# 階躍響應

## 一階線性電路的階躍響應

${\displaystyle V_{c}(s)={\frac {1/Cs}{R+1/Cs}}V_{in}(s)={\frac {1}{1+RCs}}V_{in}(s)={\frac {1}{1+\tau s}}V_{in}(s)}$

${\displaystyle V_{in}(t)=0,t\leq 0}$
${\displaystyle V_{in}(t)=V_{in},t>0}$

${\displaystyle V_{c}(t)=V_{in}(1-e^{-{\frac {t}{\tau }}})}$

## 反馈放大器的阶跃响应

${\displaystyle A_{FB}={\frac {A_{OL}}{1+\beta A_{OL}}}}$

AOL = 开环增益
AFB = 闭环增益（存在负反馈时的增益）
β = 反馈因子。

### 有一个主导极点

${\displaystyle A_{OL}={\frac {A_{0}}{1+j\omega \tau }}}$

${\displaystyle S_{OL}(t)=A_{0}(1-e^{-t/\tau })}$

${\displaystyle A_{FB}={\frac {A_{0}}{1+\beta A_{0}}}\times {\frac {1}{1+j\omega {\frac {\tau }{1+\beta A_{0}}}}}}$

${\displaystyle S_{FB}(t)={\frac {A_{0}}{1+\beta A_{0}}}(1-e^{-t(1+\beta A_{0})/\tau })}$

### 双极点放大器

${\displaystyle A_{OL}={\frac {A_{0}}{(1+j\omega \tau _{1})(1+j\omega \tau _{2})}},}$

#### 分析

${\displaystyle A_{FB}={\frac {A_{0}}{1+\beta A_{0}}}\times {\frac {1}{1+j\omega {\frac {\tau _{1}+\tau _{2}}{1+\beta A_{0}}}+(j\omega )^{2}{\frac {\tau _{1}\tau _{2}}{1+\beta A_{0}}}}}}$

${\displaystyle A_{FB}={\frac {A_{0}}{\tau _{1}\tau _{2}}}\times {\frac {1}{s^{2}+s\left({\frac {1}{\tau _{1}}}+{\frac {1}{\tau _{2}}}\right)+{\frac {1+\beta A_{0}}{\tau _{1}\tau _{2}}}}}}$

${\displaystyle 2s=-\left({\frac {1}{\tau _{1}}}+{\frac {1}{\tau _{2}}}\right)}$
${\displaystyle \pm {\sqrt {\left({\frac {1}{\tau _{1}}}-{\frac {1}{\tau _{2}}}\right)^{2}-{\frac {4\beta A_{0}}{\tau _{1}\tau _{2}}}}},}$

${\displaystyle s_{\pm }=-\rho \pm j\mu ,\,}$

${\displaystyle \rho ={\frac {1}{2}}\left({\frac {1}{\tau _{1}}}+{\frac {1}{\tau _{2}}}\right),}$

${\displaystyle \mu ={\frac {1}{2}}{\sqrt {{\frac {4\beta A_{0}}{\tau _{1}\tau _{2}}}-\left({\frac {1}{\tau _{1}}}-{\frac {1}{\tau _{2}}}\right)^{2}}}}$

${\displaystyle |s|=|s_{\pm }|={\sqrt {\rho ^{2}+\mu ^{2}}},}$

${\displaystyle \cos \phi ={\frac {\rho }{|s|}},\sin \phi ={\frac {\mu }{|s|}}}$

${\displaystyle e^{-\rho t}\sin(\mu t)\quad {\text{and}}\quad }$ ${\displaystyle e^{-\rho t}\cos(\mu t),}$

${\displaystyle S(t)=\left({\frac {A_{0}}{1+\beta A_{0}}}\right)\left(1-e^{-\rho t}\ {\frac {\sin \left(\mu t+\phi \right)}{\sin(\phi )}}\right)\ ,}$

${\displaystyle S(t)=1-e^{-\rho t}\ {\frac {\sin \left(\mu t+\phi \right)}{\sin(\phi )}}\ }$

A0 趋于无穷大时，反馈系数 β 为1。

#### 结果

1. 反馈控制了给定开环放大器并给定开环时间常数 τ1 与 τ2 时终值上下振荡的幅度。
2. 开环放大器决定了稳定时间。它确定了图3中的时标，开环放大器越快，时标越快。

#### 控制过冲

${\displaystyle S_{\max }=1+\exp \left(-\pi {\frac {\rho }{\mu }}\right)}$

${\displaystyle {\frac {4\beta A_{0}}{\tau _{1}\tau _{2}}}=\left({\frac {1}{\tau _{1}}}-{\frac {1}{\tau _{2}}}\right)^{2}}$

x = ( τ1 / τ2 )1 / 2 ，可以求解二個時間常數之間的比例，結果為

${\displaystyle x={\sqrt {\beta A_{0}}}+{\sqrt {\beta A_{0}+1}}\,}$

${\displaystyle {\frac {\tau _{1}}{\tau _{2}}}=4\beta A_{0}}$

${\displaystyle {\frac {\tau _{1}}{\tau _{2}}}=\alpha \beta A_{0},}$

α可以依允許的過沖量來設計。

#### 稳定时间控制

${\displaystyle S(t)\leq 1+\Delta ,\,}$

${\displaystyle \Delta =e^{-\rho t_{S}}{\text{ or }}t_{S}={\frac {\ln \left({\frac {1}{\Delta }}\right)}{\rho }}=\tau _{2}{\frac {2\ln \left({\frac {1}{\Delta }}\right)}{1+{\frac {\tau _{2}}{\tau _{1}}}}}\approx 2\tau _{2}\ln \left({\frac {1}{\Delta }}\right),}$

#### 相位裕度

${\displaystyle |\beta A_{\text{OL}}(f_{\text{0 db}})|=1\ }$

${\displaystyle f_{\text{0 dB}}=\beta A_{0}f_{1}\,}$

${\displaystyle \phi _{m}=180^{\circ }-\arctan(f_{\text{0 dB}}/f_{1})-\arctan(f_{\text{0 dB}}/f_{2})\ }$

${\displaystyle \phi _{m}=90^{\circ }-\arctan(f_{\text{0 dB}}/f_{2})\,}$
${\displaystyle =90^{\circ }-\arctan \left({\frac {\beta A_{0}f_{1}}{\alpha \beta A_{0}f_{1}}}\right)}$
${\displaystyle =90^{\circ }-\arctan \left({\frac {1}{\alpha }}\right)=\arctan \left(\alpha \right)}$

## 數學定義

• ${\displaystyle \textstyle t\in T}$是系統的，為方便說明，簡稱為時間
• ${\displaystyle \textstyle {\boldsymbol {x}}|_{t}\in M}$是系統在時間${\displaystyle t\,}$時的，為方便說明，稱為輸出。
• ${\displaystyle \textstyle \Phi :T\times M\longrightarrow M}$是動態系統的。
• ${\displaystyle \textstyle \Phi (0,{\boldsymbol {x}})={\boldsymbol {x}}_{0}\in M}$是動態系統的。
• ${\displaystyle \textstyle H(t)\,}$单位阶跃函数

### 非線性動態系統

${\displaystyle {\boldsymbol {x}}|_{t}=\Phi _{\{H(t)\}\left(t,{{\boldsymbol {x}}_{0}}\right)}\,}$

### 線性動態系統

${\displaystyle a(t)={h*H}(t)={H*h}(t)=\int _{-\infty }^{+\infty }h(\tau )H(t-\tau )\,d\tau =\int _{-\infty }^{t}h(\tau )\,d\tau }$

${\displaystyle h(t)={\frac {d}{dt}}\,a(t)}$

${\displaystyle a(t)={h*H}(t)={H*h}(t)=\int _{-\infty }^{+\infty }h(\tau )H(t-\tau )\,d\tau =\int _{-\infty }^{t}h(\tau )\,d\tau }$

## 参考文献与注释

1. Yuriy Shmaliy. Continuous-Time Systems. Springer Science & Business Media. 23 September 2007: 46–. ISBN 978-1-4020-6272-8.
2. ^ Benjamin C Kuo & Golnaraghi F. Automatic control systems Eighth. New York: Wiley. 2003: 253 [2015-12-20]. ISBN 0-471-13476-7. （原始内容存档于2009-05-30）.
3. ^ Benjamin C Kuo & Golnaraghi F. p. 259. [2015-12-20]. ISBN 0-471-13476-7. （原始内容存档于2009-05-30）.
4. ^ 這個估計有些保守，因為在過沖量S ( t )中，1 /sin(φ)用1 /sin(φ) ≈ 1取代
5. ^ David A. Johns & Martin K W. Analog integrated circuit design. New York: Wiley. 1997: 234–235 [2015-12-24]. ISBN 0-471-14448-7. （原始内容存档于2009-05-30）.
6. ^ Willy M C Sansen. Analog design essentials. Dordrecht, The Netherlands: Springer. 2006: §0528 p. 163 [2015-12-24]. ISBN 0-387-25746-2. （原始内容存档于2009-05-30）.