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

656 Chapter 21 The Laplace transform
Engineeringapplication21.5
Electronicthermometermeasuringoventemperature
Many engineering systems can be modelled by a first-order differential equation.
Thetimeconstantisameasureoftherapiditywithwhichthesesystemsrespondto
a change in input. Suppose an electronic thermometer is used to measure the temperature
of an oven. The sensing element does not respond instantly to changes in
the oven temperature because it takes time for the element to heat up or cool down.
Provided the electronic circuitry does not introduce further time delays then the differential
equation thatmodels the thermometer isgiven by
τ dv m
+v
dt m
=v o
where v m
=measuredtemperature, v o
=oventemperature, τ =timeconstantofthe
sensor. For convenience the temperature is measured relative to the ambient room
temperature, which forms a ‘baseline’ for temperature measurement.
Suppose the sensing element of an electronic thermometer has a time constant
of 2 seconds. If the temperature of the oven increases linearly at the rate of 3 ◦ Cs −1
startingfromanambientroomtemperatureof20 ◦ Catt = 0,calculatetheresponseof
the thermometer to the changing oven temperature. State the maximum temperature
error.
Solution
Taking Laplace transforms ofthe equation gives
τ(sV m
(s) − v m
(0)) +V m
(s) =V o
(s)
v m
(0) = 0 as the oven temperature and sensor temperature are identical att = 0.
Therefore,
τsV m
(s) +V m
(s) =V o
(s)
V m
(s) = V o (s)
1+τs
(21.1)
Forthisexample,theinputtothethermometerisatemperaturerampwithaslopeof
3 ◦ Cs −1 . Therefore, v o
= 3t fort 0:
V o
(s) = L{v o
(t)} = 3 s 2 (21.2)
Combining Equations (21.1) and (21.2) yields
V m
(s) =
3
s 2 (1+τs) = 3
s 2 (1 +2s)
Then using partialfractions, wehave
V m
(s) = 3 s 2 − 6 s + 12
1+2s
Taking the inverse Laplace transformyields
since τ = 2
v m
=3t−6+6e −0.5t
t0