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Chapter 2

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.

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}
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.