数字特征相关
| X ( t ) X(t) X(t) | R X ( τ ) R_X(\tau) RX(τ) | S X ( ω ) S_X(\omega) SX(ω) |
|---|---|---|
| a X ( t ) aX(t) aX(t) | ∣ a ∣ 2 R X ( τ ) \lvert{a}\rvert^2R_X(\tau) ∣a∣2RX(τ) | ∣ a ∣ 2 S X ( ω ) \lvert{a}\rvert^2S_X(\omega) ∣a∣2SX(ω) |
| d X ( t ) d t \frac{dX(t)}{dt} dtdX(t) | − d 2 R X ( τ ) d τ 2 -\frac{d^2R_X(\tau)}{d\tau^2} −dτ2d2RX(τ) | ω 2 S X ( ω ) \omega^2S_X(\omega) ω2SX(ω) |
| d n X ( t ) d t n \frac{d^nX(t)}{dt^n} dtndnX(t) | ( − 1 ) n d 2 n R X ( τ ) d τ 2 n (-1)^n\frac{d^{2n}R_X(\tau)}{d\tau^{2n}} (−1)ndτ2nd2nRX(τ) | ω 2 n S X ( ω ) \omega^{2n}S_X(\omega) ω2nSX(ω) |
| X ( t ) e ± j ω 0 t X(t)e^{\pm j\omega_0t} X(t)e±jω0t | R X ( τ ) e ± j ω 0 τ R_X(\tau)e^{\pm j\omega_0\tau} RX(τ)e±jω0τ | S ( ω ∓ ω 0 ) S(\omega\mp\omega_0) S(ω∓ω0) |
积分相关
一个积分的计算
求 ∫ − ∞ + ∞ e − x 2 d x \int^{+\infty}_{-\infty}e^{-x^2}{\rm d}x ∫−∞+∞e−x2dx 的值
傅里叶变换相关
常用连续傅里叶变换对
f ( n ) ( t ) ⟺ ( j ω ) n F ( j ω ) f^{(n)}(t)\Longleftrightarrow(j\omega)^nF(j\omega) f(n)(t)⟺(jω)nF(jω)
1 ⟺ 2 π δ ( ω ) 1\Longleftrightarrow2\pi\delta(\omega) 1⟺2πδ(ω)
δ ( ω ) ⟺ 1 \delta(\omega)\Longleftrightarrow 1 δ(ω)⟺1
cos ( ω 0 t ) ⟺ π [ δ ( ω + ω 0 ) + δ ( ω − ω 0 ) ] \cos(\omega_0t)\Longleftrightarrow \pi[\delta(\omega+\omega_0)+\delta(\omega-\omega_0)] cos(ω0t)⟺π[δ(ω+ω0)+δ(ω−ω0)]
e j ω 0 t ⟺ 2 π δ ( ω − ω 0 ) e^{j\omega_0 t}\Longleftrightarrow 2\pi\delta(\omega-\omega_0) ejω0t⟺2πδ(ω−ω0)
e − a ∣ t ∣ , R e { a } > 0 ⟺ 2 a ω 2 + a 2 e^{-a|t|}, \rm{Re}\{a\}>0\Longleftrightarrow \frac{2a}{\omega^2+a^2} e−a∣t∣,Re{ a}>0⟺ω2+a22a
连续傅里叶变换性质
三角函数相关
积化和差公式
cos α ⋅ cos β = 1 2 [ cos ( α + β ) + cos ( α − β ) ] \cos\alpha\cdot\cos\beta=\frac{1}{2}[\cos(\alpha+\beta)+\cos(\alpha-\beta)] cosα⋅cosβ=21[cos(α+β)+cos(α−β)]
sin α ⋅ sin β = 1 2 [ cos ( α + β ) − cos ( α − β ) ] \sin\alpha\cdot\sin\beta=\frac{1}{2}[\cos(\alpha+\beta)-\cos(\alpha-\beta)] sinα⋅sinβ=21[cos(α+β)−cos(α−β)]
sin α ⋅ cos β = 1 2 [ sin ( α + β ) + sin ( α − β ) ] \sin\alpha\cdot\cos\beta=\frac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)] sinα⋅cosβ=21[sin(α+β)+sin(α−β)]
cos α ⋅ sin β = 1 2 [ sin ( α + β ) + sin ( α − β ) ] \cos\alpha\cdot\sin\beta=\frac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)] cosα⋅sinβ=21[sin(α+β)+sin(α−β)]
来源：oschina
链接：https://my.oschina.net/u/4266314/blog/4703354