传递函数幅频特性计算
对于这个知识点首先需要回顾一下复数的相关知识
复数的辐角
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
开环传递函数
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)=Aiejϕi(ω)Aoejϕo(ω)=A(ω)ejϕ(ω)
A
(
ω
)
=
A
o
A
i
A(\omega) = \frac{A_o}{A_i}
A(ω)=AiAo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
例题
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+11
∠
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))