状态空间平均建模——Flyback

状态空间平均建模——Flyback

假设变换器工作再CCM模式下，电路如下图所示：

求平均变量

开关变换器再一个周期内有多个状态，如在CCM模式下根据开关是否闭合可分为两个状态：开关闭合状态1、开关打开状态2。针对第n个状态的等效电路列写对应的状态方程和输出方程如下：

${\dot{\text{X}}}_n(t)={\text{A}}_n\text{x}(t)+{\text{B}}_n\text{u}(t)$

${\text{y}}_n(t)={\text{C}}_n\text{x}(t)+{\text{E}}_n\text{u}(t)$

其中：

1.$\text{x}(t)$为状态向量，选取电感电流和电容电压为状态变量：$i(t)$,$v(t)$

2.$\text{u}(t)$为输入向量，为网络的独立输入量，如电源电压，二极管压降（可以看成电压源）这里为：$v_g(t)$,$v_D(t)$

3.$\dot{\text{x}}(t)$为状态向量的一阶导数：$\frac{di(t)}{dt}$,$\frac{dv(t)}{dt}$

4.$\text{y}(t)$为输出向量，可以自由选择受控变量，主要是输出电感电流和输出电容电压。

5.A，B为状态矩阵和输入矩阵

6.C，E为输出矩阵和传输矩阵

为了消除开关纹波影响，对状态变量${\text{x}}_n(t)$在一个开关周期内取平均，建立平均状态变量的状态方程如下：

$<\text{x}(t){>}_{T_s}=\frac{1}{T_s}{\int }_t^{t+T_s}x(\tau )d\tau$

因此得到：
$$\begin{array}{rl}<\dot{\text{x}}(t){>}_{T_s}& =\frac{d<\text{x}(t){>}_{T_s}}{dt}\\ & =\frac{d}{dt}(\frac{1}{T_s}{\int }_t^{t+T_s}x(\tau )d\tau )\\ & =\frac{1}{T_s}{\int }_t^{t+T_s}(\frac{d\text{x}(\tau )}{d\tau })d\tau \\ & =\frac{1}{T_s}{\int }_t^{t+T_s}\dot{\text{x}}(\tau )d\tau \\ & =\frac{1}{T_s}{\int }_t^{t+d{T}_s}\dot{\text{x}}(\tau )d\tau +\frac{1}{T_s}{\int }_{t+d{T}_s}^{t+T_s}\dot{\text{x}}(\tau )d\tau \\ & =\frac{1}{T_s}{\int }_t^{t+d{T}_s}[{\text{A}}_1\text{x}(\tau )+{\text{B}}_1\text{u}(\tau )]d\tau +\frac{1}{T_s}{\int }_{t+d{T}_s}^{t+T_s}[{\text{A}}_2\text{x}(\tau )+{\text{B}}_2\text{u}(\tau )]d\tau \\ & \approx \frac{1}{T_s}{\int }_t^{t+d{T}_s}[{\text{A}}_1<\text{x}(\tau ){>}_{T_s}+{\text{B}}_1<\text{u}(\tau ){>}_{T_s}]d\tau +\\ & \frac{1}{T_s}{\int }_{t+d{T}_s}^{t+T_s}[{\text{A}}_2<\text{x}(\tau ){>}_{T_s}+{\text{B}}_2<\text{u}(\tau ){>}_{T_s}]d\tau \\ & \approx \frac{1}{T_s}\{[{\text{A}}_1<\text{x}(\tau ){>}_{T_s}+{\text{B}}_1<\text{u}(\tau ){>}_{T_s}]d{T}_s+[{\text{A}}_2<\text{x}(\tau ){>}_{T_s}+{\text{B}}_2<\text{u}(\tau ){>}_{T_s}](1-d)T_s\}\\ & =[d(t){\text{A}}_1+d^{\prime }(t){\text{A}}_2]<\text{x}(t){>}_{T_s}+[d(t){\text{B}}_1+d^{\prime }(t){\text{B}}_2]<\text{u}(t){>}_{T_s}\end{array}$$
同理可得：

$<\text{y}(t){>}_{T_s}=[d(t){\text{C}}_1+d^{\prime }{\text{C}}_2]<\text{x}(t){>}_{T_s}+[d(t){\text{E}}_1+d^{\prime }{\text{E}}_2]<\text{u}(t){>}_{T_s}$

在Flyback电路中有：

$\text{x}(t)=\left[\begin{array}{c}i(t)\\ v(t)\end{array}\right]$

$\text{u}(t)=\left[\begin{array}{c}{v}_g(t)\\ v_D(t)\end{array}\right]$

$\text{y}(t)=\left[\begin{array}{c}{i}_g(t)\\ v(t)\end{array}\right]$

其中，$i(t)$为电感流过的电流，$V(t)$为输出电压，$v_g(t)$为输入电压，$i_g(t)$为输入电流，$V_D(t)$为二极管的压降

工作状态一，开关闭合

开关闭合示意图如下所示：

此时开关关闭，反激变压器副边侧电压为上负下正，二极管处于反向截至状态则：$i_g(t)=i(t)$对电感和电容列些微分方程，有：

$v_L=v_g-i\ast R_L=L\frac{di}{dt}⟶\frac{di}{dt}=\frac{v_g}{L}-\frac{R_L}{L}\ast i$

$i_C=-\frac{V}{R}=c\frac{dv}{dt}⟶\frac{dv}{dt}=-\frac{V}{RC}$

由此可得状态方程：

$\left[\begin{array}{c}\frac{di}{dt}\\ \frac{dv}{dt}\end{array}\right]=\left[\begin{array}{cc}-\frac{R_L}{L}& 0\\ 0& -\frac{1}{RC}\end{array}\right]\left[\begin{array}{c}i(t)\\ v(t)\end{array}\right]+\left[\begin{array}{cc}\frac{1}{L}& 0\\ 0& 0\end{array}\right]\left[\begin{array}{c}{v}_g(t)\\ v_D(t)\end{array}\right]$

输出方程：

$\left[\begin{array}{c}{i}_g\\ v\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}i(t)\\ v(t)\end{array}\right]+\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\left[\begin{array}{c}{v}_g(t)\\ v_D(t)\end{array}\right]$

由此可得矩阵${\text{A}}_1$、${\text{B}}_1$、${\text{C}}_1$、${\text{E}}_1$为：

${\text{A}}_1=\left[\begin{array}{cc}-\frac{R_L}{L}& 0\\ 0& -\frac{1}{RC}\end{array}\right]$

${\text{B}}_1=\left[\begin{array}{cc}\frac{1}{L}& 0\\ 0& 0\end{array}\right]$

${\text{C}}_1=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

${\text{E}}_1=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

工作状态二，开关断开有：

开关闭合电路示意图如下所示：

此时，开关断开，电感电流维持不变，电感电压方向突变，上负下正，反激变压器原边侧电压上负下正，副边侧电压上正下负，二极管导通，导通压降为$V_D$，$i_g=0$。

对变压器有原边侧电压：$V_s=\frac{1}{n}\ast V_p=\frac{1}{n}(v-v_D)$

电流关系：$i_s=i=n{i}_p$

对电感和电容列微分方程：

$v_L=V_s+R_L\ast i=\frac{1}{n}(v-v_D)+R_L\ast i=-L\frac{di}{dt}⟶\frac{di}{dt}=-\frac{R_L}{L}\ast i-\frac{V}{nL}+\frac{V_D}{nL}$

$i_C=\frac{1}{n}i-\frac{v}{R}=C\frac{dv}{dt}⟶\frac{dv}{dt}=\frac{1}{nC}i-\frac{1}{RC}v$

由此可得状态方程：

$\left[\begin{array}{c}\frac{di}{dt}\\ \frac{dv}{dt}\end{array}\right]=\left[\begin{array}{cc}-\frac{R_L}{L}& -\frac{1}{nL}\\ \frac{1}{nC}& -\frac{1}{RC}\end{array}\right]\left[\begin{array}{c}i(t)\\ v(t)\end{array}\right]+\left[\begin{array}{cc}0& \frac{1}{nL}\\ 0& 0\end{array}\right]\left[\begin{array}{c}{v}_g(t)\\ v_D(t)\end{array}\right]$

输出方程：

$\left[\begin{array}{c}{i}_g\\ v\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}i(t)\\ v(t)\end{array}\right]+\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\left[\begin{array}{c}{v}_g(t)\\ v_D(t)\end{array}\right]$

因此得到矩阵${\text{A}}_2$、${\text{B}}_2$、${\text{C}}_2$、${\text{E}}_2$

${\text{A}}_2=\left[\begin{array}{cc}-\frac{R_L}{L}& -\frac{1}{nL}\\ \frac{1}{nC}& -\frac{1}{RC}\end{array}\right]$

$\textbf{B}_2= \begin{bmatrix} 0 & \frac{1}{nL} \0 & 0 \end{bmatrix} $

${\text{C}}_2=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$

${\text{E}}_2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

求静态工作点

$\text{A}=d{\text{A}}_1+d^{\prime }{\text{A}}_2=\left[\begin{array}{cc}\frac{R_L}{L}& -\frac{1-d}{nL}\\ \frac{1-d}{nC}& \frac{1}{RC}\end{array}\right]$

$\text{B}=d{\text{B}}_1+d^{\prime }{\text{B}}_2=\left[\begin{array}{cc}\frac{d}{L}& \frac{1-d}{nL}\\ 0& 0\end{array}\right]$

$\text{C}=d{\text{C}}_1+d^{\prime }{\text{C}}_2=\left[\begin{array}{cc}d& 0\\ 0& 1\end{array}\right]$

$\text{E}=d{\text{E}}_1+d^{\prime }{\text{E}}_2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

等效电路

电感的平均值方程：

$<v_L{>}_{T_S}=L\frac{d<i(t){>}_{T_S}}{dt}=\frac{(1-d)(V_D-v)}{n}+d{v}_g-R_{L}i$

电容平均值方程：

$<i_C{>}_{T_S}=C\frac{d<v(t){>}_{T_s}}{dt}=\frac{1-d}{n}i-\frac{v}{R}$

稳态时一个周期内$<v_L{>}_{T_S}=0$,$<i_C{>}_{T_S}=0$得到：

$\frac{(1-d)(V_D-v)}{n}+d{v}_g-R_{L}i=0$

$\frac{1-d}{n}i-\frac{v}{R}=0$

可得等效电路：

进一步用变压器等效得：

分离扰动

为了得到开关变换器的等效电路和传递函数，需要分离出直流分量和扰动分量。小信号的传递函数实际上是对于扰动量的传递函数。

对平均向量做分解：

$<\text{x}(t){>}_{T_S}=\text{x}+\hat{\text{x}}(t)$

$<\dot{\text{x}}(t){>}_{T_S}=\dot{\text{x}}+\widehat{\dot{\text{x}}}(t)$

$<\text{u}(t){>}_{T_S}=\text{u}+\hat{\text{u}}(t)$

$<\text{y}(t){>}_{T_S}=\text{y}+\hat{\text{y}}(t)$

$d(t)=D+\hat{d}(t)$

$d^{\prime }(t)=1-d(t)=D^{\prime }-\hat{d}(t)$

带入$<\dot{\text{x}}(t){>}_{T_S}$和$<\text{y}(t){>}_{T_s}$表达式，得到：

< x ˙ ( t ) > T S = x ˙ + x ˙ ^ ( t ) = { [ D + d ^ ( t ) ] A 1 + [ 1 − ( D + d ^ ( t ) ) ] A 2 } { X + x ^ ( t ) } + { [ D + d ^ ( t ) ] B 1 + [ 1 − ( D + d ^ ( t ) ) ] B 2 } { U + u ^ ( t ) } = Ax + Bu + A x ^ ( t ) + B u ^ ( t ) + [ ( A 1 − A 2 ) x + ( B 1 − B 2 ) u ] d ^ ( t ) + ( A 1 − A 2 ) x ^ ( t ) d ^ ( t ) + ( B 1 − B 2 ) u ^ ( t ) d ^ ( t ) \begin{align} \nonumber <\dot{\textbf{x}}{(t)}>_{T_S} = & \dot{\textbf{x}} +\hat{\dot{\textbf{x}}}(t) \\ =& \nonumber \{ [D + \hat{d}(t)]A_1 +[1-(D+\hat d (t))]A_2\}\{ \textbf X + \hat{\textbf x}(t) \} \\ \nonumber & + \{ [D + \hat{d}(t)]B_1 +[1-(D+\hat d (t))]B_2\}\{ \textbf U + \hat{\textbf u}(t) \} \\ \nonumber =& \textbf{A}\textbf{x}+\textbf{B}\textbf{u} +\textbf{A} \hat{\textbf{x}}(t)+ \textbf{B} \hat{\textbf{u}}(t)+[(\textbf{A}_1-\textbf{A}_2)\textbf{x}+(\textbf{B}_1-\textbf{B}_2)\textbf{u}]\hat{d}(t) \\ \nonumber & + (\textbf{A}_1-\textbf{A}_2)\hat{\textbf{x}}(t)\hat{d}(t)+(\textbf{B}_1-\textbf{B}_2)\hat{\textbf{u}}(t)\hat{d}(t) \end{align} <x˙(t)>TS​​===​x˙+x˙^(t){[D+d^(t)]A1​+[1−(D+d^(t))]A2​}{X+x^(t)}+{[D+d^(t)]B1​+[1−(D+d^(t))]B2​}{U+u^(t)}Ax+Bu+Ax^(t)+Bu^(t)+[(A1​−A2​)x+(B1​−B2​)u]d^(t)+(A1​−A2​)x^(t)d^(t)+(B1​−B2​)u^(t)d^(t)​

< y ( t ) > T S = y + y ^ ( t ) = { [ D + d ^ ( t ) ] C 1 + [ 1 − ( D + d ^ ( t ) ) ] C 2 } { X + x ^ ( t ) } + { [ D + d ^ ( t ) ] E 1 + [ 1 − ( D + d ^ ( t ) ) ] E 2 } { U + u ^ ( t ) } = Cx + Eu + C x ^ ( t ) + E u ^ ( t ) + [ ( C 1 − C 2 ) x + ( E 1 − E 2 ) u ] d ^ ( t ) + ( C 1 − C 2 ) x ^ ( t ) d ^ ( t ) + ( E 1 − E 2 ) u ^ ( t ) d ^ ( t ) \begin{align} <{\textbf{y}}{(t)}>_{T_S}= &{\textbf{y}} +\hat{{\textbf{y}}}(t) \\ \nonumber =& \{ [D + \hat{d}(t)]C_1 +[1-(D+\hat d (t))]C_2\}\{ \textbf X + \hat{\textbf x}(t) \} \\ \nonumber & + \{ [D + \hat{d}(t)]E_1 +[1-(D+\hat d (t))]E_2\}\{ \textbf U + \hat{\textbf u}(t) \} \\ \nonumber =& \textbf{C}\textbf{x}+\textbf{E}\textbf{u} +\textbf{C} \hat{\textbf{x}}(t)+ \textbf{E} \hat{\textbf{u}}(t) +[(\textbf{C}_1-\textbf{C}_2)\textbf{x}+(\textbf{E}_1-\textbf{E}_2)\textbf{u}]\hat{d}(t) \\ \nonumber & + (\textbf{C}_1-\textbf{C}_2)\hat{\textbf{x}}(t)\hat{d}(t)+(\textbf{E}_1-\textbf{E}_2)\hat{\textbf{u}}(t)\hat{d}(t) \end{align} <y(t)>TS​​===​y+y^​(t){[D+d^(t)]C1​+[1−(D+d^(t))]C2​}{X+x^(t)}+{[D+d^(t)]E1​+[1−(D+d^(t))]E2​}{U+u^(t)}Cx+Eu+Cx^(t)+Eu^(t)+[(C1​−C2​)x+(E1​−E2​)u]d^(t)+(C1​−C2​)x^(t)d^(t)+(E1​−E2​)u^(t)d^(t)​​

忽略小信号二阶分量，对上式进行线性近似:

$\dot{\text{x}}+\widehat{\dot{\text{x}}}(t)=\text{A}\text{x}+\text{B}\text{u}+\text{A}\hat{\text{x}}(t)+\text{B}\hat{\text{u}}(t)+[({\text{A}}_1-{\text{A}}_2)\text{x}+({\text{B}}_1-{\text{B}}_2)\text{u}]\hat{d}(t)$

$\text{y}+\hat{\text{y}}(t)=\text{C}\text{x}+\text{E}\text{u}+\text{C}\hat{\text{x}}(t)+\text{E}\hat{\text{u}}(t)+[({\text{C}}_1-{\text{C}}_2)\text{x}+({\text{E}}_1-{\text{E}}_2)\text{u}]\hat{d}(t)$

直流稳态方程

由此可得直流量表达式为：

$\dot{\text{x}}=\text{A}\text{x}+\text{B}\text{u}$

$\text{y}=\text{C}\text{x}+\text{E}\text{u}$

令稳态时直流分量$\text{x}$为常数，$\dot{\text{x}}=0$，则：

$\text{x}=-{\text{A}}^{-1}\text{B}\text{u}$

$\text{y}=(\text{E}-\text{C}{\text{A}}^{-1}\text{B})\text{u}$

扰动量的表达式为：
$$\begin{array}{rl}<\widehat{\dot{\text{x}}}(t){>}_{T_S}=& \text{A}\hat{\text{x}}(t)+\text{B}\hat{\text{u}}(t)+[({\text{A}}_1-{\text{A}}_2)\text{x}+({\text{B}}_1-{\text{B}}_2)\text{u}]\hat{d}(t)\\ & +({\text{A}}_1-{\text{A}}_2)\hat{\text{x}}(t)\hat{d}(t)+({\text{B}}_1-{\text{B}}_2)\hat{\text{u}}(t)\hat{d}(t)\end{array}$$

$$\begin{array}{rl}<\hat{\text{y}}(t){>}_{T_S}=& \text{C}\hat{\text{x}}(t)+\text{E}\hat{\text{u}}(t)+[({\text{C}}_1-{\text{C}}_2)\text{x}+({\text{E}}_1-{\text{E}}_2)\text{u}]\hat{d}(t)\\ & +({\text{C}}_1-{\text{C}}_2)\hat{\text{x}}(t)\hat{d}(t)+({\text{E}}_1-{\text{E}}_2)\hat{\text{u}}(t)\hat{d}(t)\end{array}$$

对Flyback变换器，我们令：

$\text{x}=\left[\begin{array}{c}i\\ v\end{array}\right]$

$\text{u}=\left[\begin{array}{c}{v}_g\\ v_D\end{array}\right]$

$\text{y}=\left[\begin{array}{c}{i}_g\\ v\end{array}\right]$

则变换器的静态工作点可以求出：

$\text{x}=\left[\begin{array}{c}I\\ V\end{array}\right]=-{\left[\begin{array}{cc}\frac{R_L}{L}& -\frac{1-d}{nL}\\ \frac{1-d}{nC}& \frac{1}{RC}\end{array}\right]}^{-1}\left[\begin{array}{cc}\frac{d}{L}& \frac{1-d}{nL}\\ 0& 0\end{array}\right]\left[\begin{array}{c}{v}_g\\ v_D\end{array}\right]$

$\text{y}=\left[\begin{array}{c}{I}_g\\ V\end{array}\right]=(\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]-\left[\begin{array}{cc}d& nL\\ 0& 1\end{array}\right]{\left[\begin{array}{cc}\frac{R_L}{L}& -\frac{1-d}{nL}\\ \frac{1-d}{nC}& \frac{1}{RC}\end{array}\right]}^{-1}\left[\begin{array}{cc}\frac{d}{L}& \frac{1-d}{nL}\\ 0& 0\end{array}\right])\left[\begin{array}{c}{v}_g\\ v_D\end{array}\right])$

解得电感电流稳态值$I$,输出电压$V$,输入电流$I_g$：

$I=\frac{n(V_D-d{V}_D+nd{V}_g)}{d^{2}R-2dR+n^2{R}_L+R}$

$V=\frac{(1-d)R(V_D-d{V}_D+nd{V}_g)}{d^{2}R-2dR+n^2{R}_L+R}$

$I_g=\frac{nd(V_D-d{V}_D+nd{V}_g)}{d^{2}R-2dR+n^2{R}_L+R}$

则电压增益：

$M=\frac{V}{V_g}=\frac{(1-d)R(V_D-d{V}_D+nd{V}_g)}{(d^{2}R-2dR+n^2{R}_L+R)V_g}$

理想情况时：$V_D=0,R_L=0$则:$M=\frac{V}{V_g}=\frac{nd}{1-d}$

线性化交流小信号方程

对于$\widehat{\dot{\text{x}}}(t)$和$\hat{\text{y}}(t)$的表达式进行线性近似得到交流小信号模型：

$\widehat{\dot{\text{x}}}(t)=\text{A}\hat{\text{x}}(t)+\text{B}\hat{\text{u}}(t)+[({\text{A}}_1-{\text{A}}_2)\text{x}+({\text{B}}_1-{\text{B}}_2)\text{u}]\hat{d}(t)$

$\hat{\text{y}}(t)=\text{C}\hat{\text{x}}(t)+\text{E}\hat{\text{u}}(t)+[({\text{C}}_1-{\text{C}}_2)\text{x}+({\text{E}}_1-{\text{E}}_2)\text{u}]\hat{d}(t)$

解得：
$$\begin{array}{rl}\widehat{\dot{\text{x}}}(t)=\left[\begin{array}{c}\frac{d\hat{i}}{dt}\\ \frac{d\hat{v}}{dt}\end{array}\right]=& \left[\begin{array}{cc}\frac{R_L}{L}& -\frac{1-d}{nL}\\ \frac{1-d}{nC}& \frac{1}{RC}\end{array}\right]\left[\begin{array}{c}\hat{i}\\ \hat{v}\end{array}\right]+\left[\begin{array}{cc}\frac{d}{L}& \frac{1-d}{nL}\\ 0& 0\end{array}\right]\left[\begin{array}{c}{\hat{v}}_g\\ {\hat{v}}_D\end{array}\right]\\ & +(\left[\begin{array}{cc}0& \frac{1}{nL}\\ \frac{-1}{nC}& 0\end{array}\right]\left[\begin{array}{c}i\\ v\end{array}\right]+\left[\begin{array}{cc}\frac{1}{L}& \frac{-1}{nL}\\ 0& 0\end{array}\right]\left[\begin{array}{c}{v}_g\\ v_D\end{array}\right])\hat{d}(t)\end{array}$$

小信号分析

设二极管压降$V_D$为确定值不发生变化，即：${\hat{V}}_D=0$

则：$\hat{\text{u}}=\left[\begin{array}{c}\widehat{v_g}\\ 0\end{array}\right]$

对线性化后得表达式做拉普拉斯变换，得到：

$s\hat{\text{x}}(s)=\text{A}\hat{\text{x}}(s)+\text{B}\hat{\text{u}}(s)+[({\text{A}}_1-{\text{A}}_2)\text{x}+({\text{B}}_1-{\text{B}}_2)\text{u}]\hat{d}(s)$

$\hat{\text{y}}(s)=\text{C}\hat{\text{x}}(s)+\text{E}\hat{\text{u}}(s)+[({\text{C}}_1-{\text{C}}_2)\text{x}+({\text{E}}_1-{\text{E}}_2)\text{u}]\hat{d}(s)$

联立方程，解得：

$\hat{\text{x}}(s)=(s\text{I}-A{)}^{-1}\text{B}\hat{\text{u}}(s)+(s\text{I}-A{)}^{-1}[({\text{A}}_1-{\text{A}}_2)\text{x}+({\text{B}}_1-{\text{B}}_2)\text{u}]\hat{d}(s)$

y ^ ( s ) = ( C ( s I − A ) − 1 B + E ) u ^ ( s ) + { C ( s I − A ) − 1 [ ( A 1 − A 2 ) x + ( B 1 − B 2 ) u ] + [ ( C 1 − C 2 ) x + ( E 1 − E 2 ) u ] } d ^ ( s ) \begin{align} \nonumber \hat{\textbf{y}}{(s)}= (\textbf{C}(s\textbf{I}-A)^{-1} \textbf{B} +\textbf{E}) \hat{\textbf{u}}(s) + \{ \textbf{C}(s\textbf{I}-A)^{-1}[(\textbf{A}_1-\textbf{A}_2)\textbf{x}+(\textbf{B}_1-\textbf{B}_2)\textbf{u}] + \\ \nonumber [(\textbf{C}_1-\textbf{C}_2)\textbf{x}+(\textbf{E}_1-\textbf{E}_2)\textbf{u}]\} \hat{d}(s) \end{align} y^​(s)=(C(sI−A)−1B+E)u^(s)+{C(sI−A)−1[(A1​−A2​)x+(B1​−B2​)u]+[(C1​−C2​)x+(E1​−E2​)u]}d^(s)​

得到四个传递函数：

当控制变量无扰动时，可得状态变量$\hat{\text{X}}(s)$对输入变量${\hat{v}}_g(s)$的传递函数：

$G(s)=\frac{\hat{\text{X}}(s)}{{\hat{v}}_g(s)}{|}_{\hat{d}(s)=0}=(S\text{I}-A{)}^{-1}\text{B}$

当输入向量无扰动时，可得状态变量$\hat{\text{X}}(s)$对控制变量$\hat{d}(s)$的传递函数：

$G(s)=\frac{\hat{\text{X}}(s)}{\hat{d}(s)}{|}_{{\hat{v}}_g(s)=0}=(S\text{I}-A{)}^{-1}[({\text{A}}_1-{\text{A}}_2)\text{x}+({\text{B}}_1-{\text{B}}_2)\text{u}]$

当控制量无扰动时，可得输出变量$\hat{\text{y}}(s)$对输入变量${\hat{v}}_g(s)$的传递函数：

$G(s)=\frac{\hat{\text{y}}(s)}{{\hat{v}}_g(s)}{|}_{\hat{d}(s)=0}=\text{C}(S\text{I}-A{)}^{-1}\text{B}+\text{E}$

当输入向量无扰动时，可得输出变量$\hat{\text{y}}(s)$对控制变量$\hat{d}(s)$的传递函数：
$$\begin{array}{r}G(s)=\frac{\hat{\text{y}}(s)}{\hat{d}(s)}{|}_{{\hat{v}}_g(s)=0}=\text{C}(S\text{I}-A{)}^{-1}[({\text{A}}_1-{\text{A}}_2)\text{x}+({\text{B}}_1-{\text{B}}_2)\text{u}]+\\ [({\text{C}}_1-{\text{C}}_2)\text{x}+({\text{E}}_1-{\text{E}}_2)\text{u}]\end{array}$$

matlab计算代码

clc
clear
close all
syms  i v RL L d n C R Vg VD  S %A1 B1 C1 E1 A2 B2 C2 E2 

A1=[-(RL)/L 0 ; 0 -1/(R*C)];
B1 = [1/L 0 ; 0 0];
C1 = [1 0 ; 0 1];
E1 = [0 0 ; 0 0];

A2 = [-RL/L -1/(n*L) ; 1/(n*C) -1/(R*C)];
B2 = [ 0 1/(n*L);0 0];
C2 = [0 0 ; 0 1];
E2= [0 0 ; 0 0 ];
u = [Vg ; VD];
x = [i ; v];

A = simplify(d*A1+(1-d)*A2)
B = simplify(d*B1+(1-d)*B2)
C = simplify(d*C1 +(1-d)*C2)
E = simplify(d*E1 + (1-d)*E2)

x2 = simplify(A*x + B*u)
x1 = simplify(- inv(A) * B *[Vg ;VD])
y=simplify((E -C * inv(A) * B )*[Vg;VD])

G = simplify( C*inv(S*eye(2) - A)* ((A1 - A2)*x1 + (B1-B2)* u ) + ((C1-C2)*x1 + (E1 - E2)*u ))

参考文献

开关变换器建模——状态空间平均法文章来源地址https://www.toymoban.com/news/detail-659605.html

到了这里，关于状态空间平均建模——Flyback的文章就介绍完了。如果您还想了解更多内容，请在右上角搜索TOY模板网以前的文章或继续浏览下面的相关文章，希望大家以后多多支持TOY模板网！