Irregular areas and volumes, and mean values 261
Problem 5. Determine the average values over
half a cycle of the periodic waveforms shown in
Figure 28.8
(c)
Voltage (V)
0
10
2 4 6 8
210
t (ms)
(a)
0
Voltage (V)
20
1 2 3 4
220
t (ms)
(b)
0
Current (A)
3
2
1
1 2 3 4
23
22
21
t (s)
5 6
Figure 28.8
(a) Area under triangular waveform (a) for a half cycle
is given by
Area =
1
2
(base)(perpendicular height)
=
1
2
(2 × 10
−3
)(20) = 20 × 10
−3 Vs
Average value of waveform
=
area under curve
length of base
=
20 × 10 −3 Vs
2 × 10 −3 s
= 10 V
(b) Area under waveform (b) for a half cycle
= (1 × 1) + (3 × 2) = 7 As
Average value of waveform =
area under curve
length of base
=
7 As
3 s
= 2.33 A
(c) A half cycle of the voltage waveform (c) is
completed in 4 ms.
Area under curve =
1
2
{(3 − 1)10
−3
}(10)
= 10 × 10
−3 Vs
Average value of waveform =
area under curve
length of base
=
10 × 10
−3 Vs
4 × 10 −3 s
= 2.5 V
Problem 6. Determine the mean value of current
over one complete cycle of the periodic waveforms
shown in Figure 28.9
0
Current (mA)
5
4 8 12 16 20 24 28 t (ms)
0
Current (A)
2
2 4 6 8 10 12 t (ms)
Figure 28.9
(a) One cycle of the trapezoidal waveform (a) is completed in 10 ms (i.e. the periodic time is 10 ms).
Area under curve = area of trapezium
=
1
2
(sum of parallel sides)(perpendicular
distance between parallel sides)
=
1
2
{(4 + 8) ×10
−3
}(5 × 10
−3
)
= 30 × 10
−6 As
Mean value over one cycle =
area under curve
length of base
=
30 × 10 −6 As
10 × 10 −3 s
= 3mA
(b) One cycle of the saw-tooth waveform (b) is completed in 5 ms.
Area under curve =
1
2
(3 × 10
−3
)(2)
= 3 × 10
−3 As
Mean value over one cycle =
area under curve
length of base
=
3 × 10 −3 As
5 × 10 −3 s
= 0.6 A