ENG434S1: Core Skills and Electronics - Assessed Exercise Problems

This document presents a comprehensive solution to an assessed exercise from the ENG434S1 Core Skills and Electronics course. The assignment is divided into two sets. Set 1 focuses on analyzing a rainwater capture and storage system, requiring the conversion of precipitation data from inches to meters, calculating total rainfall, and determining yearly water usage based on daily usage data. The solution utilizes MATLAB code to perform these calculations. Set 2 explores solar cell operation and utilizes the Taylor series to approximate the solar cell current. The solution includes MATLAB code to plot the output voltage versus time for an RC circuit with varying resistor values and calculating the solar cell current using Taylor series expansion. The document also includes an introduction to Arduino, an open-source platform for physical computing, and provides information about a Keyestudio sensor kit, outlining its components and capabilities for learning purposes. The assignment covers topics relevant to electrical engineering, including data analysis, circuit analysis, and the application of mathematical tools.

Set 1
Problem 1
clc
a = [ 4.5 4.61 3.26 1.46 0.7 0.16 0 0.06 0.12 0.12 1.12 3.16 4.56 ] ;
display ('The data in inches:');
display ( a );
b = a * 0.0254;
display ('The data in meters:');
display ( b );
s=0;
for (i= 1:12)
s = s + b(i);
end
v = s * 40.25 * 1000000;
vl = v * 1000;
vg = vl/3.78;
display (' Volume in litres:');
display (vl);
display (' Volume in gallons:');
display (vg);
vtl = s * 10 * 30 * 0.3048 * 0.3048 *1000;
vtg = vtl / 3.78;
display ('amount of rain water collected by the system in a year in
litres:');
display (vtl);
display ('amount of rain water collected by the system in a year in
gallons:');
display (vtg);
du = [ 8.8 10 8.2 0.7 1.2 4 10.8 1.6 ] ;
display ('Daily usage in gallons: ');
display (du);
dl = du * 3.78 ;
display ('Daily usage in litres: ');
display (dl);
dy = dl * 365 ;
display ('Yearly water usage in litres for each activity: ');
display (dy);
s1=0;
for (j= 1:8)
s1 = s1 + dl(j);
end
yu = s1 * 365 ;
display (' Yearly Water usage of 1 person');
display (yu);

Output:
The data in inches:
a =
Columns 1 through 12
4.5000 4.6100 3.2600 1.4600 0.7000 0.1600 0 0.0600 0.1200 0.1200 1.1200
3.1600
Column 13
4.5600
The data in meters:
b =
Columns 1 through 12
0.1143 0.1171 0.0828 0.0371 0.0178 0.0041 0 0.0015 0.0030 0.0030 0.0284
0.0803
Column 13
0.1158
Volume in litres:

vl =
1.9701e+10
Volume in gallons:
vg =
5.2118e+09
amount of rain water collected by the system in a year in litres:
vtl =
1.3642e+04
amount of rain water collected by the system in a year in gallons:
vtg =
3.6089e+03
Daily usage in gallons:
du =

8.8000 10.0000 8.2000 0.7000 1.2000 4.0000 10.8000 1.6000
Daily usage in litres:
dl =
33.2640 37.8000 30.9960 2.6460 4.5360 15.1200 40.8240 6.0480
Yearly water usage in litres for each activity:
dy =
1.0e+04 *
1.2141 1.3797 1.1314 0.0966 0.1656 0.5519 1.4901 0.2208
Yearly Water usage of 1 person
yu =
6.2500e+04

Set 1
Problem 2
clc
vs = 0.01;
c = 0.00001;
r1 = 10000;
r2 = 1000;
r3 = 100;
t = 0 : 0.001 : 0.2;
x1 = -t/(r1*c);
v1 = vs * ( 1 - exp ( x1 ));
plot (t,v1,'m');
hold on;
x2 = -t/(r2*c);
v2 = vs * ( 1 - exp ( x2 ));
plot (t,v2,'g');
hold on;
x3 = -t/(r3*c);
v3 = vs * ( 1 - exp ( x3 ));
plot (t,v3,'c');
hold on;
legend ('r1=10k','r2=1k','r3=0.1k');
title('Output Voltage versus time');
hold off;
grid on;
Output:

Set 2
Problem 1
clc
ilight = 0.5;
idark = 0.0000006;
vt = 0.026;
x = 0.3 / 0.026 ;
p = 1;
s=1;
for (i = 1:4)
p = p * x;
q = factorial(i);
r = p / q;
s = s + r;
end
isolar = ilight + idark - (idark * s);
display('Till 5 terms:');
display (isolar);
p1 = 1;
s1=1;
for (j = 1:19)
p1 = p1 * x;
q1 = factorial(j);
r1 = p1 / q1;
s1 = s1 + r1;
end
isolar = ilight + idark - (idark * s1);
display('Till 20 terms:');
display (isolar);
Output:
Till 5 terms:
isolar =
0.4994
Till 20 terms:
isolar =
0.4394

Set 2
Problem 2
clc
c = [ 0.1 , 0 , 0 , 0 ; 0.75 , 0.05 , 0 , 0 ; 0 , 0.9 , 0.05 , 0;0 , 0 ,
0.9 , 0.05 ];
y = [ 500 ; 400 ; 300 ; 280 ];
b = [ 1000 ; 200 ; 0 ; 0];
for k = 1:10
y = c*y + b;
display (y);
s = sum(y);
bar (k,s);
hold on;
end
Output
y =
1050
595
375
284
y =
1.0e+03 *
1.1050
1.0173
0.5543
0.3517

y =
1.0e+03 *
1.1105
1.0796
0.9432
0.5164
y =
1.0e+03 *
1.1110
1.0869
1.0188
0.8747
y =
1.0e+03 *
1.1111
1.0876

1.0291
0.9607
y =
1.0e+03 *
1.1111
1.0877
1.0303
0.9742
y =
1.0e+03 *
1.1111
1.0877
1.0305
0.9760
y =
1.0e+03 *

1.1111
1.0877
1.0305
0.9762
y =
1.0e+03 *
1.1111
1.0877
1.0305
0.9762
y =
1.0e+03 *
1.1111
1.0877
1.0305
0.9762