Engineering Assignment: Analytical & Computational Methods Solutions

This document presents solutions to various tasks related to Analytical & Computational Methods, likely within an Electrical Engineering context. The solutions cover several key areas, including the analysis of current waveforms, calculating instantaneous values of power signals, and working with RLC circuits and phasor diagrams. Further tasks involve vector calculations, specifically the cross product of current and magnetic fields. The assignment also delves into signal processing, addressing the combination of sinusoidal signals and the analysis of harmonic components of sound waves. Detailed step-by-step solutions are provided for each task, making it a valuable resource for students studying electrical engineering and related fields. The solutions are contributed by a student and are available on Desklib, a platform offering AI-based study tools.

Analytical & Computational
Methods

Table of Contents
TASK 1............................................................................................................................................1
TASK 2............................................................................................................................................2
TASK 3............................................................................................................................................3
PART 1........................................................................................................................................3
PART 2........................................................................................................................................3

TASK 1
a) Given equation of current waveform as –
is = 13 cos (2πft – π/4) [A]
where, f = 1 Hz and t represents time
Transforming the given equation into t as –
t = cos-1 (is/13) + π/4
2πf
when is = +10A then, t will be
t = cos-1 (10/13) + π/4
2π. 1
= 0.69 + 45
2 x 180
= 0.13 sec
b) The instantaneous value of a power signal is
12 ∟(3π/8) [W]
here, W is units
From given equation, it has been evaluated that 12 is magnitude and 3π/8 represents argument
then, horizontal component of the signal = cos 3π/8
= -0.04
while, vertical component of the signal = sin 3π/8
= -0.99
1

TASK 2
(a) For a RLC circuit, phasor diagram where RL leads RC by 90 is drawn as –⁰
VL
V
90⁰
VR
Therefore, Resultant voltage across whole RL can be calculated by using pythogoras theorem
V2 = V2R + VL2
= 302 + 402
= 900 + 1600
= 2500
or, V = 50 volts
(b) Given,
current as
I = 2𝑖 + 3𝑗 – 4k
and,
magnetic field as
B = 3𝑖 – 2𝑗 + 6k
then, cross product of current and magnetic field is –
I X B = (2𝑖 + 3𝑗 – 4k) x (3𝑖 – 2𝑗 + 6k)
i j k
2 3 -4
3 -2 6
= i (18 – 8) – j (12 + 12) + k(-4 – 9)
= 10i – 24j – 13k
2

TASK 3
PART 1
Given signals -
𝑣1 = 40 sin (4𝑡)
𝑣2 = A cos (4𝑡)
third signal –
𝑣0 = 50 sin (4𝑡 +α)
on expanding third signal as –
𝑣0 = 50 (sin 4𝑡 cos α + cos 4𝑡 sin α)
𝑣1 + 𝑣2 = 50 (sin 4𝑡 cos α + cos 4𝑡 sin α)
40 sin (4𝑡) + A cos (4𝑡) = 50 (sin 4𝑡 cos α + cos 4𝑡 sin α)
on comparing, α = 36.87 approx.⁰
then,
A cos (4𝑡) = 50 cos 4𝑡 sin α
A = sin α = 50 sin 36.87⁰
= 30 approx.
PART 2
Given,
Third harmonic of a sound wave as –
4 cos (3𝜃) – 6 sin (3𝜃)
𝑅 sin (3𝜃 +𝛽) = R (sin 3𝜃 cos 𝛽 + cos 3𝜃 sin 𝛽)
4 cos (3𝜃) – 6 sin (3𝜃) = R (sin 3𝜃 cos 𝛽 + cos 3𝜃 sin 𝛽)
on comparing –
R sin 3𝜃 cos 𝛽 = -6 sin (3𝜃)
R cos 𝛽 = -6 …(i)
R cos 3𝜃 sin 𝛽 = 4 cos (3𝜃)
R sin 𝛽 = 4 …(ii)
3

adding and squaring both equations –
R2 cos2 𝛽 + R2 sin2 𝛽 = (-6)2 + (4)2
R2 (cos2 𝛽 + sin2 𝛽) = 36 + 16
R2 = 52
or, R = √52 = 7.2
while, on dividing both equation
tan 𝛽 = 4/6
𝛽 = tan-1 (4/6) = 0.58 approx.
so,
4 cos (3𝜃) – 6 sin (3𝜃) = 7.2 sin (3𝜃 + 0.58)
4