Wave and Vector Function Analysis - Semester 1, Electrical Engineering

This assignment solution covers the analysis of wave and vector functions, focusing on two main tasks. Task 1 involves analyzing voltage waveforms (v1 and v2), determining their amplitudes, phases, and frequencies. It includes calculating the time for maximum amplitude, using compound angle formulas to expand the functions, converting them to polar form, and graphically representing the results to understand constructive interference. Task 2 focuses on vector analysis, calculating the distance between two points (A and B), determining the angle between two vectors (AB and BC), and deriving the vector equation of a line (BC). The solution provides detailed steps and calculations for each part, including the application of relevant formulas and concepts in electrical engineering.

Wave and vector function

TABLE OF CONTENT
TASK 1 ...........................................................................................................................................1
A. Phase, frequency and amplitude .............................................................................................1
B. Time taken for maximum amplitude ......................................................................................1
C. Calculating time .....................................................................................................................2
D. Compound angle formula .......................................................................................................3
E. Conversion to polar form ........................................................................................................3
F. Using spread sheet software to complete table........................................................................4
G. Graphical representation ........................................................................................................4
H. Conclusion about v1 + v2 .....................................................................................................5
TASK 2............................................................................................................................................5
A. Distance AB ...........................................................................................................................6
B. Angle between AB and BC ....................................................................................................6
C. Vector equation of line BC .....................................................................................................6

TASK 1
v1 = 128 sin (100 πt + 0.6984 ) volts
v2 = 188 sin (100 πt - 0.8730 ) volts
A. Phase, frequency and amplitude
v1 = 128 sin (100 πt + 0.6984 ) volts
Solution
Amplitude: 128
Phase: +0.6984
Frequency: 50 Hz
v2 = 188 sin (100 πt - 0.8730 ) volts
Solution
Amplitude: 188
Phase: -0.8730
Frequency: 50 Hz
B. Time taken for maximum amplitude
Solution
The maximum value of amplitude will result in maximum value of voltages.
i) v1 = 128 sin (100 πt + 0.6984 ) volts
dv1 / dt = (128* 100 π) cos (100 πt + 0.6984 )
Equating dv1 / dt to zero we have :
(128* 100 π) cos (100 πt + 0.6984 ) = 0
cos (π / 2 ) = 0 so
100 πt + 0.6984 = π / 2
200 πt + 1.396 = π
π (200 t – 1) = -1.396
(200 t – 1) = - 0.444
time t = 0.00278 seconds
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ii) v2 = 188 sin (100 πt - 0.8730 ) volts
Differentiating both the sides
dv2/dt = (188*100 π) cos (100 πt - 0.8730 )
Equating it to zero
(188*100 π) cos (100 πt - 0.8730 ) = 0
cos (100 πt - 0.8730 ) = cos (π / 2 )
(100 πt - 0.8730 ) = (π / 2 )
200 πt – 1.746 = π
π (200t – 1 ) = 1.746
(200t – 1 ) = 0.556
time t = 0.007 seconds
C. Calculating time
i) v1 = 128 sin (100 πt + 0.6984 ) volts
At v1 = -50 volts time = ?
Solution
-50 = 128 sin (100 πt + 0.6984 ) volts
-0.39 = sin (100 πt + 0.6984 ) Now using inverse sin function values we have
(100 πt + 0.6984 ) = -0.4
time t = 0.0034 seconds
ii) v2 = 188 sin (100 πt - 0.8730 ) volts
at v2 = -50 volts time = ?
Solution
-50 = 188 sin (100 πt - 0.8730 )
-0.265 = sin (100 πt - 0.8730 )
-0.268 = (100 πt - 0.8730 )
100 πt = 0.605
time t = 0.0019 seconds
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D. Compound angle formula
Expansion of v1 and v2 in the form A sin 100 πt +- B cos 100 πt can be done by using
compound angle formulae:
i) v1 = 128 sin (100 πt + 0.6984 ) volts
Solution
Compound angle formula: sin (a+b) = sin a cos b + cos a sin b
128 sin (100 πt + 0.6984 ) = 128 [sin 100 πt cos 0.6984 + cos 100 πt sin 0.6984]
(cos 0.6984 = 0.765 sin 0.6984 = 0.642)
= 128 [0.765 sin 100 πt + 0.642 cos 100 πt ]
so v1 = 97.92 sin 100 πt + 82.17 cos 100 πt
On comparing this equation with given equation we can find value of A and B as follows:
A sin 100 πt (+-) B cos 100 πt = 97.92 sin 100 πt + 82.17 cos 100 πt
A = 97.92
B = 82.17
ii) v2 = 188 sin (100 πt - 0.8730 ) volts
Solution
Using compound angle formula : sin (a- b) = sin a cos b - cos a sin b
188 sin (100 πt - 0.8730 ) = 188 [sin 100 πt cos 0.8730 - cos 100 πt sin 0.8730]
(cos 0.8730 = 0.642 sin 0.8730 = 0.766)
= 188 [0.642 sin 100 πt – 0.766 cos 100 πt]
v2 = 120.69 sin 100 πt – 144 cos 100 πt
A sin 100 πt (+-) B cos 100 πt = 120.69 sin 100 πt – 144 cos 100 πt
A = 120.69
B = 144
E. Conversion to polar form
Solution
From above we have v1 and v2 as follows:
v1 = 97.92 sin 100 πt + 82.17 cos 100 πt
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v2 = 120.69 sin 100 πt – 144 cos 100 πt
v1 + v2 = 97.92 sin 100 πt + 82.17 cos 100 πt +120.69 sin 100 πt – 144 cos 100 πt
= 218.61 sin 100 πt – 61.83 cos 100 πt
v1 + v2 = 218.61 sin 100 πt – 61.83 cos 100 πt
The expression for v1 and v2 is of the form A sin θ + B cos θ where θ = 100 πt ,A =
218.61 and B = -61.83 . The expression can be converted into polar form Rsin (θ + α) by using
below conversion relations:
R = √(A² + B² ) and Tan α = B/A
From the given equations:
R = √((218.61)² + (-61.83)² ) = 227.185
Tan α = (-61.83) / 218.61
α = -0.27 radians
227.185 sin (100 πt – 0.27)
v1 +v2 = 218.61 sin 100 πt – 61.83 cos 100 πt = 227.185 sin (100 πt – 0.27)
F. Using spread sheet software to complete table
G. Graphical representation
Solution
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The new form v1 + v2 is expressed as: 227.185 sin (100 πt – 0.27)
The amplitude of this wave is 227.18 volts
Phase = -0.27 radians
H. Conclusion about v1 + v2
From the above graph it can be concluded that when two waveforms of same nature of
amplitude (positive or negative amplitude) are added then the resultant wave form has greater
amplitude. In the given graph it is observed that since v1 and v2 both have positive amplitude
their addition results in another positive waveform which has higher amplitude as compare to v1
and v2. Such type of interference is known as constructive interference. This type of wave
interference occurs because both v1 and v2 or adding waveforms have phase in same direction.
Though there is difference in phase magnitude of waveforms but it is also observed in terms of
its additive nature.
TASK 2
Point A (0, -3.2, 0)
Point B (4.2, 0, -2)
Point C (a, b, 0)
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A. Distance AB
Solution
A (0, -3.2, 0)
B (4.2, 0, -2)
Distance between two coordinates X (x1, y1, z1) and Y(x2, y2, z2) is given by:
XY = √ [ (x2 - x1) ² +(y2 - y1)² + (z2 - z1) ² ]
By using this formula distance AB is = √ [ (4.2 - 0) ² +(0 + 3.2)² + (-2 - 0) ² ]
AB = 5.64
B. Angle between AB and BC
Solution
Direction in which BC is drilled = 2.5i + 3j + k
Coordinates of vector AB = (4.2, 3.2, -2)
Angle between two vectors is given by formula
= Cos θ = Dot products of vector / (magnitude of first vector * magnitude of second
vector)
On substituting values we have:
Cos θ = AB. BC / mag. AB * mag. BC (mag = magnitude of vector )
AB.BC = (2.5*4.2 + 3*3.2 + 1*(-2)) = 18.1
Magnitude of AB = √ [ 4.2² +3.2² + (-2) ² ] = 5.64
Magnitude of BC = √ [ 2.5² +3² + 1 ² ] = 4.03
Cos θ = 18.1 / [4.03* 5.64] = 0.796
θ = 37.25 degree
C. Vector equation of line BC
Solution
C (a, b, 0) = vector B + rho (2.5i , 3j, 1k)
C (a, b, 0) = (4.2, 0, -2) + rho (2.5, 3, 1)
rho = 4.65
C (a, b, 0) = (4.2, 0, -2) + (11.62, 13.95, 4.65)
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C (a, b, 0) = (15.82, 13.95, 2.65 )
On comparing vectors we have:
a = 15.82
b = 13.95
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