(a) Write the state-variable equations based on the energy-storage elements L and C for the circuit shown in Fig. 7.13(a). (b) Linearize these equations for small perturbations of all variables. (c) Combine to eliminate the unwanted variable and obtain the input–output system differential equation

Question

(a) Write the state-variable equations based on the energy-storage elements L and C for the circuit shown in Fig. 7.13(a).
(b) Linearize these equations for small perturbations of all variables.
(c) Combine to eliminate the unwanted variable and obtain the input–output system differential equation relating the output [latex]\widehat{e} _{2g}[/latex] to the input [latex]\widehat{e} _{s}\left(t\right)[/latex].
(d) Draw the simulation block diagram for the linearized system.

Accepted Answer

(a) In general, the state-variable for a capacitor is its voltage [latex]e_{2g}[/latex], and the state variable for an inductor is its current [latex]i_{L}[/latex]. Thus, for the inductor L,

[latex]\frac{di_{L} }{dt} =\frac{1}{L} \left(e_{2g} -e_{23} \right)[/latex]

or

[latex]\frac{di_{L} }{dt} =\frac{-1}{L} f_{NL} \left(i_{L} \right) +\frac{1}{L} e_{2g},[/latex] (7.39)

and for the capacitor C,

[latex]\frac{de_{2g} }{dt} =\frac{1}{C} \left( i_{R1}-i_{L} \right)[/latex]

or

[latex]\frac{de_{2g} }{dt} =\frac{1}{C} \left(\frac{e_{s}-e_{2g} }{R_{1} }-i_{L} \right).[/latex]

Rearranging yields

[latex]\frac{de_{2g} }{dt} =\left(\frac{-1}{C} \right) i_{L} +\left(\frac{-1}{R_{1}C } \right) e_{2g} +\left(\frac{1}{R_{1}C } \right) e_{s}.[/latex] (7.40)

Note that Eqs. (7.39) and (7.40) comprise a set of nonlinear state-variable equations for the circuit.

Because of the nonlinear function [latex]f_{NL} \left(i_{L} \right)[/latex], Eqs. (7.39) and (7.40) cannot be combined to eliminate [latex]i_{L}[/latex] until they are linearized for small perturbations of all variables.

(b) After linearizing, Eqs. (7.39) and (7.40) become

[latex]\frac{d\widehat{i} _{L} }{dt} =\left(\frac{-1}{L} \right) \frac{de_{23} }{di_{L} } \mid _{\bar{i} L} \widehat{i} _{L} + \left(\frac{1}{L }\right) \widehat{e} _{2g},[/latex] (7.41)

[latex]\frac{d\widehat{e} _{2g} }{dt} =\left(\frac{-1}{C} \right) \widehat{i} _{L} +\left(\frac{-1}{R_{1}C } \right) \widehat{e} _{2g}+\left(\frac{1}{R_{1}C } \right) \widehat{e} _{2g}.[/latex] (7.42)

(c) Equations (7.41) and (7.42) may now be combined as follows to eliminate [latex]\widehat{i} _{L}[/latex]. Solving Eq. (7.42) for [latex]\widehat{i} _{L}[/latex] yields

[latex]\widehat{i} _{L} =-C\frac{d\widehat{e} _{2g} }{dt} -\left(\frac{1}{R_{1} } \right) \widehat{e} _{2g}+\left(\frac{1}{R_{1} } \right) \widehat{e} _{s}[/latex]

Substituting the expression for [latex]\widehat{i} _{L}[/latex] in Eq. (7.41) and using [latex]R_{inc} =\frac{de_{2g} }{di_{L} } \mid _{\bar{i} L}[/latex] yields

[latex]-C\frac{d^{2}\widehat{e} _{2g}}{dt^{2} }-\frac{1}{R_{1} } \frac{d\widehat{e} _{2g} }{dt} +\frac{1}{R_{1} } \frac{d\widehat{e} _{s} }{dt}=-\frac{R_{inc} }{L} \left(-C\frac{d\widehat{e} _{2g} }{dt}-\frac{\widehat{e} _{2g} }{R_{1} } -\frac{\widehat{e} _{s} }{R_{1} } \right) +\frac{\widehat{e} _{2g} }{L}.[/latex]

Collecting terms, we have

[latex]C\frac{d^{2}\widehat{e} _{2g}}{dt^{2} }+\left(\frac{L+R_{1}R_{inc}C }{R_{1} L} \right) \frac{d\widehat{e} _{2g} }{dt}+\left(\frac{R_{inc}+R_{1} }{R_{1} L} \right)\widehat{e} _{2g} =\frac{1}{R_{1} } \frac{d\widehat{e} _{s} }{dt}+\frac{R_{inc} }{LR_{1} } \widehat{e} _{s} .[/latex] (7.43)

(d) The simulation block diagram for the linearized system appears in Fig. 7.13(b).

Question 7.3: (a) Write the state-variable equations based on the energy-s...

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