Electrical Engineering Assignment Solutions: Analysis and Plots

This document presents a comprehensive solution to an electrical engineering assignment, addressing various problems and concepts. The solution includes calculations for error detection, analysis of sinusoidal waves, and high-pass filter responses. It further provides z-score calculations, plots for different scenarios, and analysis of system responses. The assignment also covers statistical analysis, numerical integration, and root finding. The solutions are detailed and include plots where necessary, offering a complete understanding of the problems and their solutions. This resource aims to aid students in grasping complex electrical engineering concepts and improving their problem-solving skills.

Q1:
Detecting and removing errors from the data given
0 1 2 3 4 5 6 7 8 9 10
0
50
100
150
200
250
f(x) = 1.18938107732675 x³ − 11.0928641251222 x² + 27.8009466481452 x
R² = 0.993649446545842
Plot with error
My Data Curve
Error = (1-0.9895) x 100% = 1.05%

0 1 2 3 4 5 6 7 8 9 10
0
50
100
150
200
250
f(x) = 1.002331002331 x³ − 8.04662004662004 x² + 13.2377622377622 x + 18.8867132867134
R² = 0.999981560792913
Plot with error removed
My Data Curve
Error = (1-1) x 100% = 0%

Q2:
Plot for i=32+50 sin(314.2 t)+20 sin(628.4 t − π
2 )
The expected response was a sinusoidal wave centered at 32. However, due two two
different frequency components with different magnitudes, the current results in the
above waveform. The double frequency component lags by 180o which is the reason
why the frequency components do not superimpose each other in every half a cycle.
Q3:
High-Pass filter magnitude and phase response
The transfer function of an analog high-pass filter is:
sRC
1+ sRC = s
s+ 1
RC

Q4:
The z-score is defined as: z= x − ~x
σ
In this case, σ =0.9, x=13.5
(a) ~x < 12.0A, z= 13.5− 12
0.9 =1.67
The p-value for z < 1.67 is: 0.4525
(b) ~x between 12.5A and 13.0A
z 1=13.5 −12.5
0.9 =1.11
z 2=13.5 −13
0.9 =0.56
P = 0.3643 - 0.2123 = 0.1520
(c) ~x > 14
z= 13.5− 14
0.9 =− 0.56

Since we are interested in the area to the right of the z-value and our z-score is
negative, we use z = 1-0.56 = 0.44
P = 0.1700
(d)

Q5:
Plots for
Point of contraflexure: x = 2.0428
Q6:
From the system given,
R = 0.1V, K = 50, KG = 500rpm/V x 0.01ms-1/rpm = 5ms-1/V = 0.2 V/ms-1
Hence, KG = 0.2 V/ms-1
H = 0.9 V/ms-1
For closed loop systems, T = KG
1+ KGH = 0.2
1+0.2 =0.6V /ms−1
Input, R = 0.1 V
Output, U = 0.1x50x10x0.01 = 0.5 ms-1
Q7:
Range f x fx mean
(x') x-x' (x-x')^2 f(x-x')^2 σ
34.5-36.5 7 35.5 248.5
39.167
-3.667 13.444 94.111
2.134
36.5-38.5 15 37.5 562.5 -1.667 2.778 41.667
38.5-40.5 23 39.5 908.5 0.333 0.111 2.556
40.5-42.5 11 41.5 456.5 2.333 5.444 59.889
42.5-44.5 4 43.5 174 4.333 18.778 75.111

60 2350 273.333
(a) Mean = 39.167
(b) Standard deviation = 2.134

Q8:
(a) Plot for 1 00 e− x cos(2 x )
(b) solve for ∫
A
B
100 e− x cos (2 x) dx , A=2.5 , B=3.1
(c) Answer: 2.56662

Q9:
Numerical integration
Since a curve is given without the equation, we use trapezoidal approximation to
estimate the area as follows:
A rea=0.5 ×[(5 ×5)+((5+7)×5)+((7+17) ×25)+((17+10) ×20)+(15 × 15)]=2,175 m2
Q10:
Largest positive root: x = 1.947

Q11:
Simplify and plot −15 x2 − 60 x2 +[300 −( 900 −300 Y
8− Y )](x −Y )
Simplifying:

Smallest positive root = 4.857