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

556 Chapter 19 Ordinary differential equations I
(c) To find the particular solution subject to the condition v C
(0) = 0 we need only
replaceV 0
by0intheparticularsolutionobtainedinpart(b).Hencethezero-state
response is
v C
(t) =
VRC
[
ωsin ωt +
R 2 C 2 ω 2 +1
cos ωt
]
−
RC
V
R 2 C 2 ω 2 +1 e−t/(RC)
(d) In the case of zero input we have v S
(t) = 0. In part (b) we solved the equation
with v S
(t) =V cos ωt.Hence,puttingV = 0inthesolutiontopart(b)willyield
the solution when v S
(t) = 0.So,
v C
(t) =V 0
e −t/(RC)
Alternatively we can note thatwhen v S
(t) = 0 the originalequation becomes
dv C
+ 1
dt RC v C =0
subject to v C
(0) =V 0
. This can be solved usingseparation of variables. So
1 dv C
= − 1
v C
dt RC
Integration yields
∫ 1
v C
dv C
=−∫
1
RC dt
lnv C
=− t
RC +k
v C
= e −t/(RC)+k
= e k e −t/(RC)
=Ae −t/(RC)
(k a constant)
where A is the constant e k . Applying the initial condition v C
(0) = V 0
we find
V 0
=Aand so finally
as before.
v C
(t) =V 0
e −t/(RC)
(e) Inspectionofpart(b)showsthatitisthesumofthezero-inputresponseandthe
zero-state response:
zero-input zero-state response
response
{ }} { { }} {
v C
(t)=V 0
e −t/(RC) +
VRC [
cos ωt
]
V
ωsinωt+ −
R 2 C 2 ω 2 +1 RC R 2 C 2 ω 2 +1 e−t/(RC)
(f) In this example, the terms involving e −t/(RC) tend to zero ast increases, and are
termed transients. Once the system has settled down their contribution will not
be important. The remaining terms represent the longer term behaviour of the
systemor the so-calledsteady state.
Hence the transient termsare
V 0
e −t/(RC) V
−
R 2 C 2 ω 2 +1 e−t/(RC)