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.
Chapter 2
Q. 2.6
Step-by-Step
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} |
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.
