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© 2014, John Bird
661
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 1sin (0.6792)θ −= = 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 1sin (0.1483)θ −= = 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 = 1cos (0.8739)− = 29.08° or 360° –29.08° = 330.92° as shown in diagram (i) below.
© 2014, John Bird
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(i) (ii) (b) Cosine is negative in the 2nd and 3rd quadrants x = 1cos (0.5572)− = 56.14° hence, from diagram (ii) shown above, the two values of x for which x = 1cos ( 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 1tan 0.9728θ −= = 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 1tan 2.3418θ −= = 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.
1cos (0.5316)− = 57.886° or 57°53′ as shown in the diagram below.
© 2014, John Bird
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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: ( )1sin 0.6250 α− − =
Sine is negative in the 3rd and 4th quadrants If sin θ = 0.6250, then 1sin (0.6250)α −= = 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
1tan 0.8314θ −= = 39.74° or 39°44′
© 2014, John Bird
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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′
© 2014, John Bird
665
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 = 3602° = 180°
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 = 3603° = 120°
5. State the amplitude and period of y = cos 3A and sketch the curve between 0° and 360°
Amplitude = 1 and period = 3603° = 120°
A sketch of y = cos 3A is shown below.
6. State the amplitude and period of y = 2 sin 52x and sketch the curve between 0° and 360°
© 2014, John Bird
666
If y = 2 5sin2x , amplitude = 2 and period = 360
52
° = 144°
A sketch y = 2 5sin2x is shown below
7. State the amplitude and period of y = 3 sin 4t and sketch the curve between 0° and 360°
Amplitude = 3 and period = 3604° = 90°
A sketch of y = 3 sin 4t is shown below
8. State the amplitude and period of y = 5 cos2θ and sketch the curve between 0° and 360°
Amplitude = 5 and period = 36012
° = 720°
A sketch of y = 5 cos2θ is shown below
© 2014, John Bird
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9. State the amplitude and period of y = 72
sin 38x and sketch the curve between 0° and 360°
Amplitude = 72
= 3.5 and period = 36038
° = 960°
A sketch of y = 72
sin 38x is shown below
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 = 3601° = 360°
A sketch y = 6 sin(t – 45°) is shown below.
© 2014, John Bird
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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 = 3602° = 180°
A sketch y = 4 cos(2θ + 30°) is shown below
(Note that y = 4 cos(2θ + 30°) leads y = 4 cos 2θ by 302° = 15°)
12. The frequency of a sine wave is 200 Hz. Calculate the periodic time.
Periodic time T = 1 1200f
= s = 5 ms
13. Calculate the frequency of a sine wave that has a periodic time of 25 ms.
Frequency f = 3
1 125 10T −
=×
= 40 Hz
14. Calculate the periodic time for a sine wave having a frequency of 10 kHz.
Periodic time T = 3
1 110 10f
=×
s = 100 μs or 0.1 ms
15. An alternating current completes 15 cycles in 24 ms. Determine its frequency?
© 2014, John Bird
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One cycle is completed in 2415
ms = 1.6 ms Hence, periodic time T = 1.6 ms
Frequency f = 3
1 11.6 10T −
=×
= 625 Hz
16. Graphs of 1y = 2 sin x and 2y = 3 sin(x + 50°) are drawn on the same axes. Is 2y lagging or
leading 1y ?
2y is leading 1y by 50°
17. Graphs of 1y = 6 sin x and 2y = 5 sin(x – 70°) are drawn on the same axes. Is 1y lagging or
leading 2y ?
2y is lagging 1y by 70° hence, 1y is leading 2y by 70°
© 2014, John Bird
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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 = 502ππ
= 25 Hz
periodic time, T = 1 125f
= = 0.040 s or 40 ms
phase angle = 0.29 rad leading or 1800.29π°
× = 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
ω = 40 rad/s = 2πf from which, frequency, f = 402π
= 6.37 Hz
periodic time, T = 1 16.37f
= = 0.157 s
phase angle = 0.54 rad lagging or 1800.54π°
× = 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
ω = 200π rad/s = 2πf from which, frequency, f = 2002ππ
= 100 Hz
periodic time, T = 1 1100f
= = 0.010 s or 10 ms
phase angle = 0.412 rad lagging or 1800.412π°
× = 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, 0 = 120 sin(0 + φ), i.e. 0 = 120 sin φ from which, 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 sin120
φ= and 1 50sin 24.624 24.624 0.43rad120 180
πφ − = = °= × =
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, 3
1 125 10
fT −
= =×
= 40 Hz
Angular velocity, ω = 2πt =2π(40) = 80π rad/s Hence, current i = 20 sin(80πt + φ) When t = 0, i = –10, hence –10 = 20 sin φ
from which, sin φ = 10 0.520
− = − and 1sin ( 0.5) 30 rad6
or πφ −= − = − ° −
Thus, 20sin 806
i t Aππ = −
or ( )20sin 80 0.524i t Aπ= −
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 2 2 50 100fω π π π= = × =
© 2014, John Bird
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i.e. s = 3.2 sin(100πt + α) At time t = 0, displacement = 150 cm = 1.5 m Hence, 1.5 = 3.2 sin(100πt + α) = 3.2 sin α
i.e. 1.53.2
= sin α
and α = 1 1.5sin3.2
−
= 27.953° = 0.488 rad
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 = 1002ππ
= 50 Hz
periodic time, T = 1 150f
= = 0.020 s or 20 ms
and phase angle = 0.432 rad lagging or 1800.432π°
× = 24.75° lagging or 24°45′ lagging.
(b) When t = 0, i = 5 sin(–0.432) = –2.093 A (note that –0.432 is radians) (c) When t = 8 ms, i = 5 ( )3sin 100 8 10 0.432π − × − = 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)
© 2014, John Bird
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and 100πt – 0.432 = 1sin 1 1.5708− = (again, be sure your calculator is on radians)
from which, time t = 1.5708 0.432100π+ = 0.006375 s or 6.375 ms
(e) When i = 3 A, 3 = 5 sin(100πt – 0.432)
i.e. 35
= sin(100πt – 0.432)
and 100πt – 0.432 = 1 3sin 0.64355
− =
from which, time t = 0.6435 0.432100π+ = 0.003423 s or 3.423 ms
A sketch of one cycle of the waveform is shown below.
Note that since phase angle φ = 24.75°, in terms of time tφ then
24.75360 20
tφ≡ from which, tφ = 1.375 ms
Alternatively, tφ = 0.432100
φω π= = 1.375 ms, as shown in the sketch.
© 2014, John Bird
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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. = 12
× maximum value
from which, maximum value = 2 r.m.s. 2 50× = × = 70.71 A Hence, fundamental current is: 1i = 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: 3i = 16.97 sin 3(628.3)t = 16.97 sin 1885t A Thus, current i = 1i + 3i = (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.
© 2014, John Bird
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(a) Voltage,
v = 212.1 212.1 212.1sin 2 (50) (0.30) sin 2(2 )(50) (0.1) sin 4(2 )(50)0.707 0.707 2 0.707 3
t t tπ ππ π π + − + +
volts
i.e. v = 300 sin 314.2 t + 90sin 628.3 30sin 1256.62 3
t tπ π − + +
volts
(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(2ωt – π/2) volts is shown sketched below using
harmonic synthesis.
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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 1i = 16 sin 2π(1000)t = 16 sin 2000πt = 16 sin 2π 310 t
Third harmonic 3i = 15
(16) sin 3(2000πt) = 3.2 sin 6000πt = 3.2 sin 6π 310 t
Fifth harmonic 5i = 110
(16) sin 5(2000πt) = 1.6 sin 10 000πt = 1.6 sin π 410 t
Hence, current, i = 16 sin 2π 310 t + 3.2 sin 6π 310 t + 1.6 sin π 410 t
© 2014, John Bird
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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 + 4π ) + 20 sin(1885t –
3π ) 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 = 1 12 fω π= i.e. fundamental frequency, 13772
fπ
= = 60 Hz
From the 3rd harmonic voltage, 1131 = 3 32 fω π=
i.e. 3rd harmonic frequency, 311312
fπ
= = 180 Hz