Boost電路：穩態和小信號分析

穩態分析

參考張衛平的《開關變換器的建模與控制》的1.3章節內容；伏秒平衡：在穩態下，一個開關周期內電感電流的增量是0，即
$dIL(t)dt=0\frac{dI_{L}(t)}{dt} = 0$ 。電荷平衡：在穩態下，一個開關周期內電容電壓的增量是0，即
$dVC(t)dt=0\frac{dV_{C}(t)}{dt} = 0$ 。 $Vg(t)=VgV_{g}\left. (t \right.) = V_{g}$ + $v^g(t){\widehat{v}}_{g}\left. (t \right.)$ ， $IL(t)=IL+i^L(t)I_{L}\left. (t \right.) = I_{L} + {\widehat{i}}_{L}\left. (t \right.)$ ， $VC(t)=VC+v^c(t)V_{C}\left. (t \right.) = V_{C} + {\widehat{v}}_{c}\left. (t \right.)$ ， $V(t)=V+v^(t)V\left. (t \right.) = V + \widehat{v}\left. (t \right.)$ ， $Iload(t)=Iload+i^load(t)I_{load}\left. (t \right.) = I_{load} + {\widehat{i}}_{load}\left. (t \right.)$ ,
$D(t)=D+d^(t)D\left. (t \right.) = D + \widehat{d}\left. (t \right.)$ 。上式中前者表示直流量，后者表示擾動。

在穩態下，應該用直流量 $?X,U,Y?\ \mathbf{X,U,Y\ }$ 來代替 $?x(t),u(t),y(t)?\ \mathbf{x}(\mathbf{t}),\mathbf{u(t),y(t)\ }$ 來代替狀態變量，然后得到新的穩態下的狀態空間方程，如下所示：
$[00]=A?X+B?UY=CX+EU{\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \mathbf{A\ X + B\ U} }\mathbf{Y = CX + EU}$

$[00]=A?X+B?U[00]=[?RL1?DRon?(1?D)RC1L1?(1?D)L1(1?D)C10][ILVC]+[1L1RC1L1(1?D)?(1?D)L10?1C10][VgIloadVf]{\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \mathbf{A\ X + B\ U} }{\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{- R_{L1} - DR_{on} - (1 - D)R_{C1}}{L_{1}} & \frac{- (1 - D)}{L_{1}} \\ \frac{(1 - D)}{C_{1}} & 0 \end{bmatrix}\begin{bmatrix} I_{L} \\ V_{C} \end{bmatrix} + \begin{bmatrix} \frac{1}{L_{1}} & \frac{R_{C1}}{L_{1}}(1 - D) & \frac{- (1 - D)}{L_{1}} \\ 0 & \frac{- 1}{C_{1}} & 0 \end{bmatrix}\begin{bmatrix} V_{g} \\ I_{load} \\ V_{f} \end{bmatrix}}$

將其拆分為兩項，分別是：
$\frac{- R_{L1} - DR_{on} - (1 - D)R_{C1}}{L_{1}}I_{L} - \frac{(1 - D)}{L_{1}}V_{C} + \frac{1}{L_{1}}V_{g} + \frac{R_{C1}(1 - D)}{L_{1}}I_{load} - \frac{(1 - D)}{L_{1}}V_{f} }{0 = \frac{(1 - D)}{C_{1}}I_{L} - \frac{1}{C_{1}}I_{load}}$

后一項者可得
$IL=Iload(1?D)I_{L} = \frac{I_{load}}{(1 - D)}$

代入前一項可得
$\frac{- R_{L1} - DR_{on} - (1 - D)R_{C1}}{L_{1}(1 - D)}I_{load} - \frac{(1 - D)}{L_{1}}V_{C} + \frac{1}{L_{1}}V_{g} + \frac{R_{C1}}{L_{1}}(1 - D)\ I_{load} - \frac{(1 - D)}{L_{1}}V_{f} }{\frac{(1 - D)}{L_{1}}V_{C} = \frac{- R_{L1} - DR_{on} - (1 - D)R_{C1}}{L_{1}(1 - D)}I_{load} - \frac{(1 - D)}{L_{1}}V_{C} + \frac{1}{L_{1}}V_{g} + \frac{R_{C1}}{L_{1}}(1 - D)\ I_{load} - \frac{(1 - D)}{L_{1}}V_{f}}$

可得
$VC=?RL1?DRon?(1?D)RC1(1?D)2Iload+Vg(1?D)+RC1Iload?VfV_{C} = \frac{- R_{L1} - DR_{on} - (1 - D)R_{C1}}{{(1 - D)}^{2}}I_{load} + \frac{V_{g}}{(1 - D)} + R_{C1}I_{load} - V_{f}$

輸出方程的穩態表達式如下
$[V]=[(1?D)RC11][ILVC]+[0?RC10][VgIloadVf]\lbrack V\rbrack = \left\lbrack (1 - D)\begin{matrix} R_{C1} & 1 \end{matrix} \right\rbrack\begin{bmatrix} I_{L} \\ V_{C} \end{bmatrix} + \begin{bmatrix} 0 & - R_{C1} & 0 \end{bmatrix}\begin{bmatrix} V_{g} \\ I_{load} \\ V_{f} \end{bmatrix}$

可得
$D)R_{C1}I_{L} + V_{C} - R_{C1}I_{load} }{\ = (1 - D)R_{C1}\frac{I_{load}}{(1 - D)} + \frac{- R_{L1} - DR_{on} - (1 - D)R_{C1}}{(1 - D)^{2}}I_{load} + \frac{V_{g}}{(1 - D)} + R_{C1}I_{load} - V_{f} - R_{C1}I_{load} }{= R_{C1}I_{load} + \frac{- R_{L1} - DR_{on} - (1 - D)R_{C1}}{(1 - D)^{2}}I_{load} + \frac{V_{g}}{(1 - D)} + R_{C1}I_{load} - V_{f} - R_{C1}I_{load} }{= R_{C1}I_{load} - \frac{R_{C1}}{(1 - D)}I_{load} - \frac{R_{L1} + DR_{on}}{(1 - D)^{2}}I_{load} + \frac{V_{g}}{(1 - D)} - V_{f} }{= \frac{V_{g}}{(1 - D)} - V_{f} - \frac{D}{(1 - D)}R_{C1}I_{load} - \frac{1}{(1 - D)^{2}}R_{L1}I_{load} - \frac{D}{(1 - D)^{2}}R_{on}I_{load}}$

由上式可知，由于 $V_{f}$ ， $R_{C1}$ ， $R_{L1}$ ， $R_{on}$ 的存在，使得輸出電壓穩態值是和輸出電流 $I_{load}$ 相關聯的。換言之，不同的輸出電流，會讓D發生調整，也會影響后續的各類小信號傳遞函數。

也可以用矩陣的形式直接計算，如下所示：
$X=?A?1B?UY=CX+EU=(E?CA?1B)?U{\mathbf{X =} - \mathbf{A}^{\mathbf{- 1}}\mathbf{B\ U} }{\mathbf{Y = CX + EU =}\left( \mathbf{E} - \mathbf{C}\mathbf{A}^{\mathbf{- 1}}\mathbf{B} \right)\mathbf{\ U}}$

對應的MATLAB腳本如下

小信號模型分析

參考張衛平的《開關變換器的建模與控制》的1.3章節內容；完成穩態下的狀態空間方程后，可以繼續推導小信號模型，此時不僅用擾動量 $?x^(t),?u^(t),y^(t)?\ \widehat{\mathbf{x}}\mathbf{(t),\ }\widehat{\mathbf{u}}\mathbf{(t),}\widehat{\mathbf{y}}\mathbf{(t)\ }$ 來代替 $?x(t),u(t),y(t)?\ \mathbf{x}(\mathbf{t}),\mathbf{u(t),y(t)\ }$ 來代替狀態變量，具體如下：
$dx^(t)dt=A?x^(t)+B?u^(t)+{(A1?A2)X?(B1?B2)U}d^(t)y^(t)=Cx^(t)+Eu^(t)+{(C1?C2)X?(E1?E2)U}d^(t){\frac{d\widehat{\mathbf{x}}\mathbf{(t)}}{dt} = \mathbf{A\ }\widehat{\mathbf{x}}\mathbf{(t) + B\ }\widehat{\mathbf{u}}\mathbf{(t) +}\left\{ \left( \mathbf{A}_{\mathbf{1}}\mathbf{-}\mathbf{A}_{\mathbf{2}} \right)\mathbf{X}\mathbf{-}\left( \mathbf{B}_{\mathbf{1}}\mathbf{-}\mathbf{B}_{\mathbf{2}} \right)\mathbf{U} \right\}\widehat{\mathbf{d}}\mathbf{(t)} }{\widehat{\mathbf{y}}\mathbf{(t) = C}\widehat{\mathbf{x}}\mathbf{(t) + E}\widehat{\mathbf{u}}\mathbf{(t) +}\left\{ \left( \mathbf{C}_{\mathbf{1}}\mathbf{-}\mathbf{C}_{\mathbf{2}} \right)\mathbf{X}\mathbf{-}\left( \mathbf{E}_{\mathbf{1}}\mathbf{-}\mathbf{E}_{\mathbf{2}} \right)\mathbf{U} \right\}\widehat{\mathbf{d}}\mathbf{(t)}}$

其中， $u^(t)\widehat{\mathbf{u}}(\mathbf{t})$ 擾動信號， $y^(t)\widehat{\mathbf{y}}(\mathbf{t})$ 是輸出信號， $d^(t)\widehat{\mathbf{d}}\mathbf{(t)}$ 是控制信號。接下來對上式進行拉普拉斯變換，可得( $?I\ \mathbf{I}$ 為單位矩陣)：
$sIx^(t)=A?x^(s)+B?u^(s)+[(A1?A2)X?(B1?B2)U]d^(s)y^(s)=Cx^(s)+Eu^(s)+[(C1?C2)X?(E1?E2)U]d^(s){\mathbf{sI}\widehat{\mathbf{x}}\mathbf{(t)} = \mathbf{A\ }\widehat{\mathbf{x}}\mathbf{(s) + B\ }\widehat{\mathbf{u}}\mathbf{(s) +}\left\lbrack \left( \mathbf{A}_{\mathbf{1}}\mathbf{-}\mathbf{A}_{\mathbf{2}} \right)\mathbf{X}\mathbf{-}\left( \mathbf{B}_{\mathbf{1}}\mathbf{-}\mathbf{B}_{\mathbf{2}} \right)\mathbf{U} \right\rbrack\widehat{\mathbf{d}}\mathbf{(s)} }{\widehat{\mathbf{y}}\mathbf{(s) = C}\widehat{\mathbf{x}}\mathbf{(s) + E}\widehat{\mathbf{u}}\mathbf{(s) +}\left\lbrack \left( \mathbf{C}_{\mathbf{1}}\mathbf{-}\mathbf{C}_{\mathbf{2}} \right)\mathbf{X}\mathbf{-}\left( \mathbf{E}_{\mathbf{1}}\mathbf{-}\mathbf{E}_{\mathbf{2}} \right)\mathbf{U} \right\rbrack\widehat{\mathbf{d}}\mathbf{(s)}}$

解得(下面會繼續化簡)：
$x^(t)=(sI?A)?1?B?u^(s)+(sI?A)?1[(A1?A2)X?(B1?B2)U]d^(s)\widehat{\mathbf{x}}\mathbf{(t)} = \left( \mathbf{s}\mathbf{I} - \mathbf{A} \right)^{\mathbf{- 1}}\mathbf{\ B\ }\widehat{\mathbf{u}}\mathbf{(s) +}\left( \mathbf{s}\mathbf{I} - \mathbf{A} \right)^{\mathbf{- 1}}\left\lbrack \left( \mathbf{A}_{\mathbf{1}}\mathbf{-}\mathbf{A}_{\mathbf{2}} \right)\mathbf{X}\mathbf{-}\left( \mathbf{B}_{\mathbf{1}}\mathbf{-}\mathbf{B}_{\mathbf{2}} \right)\mathbf{U} \right\rbrack\widehat{\mathbf{d}}\mathbf{(s)}$

將上式代入輸出方程，得到
$y^(s)=Cx^(s)+Eu^(s)+[(C1?C2)X?(E1?E2)U]d^(s)=C{(sI?A)?1Bu^(s)+(sI?A)?1[(A1?A2)X?(B1?B2)U]d^(s)}+Eu^(s)+[(C1?C2)X?(E1?E2)U]d^(s)={C[(sI?A)?1?B]+E}u^(s)+{C(sI?A)?1?[(A1?A2)X?(B1?B2)U]+[(C1?C2)X?(E1?E2)U]}d^(s)?{\widehat{\mathbf{y}}\left( \mathbf{s} \right)\mathbf{= C}\widehat{\mathbf{x}}\left( \mathbf{s} \right)\mathbf{+ E}\widehat{\mathbf{u}}\left( \mathbf{s} \right)\mathbf{+}\left\lbrack \left( \mathbf{C}_{\mathbf{1}}\mathbf{-}\mathbf{C}_{\mathbf{2}} \right)\mathbf{X}\mathbf{-}\left( \mathbf{E}_{\mathbf{1}}\mathbf{-}\mathbf{E}_{\mathbf{2}} \right)\mathbf{U} \right\rbrack\widehat{\mathbf{d}}\left( \mathbf{s} \right) }{\mathbf{= C}\left\{ \left( \mathbf{s}\mathbf{I} - \mathbf{A} \right)^{\mathbf{- 1}}\mathbf{B}\widehat{\mathbf{u}}\left( \mathbf{s} \right)\mathbf{+}\left( \mathbf{s}\mathbf{I} - \mathbf{A} \right)^{\mathbf{- 1}}\left\lbrack \left( \mathbf{A}_{\mathbf{1}}\mathbf{-}\mathbf{A}_{\mathbf{2}} \right)\mathbf{X}\mathbf{-}\left( \mathbf{B}_{\mathbf{1}}\mathbf{-}\mathbf{B}_{\mathbf{2}} \right)\mathbf{U} \right\rbrack\widehat{\mathbf{d}}\left( \mathbf{s} \right) \right\}\mathbf{+ E}\widehat{\mathbf{u}}\left( \mathbf{s} \right)\mathbf{+}\left\lbrack \left( \mathbf{C}_{\mathbf{1}}\mathbf{-}\mathbf{C}_{\mathbf{2}} \right)\mathbf{X}\mathbf{-}\left( \mathbf{E}_{\mathbf{1}}\mathbf{-}\mathbf{E}_{\mathbf{2}} \right)\mathbf{U} \right\rbrack\widehat{\mathbf{d}}\left( \mathbf{s} \right) }{\mathbf{=}\left\{ \mathbf{C}\left\lbrack \left( \mathbf{s}\mathbf{I} - \mathbf{A} \right)^{\mathbf{- 1}}\mathbf{\ B} \right\rbrack\mathbf{+}\mathbf{E} \right\}\widehat{\mathbf{u}}\left( \mathbf{s} \right)\mathbf{+}\left\{ \mathbf{C}\left( \mathbf{s}\mathbf{I} - \mathbf{A} \right)^{\mathbf{- 1}}\mathbf{\ }\left\lbrack \left( \mathbf{A}_{\mathbf{1}}\mathbf{-}\mathbf{A}_{\mathbf{2}} \right)\mathbf{X}\mathbf{-}\left( \mathbf{B}_{\mathbf{1}}\mathbf{-}\mathbf{B}_{\mathbf{2}} \right)\mathbf{U} \right\rbrack\mathbf{+}\left\lbrack \left( \mathbf{C}_{\mathbf{1}}\mathbf{-}\mathbf{C}_{\mathbf{2}} \right)\mathbf{X}\mathbf{-}\left( \mathbf{E}_{\mathbf{1}}\mathbf{-}\mathbf{E}_{\mathbf{2}} \right)\mathbf{U} \right\rbrack \right\}\widehat{\mathbf{d}}\left( \mathbf{s} \right)\mathbf{\ }}$

將其進一步的化簡為：
$y^(s)=︸C[(sI?A)?1?B+E]Ru^(s)+︸{C(sI?A)?1?[(A1?A2)X?(B1?B2)U]+[(C1?C2)X?(E1?E2)U]}Kd^(s)\widehat{\mathbf{y}}\left( \mathbf{s} \right)\mathbf{=}\underset{\mathbf{R}}{\overset{\mathbf{C}\left\lbrack \left( \mathbf{s}\mathbf{I} - \mathbf{A} \right)^{\mathbf{- 1}}\mathbf{\ B}\mathbf{+}\mathbf{E} \right\rbrack}{︸}}\widehat{\mathbf{u}}\left( \mathbf{s} \right)\mathbf{+}\underset{\mathbf{K}}{\overset{\left\{ \mathbf{C}\left( \mathbf{s}\mathbf{I} - \mathbf{A} \right)^{\mathbf{- 1}}\mathbf{\ }\left\lbrack \left( \mathbf{A}_{\mathbf{1}}\mathbf{-}\mathbf{A}_{\mathbf{2}} \right)\mathbf{X}\mathbf{-}\left( \mathbf{B}_{\mathbf{1}}\mathbf{-}\mathbf{B}_{\mathbf{2}} \right)\mathbf{U} \right\rbrack\mathbf{+}\left\lbrack \left( \mathbf{C}_{\mathbf{1}}\mathbf{-}\mathbf{C}_{\mathbf{2}} \right)\mathbf{X}\mathbf{-}\left( \mathbf{E}_{\mathbf{1}}\mathbf{-}\mathbf{E}_{\mathbf{2}} \right)\mathbf{U} \right\rbrack \right\}}{︸}}\widehat{\mathbf{d}}\left( \mathbf{s} \right)$

化簡得到：
$y^(s)=Ru^(s)+Kd^(s)[v^(s)]=[R11(s)R12(s)0][v^g(s)i^load(s)0]+[K11(s)]d^(s){\widehat{\mathbf{y}}\left( \mathbf{s} \right)\mathbf{= R}\widehat{\mathbf{u}}\left( \mathbf{s} \right)\mathbf{+ K}\widehat{\mathbf{d}}\left( \mathbf{s} \right) }{\left\lbrack \widehat{v}(s) \right\rbrack\mathbf{=}\begin{bmatrix} \mathbf{R}_{\mathbf{11}}\mathbf{(s)} & \mathbf{R}_{\mathbf{12}}\mathbf{(s)} & \mathbf{0} \end{bmatrix}\begin{bmatrix} {\widehat{v}}_{g}\left. (s \right.) \\ {\widehat{i}}_{load}\left. (s \right.) \\ 0 \end{bmatrix}\mathbf{+}\begin{bmatrix} \mathbf{K}_{\mathbf{11}}\mathbf{(s)} \end{bmatrix}\widehat{\mathbf{d}}\left( \mathbf{s} \right)}$

對應的MATLAB腳本如下

將其展開可得 $Gvd(s)G_{vd}\left. (s \right.)$ ， $Gvi(s)G_{vi}\left. (s \right.)$ ， $Gvg(s)G_{vg}\left. (s \right.)$ 分別是占空比對輸出電壓的控制傳遞函數，輸出電流對輸出電壓的擾動傳遞函數，輸入電壓對輸出電壓的擾動傳遞函數。
$Gvd(s)=v^(s)d^(s)∣v^g(s)=0i^load(s)=0=K11(s){G_{vd}\left. (s \right.) = \frac{\widehat{v}(s)}{\widehat{\mathbf{d}}\left( \mathbf{s} \right)}|_{\begin{matrix} {\widehat{v}}_{g}\left. (s \right.) = 0 \\ {\widehat{i}}_{load}\left. (s \right.) = 0 \end{matrix}} = \mathbf{K}_{\mathbf{11}}\mathbf{(s)} }$
$Gvi(s)=v^(s)i^load(s)∣v^g(s)=0d^(s)=0=R12(s){G_{vi}\left. (s \right.) = \frac{\widehat{v}(s)}{{\widehat{i}}_{load}\left. (s \right.)}|_{\begin{matrix} {\widehat{v}}_{g}\left. (s \right.) = 0 \\ \widehat{\mathbf{d}}\left( \mathbf{s} \right) = 0 \end{matrix}} = \mathbf{R}_{\mathbf{12}}\mathbf{(s)} }$
$Gvg(s)=v^(s)v^g(s)∣i^load(s)=0d^(s)=0=R11(s){G_{vg}\left. (s \right.) = \frac{\widehat{v}(s)}{{\widehat{v}}_{g}\left. (s \right.)}|_{\begin{matrix} {\widehat{i}}_{load}\left. (s \right.) = 0 \\ \widehat{\mathbf{d}}\left( \mathbf{s} \right) = 0 \end{matrix}} = \mathbf{R}_{\mathbf{11}}\mathbf{(s)}}$

對應的MATLAB腳本為

輸入電壓擾動，輸出電流擾動以及占空比擾動組合在一起，輸出電壓擾動為：
$v^(s)=Gvg(s)v^g(s)+Gvi(s)i^load(s)+Gvd(s)d^(s)\widehat{v}(s) = G_{vg}\left. (s \right.){\widehat{v}}_{g}\left. (s \right.) + G_{vi}\left. (s \right.){\widehat{i}}_{load}\left. (s \right.) + G_{vd}\left. (s \right.)\widehat{d}(s)$

正如下圖所示：

重新化簡下式：
$x^(t)=︸(sI?A)?1?BM?u^(s)+︸(sI?A)?1[(A1?A2)X?(B1?B2)U]Nd^(s)\widehat{x}(t) = \underset{M}{\overset{(sI - A)^{- 1}\ B}{︸}}\ \widehat{u}(s) + \underset{N}{\overset{(sI - A)^{- 1}\left\lbrack \left( A_{1} - A_{2} \right)X - \left( B_{1} - B_{2} \right)U \right\rbrack}{︸}}\widehat{d}(s)$

化簡得到：
$x^(s)=Mu^(s)+Nd^(s)[i^L(s)v^C(s)]=[M11(s)M12(s)0M21(s)M22(s)0][v^g(s)i^load(s)0]+[N11(s)N21(s)]d^(s){\widehat{\mathbf{x}}\left( \mathbf{s} \right)\mathbf{= M}\widehat{\mathbf{u}}\left( \mathbf{s} \right)\mathbf{+ N}\widehat{\mathbf{d}}\left( \mathbf{s} \right) }{\begin{bmatrix} {\widehat{i}}_{L}(s) \\ {\widehat{v}}_{C}(s) \end{bmatrix} = \begin{bmatrix} M_{11}(s) & M_{12}(s) & 0 \\ M_{21}(s) & M_{22}(s) & 0 \end{bmatrix}\begin{bmatrix} {\widehat{v}}_{g}\left. (s \right.) \\ {\widehat{i}}_{load}\left. (s \right.) \\ 0 \end{bmatrix} + \begin{bmatrix} N_{11}(s) \\ N_{21}(s) \end{bmatrix}\widehat{d}(s)}$

MATLAB腳本為：

將其展開可得 $Gild(s)G_{ild}\left. (s \right.)$ ， $Gili(s)G_{ili}\left. (s \right.)$ ， $Gilg(s)G_{ilg}\left. (s \right.)$ 分別是占空比對電感電流的控制傳遞函數，輸出電流對電感電流的擾動傳遞函數，輸入電壓對電感電流的擾動傳遞函數。

$Gild(s)=i^L(s)d^(s)∣v^g(s)=0i^load(s)=0=N11(s){G_{ild}\left. (s \right.) = \frac{{\widehat{i}}_{L}(s)}{\widehat{\mathbf{d}}\left( \mathbf{s} \right)}|_{\begin{matrix} {\widehat{v}}_{g}\left. (s \right.) = 0 \\ {\widehat{i}}_{load}\left. (s \right.) = 0 \end{matrix}} = \mathbf{N}_{\mathbf{11}}\mathbf{(s)} }$ $Gili(s)=i^L(s)i^load(s)∣v^g(s)=0d^(s)=0=M12(s){G_{ili}\left. (s \right.) = \frac{{\widehat{i}}_{L}(s)}{{\widehat{i}}_{load}\left. (s \right.)}|_{\begin{matrix} {\widehat{v}}_{g}\left. (s \right.) = 0 \\ \widehat{\mathbf{d}}\left( \mathbf{s} \right) = 0 \end{matrix}} = \mathbf{M}_{\mathbf{12}}\mathbf{(s)} }$ $Gilg(s)=i^L(s)v^g(s)∣i^load(s)=0d^(s)=0=M11(s){G_{ilg}\left. (s \right.) = \frac{{\widehat{i}}_{L}(s)}{{\widehat{v}}_{g}\left. (s \right.)}|_{\begin{matrix} {\widehat{i}}_{load}\left. (s \right.) = 0 \\ \widehat{\mathbf{d}}\left( \mathbf{s} \right) = 0 \end{matrix}} = \mathbf{M}_{\mathbf{11}}\mathbf{(s)}}$

對應的MATLAB腳本為

輸入電壓擾動，輸出電流擾動以及占空比擾動組合在一起，電感電流的擾動為：

$i^L(s)=Gilg(s)v^g(s)+Gili(s)i^load(s)+Gild(s)d^(s){\widehat{i}}_{L}(s) = G_{ilg}\left. (s \right.){\widehat{v}}_{g}\left. (s \right.) + G_{ili}\left. (s \right.){\widehat{i}}_{load}\left. (s \right.) + G_{ild}\left. (s \right.)\widehat{d}(s)$

如下圖所示

總結，這里得到了影響電感電流和輸出電壓的控制/擾動傳遞函數，重新羅列如下：

$i^L(s)=︸Gilg(s)v^g(s)+Gili(s)i^load(s)disturbance+︸Gild(s)d^(s)control{\widehat{i}}_{L}(s) = \underset{disturbance}{\overset{G_{ilg}\left. (s \right.){\widehat{v}}_{g}\left. (s \right.) + G_{ili}\left. (s \right.){\widehat{i}}_{load}\left. (s \right.)}{︸}} + \underset{control}{\overset{G_{ild}\left. (s \right.)\widehat{d}(s)}{︸}}$

$v^(s)=︸Gvg(s)v^g(s)+Gvi(s)i^load(s)disturbance+︸Gvd(s)d^(s)control\widehat{v}(s) = \underset{disturbance}{\overset{G_{vg}\left. (s \right.){\widehat{v}}_{g}\left. (s \right.) + G_{vi}\left. (s \right.){\widehat{i}}_{load}\left. (s \right.)}{︸}} + \underset{control}{\overset{G_{vd}\left. (s \right.)\widehat{d}(s)}{︸}}$

同時，也可以得到電感電流到輸出電壓的傳遞函數，我們可以認為，外部的輸入電壓和輸出電流擾動是先轉化為對應的電感電流，然后再轉化為對輸出電壓的影響。

$Gvil(s)=v^(s)d^(s)∣v^g(s)=0i^load(s)=0i^L(s)d^(s)∣v^g(s)=0i^load(s)=0=v^(s)i^L(s)G_{vil}\left. (s \right.) = \frac{\frac{\widehat{v}(s)}{\widehat{\mathbf{d}}\left( \mathbf{s} \right)}|_{\begin{matrix} {\widehat{v}}_{g}\left. (s \right.) = 0 \\ {\widehat{i}}_{load}\left. (s \right.) = 0 \end{matrix}}}{\frac{{\widehat{i}}_{L}(s)}{\widehat{\mathbf{d}}\left( \mathbf{s} \right)}|_{\begin{matrix} {\widehat{v}}_{g}\left. (s \right.) = 0 \\ {\widehat{i}}_{load}\left. (s \right.) = 0 \end{matrix}}} = \frac{\widehat{v}(s)}{{\widehat{i}}_{L}(s)}$