Engineering Math: Differential Equations

Running head: ENGINEERING MATHS
ENGINEERING MATHS
Name of the Student
Name of the University
Author Note

ENGINEERING MATHS1
Task 1:
Given, the composition of the AC voltage are two sinusoidal signals.
v1 = 122sin(100πt + 0.3492)
v2 = 178sin(100πt – 0.5238)
a. amplitude of v1 = 122
Phase of v1 = 0.3492 rad
Frequency = 100π/2π = 50 Hz.
Periodic time = 1/frequency = 1/50 = 20 msec.
Amplitude of v2 = 178
Phase of v2 = -0.5238 rad
Frequency = 100π/2π = 50 Hz.
Periodic time = 1/frequency = 1/50 = 20 msec.
b. The maximum amplitude for v1 is 122.
The time taken to reach maximum amplitude by v1 for first time
122sin(100πt + 0.3492) = 122
 Sin(100πt + 0.3492) = 1
 100πt + 0.3492 = 1.5708
 100πt = 1.2216
 t = 1.2216/(100π) = 0.0039 sec or 3.9 msec.
Similarly the time taken for v2 to reach 178 is

ENGINEERING MATHS2
178sin(100πt – 0.5238) = 178
 sin(100πt – 0.5238) = 1
 100πt – 0.5238 = 1.5708
 100πt = 1.5708 + 0.5238
 t = 0.0067 sec or 6.7 msec.
c. Now, voltage v1 first reach -50 volts at
122sin(100πt + 0.3492) = -50
 sin(100πt + 0.3492) = -50/122
 100πt + 0.3492 = asin(-50/122)
 100πt + 0.3492 = π +0.42227
 100πt = 3.56386
 t = 0.0113 sec or 11.3 msec.
Now, voltage v2 first reach -50 volts at
178sin(100πt – 0.5238) = -50
 sin(100πt – 0.5238) = -50/178
 100πt – 0.5238 = asin(-50/178)
 100πt – 0.5238 = π+0.28473
 100πt = 3.95
 t = 0.0126 or 12.6 msec.
d.
Now by compound angle formula
v1 = 122sin(100πt + 0.3492) = 122(sin(100πt)cos(0.3492) + cos(100πt)*sin(0.3492))