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

21.10 Solving linear constant coefficient differential equations 657
This response consists of threeparts:
(1) a decaying transient, 6e −0.5t , which disappears with time;
(2) a ramp,3t, with the same slope as the oven temperature;
(3) a fixed negative temperature error, −6.
Therefore,afterthetransienthasdecayedthemeasuredtemperaturefollowstheoven
temperaturewithafixednegativeerror.Itisinstructivetoobtainthetemperatureerror
by an alternative method. Given thatthe temperature error is v e
, then
and
v e
=v m
−v o
V e
(s) =V m
(s) −V o
(s) (21.3)
Combining Equations (21.1) and (21.3) yields
V e
(s) = V o (s)
1+τs −V o (s)= −τsV o (s)
1+τs
Combining Equations (21.2) and (21.4) yields
V e
(s) = −τs 3
1+τss = −3τ
2 s(1+τs)
The final value theorem can beused tofind the steady-state error:
[ ] −3τs
lim v
t→∞
e
(t) =lim sV e
(s) = lim =−3τ=−6
s→0 s→0 s(1+τs)
(21.4)
that is, the steady-state temperature error is −6 ◦ C. It is important to note that the
final value theorem can only be used if it is known that the time function tends to a
limitast → ∞. Inmany cases engineers know thisisthe case fromexperience.
The Laplace transform technique can also be used to solve simultaneous differential
equations.
Example21.27 Solve the simultaneousdifferential equations
x ′ +x+ y′
=1 x(0)=y(0)=0
2
x ′
2 +y′ +y=0
Solution Take the Laplace transforms of both equations:
sY(s) −y(0)
sX(s) −x(0) +X(s) + = 1 2 s
sX(s) −x(0)
+sY(s)−y(0)+Y(s)=0
2
These arerearranged togive
(s +1)X(s) + sY(s) = 1 2 s
sX(s)
+(s+1)Y(s)=0
2