传递函数的幅频特性计算方法

对于这个知识点首先需要回顾一下复数的相关知识

对于任意一个不为零的复数z=a+bi的辐角有无限多个值，且这些值相差为 2 π 2\pi 2π的整数倍。将适合于 − π ≤ θ ≤ π -\pi\le\theta\le\pi −π≤θ≤π的辐角的值称为辐角的主值。其指数形式记作： z = r ( c o s θ + i s i n θ ) = r e i θ z = r(cos\theta +isin\theta) = re^{i\theta} z=r(cosθ+isinθ)=reiθ

对于一个开环传递函数 G ( j ω ) G(j\omega) G(jω)那么此时就可以有如下表达式 G ( j ω ) = X o ( j ω ) X i ( j ω ) = A o e j ϕ o ( ω ) A i e j ϕ i ( ω ) = A ( ω ) e j ϕ ( ω ) G(j\omega) = \frac{X_o(j\omega)}{X_i(j\omega)}= \frac {A_o e^{j\phi_o(\omega)}}{A_i e^{j\phi_i(\omega)}} = A(\omega)e^{j\phi(\omega)} G(jω)=Xi​(jω)Xo​(jω)​=Ai​ejϕi​(ω)Ao​ejϕo​(ω)​=A(ω)ejϕ(ω)

A ( ω ) = A o A i A(\omega) = \frac{A_o}{A_i} A(ω)=Ai​Ao​​为幅频特性 ϕ ( ω ) = ϕ o ( ω ) − ϕ i ( ω ) \phi(\omega) = \phi_o(\omega)-\phi_i(\omega) ϕ(ω)=ϕo​(ω)−ϕi​(ω)为相频特性

对于传递函数 G 1 ( s ) = 1 s ( s + 1 ) G_1(s) = \frac{1}{s(s+1)} G1​(s)=s(s+1)1​ G 1 ( j ω ) = 1 j ω ( j ω + 1 ) G_1(j\omega) = \frac{1}{j\omega (j\omega +1)} G1​(jω)=jω(jω+1)1​ ∣ G 1 ( s ) ∣ = 1 ω ω 2 + 1 |G_1(s)| = \frac{1}{\omega \sqrt{\omega^2+1}} ∣G1​(s)∣=ωω2+1 ​1​ ∠ G 1 ( s ) = 0 − ( 90 ° + a r c t a n ( ω 1 ) ) \angle G_1(s) = 0 - (90°+arctan(\frac{\omega}{1})) ∠G1​(s)=0−(90°+arctan(1ω​))