣Φe(jω)∣2Su(ω)=
T(jω)+1+KT(jω)+1
2α2+ω2Du=∣(T(jω)+1+K)(jω+α)∣2Du(T2ω2+1)=∣T(jω)2+(αT+1+K)jω+(1+K)α∣2Du(T2ω2+1)均方差为(类比统计动力学笔记(二)频谱密度与线性随机系统的动态准确性(自留用)一文式(5)):
e
2
‾
=
1
2
π
∫
−
∞
∞
S
e
(
ω
)
d
ω
=
D
u
I
2
\overline{e^2} = \frac{1}{2\pi} \int_{-\infty} ^\infty S_e (\omega) {\rm d} \omega = D_u I_2
e2=2π1∫−∞∞Se(ω)dω=DuI2其中
I
2
=
1
2
π
∫
−
∞
∞
(
T
2
ω
2
+
1
)
d
ω
∣
T
(
j
ω
)
2
+
(
α
T
+
1
+
K
)
j
ω
+
(
1
+
K
)
α
∣
2
I_2 = \frac{1}{2\pi} \int_{-\infty} ^\infty \frac{ \left( T^2 \omega^2 + 1 \right) {\rm d} \omega }{ \left| T(j\omega)^2 + \left( \alpha T + 1 +K \right) j\omega + (1 + K) \alpha \right|^2 }
I2=2π1∫−∞∞∣T(jω)2+(αT+1+K)jω+(1+K)α∣2(T2ω2+1)dω可见
G
2
(
j
ω
)
=
T
2
⏟
b
0
ω
2
+
1
⏟
b
1
,
G_2 (j\omega) = \underbrace{T^2}_{b_0} \omega^2 + \underbrace{1}_{b_1},
G2(jω)=b0
T2ω2+b1
1,
H
2
(
j
ω
)
=
T
⏟
a
0
(
j
ω
)
2
+
(
α
T
+
1
+
K
)
⏟
a
1
j
ω
+
(
1
+
K
)
α
⏟
a
2
H_2 (j\omega) = \underbrace{T}_{a_0} (j\omega)^2 + \underbrace{\left( \alpha T + 1 +K \right)}_{a_1} j\omega + \underbrace{(1 + K) \alpha}_{a_2}
H2(jω)=a0
T(jω)2+a1
(αT+1+K)jω+a2
(1+K)α计算两个行列式
D
2
=
∣
a
1
a
0
a
3
a
2
∣
=
∣
α
T
+
1
+
K
T
0
(
1
+
K
)
α
∣
=
α
(
α
T
+
1
+
K
)
(
1
+
K
)
,
D_2 = \begin{vmatrix} a_1 & a_0 \\ a_3 & a_2 \end{vmatrix} = \begin{vmatrix} \alpha T + 1 +K & T \\ 0 & (1 + K) \alpha \end{vmatrix} = \alpha \left( \alpha T + 1 +K \right) (1 + K),
D2=
a1a3a0a2
=
αT+1+K0T(1+K)α
=α(αT+1+K)(1+K),
N
2
=
∣
b
0
a
0
b
1
a
2
∣
=
∣
T
2
T
1
(
1
+
K
)
α
∣
=
α
T
2
(
1
+
K
)
−
T
N_2 = \begin{vmatrix} b_0 & a_0 \\ b_1 & a_2 \end{vmatrix} = \begin{vmatrix} T^2 & T \\ 1 & (1 + K) \alpha \end{vmatrix} = \alpha T^2 (1 + K) - T
N2=
b0b1a0a2
=
T21T(1+K)α
=αT2(1+K)−T故
I
2
=
(
−
1
)
2
+
1
N
2
2
a
0
D
2
=
−
α
T
2
(
1
+
K
)
−
T
2
T
α
(
α
T
+
1
+
K
)
(
1
+
K
)
I_2 = (-1) ^{2+1} \frac{N_2}{2a_0 D_2} = - \frac{ \alpha T^2 (1 + K) - T }{2 T \alpha \left( \alpha T + 1 +K \right) (1 + K) }
I2=(−1)2+12a0D2N2=−2Tα(αT+1+K)(1+K)αT2(1+K)−T则
e
2
‾
=
D
u
I
2
=
D
u
[
T
−
α
T
2
(
1
+
K
)
]
2
T
α
(
α
T
+
1
+
K
)
(
1
+
K
)
=
D
u
[
1
−
α
T
(
1
+
K
)
]
2
α
(
α
T
+
1
+
K
)
(
1
+
K
)
\overline{e^2} = D_u I_2 = \frac{ D_u \left[ T - \alpha T^2 (1 + K) \right] }{2 T \alpha \left( \alpha T + 1 +K \right) (1 + K) } = \frac{ D_u \left[ 1 - \alpha T (1 + K) \right] }{2 \alpha \left( \alpha T + 1 +K \right) (1 + K) }
e2=DuI2=2Tα(αT+1+K)(1+K)Du[T−αT2(1+K)]=2α(αT+1+K)(1+K)Du[1−αT(1+K)]故均方差为
e
2
‾
=
D
u
[
1
−
α
T
(
1
+
K
)
]
2
α
(
α
T
+
1
+
K
)
(
1
+
K
)
\sqrt{\overline{e^2}} = \sqrt{ \frac{ D_u \left[ 1 - \alpha T (1 + K) \right] }{2 \alpha \left( \alpha T + 1 +K \right) (1 + K) } }
e2=2α(αT+1+K)(1+K)Du[1−αT(1+K)]文章来源地址https://www.toymoban.com/news/detail-675134.html
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