CHAPTER 39 TRIGONOMETRIC WAVEFORMS
CHAPTER 39 TRIGONOMETRIC WAVEFORMS EXERCISE 165 Page 448
1. Determine all of the angles between 0° and 360° whose sine is: (a) 0.6792
(b) –0.1483
(a) Sine is positive in the 1st and 2nd quadrants If sin θ = 0.6792, then θ = sin −1 (0.6792) = 42.78° or 180° – 42.78° = 137.22° as shown in diagram (i) below.
(i)
(ii)
(b) Sine is negative in the 3rd and 4th quadrants If sin θ = 0.1483, then θ = sin −1 (0.1483) = 8.53° From diagram (ii) above, the two values where sin θ = –0.1483 are 180° + 8.53° = 188.53° and 360° – 8.53° = 351.47°
2. Solve the following equations for values of x between 0° and 360°: (a) x = cos −1 0.8739
(b) x = cos −1 (–0.5572)
(a) Cosine is positive in the 1st and 4th quadrants x = cos −1 (0.8739) = 29.08° or 360° –29.08° = 330.92° as shown in diagram (i) below.
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(i)
(ii)
(b) Cosine is negative in the 2nd and 3rd quadrants x = cos −1 (0.5572) = 56.14° hence, from diagram (ii) shown above, the two values of x for which x = cos −1 (−0.5572) are:
180° – 56.14° = 123.86
and
180° + 56.14° = 236.14°
3. Find the angles between 0° to 360° whose tangent is: (a) 0.9728
(b) –2.3420
(a) Tangent is positive in the 1st and 3rd quadrants
θ = tan −1 0.9728 = 44.21° or 180° + 44.21° = 224.21° as shown in diagram (i) below.
(i)
(ii)
(b) Tangent is negative in the 2nd and 4th quadrants
θ = tan −1 2.3418 = 66.88° and from diagram (ii) shown above, θ = 180° – 66.88° = 113.12°
and
360° – 66.88° = 293.12°
4. Solve, in the range 0° to 360°, giving the answers in degrees and minutes: cos −1 (–0.5316) = t
Cosine is negative in the 2nd and 3rd quadrants. cos −1 (0.5316) = 57.886° or 57°53′ as shown in the diagram below.
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From the diagram, t = 180° – 57°53′ = 122°7′
and
t = 180° + 57°53′ = 237°53′
5. Solve, in the range 0° to 360°, giving the answers in degrees and minutes:
sin −1 ( −0.6250 ) = α
Sine is negative in the 3rd and 4th quadrants If sin θ = 0.6250, then α = sin −1 (0.6250) = 38.682° = 38°41′
From the diagram above, the two values where sin α = –0.6250 are 180° + 38°41′ = 218°41′ and 360° – 38°41′ = 321°19′
6. Solve, in the range 0° to 360°, giving the answers in degrees and minutes:
tan −1 0.8314 = θ
Tangent is positive in the 1st and 3rd quadrants
θ = tan −1 0.8314 = 39.74° or 39°44′
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From the diagram, the two values of θ between 0° and 360° are: 39°44′
and 180° + 39°44′ = 219°44′
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EXERCISE 166 Page 452
1. A sine wave is given by y = 5 sin 3x. State its peak value.
Peak value = amplitude = maximum value = 5
2. A sine wave is given by y = 4 sin 2x. State its period in degrees.
Period =
360° = 180° 2
3. A periodic function is given by y = 30 cos 5x. State its maximum value.
Maximum value = amplitude = peak value = 30
4. A periodic function is given by y = 25 cos 3x. State its period in degrees.
Period =
360° = 120° 3
5. State the amplitude and period of y = cos 3A and sketch the curve between 0° and 360°
Amplitude = 1 and period =
360° = 120° 3
A sketch of y = cos 3A is shown below.
6. State the amplitude and period of y = 2 sin
5x and sketch the curve between 0° and 360° 2 665
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5x 360° , amplitude = 2 and period = = 144° 5 2 2 5x A sketch y = 2 sin is shown below 2
If y = 2 sin
7. State the amplitude and period of y = 3 sin 4t and sketch the curve between 0° and 360°
Amplitude = 3 and period =
360° = 90° 4
A sketch of y = 3 sin 4t is shown below
8. State the amplitude and period of y = 5 cos
Amplitude = 5 and period =
A sketch of y = 5 cos
θ 2
θ 2
and sketch the curve between 0° and 360°
360° = 720° 1 2
is shown below
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9. State the amplitude and period of y =
7 3x sin and sketch the curve between 0° and 360° 8 2
7 360° = 3.5 and period = = 960° 3 2 8 7 3x A sketch of y = sin is shown below 8 2
Amplitude =
10. State the amplitude and period of y = 6 sin(t – 45°) and sketch the curve between 0° and 360°
If y = 6 sin(t – 45°), amplitude = 6 and period =
360° = 360° 1
A sketch y = 6 sin(t – 45°) is shown below.
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11. State the amplitude and period of y = 4 cos(2θ + 30°) and sketch the curve between 0° and 360°
If y = 4 cos(2θ + 30°), amplitude = 4 and period =
360° = 180° 2
A sketch y = 4 cos(2θ + 30°) is shown below (Note that y = 4 cos(2θ + 30°) leads y = 4 cos 2θ by
30° = 15°) 2
12. The frequency of a sine wave is 200 Hz. Calculate the periodic time.
Periodic time T =
1 1 = s = 5 ms f 200
13. Calculate the frequency of a sine wave that has a periodic time of 25 ms.
Frequency f =
1 1 = 40 Hz = T 25 ×10−3
14. Calculate the periodic time for a sine wave having a frequency of 10 kHz.
Periodic time T =
1 1 = s = 100 μs or 0.1 ms f 10 ×103
15. An alternating current completes 15 cycles in 24 ms. Determine its frequency?
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One cycle is completed in
Frequency f =
24 ms = 1.6 ms 15
Hence, periodic time T = 1.6 ms
1 1 = 625 Hz = T 1.6 ×10−3
16. Graphs of y1 = 2 sin x and y2 = 3 sin(x + 50°) are drawn on the same axes. Is y2 lagging or leading y1 ?
y2 is leading y1 by 50°
17. Graphs of y1 = 6 sin x and y2 = 5 sin(x – 70°) are drawn on the same axes. Is y1 lagging or leading y2 ?
y2 is lagging y1 by 70°
hence,
y1 is leading y2 by 70°
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EXERCISE 167 Page 454
1. Find the (a) amplitude, (b) frequency, (c) periodic time and (d) phase angle (stating whether it is leading or lagging sin ωt) for: i = 40 sin(50πt + 0.29) mA. If i = 40 sin(50πt + 0.29) mA, then
amplitude = 40 mA,
ω = 50π rad/s = 2πf from which, frequency, f =
50π = 25 Hz 2π
1 1 = = 0.040 s or 40 ms f 25
periodic time, T = phase angle = 0.29 rad leading or 0.29 ×
180°
π
= 16.62° leading or 16°37′ leading
2. Find the (a) amplitude, (b) frequency, (c) periodic time and (d) phase angle (stating whether it is leading or lagging sin ωt) for: y = 75 sin(40t – 0.54) cm.
If y = 75 sin(40t – 0.54) cm, then
amplitude = 75 cm 40 = 6.37 Hz 2π
ω = 40 rad/s = 2πf from which, frequency, f =
1 1 = = 0.157 s f 6.37
periodic time, T = phase angle = 0.54 rad lagging or 0.54 ×
180°
π
= 30.94° lagging or 30°56′ lagging
3. Find the (a) amplitude, (b) frequency, (c) periodic time and (d) phase angle (stating whether it is leading or lagging sin ωt) for: v = 300 sin(200πt – 0.412) V. If v = 300 sin(200πt – 0.412) V, then
amplitude = 300 V 200π = 100 Hz 2π
ω = 200π rad/s = 2πf from which, frequency, f = periodic time, T = phase angle = 0.412 rad lagging or 0.412 ×
180°
π
1 1 = = 0.010 s or 10 ms f 100 = 23.61° lagging or 23°36′ lagging
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4. A sinusoidal voltage has a maximum value of 120 V and a frequency of 50 Hz. At time t = 0, the voltage is (a) zero, and (b) 50 V. Express the instantaneous voltage v in the form v = A sin( ωt ± α). Let v = A sin( ωt ± α) = 120 sin(2πft + φ) = 120 sin(100πt + φ) volts, since f = 50 Hz (a) When t = 0, v = 0 hence, from which,
0 = 120 sin(0 + φ), i.e. 0 = 120 sin φ
sin φ = 0 and
φ=0
Hence, if v = 0 when t = 0, then v = 120 sin 100πt volts (b) When t = 0, v = 50 V hence, 50 = 120 sin(0 + φ) from which,
50 = sin φ 120
π 50 φ sin −1 = and = = ° 24.624 ×= 0.43rad 24.624 180 120
Hence, if v = 50 when t = 0, then v = 120 sin(100πt + 0.43) volts
5. An alternating current has a periodic time of 25 ms and a maximum value of 20 A. When time = 0, current i = –10 amperes. Express the current i in the form i = A sin(ωt ± α).
If periodic time T = 25 ms, then frequency, f=
1 1 = 40 Hz = T 25 ×10−3
Angular velocity, ω = 2πt =2π(40) = 80π rad/s Hence, current i = 20 sin(80πt + φ) When t = 0, i = –10, hence from which,
sin φ = −
–10 = 20 sin φ 10 = −0.5 20
and
sin −1 (−0.5) = φ= −30° or −
π 6
rad
π Thus, i 20sin 80π t − = = A or i 20sin ( 80π t − 0.524 ) A 6
6. An oscillating mechanism has a maximum displacement of 3.2 m and a frequency of 50 Hz. At time t = 0 the displacement is 150 cm. Express the displacement in the general form A sin( ωt ± α). Displacement, s = A sin(ωt ± α) where A = 3.2 m and ω = 2π f = 2π × 50 = 100π
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s = 3.2 sin(100πt + α)
i.e.
At time t = 0, displacement = 150 cm = 1.5 m Hence,
1.5 = 3.2 sin(100πt + α) = 3.2 sin α
i.e.
1.5 = sin α 3.2
1.5 α = sin −1 = 27.953° = 0.488 rad 3.2
and
Hence, displacement = 3.2 sin(100πt + 0.488) m
7. The current in an a.c. circuit at any time t seconds is given by: i = 5 sin(100πt – 0.432) amperes Determine the (a) amplitude, frequency, periodic time and phase angle (in degrees) (b) value of current at t = 0 (c) value of current at t = 8 ms (d) time when the current is first a maximum (e) time when the current first reaches 3 A. Sketch one cycle of the waveform showing relevant points. (a) If i = 5 sin(100πt – 0.432) mA, then
amplitude = 5 A,
ω = 100π rad/s = 2πf from which, frequency, f = periodic time, T = and phase angle = 0.432 rad lagging or 0.432 × (b) When t = 0, i = 5 sin(–0.432) = –2.093 A
100π = 50 Hz 2π
1 1 = = 0.020 s or 20 ms f 50
180°
π
= 24.75° lagging or 24°45′ lagging. (note that –0.432 is radians)
(c) When t = 8 ms, i = 5 sin 100π ( 8 ×10−3 ) − 0.432 = 5 sin (2.081274) = 4.363 A (d) When the current is first a maximum, 5 = 5 sin(100πt – 0.432) i.e.
1 = sin(100πt – 0.432) 672
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100πt – 0.432 = sin −1 1 = 1.5708
and
from which,
time t =
(again, be sure your calculator is on radians)
1.5708 + 0.432 = 0.006375 s or 6.375 ms 100π
3 = 5 sin(100πt – 0.432)
(e) When i = 3 A,
3 = sin(100πt – 0.432) 5
i.e.
100πt – 0.432 = sin −1
and
from which,
time t =
3 = 0.6435 5
0.6435 + 0.432 = 0.003423 s or 3.423 ms 100π
A sketch of one cycle of the waveform is shown below.
Note that since phase angle φ = 24.75°, in terms of time φt then 24.75 φt from which, φt = 1.375 ms ≡ 360 20
Alternatively,
φt =
φ 0.432 = 1.375 ms, as shown in the sketch. = ω 100π
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EXERCISE 168 Page 459
1. A complex current waveform i comprises a fundamental current of 50 A r.m.s. and frequency 100 Hz, together with a 24% third harmonic, both being in phase with each other at zero time. (a) Write down an expression to represent current i. (b) Sketch the complex waveform of current using harmonic synthesis over one cycle of the fundamental.
(a) Fundamental current:
r.m.s. =
from which, maximum value =
1 × maximum value 2 2 × r.m.s. =2 × 50 = 70.71 A
Hence, fundamental current is: i1 = 70.71 sin 2π(100)t = 70.71 sin 628.3t A Third harmonic:
amplitude = 24% of 70.71 = 16.97 A
Hence, third harmonic current is: i3 = 16.97 sin 3(628.3)t = 16.97 sin 1885t A Thus, current i = i1 + i3 = (70.71 sin 628.3t + 16.97 sin 1885t) amperes (b) The complex waveform for current i is shown sketched below:
2. A complex voltage waveform v comprises a 212.1 V r.m.s. fundamental voltage at a frequency of 50 Hz, a 30% second harmonic component lagging the fundamental voltage at zero time by π/2 rad, and a 10% fourth harmonic component leading the fundamental at zero time by π/3 rad. (a) Write down an expression to represent voltage v. (b) Sketch the complex voltage waveform using harmonic synthesis over one cycle of the fundamental waveform. 674
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(a) Voltage, v=
212.1 212.1 π 212.1 π sin 2π (50)t + (0.30) sin 2(2π )(50)t − + (0.1) sin 4(2π )(50)t + 0.707 0.707 2 0.707 3
volts i.e.
π π v = 300 sin 314.2 t + 90 sin 628.3t − + 30sin 1256.6t + volts 2 3
(b) The complex waveform representing v is shown sketched below
3. A voltage waveform is represented by:
v = 20 + 50 sin ωt + 20 sin(2ωt – π/2) volts.
Draw the complex waveform over one cycle of the fundamental by using harmonic synthesis. One waveform of v = 20 + 50 sin ωt + 20 sin(2 ωt – π/2) volts is shown sketched below using harmonic synthesis. 675
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4. Write down an expression representing a current i having a fundamental component of amplitude 16 A and frequency 1 kHz, together with its third and fifth harmonics being respectively one-fifth and one-tenth the amplitude of the fundamental, all components being in phase at zero time. Sketch the complex current waveform for one cycle of the fundamental using harmonic synthesis.
Fundamental current i1 = 16 sin 2π(1000)t = 16 sin 2000πt = 16 sin 2π 103 t Third harmonic i3 =
1 (16) sin 3(2000πt) = 3.2 sin 6000πt = 3.2 sin 6π 103 t 5
Fifth harmonic i5 =
1 (16) sin 5(2000πt) = 1.6 sin 10 000πt = 1.6 sin π 104 t 10
Hence,
current, i = 16 sin 2π 103 t + 3.2 sin 6π 103 t + 1.6 sin π 104 t 676
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A sketch of current is shown below
5. A voltage waveform is described by: v = 200 sin 377t + 80 sin(1131t +
π 4
) + 20 sin(1885t –
π 3
) volts
Determine (a) the fundamental and harmonic frequencies of the waveform, (b) the percentage third harmonic and (c) the percentage fifth harmonic. Sketch the voltage waveform using harmonic synthesis over one cycle of the fundamental.
(a) From the fundamental voltage, 377 = ω1 = 2π f1 i.e. fundamental frequency, f1 =
377 = 60 Hz 2π
From the 3rd harmonic voltage, 1131 = ω3 = 2π f3 i.e.
3rd harmonic frequency, f3 =
1131 = 180 Hz 2π
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From the 5th harmonic voltage, 1885 = ω5 = 2π f5 i.e.
5th harmonic frequency, f5 =
1885 = 300 Hz 2π
(b) Percentage 3rd harmonic =
80 ×100% = 40% 200
(c) Percentage 5th harmonic =
20 ×100% = 10% 200
The complex waveform representing v is shown sketched below:
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1. Determine all of the angles between 0° and 360° whose sine is: (a) 0.6792
(b) –0.1483
(a) Sine is positive in the 1st and 2nd quadrants If sin θ = 0.6792, then θ = sin −1 (0.6792) = 42.78° or 180° – 42.78° = 137.22° as shown in diagram (i) below.
(i)
(ii)
(b) Sine is negative in the 3rd and 4th quadrants If sin θ = 0.1483, then θ = sin −1 (0.1483) = 8.53° From diagram (ii) above, the two values where sin θ = –0.1483 are 180° + 8.53° = 188.53° and 360° – 8.53° = 351.47°
2. Solve the following equations for values of x between 0° and 360°: (a) x = cos −1 0.8739
(b) x = cos −1 (–0.5572)
(a) Cosine is positive in the 1st and 4th quadrants x = cos −1 (0.8739) = 29.08° or 360° –29.08° = 330.92° as shown in diagram (i) below.
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(i)
(ii)
(b) Cosine is negative in the 2nd and 3rd quadrants x = cos −1 (0.5572) = 56.14° hence, from diagram (ii) shown above, the two values of x for which x = cos −1 (−0.5572) are:
180° – 56.14° = 123.86
and
180° + 56.14° = 236.14°
3. Find the angles between 0° to 360° whose tangent is: (a) 0.9728
(b) –2.3420
(a) Tangent is positive in the 1st and 3rd quadrants
θ = tan −1 0.9728 = 44.21° or 180° + 44.21° = 224.21° as shown in diagram (i) below.
(i)
(ii)
(b) Tangent is negative in the 2nd and 4th quadrants
θ = tan −1 2.3418 = 66.88° and from diagram (ii) shown above, θ = 180° – 66.88° = 113.12°
and
360° – 66.88° = 293.12°
4. Solve, in the range 0° to 360°, giving the answers in degrees and minutes: cos −1 (–0.5316) = t
Cosine is negative in the 2nd and 3rd quadrants. cos −1 (0.5316) = 57.886° or 57°53′ as shown in the diagram below.
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From the diagram, t = 180° – 57°53′ = 122°7′
and
t = 180° + 57°53′ = 237°53′
5. Solve, in the range 0° to 360°, giving the answers in degrees and minutes:
sin −1 ( −0.6250 ) = α
Sine is negative in the 3rd and 4th quadrants If sin θ = 0.6250, then α = sin −1 (0.6250) = 38.682° = 38°41′
From the diagram above, the two values where sin α = –0.6250 are 180° + 38°41′ = 218°41′ and 360° – 38°41′ = 321°19′
6. Solve, in the range 0° to 360°, giving the answers in degrees and minutes:
tan −1 0.8314 = θ
Tangent is positive in the 1st and 3rd quadrants
θ = tan −1 0.8314 = 39.74° or 39°44′
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From the diagram, the two values of θ between 0° and 360° are: 39°44′
and 180° + 39°44′ = 219°44′
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EXERCISE 166 Page 452
1. A sine wave is given by y = 5 sin 3x. State its peak value.
Peak value = amplitude = maximum value = 5
2. A sine wave is given by y = 4 sin 2x. State its period in degrees.
Period =
360° = 180° 2
3. A periodic function is given by y = 30 cos 5x. State its maximum value.
Maximum value = amplitude = peak value = 30
4. A periodic function is given by y = 25 cos 3x. State its period in degrees.
Period =
360° = 120° 3
5. State the amplitude and period of y = cos 3A and sketch the curve between 0° and 360°
Amplitude = 1 and period =
360° = 120° 3
A sketch of y = cos 3A is shown below.
6. State the amplitude and period of y = 2 sin
5x and sketch the curve between 0° and 360° 2 665
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5x 360° , amplitude = 2 and period = = 144° 5 2 2 5x A sketch y = 2 sin is shown below 2
If y = 2 sin
7. State the amplitude and period of y = 3 sin 4t and sketch the curve between 0° and 360°
Amplitude = 3 and period =
360° = 90° 4
A sketch of y = 3 sin 4t is shown below
8. State the amplitude and period of y = 5 cos
Amplitude = 5 and period =
A sketch of y = 5 cos
θ 2
θ 2
and sketch the curve between 0° and 360°
360° = 720° 1 2
is shown below
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9. State the amplitude and period of y =
7 3x sin and sketch the curve between 0° and 360° 8 2
7 360° = 3.5 and period = = 960° 3 2 8 7 3x A sketch of y = sin is shown below 8 2
Amplitude =
10. State the amplitude and period of y = 6 sin(t – 45°) and sketch the curve between 0° and 360°
If y = 6 sin(t – 45°), amplitude = 6 and period =
360° = 360° 1
A sketch y = 6 sin(t – 45°) is shown below.
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11. State the amplitude and period of y = 4 cos(2θ + 30°) and sketch the curve between 0° and 360°
If y = 4 cos(2θ + 30°), amplitude = 4 and period =
360° = 180° 2
A sketch y = 4 cos(2θ + 30°) is shown below (Note that y = 4 cos(2θ + 30°) leads y = 4 cos 2θ by
30° = 15°) 2
12. The frequency of a sine wave is 200 Hz. Calculate the periodic time.
Periodic time T =
1 1 = s = 5 ms f 200
13. Calculate the frequency of a sine wave that has a periodic time of 25 ms.
Frequency f =
1 1 = 40 Hz = T 25 ×10−3
14. Calculate the periodic time for a sine wave having a frequency of 10 kHz.
Periodic time T =
1 1 = s = 100 μs or 0.1 ms f 10 ×103
15. An alternating current completes 15 cycles in 24 ms. Determine its frequency?
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One cycle is completed in
Frequency f =
24 ms = 1.6 ms 15
Hence, periodic time T = 1.6 ms
1 1 = 625 Hz = T 1.6 ×10−3
16. Graphs of y1 = 2 sin x and y2 = 3 sin(x + 50°) are drawn on the same axes. Is y2 lagging or leading y1 ?
y2 is leading y1 by 50°
17. Graphs of y1 = 6 sin x and y2 = 5 sin(x – 70°) are drawn on the same axes. Is y1 lagging or leading y2 ?
y2 is lagging y1 by 70°
hence,
y1 is leading y2 by 70°
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EXERCISE 167 Page 454
1. Find the (a) amplitude, (b) frequency, (c) periodic time and (d) phase angle (stating whether it is leading or lagging sin ωt) for: i = 40 sin(50πt + 0.29) mA. If i = 40 sin(50πt + 0.29) mA, then
amplitude = 40 mA,
ω = 50π rad/s = 2πf from which, frequency, f =
50π = 25 Hz 2π
1 1 = = 0.040 s or 40 ms f 25
periodic time, T = phase angle = 0.29 rad leading or 0.29 ×
180°
π
= 16.62° leading or 16°37′ leading
2. Find the (a) amplitude, (b) frequency, (c) periodic time and (d) phase angle (stating whether it is leading or lagging sin ωt) for: y = 75 sin(40t – 0.54) cm.
If y = 75 sin(40t – 0.54) cm, then
amplitude = 75 cm 40 = 6.37 Hz 2π
ω = 40 rad/s = 2πf from which, frequency, f =
1 1 = = 0.157 s f 6.37
periodic time, T = phase angle = 0.54 rad lagging or 0.54 ×
180°
π
= 30.94° lagging or 30°56′ lagging
3. Find the (a) amplitude, (b) frequency, (c) periodic time and (d) phase angle (stating whether it is leading or lagging sin ωt) for: v = 300 sin(200πt – 0.412) V. If v = 300 sin(200πt – 0.412) V, then
amplitude = 300 V 200π = 100 Hz 2π
ω = 200π rad/s = 2πf from which, frequency, f = periodic time, T = phase angle = 0.412 rad lagging or 0.412 ×
180°
π
1 1 = = 0.010 s or 10 ms f 100 = 23.61° lagging or 23°36′ lagging
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4. A sinusoidal voltage has a maximum value of 120 V and a frequency of 50 Hz. At time t = 0, the voltage is (a) zero, and (b) 50 V. Express the instantaneous voltage v in the form v = A sin( ωt ± α). Let v = A sin( ωt ± α) = 120 sin(2πft + φ) = 120 sin(100πt + φ) volts, since f = 50 Hz (a) When t = 0, v = 0 hence, from which,
0 = 120 sin(0 + φ), i.e. 0 = 120 sin φ
sin φ = 0 and
φ=0
Hence, if v = 0 when t = 0, then v = 120 sin 100πt volts (b) When t = 0, v = 50 V hence, 50 = 120 sin(0 + φ) from which,
50 = sin φ 120
π 50 φ sin −1 = and = = ° 24.624 ×= 0.43rad 24.624 180 120
Hence, if v = 50 when t = 0, then v = 120 sin(100πt + 0.43) volts
5. An alternating current has a periodic time of 25 ms and a maximum value of 20 A. When time = 0, current i = –10 amperes. Express the current i in the form i = A sin(ωt ± α).
If periodic time T = 25 ms, then frequency, f=
1 1 = 40 Hz = T 25 ×10−3
Angular velocity, ω = 2πt =2π(40) = 80π rad/s Hence, current i = 20 sin(80πt + φ) When t = 0, i = –10, hence from which,
sin φ = −
–10 = 20 sin φ 10 = −0.5 20
and
sin −1 (−0.5) = φ= −30° or −
π 6
rad
π Thus, i 20sin 80π t − = = A or i 20sin ( 80π t − 0.524 ) A 6
6. An oscillating mechanism has a maximum displacement of 3.2 m and a frequency of 50 Hz. At time t = 0 the displacement is 150 cm. Express the displacement in the general form A sin( ωt ± α). Displacement, s = A sin(ωt ± α) where A = 3.2 m and ω = 2π f = 2π × 50 = 100π
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s = 3.2 sin(100πt + α)
i.e.
At time t = 0, displacement = 150 cm = 1.5 m Hence,
1.5 = 3.2 sin(100πt + α) = 3.2 sin α
i.e.
1.5 = sin α 3.2
1.5 α = sin −1 = 27.953° = 0.488 rad 3.2
and
Hence, displacement = 3.2 sin(100πt + 0.488) m
7. The current in an a.c. circuit at any time t seconds is given by: i = 5 sin(100πt – 0.432) amperes Determine the (a) amplitude, frequency, periodic time and phase angle (in degrees) (b) value of current at t = 0 (c) value of current at t = 8 ms (d) time when the current is first a maximum (e) time when the current first reaches 3 A. Sketch one cycle of the waveform showing relevant points. (a) If i = 5 sin(100πt – 0.432) mA, then
amplitude = 5 A,
ω = 100π rad/s = 2πf from which, frequency, f = periodic time, T = and phase angle = 0.432 rad lagging or 0.432 × (b) When t = 0, i = 5 sin(–0.432) = –2.093 A
100π = 50 Hz 2π
1 1 = = 0.020 s or 20 ms f 50
180°
π
= 24.75° lagging or 24°45′ lagging. (note that –0.432 is radians)
(c) When t = 8 ms, i = 5 sin 100π ( 8 ×10−3 ) − 0.432 = 5 sin (2.081274) = 4.363 A (d) When the current is first a maximum, 5 = 5 sin(100πt – 0.432) i.e.
1 = sin(100πt – 0.432) 672
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100πt – 0.432 = sin −1 1 = 1.5708
and
from which,
time t =
(again, be sure your calculator is on radians)
1.5708 + 0.432 = 0.006375 s or 6.375 ms 100π
3 = 5 sin(100πt – 0.432)
(e) When i = 3 A,
3 = sin(100πt – 0.432) 5
i.e.
100πt – 0.432 = sin −1
and
from which,
time t =
3 = 0.6435 5
0.6435 + 0.432 = 0.003423 s or 3.423 ms 100π
A sketch of one cycle of the waveform is shown below.
Note that since phase angle φ = 24.75°, in terms of time φt then 24.75 φt from which, φt = 1.375 ms ≡ 360 20
Alternatively,
φt =
φ 0.432 = 1.375 ms, as shown in the sketch. = ω 100π
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© 2014, John Bird
EXERCISE 168 Page 459
1. A complex current waveform i comprises a fundamental current of 50 A r.m.s. and frequency 100 Hz, together with a 24% third harmonic, both being in phase with each other at zero time. (a) Write down an expression to represent current i. (b) Sketch the complex waveform of current using harmonic synthesis over one cycle of the fundamental.
(a) Fundamental current:
r.m.s. =
from which, maximum value =
1 × maximum value 2 2 × r.m.s. =2 × 50 = 70.71 A
Hence, fundamental current is: i1 = 70.71 sin 2π(100)t = 70.71 sin 628.3t A Third harmonic:
amplitude = 24% of 70.71 = 16.97 A
Hence, third harmonic current is: i3 = 16.97 sin 3(628.3)t = 16.97 sin 1885t A Thus, current i = i1 + i3 = (70.71 sin 628.3t + 16.97 sin 1885t) amperes (b) The complex waveform for current i is shown sketched below:
2. A complex voltage waveform v comprises a 212.1 V r.m.s. fundamental voltage at a frequency of 50 Hz, a 30% second harmonic component lagging the fundamental voltage at zero time by π/2 rad, and a 10% fourth harmonic component leading the fundamental at zero time by π/3 rad. (a) Write down an expression to represent voltage v. (b) Sketch the complex voltage waveform using harmonic synthesis over one cycle of the fundamental waveform. 674
© 2014, John Bird
(a) Voltage, v=
212.1 212.1 π 212.1 π sin 2π (50)t + (0.30) sin 2(2π )(50)t − + (0.1) sin 4(2π )(50)t + 0.707 0.707 2 0.707 3
volts i.e.
π π v = 300 sin 314.2 t + 90 sin 628.3t − + 30sin 1256.6t + volts 2 3
(b) The complex waveform representing v is shown sketched below
3. A voltage waveform is represented by:
v = 20 + 50 sin ωt + 20 sin(2ωt – π/2) volts.
Draw the complex waveform over one cycle of the fundamental by using harmonic synthesis. One waveform of v = 20 + 50 sin ωt + 20 sin(2 ωt – π/2) volts is shown sketched below using harmonic synthesis. 675
© 2014, John Bird
4. Write down an expression representing a current i having a fundamental component of amplitude 16 A and frequency 1 kHz, together with its third and fifth harmonics being respectively one-fifth and one-tenth the amplitude of the fundamental, all components being in phase at zero time. Sketch the complex current waveform for one cycle of the fundamental using harmonic synthesis.
Fundamental current i1 = 16 sin 2π(1000)t = 16 sin 2000πt = 16 sin 2π 103 t Third harmonic i3 =
1 (16) sin 3(2000πt) = 3.2 sin 6000πt = 3.2 sin 6π 103 t 5
Fifth harmonic i5 =
1 (16) sin 5(2000πt) = 1.6 sin 10 000πt = 1.6 sin π 104 t 10
Hence,
current, i = 16 sin 2π 103 t + 3.2 sin 6π 103 t + 1.6 sin π 104 t 676
© 2014, John Bird
A sketch of current is shown below
5. A voltage waveform is described by: v = 200 sin 377t + 80 sin(1131t +
π 4
) + 20 sin(1885t –
π 3
) volts
Determine (a) the fundamental and harmonic frequencies of the waveform, (b) the percentage third harmonic and (c) the percentage fifth harmonic. Sketch the voltage waveform using harmonic synthesis over one cycle of the fundamental.
(a) From the fundamental voltage, 377 = ω1 = 2π f1 i.e. fundamental frequency, f1 =
377 = 60 Hz 2π
From the 3rd harmonic voltage, 1131 = ω3 = 2π f3 i.e.
3rd harmonic frequency, f3 =
1131 = 180 Hz 2π
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From the 5th harmonic voltage, 1885 = ω5 = 2π f5 i.e.
5th harmonic frequency, f5 =
1885 = 300 Hz 2π
(b) Percentage 3rd harmonic =
80 ×100% = 40% 200
(c) Percentage 5th harmonic =
20 ×100% = 10% 200
The complex waveform representing v is shown sketched below:
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