082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

X(s)= s3 −3s 2 −4s+6
s 2 (s 2 −5s+6)
21.10 Solving linear constant coefficient differential equations 655
= A s + B s + C
2 s −2 + D
s −3
= s3 −3s 2 −4s+6
s 2 (s −2)(s −3)
The constantsA,B,C andDare evaluated inthe usual way:
X(s)= 1 6s + 1 s 2 + 3
2(s −2) − 2
3(s −3)
Taking the inverse Laplace transform yields
x(t) = 1 6 +t + 3 2 e2t − 2 3 e3t
Engineeringapplication21.4
Dischargeofacapacitorthroughaloadresistor
InEngineeringapplication2.8,weexaminedthevariationinvoltageacrossacapacitor,C,
when it was switched in series with a resistor,R, at timet = 0. We stated a
relationshipforthetime-varyingvoltage, v,acrossthecapacitor.Provethisrelationship.
Refer tothe example fordetails ofthe circuit.
Solution
Firstwemustderiveadifferentialequationforthecircuit.UsingKirchhoff’svoltage
law and denoting the voltageacrossthe resistorby v R
weobtain
v+v R
=0
Using Ohm’slaw and denoting the currentinthe circuitbyiweobtain
v+iR=0
For the capacitor,
i =C dv
dt
Combining these equations gives
v+RC dv = 0
dt
We now take the Laplace transformof this equation. Using L{v} =V(s)weobtain
V(s) +RC(sV(s) − v(0)) =0
V(s)(1 +RCs) =RCv(0)
V(s)= RCv(0)
1+RCs = v(0)
1
RC +s
Taking the inverse Laplace transform ofthe equation yields
v = v(0)e −t/(RC)
t 0
This isequivalent tothe relationship statedinEngineering application 2.8.