工程数学基础
参考：DR_CAN
1.特征值与特征向量
在数学上，特别是线性代数中，对于一个给定的线性变换
A
A
A，它的特征向量
v
v
v经过这个线性变换的作用之后，得到的新向量仍然与原来的
v
v
v保持在同一条直线上。但其长度或方向也许会改变。即
A
v
=
λ
v
Av=\lambda v
Av=λv
其中
λ
\lambda
λ为标量，即特征向量的长度在该线性变换下缩放的比例，称为其特征值。
线性变化
A
=
[
1
1
4
−
2
]
,
v
1
=
[
1
2
]
A=\begin{bmatrix} 1 & 1\\4 &-2\end{bmatrix}, v_1=\begin{bmatrix}1 \\2\end{bmatrix}
A=[141−2],v1=[12]
A
v
1
=
[
3
0
]
Av_1=\begin{bmatrix}3 \\0\end{bmatrix}
Av1=[30]
v
1
→
A
v
1
v_1 \to Av_1
v1→Av1
v
2
=
[
1
1
]
v_2=\begin{bmatrix}1 \\1\end{bmatrix}
v2=[11]
A
v
2
=
2
[
1
1
]
=
2
v
2
Av_2=2\begin{bmatrix}1 \\1\end{bmatrix}=2v_2
Av2=2[11]=2v2
A
v
2
Av_2
Av2与
v
2
v_2
v2在一条直线上，
v
2
v_2
v2为特征向量，
λ
=
2
\lambda =2
λ=2为特征值
如何求解特征值和特征向量？
A
v
=
λ
v
Av=\lambda v
Av=λv
A
v
−
λ
v
=
0
Av-\lambda v=0
Av−λv=0
(
A
−
λ
I
)
v
=
0
(A-\lambda I)v=0
(A−λI)v=0
其中
I
I
I为单位矩阵
[
1
1
⋯
1
]
\begin{bmatrix} 1 & & & \\ & 1 & & \\ & & \cdots & \\ & & &1\end{bmatrix}
⎣⎢⎢⎡11⋯1⎦⎥⎥⎤
式子有非零解，则有
∣
A
−
λ
I
∣
=
0
|A-\lambda I|=0
∣A−λI∣=0
A
=
[
1
1
4
−
2
]
A=\begin{bmatrix} 1 & 1\\4 &-2\end{bmatrix}
A=[141−2]
A
−
λ
I
=
[
1
−
λ
1
4
−
2
−
λ
]
A-\lambda I=\begin{bmatrix}1-\lambda & 1\\ 4 &-2-\lambda \end{bmatrix}
A−λI=[1−λ41−2−λ]
∣
A
−
λ
I
∣
=
0
|A-\lambda I|=0
∣A−λI∣=0
⇒
∣
1
−
λ
1
4
−
2
−
λ
∣
=
0
\Rightarrow \begin{vmatrix}1-\lambda & 1\\ 4 &-2-\lambda \end{vmatrix}=0
⇒∣∣∣∣1−λ41−2−λ∣∣∣∣=0
⇒
(
λ
−
2
)
(
λ
+
3
)
=
0
\Rightarrow (\lambda -2)(\lambda +3)=0
⇒(λ−2)(λ+3)=0
λ
1
=
2
,
λ
2
=
−
3
\lambda_1=2,\lambda_2=-3
λ1=2,λ2=−3
将计算得到的代回去,
λ
1
=
2
\lambda_1=2
λ1=2时
[
1
−
2
1
4
−
2
−
2
]
v
1
=
0
\begin{bmatrix}1-2 & 1\\ 4 &-2-2 \end{bmatrix}v_1=0
[1−241−2−2]v1=0
[
−
1
1
4
−
4
]
[
v
1
v
2
]
=
0
\begin{bmatrix}-1 & 1\\ 4 &-4 \end{bmatrix}\begin{bmatrix}v_1 \\v_2\end{bmatrix}=0
[−141−4][v1v2]=0
−
v
11
+
v
12
=
0
4
v
11
−
4
v
12
=
0
-v_{11}+v_{12}=0\\ 4v_{11}-4v_{12}=0
−v11+v12=04v11−4v12=0
⇒
v
11
=
v
12
\Rightarrow v_{11}=v_{12}
⇒v11=v12
可任取一组解： v 11 = 1 , v 12 = 1 v_{11}=1,v_{12}=1 v11=1,v12=1
v 1 = [ 1 1 ] v_1=\begin{bmatrix}1 \\1\end{bmatrix} v1=[11]
同理求得
λ
1
=
−
3
\lambda_1=-3
λ1=−3时
4
v
21
+
v
22
=
0
4v_{21}+v_{22}=0
4v21+v22=0
可取
v
21
=
1
,
v
22
=
−
4
v_{21}=1,v_{22}=-4
v21=1,v22=−4
v 2 = [ 1 − 4 ] v_2=\begin{bmatrix}1 \\-4\end{bmatrix} v2=[1−4]
应用：化对角矩阵，解耦，Decouple
设一个新的矩阵(过度矩阵 corrdinate transformation matrix)
P
=
[
v
1
v
2
]
P=\begin{bmatrix}v_1 &v_2\end{bmatrix}
P=[v1v2]
A P = A [ v 1 v 2 ] = A [ v 11 v 12 v 21 v 22 ] = [ A [ v 11 v 12 ] A [ v 21 v 22 ] ] = [ λ 1 [ v 11 v 12 ] λ 2 [ v 21 v 22 ] ] = [ λ 1 v 11 λ 2 v 21 λ 1 v 21 λ 2 v 22 ] = [ v 11 v 21 v 21 v 22 ] [ λ 1 0 0 λ 2 ] = P Λ \begin{aligned} AP & = A\begin{bmatrix}v_1 &v_2\end{bmatrix} = A\begin{bmatrix}v_{11} &v_{12}\\v_{21} &v_{22}\end{bmatrix}\\ & = \begin{bmatrix} A\begin{bmatrix}v_{11} \\v_{12}\end{bmatrix} &A\begin{bmatrix}v_{21} \\v_{22}\end{bmatrix} \end{bmatrix} \\ & = \begin{bmatrix} \lambda_1 \begin{bmatrix}v_{11} \\v_{12}\end{bmatrix} &\lambda_2 \begin{bmatrix}v_{21} \\v_{22}\end{bmatrix} \end{bmatrix}\\ & = \begin{bmatrix} \lambda_1 v_{11} &\lambda_2v_{21}\\ \lambda_1 v_{21} &\lambda_2v_{22} \end{bmatrix} \\ & = \begin{bmatrix} v_{11} &v_{21}\\ v_{21} &v_{22} \end{bmatrix}\begin{bmatrix} \lambda_1 & 0\\ 0 &\lambda_2 \end{bmatrix}\\ &=P \Lambda \end{aligned} AP=A[v1v2]=A[v11v21v12v22]=[A[v11v12]A[v21v22]]=[λ1[v11v12]λ2[v21v22]]=[λ1v11λ1v21λ2v21λ2v22]=[v11v21v21v22][λ100λ2]=PΛ
P
−
1
A
P
=
P
−
1
P
Λ
P^{-1}AP =P^{-1}P \Lambda
P−1AP=P−1PΛ
P
−
1
A
P
=
Λ
P^{-1}AP = \Lambda
P−1AP=Λ
微分方程，状态空间表达式
d x 1 d t = x 1 + x 2 d x 2 d t = 4 x 1 − 2 x 2 \frac{\mathrm{d} x_1}{\mathrm{d} t} =x_1+x_2\\ \frac{\mathrm{d} x_2}{\mathrm{d} t} =4x_1-2x_2 dtdx1=x1+x2dtdx2=4x1−2x2
d d t [ x 1 x 2 ] = [ 1 1 4 − 2 ] [ x 1 x 2 ] \frac{\mathrm{d} }{\mathrm{d} t} \begin{bmatrix}x_1 \\x_2\end{bmatrix} =\begin{bmatrix} 1 & 1\\4 &-2\end{bmatrix} \begin{bmatrix}x_1 \\x_2\end{bmatrix} dtd[x1x2]=[141−2][x1x2]
x ˙ = A x \dot x=Ax x˙=Ax
令 x = P y , x ˙ = P y ˙ = A x = A P y , P = [ 1 1 1 − 4 ] x=Py,\dot x=P\dot y=Ax=APy,P=\begin{bmatrix} 1 & 1\\ 1 &-4\end{bmatrix} x=Py,x˙=Py˙=Ax=APy,P=[111−4]
P
−
1
P
y
˙
=
P
−
1
A
P
y
P^{-1}P\dot y=P^{-1}APy
P−1Py˙=P−1APy
y
˙
=
P
−
1
A
P
y
=
Λ
y
\dot y=P^{-1}APy=\Lambda y
y˙=P−1APy=Λy
Λ = [ λ 1 0 0 λ 2 ] = [ 2 0 0 − 3 ] \Lambda=\begin{bmatrix} \lambda_1 & 0\\0 &\lambda_2 \end{bmatrix}=\begin{bmatrix} 2& 0\\0 &-3\end{bmatrix} Λ=[λ100λ2]=[200−3]
y ˙ = [ 2 0 0 − 3 ] y \dot y= \begin{bmatrix} 2& 0\\0 &-3\end{bmatrix}y y˙=[200−3]y
y ˙ 1 = 2 y 1 y ˙ 2 = − 3 y 2 \dot y_1=2y_1\\\dot y_2=-3y_2 y˙1=2y1y˙2=−3y2
解微分方程
y
1
=
C
1
e
2
t
y
˙
2
=
C
2
e
−
3
t
y_1=C_1 e^{2t}\\\dot y_2=C_2 e^{-3t}
y1=C1e2ty˙2=C2e−3t
C
1
,
C
2
C_1,C_2
C1,C2为常数
x = P y = [ 1 1 1 − 4 ] [ C 1 e 2 t C 2 e − 3 t ] = [ C 1 e 2 t + C 2 e − 3 t C 1 e 2 t − 4 C 2 e − 3 t ] x=Py=\begin{bmatrix} 1 & 1\\ 1 &-4\end{bmatrix}\begin{bmatrix}C_1 e^{2t}\\C_2 e^{-3t}\end{bmatrix}\\=\begin{bmatrix}C_1 e^{2t}+C_2 e^{-3t}\\C_1 e^{2t}-4C_2 e^{-3t}\end{bmatrix} x=Py=[111−4][C1e2tC2e−3t]=[C1e2t+C2e−3tC1e2t−4C2e−3t]
2.泰勒级数&泰勒公式
对于一个线性系统，它应该符合叠加原理，如
x
˙
=
f
(
x
)
\dot x=f(x)
x˙=f(x)
- x 1 , x 2 x_1,x_2 x1,x2是解
- x 3 = k 1 x 1 + k 2 x 2 , k 1 , k 2 x_3=k_1x_1+k_2x_2,\quad k_1,k_2 x3=k1x1+k2x2,k1,k2为常数
- x 3 x_3 x3是解
线性化：泰勒级数
在导数附近线性化
3.线性时不变系统的冲击响应与卷积
linear time invariant
o
{
f
(
t
)
}
=
x
(
t
)
o\left \{ f(t) \right \} =x(t)
o{f(t)}=x(t)
线性
o
{
f
1
(
t
)
+
f
2
(
t
)
}
=
x
1
(
t
)
+
x
2
(
t
)
o\left \{ f_1(t)+ f_2(t) \right \} =x_1(t)+ x_2(t)
o{f1(t)+f2(t)}=x1(t)+x2(t)
o
{
a
f
(
t
)
}
=
a
x
(
t
)
o\left \{ af(t) \right \} =ax(t)
o{af(t)}=ax(t)
叠加原理
o
{
a
f
1
(
t
)
+
b
f
2
(
t
)
}
=
a
x
1
(
t
)
+
b
x
2
(
t
)
o\left \{ af_1(t)+b f_2(t) \right \} =ax_1(t)+ bx_2(t)
o{af1(t)+bf2(t)}=ax1(t)+bx2(t)
时不变
o
{
f
(
t
−
τ
)
}
=
x
(
t
−
τ
)
o\left \{ f(t-\tau) \right \} =x(t-\tau)
o{f(t−τ)}=x(t−τ)
弹簧质量阻尼系统
对小块施加短暂的外力，及其响应
系统的传递方程作拉普拉斯逆变换等于卷积
L
−
1
[
F
(
s
)
H
(
s
)
]
=
L
−
1
[
X
(
s
)
]
⇒
f
(
t
)
∗
h
(
t
)
=
x
(
t
)
\mathcal{L}^{-1}\left [ F(s)H(s) \right ] =\mathcal{L}^{-1}\left [ X(s) \right ]\\ \Rightarrow f(t)*h(t)=x(t)
L−1[F(s)H(s)]=L−1[X(s)]⇒f(t)∗h(t)=x(t)
现在输入连续的外力，为了便于分析将其离散化
连续化
lim
Δ
T
→
0
∑
→
∫
\lim_{\Delta T \to 0} \sum \to \int
ΔT→0lim∑→∫
单位冲激响应（狄拉克函数）
宽度为0，面积为1
δ
(
t
)
=
{
0
,
t
≠
0
∫
−
∞
∞
δ
(
t
)
d
t
=
1
\delta(t)=\begin{cases}0,t\ne 0\\\int_{-\infty} ^\infty \delta(t)dt=1\end{cases}
δ(t)={0,t=0∫−∞∞δ(t)dt=1
它是纯数学上的意义，现实生活不存在，通过另外一个辅助函数来理解它
δ
Δ
(
t
)
=
{
1
Δ
T
,
0
<
t
<
Δ
T
0
,
e
l
s
e
\delta_\Delta(t)=\begin{cases}\frac{1}{\Delta T},0<t < \Delta T\\0,else\end{cases}
δΔ(t)={ΔT1,0<t<ΔT0,else
lim Δ T → 0 δ Δ = δ ( t ) \lim_{\Delta T \to 0} \delta_\Delta = \delta(t) ΔT→0limδΔ=δ(t)
对小块施加单位冲激响应
| f ( t ) f(t) f(t) | x ( t ) x(t) x(t) |
|---|---|
| δ Δ ( t ) \delta_\Delta (t) δΔ(t) | h Δ ( t ) h_\Delta (t) hΔ(t) |
| δ Δ ( t − i Δ T ) \delta_\Delta (t -i\Delta T) δΔ(t−iΔT) | h Δ ( t − i Δ T ) h_\Delta (t -i\Delta T) hΔ(t−iΔT) |
| A δ Δ ( t − i Δ T ) A\delta_\Delta (t -i\Delta T) AδΔ(t−iΔT) | A h Δ ( t − i Δ T ) Ah_\Delta (t -i\Delta T) AhΔ(t−iΔT) |
| Δ T f ( i Δ T ) δ Δ ( t − i Δ T ) \Delta Tf(i\Delta T)\delta_\Delta (t -i\Delta T) ΔTf(iΔT)δΔ(t−iΔT) | Δ T f ( i Δ T ) h Δ ( t − i Δ T ) \Delta Tf(i\Delta T)h_\Delta (t -i\Delta T) ΔTf(iΔT)hΔ(t−iΔT) |
t
=
i
Δ
T
t=i\Delta T
t=iΔT时刻冲激响应:
x
(
t
)
=
∑
i
=
0
I
Δ
T
f
(
i
Δ
T
)
h
Δ
(
t
−
i
Δ
T
)
x(t)=\sum_{i=0}^I \Delta Tf(i\Delta T)h_\Delta (t -i\Delta T)
x(t)=i=0∑IΔTf(iΔT)hΔ(t−iΔT)
lim
Δ
T
→
0
,
h
Δ
(
t
)
→
h
(
t
)
,
Δ
T
=
d
τ
,
i
Δ
T
=
τ
\lim_{\Delta T \to 0},h_\Delta (t) \to h(t),\Delta T=d\tau,i\Delta T=\tau
limΔT→0,hΔ(t)→h(t),ΔT=dτ,iΔT=τ:
x
(
t
)
=
∫
0
t
f
(
τ
)
h
(
t
−
τ
)
d
τ
=
f
(
t
)
∗
h
(
t
)
x(t)=\int_{0}^t f(\tau)h (t -\tau) d \tau=f(t)*h(t)
x(t)=∫0tf(τ)h(t−τ)dτ=f(t)∗h(t)
对冲激响应作拉普拉斯变换
L
[
δ
(
t
)
]
=
1
\mathcal{L}[\delta(t)]=1
L[δ(t)]=1
X
(
s
)
=
1
⋅
H
(
s
)
=
H
(
s
)
X(s)=1\cdot H(s)=H(s)
X(s)=1⋅H(s)=H(s)
4.卷积的拉普拉斯变换
拉普拉斯变换
X
(
s
)
=
L
[
x
(
t
)
]
=
∫
0
∞
x
(
t
)
e
−
s
t
d
t
X(s)=\mathcal{L}[x(t)]=\int_0^\infty x(t)e^{-st}dt
X(s)=L[x(t)]=∫0∞x(t)e−stdt
卷积
x
(
t
)
∗
g
(
t
)
=
∫
0
t
x
(
τ
)
g
(
t
−
τ
)
d
t
x(t)*g(t)=\int_0^t x(\tau)g(t-\tau)dt
x(t)∗g(t)=∫0tx(τ)g(t−τ)dt
证明：
L
[
x
(
t
)
∗
g
(
t
)
]
=
X
(
s
)
G
(
s
)
\mathcal{L}[x(t)*g(t)]=X(s)G(s)
L[x(t)∗g(t)]=X(s)G(s)
L
[
x
(
t
)
∗
g
(
t
)
]
=
∫
0
∞
∫
0
t
x
(
τ
)
g
(
t
−
τ
)
d
τ
e
−
s
t
d
t
=
∫
0
∞
∫
τ
∞
x
(
τ
)
g
(
t
−
τ
)
e
−
s
t
d
t
d
τ
=
∫
0
∞
∫
0
∞
x
(
τ
)
g
(
u
)
e
−
s
(
u
+
τ
)
d
u
d
τ
=
∫
0
∞
x
(
τ
)
e
−
s
τ
d
τ
∫
0
∞
g
(
u
)
e
−
s
u
d
u
=
X
(
s
)
G
(
s
)
\begin{aligned} \mathcal{L}[x(t)*g(t)] & = \int_0^\infty \int_0^t x(\tau)g(t-\tau)d\tau e^{-st}dt\\ & = \int_0^\infty \int_\tau^\infty x(\tau)g(t-\tau)e^{-st}dtd\tau \\ & = \int_0^\infty \int_0^\infty x(\tau)g(u)e^{-s(u+\tau)}dud\tau \\ & = \int_0^\infty x(\tau)e^{-s\tau} d\tau \int_0^\infty g(u)e^{-su}du\\ & = X(s)G(s) \end{aligned}
L[x(t)∗g(t)]=∫0∞∫0tx(τ)g(t−τ)dτe−stdt=∫0∞∫τ∞x(τ)g(t−τ)e−stdtdτ=∫0∞∫0∞x(τ)g(u)e−s(u+τ)dudτ=∫0∞x(τ)e−sτdτ∫0∞g(u)e−sudu=X(s)G(s)
其中，
u
=
t
−
τ
,
d
t
=
d
u
+
d
τ
=
d
u
,
t
∈
[
τ
,
∞
)
,
u
=
t
−
τ
∈
[
0
,
∞
)
u=t-\tau,dt=du+d\tau=du,t \in [\tau,\infty),u=t-\tau \in[0,\infty)
u=t−τ,dt=du+dτ=du,t∈[τ,∞),u=t−τ∈[0,∞)
这是一个二重积分，就是去求体积，改变积分次序
L [ x ( t ) ∗ g ( t ) ] = L [ x ( t ) ] L [ g ( t ) ] = X ( s ) G ( s ) \mathcal{L}[x(t)*g(t)] =\mathcal{L}[x(t)]\mathcal{L}[g(t)] =X(s)G(s) L[x(t)∗g(t)]=L[x(t)]L[g(t)]=X(s)G(s)
5.欧拉公式证明
e i θ = cos θ + i sin θ e^{i\theta}=\cos \theta +i\sin \theta eiθ=cosθ+isinθ
令 f ( θ ) = e i θ cos θ + i sin θ f(\theta)=\frac{e^{i\theta}}{\cos \theta +i\sin \theta} f(θ)=cosθ+isinθeiθ
f
′
(
θ
)
=
i
e
i
θ
(
cos
θ
+
i
sin
θ
)
−
e
i
θ
(
−
sin
θ
+
i
cos
θ
)
(
cos
θ
+
i
sin
θ
)
2
=
0
f^{'}(\theta)=\frac{ie^{i\theta}(\cos \theta +i\sin \theta)-e^{i\theta}(-\sin \theta +i\cos \theta)}{(\cos \theta +i\sin \theta)^2}=0
f′(θ)=(cosθ+isinθ)2ieiθ(cosθ+isinθ)−eiθ(−sinθ+icosθ)=0
说明
f
(
θ
)
f(\theta)
f(θ)为常数
令
f
(
0
)
=
e
0
cos
0
+
i
sin
0
=
1
f(0)=\frac{e^{0}}{\cos 0 +i\sin 0}=1
f(0)=cos0+isin0e0=1
即 e i θ = cos θ + i sin θ e^{i\theta}=\cos \theta +i\sin \theta eiθ=cosθ+isinθ
6.复数的不同表达形式
x
2
−
2
x
+
2
=
0
x^2-2x+2=0
x2−2x+2=0
x
=
1
±
−
1
=
1
±
i
x=1\pm \sqrt{-1}=1\pm i
x=1±−1
=1±i
∣
z
∣
=
a
2
+
b
2
|z|=\sqrt{a^2+b^2}
∣z∣=a2+b2
θ
=
arctan
b
a
\theta =\arctan \frac{b}{a}
θ=arctanab
a
=
∣
z
∣
cos
θ
,
b
=
∣
z
∣
sin
θ
a=|z|\cos \theta,b=|z|\sin \theta
a=∣z∣cosθ,b=∣z∣sinθ
z
=
∣
z
∣
cos
θ
+
∣
z
∣
sin
θ
i
=
∣
z
∣
e
i
θ
z=|z|\cos \theta+|z|\sin \theta i=|z|e^{i\theta}
z=∣z∣cosθ+∣z∣sinθi=∣z∣eiθ
z
1
=
∣
z
1
∣
e
i
θ
z_1=|z_1|e^{i\theta}
z1=∣z1∣eiθ
z
2
=
∣
z
2
∣
e
i
θ
z_2=|z_2|e^{i\theta}
z2=∣z2∣eiθ
z
1
⋅
z
2
=
∣
z
1
∣
∣
z
2
∣
e
i
(
θ
1
+
θ
2
)
z_1\cdot z_2=|z_1||z_2|e^{i(\theta_1+\theta_2)}
z1⋅z2=∣z1∣∣z2∣ei(θ1+θ2)
共轭
z
1
=
a
+
b
i
z_1=a+bi
z1=a+bi
z
2
=
a
−
b
i
z_2=a-bi
z2=a−bi