数字滤波器设计指南
- 数学基础（一）
-
- 指数与对数
-
- 指数
- 对数
- 分贝
- 复数
-
- 复数的矩形形式
- 复数的极坐标形式
- 三角函数
- 微分
- 积分
- 狄拉克
δ
\\delta
δ函数
原书为C. Britton Rorabaugh的 Digital Filter Designer’s Book，此处是我个人的翻译以及一些理解
数学基础（一）
指数与对数
指数
无理数
e
e
e定义为：
e
≜
lim
x
→
+
∞
(
1
+
1
x
)
x
≃
2.71828…
e \\triangleq {\\lim\\limits_{x \\to +\\infty} (1+\\frac{1}{x})^x}\\simeq 2.71828…
e≜x→+∞lim(1+x1)x≃2.71828… 欧拉常数
γ
\\gamma
γ定义为：
γ
=
lim
N
→
∞
(
∑
N
=
1
n
1
n
−
log
e
N
)
≃
0.577215664…
\\gamma = \\lim\\limits_{N \\to \\infty} (\\sum\\limits_{N=1}^n \\frac{1}{n}-\\log_e N)\\simeq 0.577215664…
γ=N→∞lim(N=1∑nn1−logeN)≃0.577215664…
e
e
e的复数幂的级数展开：
e
z
=
∑
n
=
0
∞
z
n
n
!
e^z = \\sum\\limits_{n=0}^\\infty\\frac{z^n}{n!}
ez=n=0∑∞n!zn
对数
以10为底的对数定义：
y
=
log
10
x
⇔
x
=
10
y
y=\\log_{10}x \\Leftrightarrow x=10^y
y=log10x⇔x=10y 自然对数定义：
y
=
log
e
x
⇔
x
=
e
y
y=\\log_e x \\Leftrightarrow x=e^y
y=logex⇔x=ey
对数性质：
ln
x
=
∫
1
x
1
y
d
y
,
x
>
0
\\ln x = \\int_1^x \\frac{1}{y}dy, x>0
lnx=∫1xy1dy,x>0
d
d
x
(
ln
x
)
=
1
x
,
x
>
0
\\frac{d}{dx} (\\ln x)=\\frac{1}{x}, x>0
dxd(lnx)=x1,x>0
log
b
(
x
y
)
=
log
b
x
+
log
b
y
\\log_b (xy) = \\log_b x+\\log_b y
logb(xy)=logbx+logby
log
b
(
1
x
)
=
−
log
b
x
\\log_b (\\frac{1}{x}) = -\\log_b x
logb(x1)=−logbx
log
b
(
y
x
)
=
x
log
b
y
\\log_b (y^x) = x\\log_b y
logb(yx)=xlogby
log
c
x
=
(
log
b
x
)
(
log
c
b
)
=
log
b
x
log
b
c
\\log_c x = (\\log_b x)(\\log_c b)= \\frac{\\log_b x}{\\log_b c}
logcx=(logbx)(logcb)=logbclogbx
ln
(
1
+
z
)
=
∑
n
=
1
∞
(
−
1
)
n
−
1
z
n
n
,
∣
z
∣
<
1
\\ln (1+z) = \\sum\\limits_{n=1}^{\\infty} (-1)^{n-1}\\frac{z^n}{n}, |z| < 1
ln(1+z)=n=1∑∞(−1)n−1nzn,∣z∣<1
分贝
假设一个系统对于给定的输入能量
¶
i
n
\\P_{in}
¶in和输入电压
V
i
n
V_{in}
Vin，对应的输出能量为
P
o
u
t
P_{out}
Pout以及输出电压为
V
o
u
t
V_{out}
Vout，增益
G
G
G的分贝数
d
B
dB
dB定义为：
G
d
B
=
10
log
10
P
o
u
t
P
i
n
=
10
log
10
V
o
u
t
2
/
Z
o
u
t
V
i
n
2
/
Z
i
n
G_{dB} = 10\\log_{10}\\frac{P_{out}}{P_{in}}=10\\log_{10}\\frac{V^2_{out}/Z_{out}}{V^2_{in}/Z_{in}}
GdB=10log10PinPout=10log10Vin2/ZinVout2/Zout 如果输入和输出阻抗相同，则：
G
d
B
=
10
log
10
V
o
u
t
2
V
i
n
2
=
20
log
10
V
o
u
t
V
i
n
G_{dB} =10\\log_{10}\\frac{V^2_{out}}{V^2_{in}}=20\\log_{10}\\frac{V_{out}}{V_{in}}
GdB=10log10Vin2Vout2=20log10VinVout
例：如果一个功放的增益为17.0
d
B
dB
dB，对于一个3-
m
W
mW
mW的输入，输出能量是多少？
17.0
d
B
=
10
log
10
P
o
u
t
3
×
10
−
3
17.0dB =10\\log_{10}\\frac{P_{out}}{3\\times 10^{-3}}
17.0dB=10log103×10−3Pout
P
o
u
t
=
(
3
×
10
−
3
)
10
17
/
10
=
150
m
W
P_{out} = (3\\times 10^{-3})10^{17/10}=150mW
Pout=(3×10−3)1017/10=150mW
例：8位无符号整型能表示的分贝数范围是多少？ 8位无符号整型可表示范围位1~255，对应的
P
o
u
t
=
20
log
10
(
255
1
)
=
48.13
d
B
P_{out} = 20\\log_{10}(\\frac{255}{1})=48.13dB
Pout=20log10(1255)=48.13dB
复数
复数的矩形形式
复数矩形形式定义：
z
=
a
+
b
j
z = a+bj
z=a+bj，其中
a
a
a、
b
b
b为实数，
j
=
−
1
j=\\sqrt{-1}
j=−1
复数的绝对值/模：
∣
z
∣
=
∣
a
+
b
j
∣
=
a
2
+
b
2
|z|=|a+bj|=\\sqrt{a^2+b^2}
∣z∣=∣a+bj∣=a2+b2
复数的共轭：
(
z
=
a
+
b
j
)
⇔
(
z
∗
=
a
−
b
j
)
(z=a+bj)\\Leftrightarrow(z^*=a-bj)
(z=a+bj)⇔(z∗=a−bj) 共轭的加/乘/除：
(
z
1
+
z
2
)
∗
=
z
1
∗
+
z
2
∗
(
z
1
z
2
)
∗
=
z
1
∗
z
2
∗
(
z
1
z
2
)
∗
=
z
1
∗
z
2
∗
(z_1+z_2)^* = z_1^*+z_2^* \\\\ (z_1z_2)^*=z_1^*z_2^* \\\\ (\\frac{z_1}{z_2})^* = \\frac{z_1^*}{z_2^*}
(z1+z2)∗=z1∗+z2∗(z1z2)∗=z1∗z2∗(z2z1)∗=z2∗z1∗ 复数矩形形式四则运算 假设两个复数
z
1
=
a
+
b
j
z_1=a+bj
z1=a+bj、
z
2
=
c
+
d
j
z_2=c+dj
z2=c+dj，复数的矩形形式四则运算：
z
1
+
z
2
=
(
a
+
c
)
+
j
(
b
+
d
)
z
1
−
z
2
=
(
a
−
c
)
+
j
(
b
−
d
)
z
1
z
2
=
(
a
c
−
b
d
)
+
j
(
a
d
+
b
c
)
z
1
z
2
=
a
c
+
b
d
c
2
+
d
2
+
j
b
c
−
a
d
c
2
+
d
2
z_1+z_2 = (a+c)+j(b+d) \\\\ z_1-z_2 = (a-c)+j(b-d) \\\\ z_1z_2 = (ac-bd)+j(ad+bc) \\\\ \\frac{z_1}{z_2}=\\frac{ac+bd}{c^2+d^2}+j \\frac{bc-ad}{c^2+d^2}
z1+z2=(a+c)+j(b+d)z1−z2=(a−c)+j(b−d)z1z2=(ac−bd)+j(ad+bc)z2z1=c2+d2ac+bd+jc2+d2bc−ad
复数的极坐标形式
复数的极坐标形式定义：
a
=
r
c
o
s
θ
,
b
=
r
s
i
n
θ
,
z
=
r
c
o
s
θ
+
j
r
s
i
n
θ
=
r
(
c
o
s
θ
+
j
s
i
n
θ
)
a = rcos\\theta, b = rsin\\theta, z = rcos\\theta+jrsin\\theta = r(cos\\theta+jsin\\theta)
a=rcosθ,b=rsinθ,z=rcosθ+jrsinθ=r(cosθ+jsinθ) 通常的，将
(
c
o
s
θ
+
j
s
i
n
θ
)
(cos\\theta+jsin\\theta)
(cosθ+jsinθ)记作：
c
i
s
θ
cis\\theta
cisθ。因此：
z
=
r
c
i
s
θ
=
r
e
j
θ
z = r cis\\theta = re^{j\\theta}
z=rcisθ=rejθ 复数极坐标的四则运算 假设三个复数：
z
=
r
(
c
o
s
θ
+
j
s
i
n
θ
)
=
r
e
j
θ
z
1
=
r
1
(
c
o
s
θ
1
+
j
s
i
n
θ
1
)
=
r
1
e
j
θ
1
z
2
=
r
2
(
c
o
s
θ
2
+
j
s
i
n
θ
2
)
=
r
2
e
j
θ
2
z = r(cos\\theta+jsin\\theta) = re^{j\\theta} \\\\ z_1 = r_1(cos\\theta_1+jsin\\theta_1) = r_1e^{j\\theta_1} \\\\ z_2 = r_2(cos\\theta_2+jsin\\theta_2) = r_2e^{j\\theta_2}
z=r(cosθ+jsinθ)=rejθz1=r1(cosθ1+jsinθ1)=r1ejθ1z2=r2(cosθ2+jsinθ2)=r2ejθ2 乘法
z
1
z
2
=
r
1
r
2
(
c
o
s
(
θ
1
+
θ
2
)
+
j
s
i
n
(
θ
1
+
θ
2
)
)
=
r
1
r
2
e
j
(
θ
1
+
θ
2
)
z_1z_2 = r_1r_2(cos(\\theta_1+\\theta_2) + jsin(\\theta_1+\\theta_2)) = r_1r_2e^{j(\\theta_1+\\theta_2)}
z1z2=r1r2(cos(θ1+θ2)+jsin(θ1+θ2))=r1r2ej(θ1+θ2) 除法
z
1
z
2
=
r
1
r
2
(
c
o
s
(
θ
1
−
θ
2
)
+
j
s
i
n
(
θ
1
−
θ
2
)
)
=
r
1
r
2
e
j
(
θ
1
−
θ
2
)
\\frac{z_1}{z_2} = \\frac{r_1}{r_2}(cos(\\theta_1-\\theta_2)+jsin(\\theta_1-\\theta_2)) = \\frac{r_1}{r_2}e^{j(\\theta_1-\\theta_2)}
z2z1=r2r1(cos(θ1−θ2)+jsin(θ1−θ2))=r2r1ej(θ1−θ2) 幂
z
n
=
r
n
(
c
o
s
(
n
θ
)
+
j
s
i
n
(
n
θ
)
)
=
r
n
e
j
n
θ
z^n = r^n(cos(n\\theta) + jsin(n\\theta)) = r^ne^{jn\\theta}
zn=rn(cos(nθ)+jsin(nθ))=rnejnθ 根
z
n
=
z
1
/
n
=
r
1
/
n
(
c
o
s
(
θ
+
2
k
π
n
)
+
j
s
i
n
(
θ
+
2
k
π
n
)
)
=
r
1
/
n
e
j
(
θ
+
2
k
π
)
n
,
k
=
0
,
1
,
2
,
.
.
.
\\sqrt[n]{z} = z^{1/n}=r^{1/n}\\left(cos(\\frac{\\theta+2k\\pi}{n}) + jsin(\\frac{\\theta+2k\\pi}{n})\\right) = r^{1/n}e^{\\frac{j(\\theta+2k\\pi)}{n}}, \\ \\ k = 0,1,2,…
nz
=z1/n=r1/n(cos(nθ+2kπ)+jsin(nθ+2kπ))=r1/nenj(θ+2kπ), k=0,1,2,… 对数
ln
(
z
)
=
ln
(
r
e
j
θ
)
=
ln
(
r
e
j
(
θ
+
2
k
π
)
)
=
ln
(
r
)
+
j
(
θ
+
2
k
π
)
,
k
=
0
,
1
,
2
,
.
.
.
\\begin{split} \\ln(z) &= \\ln(re^{j\\theta}) \\\\ &= \\ln(re^{j(\\theta + 2k\\pi)}) \\\\ &= \\ln(r) + j(\\theta+2k\\pi), \\ \\ \\ k = 0,1,2,… \\end{split}
ln(z)=ln(rejθ)=ln(rej(θ+2kπ))=ln(r)+j(θ+2kπ), k=0,1,2,…
三角函数
如图所示，定义：
S
i
n
e
:
s
i
n
θ
=
y
r
C
o
s
i
n
e
:
c
o
s
θ
=
x
r
T
a
n
g
e
n
t
:
t
a
n
θ
=
y
x
C
o
s
e
c
a
n
t
:
c
s
c
θ
=
r
y
S
e
c
a
n
t
:
s
e
c
θ
=
r
x
C
o
t
a
n
g
e
n
t
:
c
o
t
θ
=
x
y
\\begin{split} Sine: \\ \\ sin\\theta = \\frac{y}{r} \\\\ Cosine: \\ \\ cos\\theta = \\frac{x}{r} \\\\ Tangent: \\ \\ tan\\theta = \\frac{y}{x} \\\\ Cosecant: \\ \\ csc\\theta = \\frac{r}{y} \\\\ Secant: \\ \\ sec\\theta = \\frac{r}{x} \\\\ Cotangent: \\ \\ cot\\theta = \\frac{x}{y} \\end{split}
Sine: sinθ=ryCosine: cosθ=rxTangent: tanθ=xyCosecant: cscθ=yrSecant: secθ=xrCotangent: cotθ=yx 正弦曲线的相移
c
o
s
(
ω
t
)
=
s
i
n
(
ω
t
+
π
2
)
c
o
s
(
ω
t
)
=
c
o
s
(
ω
t
+
2
n
π
)
,
n
∈
Z
s
i
n
(
ω
t
)
=
s
i
n
(
ω
t
+
2
n
π
)
,
n
∈
Z
s
i
n
(
ω
t
)
=
c
o
s
(
ω
t
−
π
2
)
c
o
s
(
ω
t
)
=
c
o
s
(
ω
t
+
(
2
n
+
1
)
π
)
,
n
∈
Z
s
i
n
(
ω
t
)
=
−
s
i
n
(
ω
t
+
(
2
n
+
1
)
π
)
,
n
∈
Z
cos(\\omega t) = sin\\left(\\omega t + \\frac{\\pi}{2}\\right) \\\\ cos(\\omega t) = cos\\left(\\omega t + 2n\\pi\\right), \\ \\ n \\in \\Zeta \\\\ sin(\\omega t) = sin\\left(\\omega t + 2n\\pi\\right), \\ \\ n \\in \\Zeta \\\\ sin(\\omega t) = cos\\left(\\omega t – \\frac{\\pi}{2} \\right) \\\\ cos(\\omega t) = cos\\left(\\omega t + (2n+1)\\pi\\right), \\ \\ n \\in \\Zeta \\\\ sin(\\omega t) = -sin\\left(\\omega t + (2n+1)\\pi\\right), \\ \\ n \\in \\Zeta
cos(ωt)=sin(ωt+2π)cos(ωt)=cos(ωt+2nπ), n∈Zsin(ωt)=sin(ωt+2nπ), n∈Zsin(ωt)=cos(ωt−2π)cos(ωt)=cos(ωt+(2n+1)π), n∈Zsin(ωt)=−sin(ωt+(2n+1)π), n∈Z 性质
s
i
n
(
−
x
)
=
−
s
i
n
(
x
)
,
c
o
s
(
−
x
)
=
c
o
s
(
x
)
,
t
a
n
(
−
x
)
=
−
t
a
n
(
x
)
,
c
o
s
2
x
+
s
i
n
2
x
=
1
,
c
o
s
(
2
x
)
=
2
c
o
s
2
x
−
1
,
s
i
n
(
x
±
y
)
=
s
i
n
x
c
o
s
y
±
c
o
s
y
s
i
n
x
,
c
o
s
(
x
±
y
)
=
c
o
s
x
c
o
s
y
∓
s
i
n
x
s
i
n
y
,
t
a
n
(
x
±
y
)
=
t
a
n
x
±
t
a
n
y
1
∓
t
a
n
x
t
a
n
y
,
s
i
n
(
2
x
)
=
2
s
i
n
x
c
o
s
x
,
c
o
s
(
2
x
)
=
c
o
s
2
x
−
s
i
n
2
x
,
t
a
n
(
2
x
)
=
2
t
a
n
x
1
−
t
a
n
2
x
,
s
i
n
x
s
i
n
y
=
1
2
(
−
c
o
s
(
x
+
y
)
+
c
o
s
(
x
−
y
)
)
,
s
i
n
x
c
o
s
y
=
1
2
(
s
i
n
(
x
+
y
)
+
s
i
n
(
x
−
y
)
)
,
c
o
s
x
c
o
s
y
=
1
2
(
c
o
s
(
x
+
y
)
+
c
o
s
(
x
−
y
)
)
,
s
i
n
x
+
s
i
n
y
=
2
s
i
n
x
+
y
2
c
o
s
x
−
y
2
,
s
i
n
x
−
s
i
n
y
=
2
s
i
n
x
−
y
2
c
o
s
x
+
y
2
,
c
o
s
x
+
c
o
s
y
=
2
c
o
s
x
+
y
2
c
o
s
x
−
y
2
,
c
o
s
x
−
c
o
s
y
=
−
2
s
i
n
x
+
y
2
s
i
n
x
−
y
2
,
A
c
o
s
(
ω
t
+
ψ
)
+
B
c
o
s
(
ω
t
+
ϕ
)
=
C
c
o
s
(
ω
t
+
θ
)
,
其中：
C
=
[
A
2
+
B
2
−
2
A
B
c
o
s
(
ϕ
−
ψ
)
]
1
/
2
θ
=
t
a
n
−
1
(
A
s
i
n
ψ
+
B
s
i
n
ϕ
A
c
o
s
ψ
+
B
c
o
s
ϕ
)
A
c
o
s
(
ω
t
+
ψ
)
+
B
s
i
n
(
ω
t
+
ϕ
)
=
C
c
o
s
(
ω
t
+
θ
)
,
其中：
C
=
[
A
2
+
B
2
−
2
A
B
s
i
n
(
ϕ
−
ψ
)
]
1
/
2
θ
=
t
a
n
−
1
(
A
s
i
n
ψ
−
B
c
o
s
ϕ
A
c
o
s
ψ
+
B
s
i
n
ϕ
)
\\begin{split} &sin(-x) = -sin(x), \\\\ &cos(-x) = cos(x), \\\\ &tan(-x) = -tan(x), \\\\ &cos^2x + sin^2x = 1, \\\\ &cos(2x) = 2cos^2x – 1, \\\\ &sin(x\\pm y) = sinxcosy\\pm cosysinx, \\\\ &cos(x\\pm y) = cosxcosy\\mp sinxsiny, \\\\ &tan(x\\pm y) = \\frac{tanx\\pm tany}{1\\mp tanxtany}, \\\\ &sin(2x) = 2sinxcosx, \\\\ &cos(2x) = cos^2x- sin^2x, \\\\ &tan(2x) = \\frac{2tanx}{1 – tan^2x}, \\\\ &sinxsiny = \\frac{1}{2}\\left(-cos(x+y) + cos(x-y) \\right), \\\\ &sinxcosy = \\frac{1}{2}\\left(sin(x+y) + sin(x-y) \\right), \\\\ &cosxcosy = \\frac{1}{2}\\left(cos(x+y) + cos(x-y) \\right), \\\\ &sinx + siny = 2sin\\frac{x+y}{2}cos\\frac{x-y}{2}, \\\\ &sinx – siny = 2sin\\frac{x-y}{2}cos\\frac{x+y}{2}, \\\\ &cosx + cosy = 2cos\\frac{x+y}{2}cos\\frac{x-y}{2}, \\\\ &cosx – cosy = -2sin\\frac{x+y}{2}sin\\frac{x-y}{2}, \\\\ &Acos(\\omega t + \\psi) + Bcos(\\omega t + \\phi) = Ccos(\\omega t + \\theta), \\\\ 其中：&C = \\left[A^2 + B^2 – 2ABcos(\\phi – \\psi) \\right]^{1/2} \\\\ &\\theta = tan^{-1}\\left(\\frac{Asin\\psi + Bsin\\phi}{Acos\\psi + Bcos\\phi}\\right) \\\\ &Acos(\\omega t + \\psi) + Bsin(\\omega t + \\phi) = Ccos(\\omega t + \\theta), \\\\ 其中：&C = \\left[A^2 + B^2 – 2ABsin(\\phi – \\psi) \\right]^{1/2} \\\\ &\\theta = tan^{-1}\\left(\\frac{Asin\\psi – Bcos\\phi}{Acos\\psi + Bsin\\phi}\\right) \\\\ \\end{split}
其中：其中：sin(−x)=−sin(x),cos(−x)=cos(x),tan(−x)=−tan(x),cos2x+sin2x=1,cos(2x)=2cos2x−1,sin(x±y)=sinxcosy±cosysinx,cos(x±y)=cosxcosy∓sinxsiny,tan(x±y)=1∓tanxtanytanx±tany,sin(2x)=2sinxcosx,cos(2x)=cos2x−sin2x,tan(2x)=1−tan2x2tanx,sinxsiny=21(−cos(x+y)+cos(x−y)),sinxcosy=21(sin(x+y)+sin(x−y)),cosxcosy=21(cos(x+y)+cos(x−y)),sinx+siny=2sin2x+ycos2x−y,sinx−siny=2sin2x−ycos2x+y,cosx+cosy=2cos2x+ycos2x−y,cosx−cosy=−2sin2x+ysin2x−y,Acos(ωt+ψ)+Bcos(ωt+ϕ)=Ccos(ωt+θ),C=[A2+B2−2ABcos(ϕ−ψ)]1/2θ=tan−1(Acosψ+BcosϕAsinψ+Bsinϕ)Acos(ωt+ψ)+Bsin(ωt+ϕ)=Ccos(ωt+θ),C=[A2+B2−2ABsin(ϕ−ψ)]1/2θ=tan−1(Acosψ+BsinϕAsinψ−Bcosϕ) 欧拉表示
e
j
x
=
c
o
s
x
+
j
s
i
n
x
,
e
−
j
x
=
c
o
s
x
−
j
s
i
n
x
,
c
o
s
x
=
e
j
x
+
e
−
j
x
2
,
s
i
n
x
=
e
j
x
−
e
−
j
x
2
j
,
e^{jx} = cosx + jsinx, \\\\ e^{-jx} = cosx – jsinx, \\\\ cosx = \\frac{e^{jx} + e^{-jx}}{2}, \\\\ sinx= \\frac{e^{jx} – e^{-jx}}{2j}, \\\\
ejx=cosx+jsinx,e−jx=cosx−jsinx,cosx=2ejx+e−jx,sinx=2jejx−e−jx, 级数与乘积展开
s
i
n
x
=
∑
n
=
0
∞
(
−
1
)
n
x
2
n
+
1
(
2
n
+
1
)
!
,
c
o
s
x
=
∑
n
=
0
∞
(
−
1
)
n
x
2
n
(
2
n
)
!
,
t
a
n
x
=
∑
n
=
1
∞
(
−
1
)
n
−
1
2
2
n
(
2
2
n
−
1
)
B
2
n
x
2
n
−
1
(
2
n
)
!
,
∣
x
∣
<
π
2
c
o
t
x
=
∑
n
=
0
∞
(
−
1
)
n
2
2
n
B
2
n
x
2
n
−
1
(
2
n
)
!
,
∣
x
∣
<
π
s
e
c
x
=
∑
n
=
0
∞
(
−
1
)
n
E
2
n
x
2
n
(
2
n
)
!
,
∣
x
∣
<
π
2
c
s
c
x
=
∑
n
=
0
∞
(
−
1
)
n
−
1
2
(
2
2
n
−
1
−
1
B
2
n
x
2
n
−
1
)
(
2
n
)
!
,
∣
x
∣
<
π
\\begin{split} sinx &= \\sum\\limits_{n=0}^\\infty \\frac{(-1)^nx^{2n+1}}{(2n+1)!}, \\\\ cosx &= \\sum\\limits_{n=0}^\\infty \\frac{(-1)^nx^{2n}}{(2n)!}, \\\\ tanx &= \\sum\\limits_{n=1}^\\infty \\frac{(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}x^{2n-1}}{(2n)!}, |x| < \\frac{\\pi}{2}\\\\ cotx &= \\sum\\limits_{n=0}^\\infty \\frac{(-1)^{n}2^{2n}B_{2n}x^{2n-1}}{(2n)!}, |x| < \\pi \\\\ secx &= \\sum\\limits_{n=0}^\\infty \\frac{(-1)^nE_{2n}x^{2n}}{(2n)!}, |x| < \\frac{\\pi}{2} \\\\ cscx &= \\sum\\limits_{n=0}^\\infty \\frac{(-1)^{n-1}2(2^{2n-1}-1B_{2n}x^{2n-1})}{(2n)!}, |x| < \\pi \\end{split}
sinxcosxtanxcotxsecxcscx=n=0∑∞(2n+1)!(−1)nx2n+1,=n=0∑∞(2n)!(−1)nx2n,=n=1∑∞(2n)!(−1)n−122n(22n−1)B2nx2n−1,∣x∣<2π=n=0∑∞(2n)!(−1)n22nB2nx2n−1,∣x∣<π=n=0∑∞(2n)!(−1)nE2nx2n,∣x∣<2π=n=0∑∞(2n)!(−1)n−12(22n−1−1B2nx2n−1),∣x∣<π 其中，
B
n
B_{n}
Bn为伯努利数，
E
n
E_n
En为欧拉数
s
i
n
x
=
x
∏
n
=
1
∞
(
1
−
x
2
n
2
π
2
)
,
c
o
s
x
=
∏
n
=
1
∞
(
1
−
4
x
2
(
2
n
−
1
)
2
π
2
)
\\begin{split} sinx &= x\\prod\\limits_{n=1}^{\\infty}\\left(1 – \\frac{x^2}{n^2\\pi^2} \\right), \\\\ cosx &= \\prod\\limits_{n=1}^{\\infty}\\left(1 – \\frac{4x^2}{(2n-1)^2\\pi^2} \\right) \\end{split}
sinxcosx=xn=1∏∞(1−n2π2x2),=n=1∏∞(1−(2n−1)2π24x2) 正余弦的正交性 假设存在两个函数分别为
ϕ
1
(
t
)
\\phi_1(t)
ϕ1(t)和
ϕ
2
(
t
)
\\phi_2(t)
ϕ2(t)，若：
∫
0
T
ϕ
1
(
t
)
ϕ
2
(
t
)
d
t
=
0
\\int_0^T \\phi_1(t)\\phi_2(t)dt = 0
∫0Tϕ1(t)ϕ2(t)dt=0 我们称这两个函数在区间
[
0
,
T
]
[0, T]
[0,T]上是正交的，
ϕ
1
(
t
)
\\phi_1(t)
ϕ1(t)和
ϕ
2
(
t
)
\\phi_2(t)
ϕ2(t)形成正交集。如果这两个函数满足：
∫
0
T
ϕ
1
2
(
t
)
d
t
=
∫
0
T
ϕ
2
2
(
t
)
d
t
=
1
\\int_0^T \\phi_1^2(t)dt = \\int_0^T \\phi_2^2(t)dt = 1
∫0Tϕ12(t)dt=∫0Tϕ22(t)dt=1 则称这两个函数是
[
0
,
T
]
[0, T]
[0,T]上的正交规范集。 假设存在两组信号：
ϕ
1
(
t
)
=
A
s
i
n
(
ω
0
t
)
,
ϕ
2
(
t
)
=
A
c
o
s
(
ω
0
t
)
\\phi_1(t) = Asin(\\omega_0t), \\\\ \\phi_2(t) = Acos(\\omega_0t)
ϕ1(t)=Asin(ω0t),ϕ2(t)=Acos(ω0t) 如果
ω
0
T
\\omega_0T
ω0T是
π
\\pi
π的整数倍，那么信号
ϕ
1
\\phi_1
ϕ1 和
ϕ
2
\\phi_2
ϕ2是区间
[
0
,
T
]
[0, T]
[0,T]上的正交集。当
A
2
=
2
/
T
A^2=2/T
A2=2/T时，这两个信号为正交规范集。当
ω
0
T
≫
1
\\omega_0T\\gg1
ω0T≫1且
A
2
=
2
/
T
A^2=2/T
A2=2/T，信号
ϕ
1
\\phi_1
ϕ1 和
ϕ
2
\\phi_2
ϕ2近似正交规范集，推导如下：
∫
0
T
ϕ
1
(
t
)
ϕ
2
(
t
)
d
t
=
A
2
∫
0
T
s
i
n
(
ω
0
t
)
c
o
s
(
ω
0
t
)
d
t
=
A
2
2
∫
0
T
s
i
n
(
ω
0
t
+
ω
0
t
)
+
s
i
n
(
ω
0
t
−
ω
0
t
)
d
t
=
A
2
2
∫
0
T
s
i
n
(
2
ω
0
t
)
=
A
2
2
(
c
o
s
(
2
ω
0
t
)
2
ω
0
)
∣
t
=
0
T
=
A
2
4
ω
0
t
(
1
−
c
o
s
2
ω
0
T
)
\\begin{split} \\int_0^T \\phi_1(t)\\phi_2(t)dt &=A^2\\int_0^T sin(\\omega_0t)cos(\\omega_0t)dt \\\\ &=\\frac{A^2}{2}\\int_0^Tsin(\\omega_0t + \\omega_0t)+sin(\\omega_0t – \\omega_0t)dt \\\\ &=\\frac{A^2}{2}\\int_0^Tsin(2\\omega_0t) = \\frac{A^2}{2}\\left(\\frac{cos(2\\omega_0t)}{2\\omega_0}\\right)|^T_{t=0} \\\\ &=\\frac{A^2}{4\\omega_0t}(1-cos2\\omega_0T) \\end{split}
∫0Tϕ1(t)ϕ2(t)dt=A2∫0Tsin(ω0t)cos(ω0t)dt=2A2∫0Tsin(ω0t+ω0t)+sin(ω0t−ω0t)dt=2A2∫0Tsin(2ω0t)=2A2(2ω0cos(2ω0t))∣t=0T=4ω0tA2(1−cos2ω0T) 因此，当
ω
0
T
\\omega_0T
ω0T是
π
\\pi
π的整数倍时，
c
o
s
2
ω
0
T
=
1
cos2\\omega_0T = 1
cos2ω0T=1，
ϕ
1
\\phi_1
ϕ1 和
ϕ
2
\\phi_2
ϕ2正交。当
ω
0
T
≫
1
\\omega_0T\\gg1
ω0T≫1时，
A
2
4
ω
0
t
(
1
−
c
o
s
2
ω
0
T
)
\\frac{A^2}{4\\omega_0t}(1-cos2\\omega_0T)
4ω0tA2(1−cos2ω0T)将无限趋近于0，
ϕ
1
\\phi_1
ϕ1 和
ϕ
2
\\phi_2
ϕ2近似正交规范集。 信号
ϕ
1
\\phi_1
ϕ1在区间
[
0
,
T
]
[0, T]
[0,T]上的能量：
E
1
=
∫
0
T
ϕ
1
2
(
t
)
d
t
=
A
2
∫
0
T
s
i
n
2
(
ω
0
t
)
d
t
=
A
2
(
T
2
−
s
i
n
(
2
ω
0
T
)
4
ω
0
)
\\begin{split} E_1 &= \\int_0^T \\phi_1^2(t)dt = A^2\\int_0^T sin^2(\\omega_0t)dt \\\\ &= A^2 \\left(\\frac{T}{2} – \\frac{sin(2\\omega_0T)}{4\\omega_0}\\right) \\end{split}
E1=∫0Tϕ12(t)dt=A2∫0Tsin2(ω0t)dt=A2(2T−4ω0sin(2ω0T)) 若
ϕ
1
\\phi_1
ϕ1具有单位能量，则
A
2
A^2
A2必定满足：
A
2
=
(
T
2
−
s
i
n
(
2
ω
0
T
)
4
ω
0
)
−
1
A^2 = \\left(\\frac{T}{2} – \\frac{sin(2\\omega_0T)}{4\\omega_0}\\right)^{-1}
A2=(2T−4ω0sin(2ω0T))−1 若
ω
0
T
=
n
π
\\omega_0T = n\\pi
ω0T=nπ，则
s
i
n
(
2
ω
0
T
)
=
0
sin(2\\omega_0T) = 0
sin(2ω0T)=0，那么有：
A
=
2
T
A = \\sqrt{\\frac{2}{T}}
A=T2
故：
E
1
=
1
−
s
i
n
(
2
ω
0
T
)
2
ω
0
E_1 = 1 – \\frac{sin(2\\omega_0T)}{2\\omega_0}
E1=1−2ω0sin(2ω0T) 对于
ϕ
2
\\phi_2
ϕ2同理可得:
E
2
=
A
2
(
T
2
+
s
i
n
(
2
ω
0
T
)
4
ω
0
)
E_2 = A^2 \\left(\\frac{T}{2} + \\frac{sin(2\\omega_0T)}{4\\omega_0}\\right)
E2=A2(2T+4ω0sin(2ω0T)) 不难看出，当
ω
0
T
≫
1
\\omega_0T\\gg1
ω0T≫1且
A
2
=
2
/
T
A^2=2/T
A2=2/T，
ϕ
1
\\phi_1
ϕ1 和
ϕ
2
\\phi_2
ϕ2近似正交规范集。
微分
常用微分：
d
d
x
s
i
n
u
=
c
o
s
u
d
u
d
x
d
d
x
c
o
s
u
=
−
s
i
n
u
d
u
d
x
d
d
x
t
a
n
u
=
s
e
c
2
u
d
u
d
x
=
1
c
o
s
2
u
d
u
d
x
d
d
x
c
o
t
u
=
c
s
c
2
u
d
u
d
x
=
1
s
i
n
2
u
d
u
d
x
d
d
x
s
e
c
u
=
s
e
c
u
t
a
n
u
d
u
d
x
=
s
i
n
u
c
o
s
2
u
d
u
d
x
d
d
x
c
s
c
u
=
−
c
s
c
u
c
o
t
u
d
u
d
x
=
−
c
o
s
u
s
i
n
2
u
d
u
d
x
d
d
x
e
u
=
e
u
d
u
d
x
d
d
x
l
n
u
=
1
u
d
u
d
x
d
d
x
l
o
g
u
=
l
o
g
e
u
d
u
d
x
d
d
x
(
u
v
)
=
1
v
2
(
v
d
u
d
x
−
u
d
v
d
x
)
\\begin{split} & \\frac{d}{dx}sinu = cosu\\frac{du}{dx} \\\\ & \\frac{d}{dx}cosu = -sinu\\frac{du}{dx} \\\\ & \\frac{d}{dx}tanu = sec^2u\\frac{du}{dx} = \\frac{1}{cos^2u}\\frac{du}{dx} \\\\ & \\frac{d}{dx}cotu = csc^2u\\frac{du}{dx} = \\frac{1}{sin^2u}\\frac{du}{dx} \\\\ & \\frac{d}{dx}secu = secu\\ tanu\\frac{du}{dx} = \\frac{sinu}{cos^2u}\\frac{du}{dx} \\\\ & \\frac{d}{dx}cscu = -cscu\\ cotu\\frac{du}{dx} = -\\frac{cosu}{sin^2u}\\frac{du}{dx} \\\\ & \\frac{d}{dx}e^u= e^u\\frac{du}{dx} \\\\ & \\frac{d}{dx}lnu= \\frac{1}{u}\\frac{du}{dx} \\\\ & \\frac{d}{dx}logu= \\frac{loge}{u}\\frac{du}{dx} \\\\ & \\frac{d}{dx}\\left(\\frac{u}{v}\\right)= \\frac{1}{v^2}\\left(v\\frac{du}{dx}-u\\frac{dv}{dx}\\right) \\end{split}
dxdsinu=cosudxdudxdcosu=−sinudxdudxdtanu=sec2udxdu=cos2u1dxdudxdcotu=csc2udxdu=sin2u1dxdudxdsecu=secu tanudxdu=cos2usinudxdudxdcscu=−cscu cotudxdu=−sin2ucosudxdudxdeu=eudxdudxdlnu=u1dxdudxdlogu=ulogedxdudxd(vu)=v21(vdxdu−udxdv)
积分
常用积分：
∫
1
x
d
x
=
l
n
x
∫
e
a
x
d
x
=
1
a
e
a
x
∫
x
e
a
x
d
x
=
a
x
−
1
a
2
e
a
x
∫
s
i
n
(
a
x
)
d
x
=
−
1
a
c
o
s
(
a
x
)
∫
c
o
s
(
a
x
)
d
x
=
1
a
s
i
n
(
a
x
)
∫
s
i
n
(
a
x
+
b
)
d
x
=
−
1
a
c
o
s
(
a
x
+
b
)
∫
c
o
s
(
a
x
+
b
)
d
x
=
1
a
s
i
n
(
a
x
+
b
)
∫
x
s
i
n
(
a
x
)
d
x
=
−
x
a
c
o
s
(
a
x
)
+
1
a
2
s
i
n
(
a
x
)
∫
x
c
o
s
(
a
x
)
d
x
=
x
a
s
i
n
(
a
x
)
+
1
a
2
c
o
s
(
a
x
)
∫
s
i
n
2
(
a
x
)
d
x
=
x
2
−
s
i
n
(
2
a
x
)
4
a
∫
c
o
s
2
(
a
x
)
d
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a
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a
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u
\\begin{split} &\\int \\frac{1}{x}dx = lnx \\\\ &\\int e^{ax}dx = \\frac{1}{a}e^{ax} \\\\ &\\int xe^{ax}dx = \\frac{ax – 1}{a^2}e^{ax} \\\\ &\\int sin(ax)dx = -\\frac{1}{a}cos(ax) \\\\ &\\int cos(ax)dx = \\frac{1}{a}sin(ax) \\\\ &\\int sin(ax+b)dx = -\\frac{1}{a}cos(ax+b) \\\\ &\\int cos(ax+b)dx = \\frac{1}{a}sin(ax+b) \\\\ &\\int xsin(ax)dx = -\\frac{x}{a}cos(ax) + \\frac{1}{a^2}sin(ax) \\\\ &\\int xcos(ax)dx = \\frac{x}{a}sin(ax) + \\frac{1}{a^2}cos(ax) \\\\ &\\int sin^2(ax)dx = \\frac{x}{2} – \\frac{sin(2ax)}{4a} \\\\ &\\int cos^2(ax)dx = \\frac{x}{2} + \\frac{sin(2ax)}{4a} \\\\ &\\int x^2sin(ax)dx = \\frac{1}{a^3}(2axsin(ax) + 2cos(ax) – a^2x^2cos(ax)) \\\\ &\\int x^2cos(ax)dx = \\frac{1}{a^3}(2axcos(ax) – 2sin(ax) + a^2x^2sin(ax)) \\\\ &\\int sin^3(ax)dx = -\\frac{1}{3}cosx(sin^2x + 2) \\\\ &\\int cos^3(ax)dx = \\frac{1}{3}sinx(cos^2x + 2) \\\\ &\\int sinxcosxdx = \\frac{1}{2}sin^2x \\\\ &\\int sin(mx)cos(nx)dx = -\\frac{cos(m-n)x}{2(m-n)} – \\frac{cos(m+n)x}{2(m+n)} \\\\ &\\int sin^2xcos^2xdx = \\frac{1}{8}\\left(x – \\frac{1}{4}sin4x\\right) \\\\ &\\int sinxcos^mxdx = -\\frac{cos^{m+1}x}{m+1} \\\\ &\\int sin^mxcosxdx = \\frac{sin^{m+1}x}{m+1} \\\\ &\\int cos^mxsin^nxdx = \\frac{cos^{m-1}xsin^{n+1}x}{m+n} + \\frac{m-1}{m+n}\\int cos^{m-2}xsin^nxdx \\ \\ (m\\neq-n) \\\\ &\\int cos^mxsin^nxdx = -\\frac{cos^{m+1}xsin^{n-1}x}{m+n} + \\frac{m-1}{m+n}\\int cos^mxsin^{n-2}xdx \\ \\ (m\\neq-n) \\\\ &\\int udv = uv – \\int vdu \\end{split}
∫x1dx=lnx∫eaxdx=a1eax∫xeaxdx=a2ax−1eax∫sin(ax)dx=−a1cos(ax)∫cos(ax)dx=a1sin(ax)∫sin(ax+b)dx=−a1cos(ax+b)∫cos(ax+b)dx=a1sin(ax+b)∫xsin(ax)dx=−axcos(ax)+a21sin(ax)∫xcos(ax)dx=axsin(ax)+a21cos(ax)∫sin2(ax)dx=2x−4asin(2ax)∫cos2(ax)dx=2x+4asin(2ax)∫x2sin(ax)dx=a31(2axsin(ax)+2cos(ax)−a2x2cos(ax))∫x2cos(ax)dx=a31(2axcos(ax)−2sin(ax)+a2x2sin(ax))∫sin3(ax)dx=−31cosx(sin2x+2)∫cos3(ax)dx=31sinx(cos2x+2)∫sinxcosxdx=21sin2x∫sin(mx)cos(nx)dx=−2(m−n)cos(m−n)x−2(m+n)cos(m+n)x∫sin2xcos2xdx=81(x−41sin4x)∫sinxcosmxdx=−m+1cosm+1x∫sinmxcosxdx=m+1sinm+1x∫cosmxsinnxdx=m+ncosm−1xsinn+1x+m+nm−1∫cosm−2xsinnxdx (m=−n)∫cosmxsinnxdx=−m+ncosm+1xsinn−1x+m+nm−1∫cosmxsinn−2xdx (m=−n)∫udv=uv−∫vdu
狄拉克
δ
\\delta
δ函数
在电力工程中，没有比狄拉克函数更能导致抽象解释的函数了。狄拉克函数，又称作
δ
\\delta
δ函数或冲激函数。单位冲激通常被宽泛地描述为：在原点处具有零宽度和无限幅值，并且其总面积为1的冲激。冲激下的面积等于 1，怎么可能说零乘以无穷等于 1 呢？我们可以给出定义：
∫
−
∞
+
∞
f
n
(
t
)
d
t
=
1
\\int^{+\\infty}_{-\\infty} f_n(t)dt = 1
∫−∞+∞fn(t)dt=1 且
lim
n
→
∞
f
n
(
t
)
=
0
,
t
≠
0
\\lim\\limits_{n\\rightarrow \\infty} f_n(t) = 0, t\\neq0
n→∞limfn(t)=0,t=0 因此，delta函数被定义为：
δ
(
t
)
=
lim
n
→
∞
f
n
(
t
)
\\delta(t) = \\lim\\limits_{n\\rightarrow \\infty} f_n(t)
δ(t)=n→∞limfn(t) 第二种定义方式为：
∫
−
∞
+
∞
f
n
(
t
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d
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=
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且
δ
(
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)
=
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≠
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\\int^{+\\infty}_{-\\infty} f_n(t)dt = 1且\\delta(t) = 0, t\\neq0
∫−∞+∞fn(t)dt=1且δ(t)=0,t=0 第三种定义方式为：
∫
−
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+
∞
δ
(
t
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f
(
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d
t
=
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(
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)
\\int^{+\\infty}_{-\\infty} \\delta(t)f(t)dt = f(0)
∫−∞+∞δ(t)f(t)dt=f(0)
