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

21.5 Laplace transform of derivatives and integrals 637
Consider the Laplace transformof theequation forthe voltage across acapacitor
of capacitanceC. Recall fromEngineering application 21.1 that
V(s)= 1 Cs I(s)
Rearrangingthisexpressiongivesaformulafortheimpedanceofthecapacitor,Z(s),
inthesdomain
Z(s) = V(s)
I(s) = 1 Cs
Thisformoftheequationcanbeusedtodeterminethebehaviourofthecapacitorin
a variety of situations where the voltages and currents can take a number of forms,
for example step inputs, sine waves, triangular waves, etc. However, if we are only
interested in the sinusoidal steady-state response of the system when all transients
have decayed, then itispossible tosubstitutes = jω into the expression forZ(s)
Z(jω) = V(jω)
I(jω) = 1
jωC
This form of the impedance is the one regularly used in a.c. circuit theory. It gives
theimpedanceofthecapacitorintermsofthecapacitanceandtheangularfrequency
ω,where ω = 2πf.(Thisresulthasalreadybeendiscussedinthecontextofphasors
on page 342).
Thesamesubstitutioncan bemade toobtain thefrequency responseofasystem,
givenitstransferfunction.Thesystemfrequencyresponseissomethingthatcanusuallybeobtainedeasilybyexperiment.Allthatisrequiredisasignalsourceofaknown
amplitude and a means of measuring the output amplitude and phase relative to the
original signal. Measurements of this sort can reveal some of the properties of the
systeminquestionreflectingthecloserelationshipbetweenthetransferfunctionand
the frequency response.
EXERCISES21.5
1 TheLaplacetransform ofy(t)isY(s),y(0) = 3,
y ′ (0) = 1. Find the Laplacetransformsofthe
followingexpressions:
(a) y ′ (b) y ′′
(c) 3y ′′ −y ′
(d) y ′′ +2y ′ +3y
(e) 3y ′′ −y ′ +2y
(f) −4y ′′ +5y ′ −3y
(g) 3 d2 y
dt 2 +6dy dt +8y
(h) 4 d2 y
dt 2 −8dy dt +6y
2 Given the Laplacetransform of f (t)isF(s),
f(0)=2,f ′ (0)=3andf ′′ (0) = −1,findthe
Laplace transformsof
(a) 3f ′ −2f
(c) f ′′′
(b) 3f ′′ − f ′ + f
(d) 2f ′′′ − f ′′ +4f ′ −2f
3 (a) IfF(s) = L{f(t)} = ∫ ∞
0 e −st f (t)dt,showusing
integration byparts that
(i) L{f ′ (t)} =sF(s)−f(0)
(ii) L{f ′′ (t)} =s 2 F(s)−sf(0)−f ′ (0)
(b) IfF(s) = L{f (t)}prove that
L{e −at f(t)} =F(s +a)
Deduce L{te −t },given L{t} = 1 s 2.