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## Q. 2.6

Differential Equation Transfer Function—Single Loop via the Differential Equation

PROBLEM: Find the transfer function relating the capacitor voltage, Vc (s),to the input voltage ,V(s) in Figure 2.3.

## Verified Solution

In any problem, the designer must ﬁrst decide what the input and outputshouldbe.Inthisnetwork,severalvariablescouldhavebeenchosentobethe output—forexample,theinductorvoltage,thecapacitorvoltage,theresistorvoltage, orthecurrent.Theproblemstatement,however,isclearinthiscase:Wearetotreatthe capacitor voltage as the output and the applied voltage as the input. Summing the voltages around the loop, assuming zero initial conditions,

yields the integro-differential equation for this network as

$L\frac{di(t)}{dt} +Ri(t)+\frac{1}{c} ∫^{t}_{0}i(\tau )(d\tau )=\nu (t)$

Changing variables from current to charge using i(t)= dq (t)/dt yields

$L\frac{d^{2}q(t) }{dt^{2} }+R\frac{dq(t)}{dt}+\frac{1}{C} q(t)=\nu (t)$

From the voltage-charge relationship for a capacitor in Table 2.3,

 Component Voltage-current Current-voltage Voltage-charge Impedance Z(s)=V(s)/I(s) Admittance Y(s)=I(s)/V(s) $\nu (t)=\frac{1}{C} ∫^{1}_{0} i(\tau )d\tau$ $i(t)=C\frac{d\nu (t)}{dt}$ $\nu (t)=\frac{1}{C}q(t)$ $\frac{1}{Cs}$ Cs $\nu (t)=Ri(t)$ $i(t)=\frac{1}{R} \nu (t)$ $\nu (t)=R\frac{dq(t)}{dt}$ R $\frac{1}{R}=G$ $\nu (t)=L\frac{di(t)}{dt}$ $i(t)=\frac{1}{L∫^{1}_{0} }\nu (\tau)d\tau$ $\nu (t)=L\frac{d^{2}q(t) }{dt^{2} }$ Ls $\frac{1}{Ls}$
$q(t)=C\nu c(t)$

Substituting Eq. (2.63) into Eq. (2.62) yields

$LC\frac{d^{2}\nu c(t) }{dt^{2} } +RC\frac{d\nu c(t)}{dt} +\nu c=\nu (t)$

Taking the Laplace transform assuming zero initial conditions, rearranging terms, and simplifying yields

$(LCs^{2}+RCs+1)Vc(s)=V(s)$

Solving for the transfer function, Vc(s)/V(s) , we obtain

$\frac{Vc(s)}{V(s)} =\frac{1/LC}{s^{2}+\frac{R}{L}s+\frac{1}{LC}}$

as shown in Figure 2.4.