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

412 Chapter 12 Applications of differentiation
= −U(−αe −αt cos(βt) −e −αt βsin(βt))
( −αe −αt )
αsin(βt)
−U
+ e−αt αβcos(βt)
β
β
(
)
= −Ue −αt −αcos(βt) − βsin(βt) − α2 sin(βt)
+ αcos(βt)
β
( )
=Ue −αt β + α2
sin(βt)
β
At a turning point dv o
= 0.Hence
dt
( β
Ue −αt 2 + α 2 )
sin(βt) =0
β
y o (t)
z < 1
a
R
L
c
U
z = 1
y i
b
Figure12.7
Asecond-order electrical system.
C
d
y o
t m
z > 1
Figure12.8
Responseofasecond-order system to a step
input.
This occurs when sin(βt) = 0, which corresponds tot = kπ/β, k = 0,1,2....
It is now straightforward to calculate t m
, once β has been calculated, using Equations
(12.2), (12.3) and (12.4) for particular values ofR,L andC. You may like to
show that the turning point corresponding to k = 1 is a maximum by calculating
d 2 v o
dt 2 and carrying outthe second-derivative test.
Itispossibletocheckwhetherornotasystemisunderdampedusingthefollowing
formulae:
ζ = R R c
damping ratio (12.5)
t
√
L
R c
=2
C
criticalresistance (12.6)
LetuslookataspecificcasewithtypicalvaluesL = 40mH,C = 1 µF,R = 200.
UsingEquations (12.5) and (12.6),wefind
√ √
L
R c
=2
C = 2 4 ×10 −2
= 400
1 ×10−6 ζ = R R c
= 200
400 = 0.5