Question 6.7: State-Variable Model of the Circuit in Example 6.4 Reconside...
State-Variable Model of the Circuit in Example 6.4
Reconsider the circuit shown in Figure 6.21.
a. Derive the state-variable model with inductor currents and capacitor voltages as states. The input is the applied voltage v_{\mathrm{a}}, and the output is the voltage across the capacitor C.
b. Based on the state-space form obtained in Part (a), determine the differential equation relating the output voltage v_{\mathrm{o}}(t) to the input voltage v_{\mathrm{a}}(t).
a. We first label the nodes and currents as we did in Example 6.4. Note that the circuit has two independent energy storage elements, L and C. This implies that two state variables are needed, and they are
x_{1}=i_{L^{\prime}} \quad x_{2}=v_{C}.
Their time derivatives are
\begin{aligned} & \dot{x}_{1}=\frac{\mathrm{d} i_{\mathrm{L}}}{\mathrm{d} t}=\frac{1}{L} v_{\mathrm{L}} \\ & \dot{x}_{2}=\frac{\mathrm{d} v_{\mathrm{C}}}{\mathrm{d} t}=\frac{1}{C} i_{\mathrm{C}} . \end{aligned}
We need to express the voltage across the inductor, v_{L}, and the current through the capacitor, i_{\mathrm{C}}, in terms of the state variables and the input. Note that
v_{\mathrm{L}}=v_{\mathrm{a}}-v_{1}=v_{\mathrm{a}}-v_{\mathrm{C}}=u-x_{2}.
Applying Kirchhoff’s current law to node 1 gives
i_{\mathrm{C}}=i_{\mathrm{L}}-i_{\mathrm{R}}=i_{\mathrm{L}}-\frac{v_{\mathrm{C}}}{R}=x_{1}-\frac{1}{R} x_{2}.
Thus, the complete set of two state-variable equations is
\begin{aligned} & \dot{x}_{1}=\frac{1}{L} v_{\mathrm{L}}=-\frac{1}{L} x_{2}+\frac{1}{L} u, \\ & \dot{x}_{2}=\frac{1}{C} i_{\mathrm{C}}=\frac{1}{C} x_{1}-\frac{1}{R C} x_{2} . \end{aligned}
The output equation is
y=v_{\mathrm{o}}=v_{\mathrm{C}}=x_{2}.
The state equation and the output equation can be written in matrix form as follows:
\begin{aligned} \left\{\begin{array}{l} \dot{x}_{1} \\ \dot{x}_{2} \end{array}\right\} & =\left[\begin{array}{rr} 0 & -\frac{1}{L} \\ \frac{1}{C} & -\frac{1}{R C} \end{array}\right]\left\{\begin{array}{l} x_{1} \\ x_{2} \end{array}\right\}+\left[\begin{array}{c} 1 / L \\ 0 \end{array}\right] u, \\ y & =\left[\begin{array}{ll} 0 & 1 \end{array}\right]\left\{\begin{array}{l} x_{1} \\ x_{2} \end{array}\right\}+0 \cdot u . \end{aligned}
b. Note that V_{\mathrm{o}}(s)=Y(s) and V_{\mathrm{a}}(s)=U(s). As presented in Section 4.4, the state-space form can be converted to a transfer function using
\frac{Y(s)}{U(s)}=\mathbf{C}(s \mathbf{I}-\mathbf{A})^{-1} \mathbf{B}+D .
Substituting A, B, C, and D matrices obtained in Part (a) gives
\frac{V_{0}(s)}{V_{\mathrm{a}}(s)}=\left[\begin{array}{ll} 0 & 1 \end{array}\right]\left[\begin{array}{cc} s & \frac{1}{L} \\ -\frac{1}{C} & s+\frac{1}{R C} \end{array}\right]^{-1}\left[\begin{array}{c} 1 / L \\ 0 \end{array}\right]=\frac{R}{R L C s^{2}+L s+R},
which returns the same input-output equation as the one obtained in Examples 6.4 and 6.6.