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

15.2 Average value of a function 473
Solution
We first need to obtain an equation for the waveform. We choose the interval
0t<2.
The general equation for a straightlineis
v=mt+c
Whent=0thenv=0.So
0=0+c
c = 0
Whent=2thenv=5.So
Hence
5 =m(2)
m=2.5
v = 2.5t
The average value isgiven by
∫ 2
[ 2.5t
2
] 2
0
v av
= 1 2.5tdt = 1 2 0 2 2
v av
= 1 ( ) 2.5×4
−0 = 10 2 2 4 =2.5V
Engineeringapplication15.2
Athyristorfiringcircuit
Figure 15.3 shows a simple circuit to control the voltage across a load resistor,R L
.
This circuit has many uses, one of which is to adjust the level of lighting in a room.
Thecircuithasana.c.powersupplywithpeakvoltage,V S
.Themaincontrolelement
is the thyristor. This device is similar in many ways to a diode. It has a very high
resistance when it is reverse biased and a low resistance when it is forward biased.
However,unlikeadiode,thislowresistancedependsonthethyristorbeing‘switched
on’ by the application of a gate current. The point at which the thyristor is switched
on can be varied by varying the resistor,R G
. Figure 15.4 shows a typical waveform
ofthe voltage, v L
, across the loadresistor.
The point at which the thyristor is turned on in each cycle is characterized by
the quantity αT, where 0 α 0.25 and T is the period of the waveform. This
restrictionon αreflectsthefactthatifthethyristorhasnotturnedonwhenthesupply
voltage has peaked inthe forward direction thenitwill never turnon.
Calculate the average value of the waveform over a period and comment on the
result.
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