Question 7.1: Find the input–output differential equation relating eo to e...

For the source,

e_{s} =e_{1g}.          (7.24)

For the resistor R_{1},

i_{R1} =\frac{1}{R_{1} } \left(e_{1g}-e_{2g}\right) .          (7.25)

For the inductor L,

e_{2g}-e_{3g}=\frac{di_{L} }{dt} .          (7.26)

For the resistor R_{2},

e_{3g}=R_{2} i_{L} .          (7.27)

For the capacitor C,

i_{c} =C\frac{de_{2g} }{dt} .          (7.28)

At node 2,

i_{R1} =i_{L} +i_{C} .          (7.29)

Equations (7.24)–(7.29) comprise a set of six equations involving six unknown variables: e_{1g}, e_{2g}, e_{3g}, i_{R1}, i_{L}, and i_{C}. (Note that using i_{L} to describe the current through both L and R_{2} satisfies Kirchhoff’s law at node 3 and eliminates one variable and one equation.)

The node method may be applied to node 2 to eliminate the unwanted variables e_{1g}, e_{3g}, i_{R1}, i_{L}, and i_{C}. Substituting for i_{R1} from Eq. (7.25) and for i_{C} from Eq. (7.28) into Eq. (7.29) yields

\frac{1}{R_{1} } \left(e_{1g} -e_{2g} \right) =i_{L} +C\frac{de_{2g} }{dt} .          (7.30)

Then, using Eq. (7.24) for e_{1g} and rearranging Eq. (7.30) yields

i_{L}=\frac{1}{R_{1} }e_{s}- \frac{1}{R_{1} }e_{2g} -C\frac{de_{2g} }{dt} .          (7.31)

Differentiating Eq. (7.31) with respect to time gives

\frac{di_{L} }{dt} =\left(\frac{1}{R_{1} } \right) \frac{de_{s} }{dt} -\left(\frac{1}{R_{1} } \right) \frac{de_{2g} }{dt}-C\frac{d^{2}e_{2g} }{dt^{2} } .          (7.32)

Combine Eqs. (7.26) and (7.27) to solve for e_{2g}:

e_{2g}=L \frac{di_{L} }{dt}+R_{2} i_{L} .          (7.33)

Substitution from Eqs. (7.31) and (7.32) into Eq. (7.33) then eliminates i_{L}:

e_{2g}=\left(\frac{L}{R_{1} } \right) \left(\frac{de_{s} }{dt}-\frac{de_{2g} }{dt} \right) -LC\frac{d^{2}e_{2g} }{dt^{2} } +\left(\frac{R_{2} }{R_{1} } \right) \left(e_{s}-e_{2g} \right) -R_{2} C\frac{de_{2g} }{dt}.          (7.34)

Collecting terms and noting that e_{o} is the same as e_{2g}, we have

LC\frac{d^{2}e_{o} }{dt^{2} }+\left(\frac{L}{R_{1} }+R_{2}C \right) \frac{de_{o} }{dt} +\left(1+\frac{R_{2} }{R_{1} } \right) e_{o} =\frac{R_{2} }{R_{1}}e_{s} +\frac{L}{R_{1} } \frac{de_{s} }{dt} .          (7.35)