Finalise lab reportHEADmaster

Signed-off-by: Warrick Lo <wlo@warricklo.net>

7 files changed, 448 insertions, 35 deletions

diff --git a/general.cls b/general.cls
index 94773e1..bef7311 100644
--- a/general.cls
+++ b/general.cls
@@ -25,6 +25,7 @@
 \RequirePackage{amsmath}
 \RequirePackage{unicode-math}
 \RequirePackage{siunitx}
+\RequirePackage{derivative}
 
 % Code listings.
 \RequirePackage{listings}
@@ -40,12 +41,16 @@
 	BoldItalicFont=*-BoldItalic
 ]
 
-\setmathfont[math-style=ISO, bold-style=ISO, partial=upright]{STIX Two Math}
+\setmathfont[
+	math-style=ISO,
+	bold-style=ISO,
+	partial=upright,
+	StylisticSet=8
+]{STIX Two Math}
 
 \captionsetup{
 	labelfont=bf,
-	labelsep=period,
-	justification=centering
+	labelsep=period
 }
 
 \sisetup{
diff --git a/images/manav.png b/images/manav.png
new file mode 100644
index 0000000..3007bce
--- /dev/null
+++ b/images/manav.png
diff --git a/images/warrick.png b/images/warrick.png
new file mode 100644
index 0000000..132006a
--- /dev/null
+++ b/images/warrick.png
diff --git a/images/yuvraj.png b/images/yuvraj.png
new file mode 100644
index 0000000..a51fd2f
--- /dev/null
+++ b/images/yuvraj.png
diff --git a/macros.sty b/macros.sty
index 5cc98b9..56854d0 100644
--- a/macros.sty
+++ b/macros.sty
@@ -38,13 +38,10 @@
 \DeclareMathOperator{\arsech}{arsech}
 \DeclareMathOperator{\arcsch}{arcsch}
 
-\makeatletter
 \let\@epsilon\epsilon
 \let\@theta\theta
 \let\@phi\phi
 \AtBeginDocument{
-	\RenewCommandCopy{\angle}{\measuredangle}
-
 	\RenewCommandCopy{\epsilon}{\varepsilon}
 	\RenewCommandCopy{\varepsilon}{\@epsilon}
 	\RenewCommandCopy{\theta}{\vartheta}
@@ -54,5 +51,6 @@
 	\undef\@epsilon
 	\undef\@theta
 	\undef\@phi
+
+	\RenewCommandCopy{\angle}{\measuredangle}
 }
-\makeatother
diff --git a/master.tex b/master.tex
index 3a79ddf..b565cba 100644
--- a/master.tex
+++ b/master.tex
@@ -3,12 +3,48 @@
 
 \title{ELEC 342 Lab 1\\AC/DC Circuits and Basic Measurements}
 \author{Manav Khangura \and Warrick Lo \and Yuvraj Chadha}
-\date{09 February 2026}
+\date{}
+
+\DeclareSIUnit{\VAR}{VAR}
 
 \begin{document}
 
+\vfill
+
 \maketitle
 
+\vfill
+
+\begin{centering}
+	\Large
+
+	Lab Experiment: 1 \\
+	Section: L2C \\
+	Bench: \#7
+
+	\vfill
+
+	\begin{tabular}{|c|c|c|c|}
+		\hline
+		Partners & Student ID & Participation & Signatures \\
+		\hline
+		\hline
+		Manav Khangura & 16725475 & \qty{33.3}{\%} & \includegraphics[height=20pt]{images/manav.png} \\
+		Warrick Lo & 47938733 & \qty{33.3}{\%} & \includegraphics[height=20pt]{images/warrick.png} \\
+		Yuvraj Chadha & 23005697 & \qty{33.3}{\%} & \includegraphics[height=20pt]{images/yuvraj.png} \\
+		\hline
+	\end{tabular}
+
+	\vfill
+
+	Date(s) Performed: 30 January, 13 February \\
+	Date Submitted: 20 February 2026 \\
+\end{centering}
+
+\vfill
+
+\newpage
+
 \tableofcontents
 
 \listoftables
@@ -599,14 +635,14 @@ and thus,
 			& $I_\text{3(PP)}$ (\si{\A}) & $P_\text{3(out)}$ (\si{\W}) \\
 		\midrule
 		No load & 15.05 $\angle$ -1.7 & 0.08 $\angle$ 36.2 & 0.8
-			& 14.19 $\angle$ -93.8 &
-			& 0.11 $\angle$ -224.5 && 1.0 \\
+			& 14.19 $\angle$ -93.8 & 20.5
+			& 0.11 $\angle$ -224.5 & 0.4 & 1.0 \\
 		Light load & 14.87 $\angle$ -1.8 & 0.84 $\angle$ -2.4 & 12.5
-			& 13.34 $\angle$ -93.9 &
-			& 0.88 $\angle$ -96.2 && 11.7 \\
+			& 13.34 $\angle$ -93.9 & 20.6
+			& 0.88 $\angle$ -96.2 & 0.9 & 11.7 \\
 		Heavy load & 14.77 $\angle$ -1.8 & 1.50 $\angle$ -2.0 & 22.0
-			& 13.09 $\angle$ -94.1 &
-			& 1.53 $\angle$ -95.6 && 19.9 \\
+			& 13.09 $\angle$ -94.1 & 20.2
+			& 1.53 $\angle$ -95.6 & 2.6 & 19.9 \\
 		\bottomrule
 	\end{tabular}
 	\caption[Single-phase full-wave rectifier, without a capacitor filter]
@@ -615,6 +651,16 @@ and thus,
 \end{adjustwidth}
 \end{table}
 
+\begin{itemize}
+	\item \textbf{What relationship between the voltage and current
+		peak/ripple do you observe?} \\
+		The current follows the same waveform shape as the voltage. Since the
+		load is resistive and there is no capacitor, the current peak is in phase
+		with the voltage peak. As the load resistance decreases (heavier load),
+		the peak to peak current increases while the voltage peak to peak
+		remains about constant.
+\end{itemize}
+
 \begin{table}[H]
 \begin{adjustwidth}{-15mm}{-15mm}
 	\centering
@@ -626,15 +672,15 @@ and thus,
 			& $I_\text{3(RMS)}$ (\si{\A}) & $I_\text{3(PP)}$ (\si{\A})
 			& $P_\text{3(out)}$ (\si{\W}) \\
 		\midrule
-		No load & 15.04 $\angle$ -1.7 & 0.34 $\angle$ -8.6 && 2.2
-			& 19.90 $\angle$ -176.3 &
-			& 0.17 $\angle$ 13.1 && 2.8 \\
-		Light load & 14.73 $\angle$ -1.8 & 2.5 $\angle$ -3.9 && 24.8
-			& 17.73 $\angle$ -165.8 &
-			& 1.23 $\angle$ 203.1 && 21.7 \\
-		Heavy load & 14.54 $\angle$ -2.1 & 3.80 $\angle$ -2.3 && 39.4
-			& 16.73 $\angle$ -161.9 &
-			& 2.01 $\angle$ -159.5 && 33.6 \\
+		No load & 15.04 $\angle$ -1.7 & 0.34 $\angle$ -8.6 & 2.8 & 2.2
+			& 19.90 $\angle$ -176.3 & 0.29
+			& 0.17 $\angle$ 13.1 & 0.16 & 2.8 \\
+		Light load & 14.73 $\angle$ -1.8 & 2.5 $\angle$ -3.9 & 13.5 & 24.8
+			& 17.73 $\angle$ -165.8 & 1.58
+			& 1.23 $\angle$ 203.1 & 0.23 & 21.7 \\
+		Heavy load & 14.54 $\angle$ -2.1 & 3.80 $\angle$ -2.3 & 18.5 & 39.4
+			& 16.73 $\angle$ -161.9 & 2.30
+			& 2.01 $\angle$ -159.5 & 0.36 & 33.6 \\
 		\bottomrule
 	\end{tabular}
 	\caption[Single-phase full-wave rectifier, with a capacitor filter]
@@ -643,6 +689,27 @@ and thus,
 \end{adjustwidth}
 \end{table}
 
+\begin{itemize}
+	\item \textbf{How do the output voltage and current ripples compare with
+		the previous case without the capacitor filter?} \\
+		Compared to the no-capacitor case, adding the capacitor filter reduces both
+		the output voltage and current ripple, because the capacitor supplies current
+		between rectified peaks. Ripple increases with heavier load due to
+		faster discharge, but it is smaller than without the filter.
+	\item \textbf{How does the presence of a capacitor impact the input current peak?} \\
+		The input current peak is increased when the capacitor is present.
+		Current flows in brief, high-amplitude pulses close to the AC peaks
+		rather than continuously throughout the cycle because the capacitor charges
+		only when the input voltage is greater than the capacitor voltage.
+		In comparison to the scenario in which the capacitor filter is not used,
+		this leads to significantly greater input current peaks.
+	\item \textbf{How does the peak of the input current change with the load current?} \\
+		The peak input current rises noticeably as the load current does.
+		A higher recharge current is needed over a brief interval when the load
+		is heavier because the capacitor discharges more between AC peaks.
+		Higher amplitude current pulses at the input are the outcome of this.
+\end{itemize}
+
 \phantomsection
 \addcontentsline{toc}{subsection}{Task 4}
 \section*{Task 4A. Three-Phase Full-Wave Rectifier}
@@ -655,12 +722,12 @@ and thus,
 			& $V_\text{2(PP)}$ (\si{\V}) & $I_\text{2(RMS)}$ (\si{\A})
 			& $I_\text{2(PP)}$ (\si{\A}) & $P_\text{3(out)}$ (\si{\W}) \\
 		\midrule
-		No load & 0.09 $\angle$ 33.3 & 12.18 $\angle$ -62.2 &
-			& 0.16 $\angle$ -148.3 && 1.5 \\
-		Light load & 0.65 $\angle$ 33.3 & 11.99 $\angle$ -50.1 &
-			& 0.86 $\angle$ -36.7 && 9.9 \\
-		Heavy load & 1.10 $\angle$ 34.2 & 11.86 $\angle$ -40.6 &
-			& 1.43 $\angle$ -32.2 && 16.2 \\
+		No load & 0.09 $\angle$ 33.3 & 12.18 $\angle$ -62.2 & 1.5
+			& 0.16 $\angle$ -148.3 & 0.14 & 1.5 \\
+		Light load & 0.65 $\angle$ 33.3 & 11.99 $\angle$ -50.1 & 1.5
+			& 0.86 $\angle$ -36.7 & 0.22 & 9.9 \\
+		Heavy load & 1.10 $\angle$ 34.2 & 11.86 $\angle$ -40.6 & 1.5
+			& 1.43 $\angle$ -32.2 & 0.30 & 16.2 \\
 		\bottomrule
 	\end{tabular}
 	\caption[Three-phase full-wave rectifier, without a capacitor filter]
@@ -668,6 +735,18 @@ and thus,
 	\label{tab:4a}
 \end{table}
 
+\begin{itemize}
+	\item \textbf{What relationship between the voltage and current peak/ripple
+		do you observe?} \\
+		The ripples are the same shape and in phase as the load is purely resistive.
+		The voltage peak to peak remains constant while the current peak to increases.
+	\item \textbf{How does the input current differ from that of the single-phase rectifier?} \\
+		The input current of a three-phase rectifier is smoother and more continuous
+		than that of a single-phase rectifier. In the single-phase case, current flows
+		in large pulses and drops to zero, in the three-phase case, conduction overlaps
+		between phases, reducing ripple.
+\end{itemize}
+
 \begin{table}[H]
 \begin{adjustwidth}{-15mm}{-15mm}
 	\centering
@@ -679,12 +758,12 @@ and thus,
 			& $I_\text{3(RMS)}$ (\si{\A}) & $I_\text{3(PP)}$ (\si{\A})
 			& $P_\text{3(out)}$ (\si{\W}) \\
 		\midrule
-		No load & 0.08 $\angle$ 22.4 && 0.20 $\angle$ -31.0 &
-			& 12.58 $\angle$ -236.0 && 0.12 $\angle$ -218.5 && 1.6 \\
-		Light load & 0.70 $\angle$ 28.6 && 1.02 $\angle$ -54.3 &
-			& 11.91 $\angle$ -95.9 && 0.83 $\angle$ 96.0 && 10.0 \\
-		Heavy load & 1.17 $\angle$ 29.0 && 1.73 $\angle$ -54.4 &
-			& 11.63 $\angle$ -96.0 && 1.40 $\angle$ -183.2 && 16.3 \\
+		No load & 0.08 $\angle$ 22.4 & 0.5 & 0.20 $\angle$ -31.0 & 0.6
+			& 12.58 $\angle$ -236.0 & 0.1 & 0.12 $\angle$ -218.5 & 0.1 & 1.6 \\
+		Light load & 0.70 $\angle$ 28.6 & 2.7 & 1.02 $\angle$ -54.3 & 2.0
+			& 11.91 $\angle$ -95.9 & 0.3 & 0.83 $\angle$ 96.0 & 0.1 & 10.0 \\
+		Heavy load & 1.17 $\angle$ 29.0 & 4.2 & 1.73 $\angle$ -54.4 & 2.7
+			& 11.63 $\angle$ -96.0 & 0.4 & 1.40 $\angle$ -183.2 & 0.1 & 16.3 \\
 		\bottomrule
 	\end{tabular}
 	\caption[Three-phase full-wave rectifier, with a capacitor filter]
@@ -693,6 +772,324 @@ and thus,
 \end{adjustwidth}
 \end{table}
 
+\begin{itemize}
+	\item \textbf{How do the output voltage and current ripples compare with
+		the previous case without the capacitor filter?} \\
+		With the capacitor filter, both output voltage and current become smoother.
+		Ripple still increases with heavier load but it remains smaller than
+		the case without the capacitor.
+	\item \textbf{How does the presence of a capacitor impact the input current peak?} \\
+		The input current peak is increased when the capacitor is present.
+		Current only flows when the input voltage is higher than the capacitor
+		voltage because the capacitor keeps the output voltage close to its
+		maximum value. This causes short pulses as opposed to continuous conduction,
+		which results in larger peak input currents.
+	\item \textbf{How do the output voltage and current ripples compare with the
+		single-phase rectifier?} \\
+		Compared to the single-phase rectifier, the three-phase rectifier produces
+		significantly smaller output voltage and current ripple. This is because
+		the three-phase rectifier has overlapping output between phases,
+		which reduces the voltage dips between peaks. As a result,
+		the output is smoother.
+	\item \textbf{How does the peak of the input current change with the load current?} \\
+		The peak input current increases as the load current increases.
+		A heavier load causes the capacitor to discharge more current between peaks,
+		requiring a larger recharge current over a short time. As a result,
+		the input current peaks become taller under heavier load conditions.
+\end{itemize}
+
+
+\phantomsection
+\addcontentsline{toc}{subsection}{Task 5}
+\section*{Task 5A. Instantaneous, Real, and Reactive Power in Parallel RLC Circuit}
+
+The instantaneous power is given by
+\begin{align}
+	p(t) = v(t) i(t),
+	\label{eq:inst-power}
+\end{align}
+where $v(t)$ and $i(t)$ are the instantaneous voltages and currents,
+respectively. The average power is
+\begin{align}
+	P = \frac{1}{T} \int_{\langle T \rangle} v(t) i(t) \odif{t}.
+	\label{eq:power}
+\end{align}
+
+The measured RMS voltages and currents are slightly different than the ones
+calculated by MATLAB, which uses the discrete version of equation \eqref{eq:power}
+(see the appendix). We will use the values calculated by MATLAB here.
+
+For a sinusoidal signal, $P = V_\text{RMS} I_\text{RMS} \cos\phi$ and
+$Q = V_\text{RMS} I_\text{RMS} \sin\phi$. We will assume that the signals
+are ``sinusoidal enough''.
+
+\begin{table}[H]
+	\centering
+	\begin{tabular}{ccc}
+		\toprule
+		& Real Power, $P$ (\si{\W}) & Reactive Power, $Q$ (\si{\VAR}) \\
+		\midrule
+		Task 3C, step 3 & 9.98 & 0.03 \\
+		Task 3C, step 4 & 10.68 & 8.90 \\
+		Task 3C, step 5 & 11.25 & -0.20 \\
+		\bottomrule
+	\end{tabular}
+	\caption[Calculations for the instantaneous, real, and reactive power
+		in a parallel RLC circuit]
+		{Calculations for the instantaneous, real, and reactive power in a
+		parallel RLC circuit. The equations for a sinusoidal signal,
+		shown above, were used. RMS values are calculated from MATLAB.}
+\end{table}
+
+The instantaneous power gives the power absorbed at any given moment in time,
+while the real power is the average power absorbed or dissipated over
+an interval. The reactive power is power that is temporarily stored in the
+magnetic and electric fields of inductors and capacitors, respectively.
+
+By adding inductors, the reactive power increases since they have positive reactance
+and will increase the power factor angle. Capacitors, on the other hand, decrease
+the reactive power since they have negative reactive and will decrease the
+power factor angle.
+
+\begin{figure}[H]
+	\centering
+	\includegraphics[width=0.5\linewidth]{images/5a-phasor.png}
+	\caption[Phasor plot of the voltage and currents for the parallel RLC circuit]
+		{Phasor plot of the applied voltage $V_1$, input current $I_1$,
+		inductor current $I_2$, and capacitor current $I_3$ for the parallel
+		RLC circuit. Here the voltage $V_1$ is normalised to $1$. Notice how
+		the phasors $I_2$ and $I_3$ effectively cancel each other.}
+\end{figure}
+
+\begin{figure}[H]
+	\centering
+	\begin{subfigure}{0.75\linewidth}
+		\includegraphics[width=\linewidth]{images/5a-1-plot.png}
+		\caption{Plot of circuit with one resistor.}
+	\end{subfigure}
+
+	\medskip
+
+	\begin{subfigure}{0.75\linewidth}
+		\includegraphics[width=\linewidth]{images/5a-2-plot.png}
+		\caption{Plot of circuit with one resistor and one inductor.}
+	\end{subfigure}
+
+	\medskip
+
+	\begin{subfigure}{0.75\linewidth}
+		\includegraphics[width=\linewidth]{images/5a-3-plot.png}
+		\caption{Plot of circuit with one resistor, one inductor,
+			and three capacitors.}
+	\end{subfigure}
+
+	\caption[Plots of instantaneous, real, and reactive power in the parallel RLC circuits]
+		{Plots of instantaneous, real, and reactive power in the parallel RLC circuits.}
+\end{figure}
+
+\section*{Task 5B. Single-Phase Phasor Diagram for the Series RLC Circuit}
+
+We see that in the phasor diagrams, the inductor and capacitor voltages are
+always \ang{180} apart from each other, as expected since each of them have
+purely reactive loads. We see that with one resistor, the voltages $V_2$
+and $V_3$ are almost exactly \ang{\pm 90} from $V_1$. This means that the
+power factor angle is near the minimum (\ang{0}) and almost all of the
+power delivered is real.
+
+When we add the other resistors in parallel and decrease the series resistance,
+the current starts to lead the voltage, meaning that less of the total power
+delivered is real and more of it is reactive. Because the voltages $V_2$ and $V_3$
+have to be \ang{90} apart from the current $I_1$, we see that the shift in the
+current also causes the angles of the reactive components to shift as well.
+
+Finally we observe that the difference between $V_1$ and $V_2+V_3$ should be
+the voltage drop across the resistor(s).
+
+\begin{figure}[H]
+	\centering
+	\begin{subfigure}{0.45\linewidth}
+		\includegraphics[width=\linewidth]{images/5b-1-phasor.png}
+		\caption{Phasor diagram of circuit with one resistor, one inductor,
+			and three capacitors.}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{0.45\linewidth}
+		\includegraphics[width=\linewidth]{images/5b-2-phasor.png}
+		\caption{Phasor diagram of circuit with three resistors, one inductor,
+			and three capacitors.}
+	\end{subfigure}
+	\caption[Phasor plots of the voltages and current for the series RLC circuit]
+		{Phasor plots of the applied voltage $V_1$, inductor voltage $V_2$,
+		capacitor voltage $V_3$, and input current $I_1$ for the series RLC circuit.}
+\end{figure}
+
+\section*{Task 5C. Three-Phase Phasor Diagrams for Wye-Connected RL Load}
+
+We see that without the neutral line connected, shorting one of the load elements
+will shift the phase voltage angles away from \ang{120}. Shorting the resistor
+caused the angle of the unbalanced load's voltage to increase, and shorting the
+inductor caused it to decrease.
+
+With the neutral line connected, the unbalanced current now has a return path
+back to the source, ensuring that the phase voltages stay relatively close to
+\ang{120} apart from each other.
+
+Looking at our measurements in table \ref{tab:2b}, we see that the unbalanced
+loads wouldn't affect the phase voltages, since each point of connection in the
+delta configuration is directly connected to a voltage source. As expected from
+having unbalanced impedances, we will see differences between the phases in
+current.
+
+\begin{figure}[H]
+	\begin{subfigure}{0.45\linewidth}
+		\includegraphics[width=\linewidth]{images/5c-1-phasor.png}
+		\caption{Phasor diagram of circuit with RL balanced load, \\
+			neutrals disconnected.}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{0.45\linewidth}
+		\includegraphics[width=\linewidth]{images/5c-2-phasor.png}
+		\caption{Phasor diagram of circuit with RL unbalanced load, \\
+			shorted resistor, neutrals disconnected.}
+	\end{subfigure}
+	\caption[Phasor plots of the applied phase voltages and input currents
+		for the wye-connected RL load]
+		{Phasor plots of the applied phase voltages, $V_1$, $V_2$, $V_3$,
+		and input currents, $I_1$, $I_2$, $I_3$, for the wye-connected RL load.
+		Note that all current magnitudes have been scaled up by $10$.}
+
+	\medskip
+
+	\begin{subfigure}{0.45\linewidth}
+		\includegraphics[width=\linewidth]{images/5c-3-phasor.png}
+		\caption{Phasor diagram of circuit with RL unbalanced load, \\
+			shorted inductor, neutrals disconnected.}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{0.45\linewidth}
+		\includegraphics[width=\linewidth]{images/5c-4-phasor.png}
+		\caption{Phasor diagram of circuit with RL unbalanced load, \\
+			shorted resistor, neutrals connected.}
+	\end{subfigure}
+
+	\caption[Phasor diagrams of the applied phase voltages and input currents
+		for the wye-connected RL load]
+		{Phasor diagrams of the applied phase voltages, $V_1$, $V_2$, $V_3$,
+		and input currents, $I_1$, $I_2$, $I_3$, for the wye-connected RL load.
+		Note that all current magnitudes have been scaled up by $10$.}
+\end{figure}
+
+\begin{figure}[H]
+	\ContinuedFloat
+	\centering
+	\begin{subfigure}{0.45\linewidth}
+		\includegraphics[width=\linewidth]{images/5c-5-phasor.png}
+		\caption{Phasor diagram of circuit with RL unbalanced load, \\
+			shorted inductor, neutrals connected.}
+	\end{subfigure}
+
+	\captionsetup{list=no}
+	\caption{Phasor diagrams of the applied phase voltages, $V_1$, $V_2$, $V_3$,
+		and input currents, $I_1$, $I_2$, $I_3$, for the wye-connected RL load.
+		Note that all current magnitudes have been scaled up by $10$.}
+	\captionsetup{list=yes}
+\end{figure}
+
+\section*{Task 5D. Harmonics in AC Mains}
+
+\textit{For this section, we will use $V_1$ from Task 2B, with a balanced load.}
+
+Harmonics are distortions that occur at integer multiples of the fundamental
+frequency. For AC mains in North America, this is \qty{60}{\Hz}. These are caused
+by non-linear loads. In our lab, these harmonics may be caused by the power supply's
+internal components being non-linear. Since our load is purely ohmic, the load should
+not be contributing to the harmonic content.
+
+\begin{figure}[H]
+	\centering
+	\includegraphics[width=0.75\linewidth]{images/5d-plot.png}
+	\caption[Frequency spectrum of a sinusoidal voltage signal]
+		{Frequency spectrum of $V_1$ from Task 2B using MATLAB's \texttt{fft()}
+		function (see the appendix).}
+\end{figure}
+
+As expected, we see a large spike at \qty{60}{\Hz} as that is our power supply's
+frequency. We also notice distortions at multiples the fundamental frequency.
+
+\section*{Task 5E. Single-Phase AC/DC}
+
+\textit{For this section, we will use $I_1$ from Task 3A,
+	under a full load with and without a capacitor filter.}
+
+We see that the output current without a capacitor filter does look close to a
+sinusoidal wave. However, the peaks are sharper than a sine wave, resembling
+more of a sawtooth wave.
+
+The output current with a capacitor filter, apart from the non-zero parts
+of the current, does not resemble a sinusoidal wave.
+
+As expected from our initial observations on the waveforms, the frequency spectrum
+of the current without a capacitor filter has a single large spike at \qty{60}{\Hz},
+with a small amount of distortion at higher frequencies. On the other hand,
+the current with a capacitor filter has strong harmonic content.
+
+\begin{figure}[H]
+	\centering
+
+	\begin{subfigure}{0.75\linewidth}
+		\includegraphics[width=\linewidth]{images/5e-1-plot.png}
+		\caption{Waveform and frequency spectrum under a full load without
+			a capacitor filter.}
+	\end{subfigure}
+
+	\medskip
+
+	\begin{subfigure}{0.75\linewidth}
+		\includegraphics[width=\linewidth]{images/5e-2-plot.png}
+		\caption{Waveform and frequency spectrum under a full load with
+			a capacitor filter.}
+	\end{subfigure}
+
+	\caption[Waveform and frequency spectrum of a rectified current]
+		{Waveform and frequency spectrum of $I_1$ from Task 3A using MATLAB's
+		\texttt{fft()} function (see the appendix).}
+\end{figure}
+
+\section*{Task 5F. Three-Phase AC/DC Rectifiers}
+
+\textit{For this section, we will use $I_1$, $I_2$, and $I_3$ from
+	Task 4A, under a full load with a capacitor filter.}
+
+The current $I_2$ is the output of the current from the full-bridge rectifier.
+The diodes ensure that all of the current flows in one single direction. For $I_3$,
+the capacitor helps maintain a steady voltage and current, making it comparable
+to an ideal DC current.
+
+\begin{figure}[H]
+	\centering
+	\includegraphics[width=0.75\linewidth]{images/5f-plot.png}
+	\caption[Waveform and frequency spectrum of the current in a
+		three-phase rectifier]
+		{Waveform and frequency spectrum of $I_1$, $I_2$, and $I_3$ from Task 4A
+		using MATLAB's \texttt{fft()} function (see the appendix).}
+\end{figure}
+
+The $I_1$ currents in Task 5E are much closer to an ideal sinusoidal wave,
+compared to here. However, we also observe that the three-phase input current
+has much less harmonic content compared to the single-phase currents.
+
+\phantomsection
+\addcontentsline{toc}{subsection}{Conclusion}
+\section*{Conclusion}
+
+In this laboratory experiment, we were able to measure and calculate different
+parameters of single-phase and three-phase circuits. We looked into real-world
+applications and ideal versus non-ideal components, such as inductors, capacitors,
+and rectifiers. We also saw the difference between balanced and unbalanced loads,
+and their different effects in the wye and delta configurations. In addition, this
+lab showed us the difference between single-phase and three-phase rectifiers.
+Finally, we looked into the harmonic content of AC voltages and currents.
+
 \newpage
 
 \appendix
@@ -725,6 +1122,19 @@ and thus,
 	stringstyle=\color{red},
 	caption={Phasor plot code.},
 	captionpos=b
+]{matlab/poweran.m}
+
+\newpage
+
+\lstinputlisting[
+	style=Matlab-editor,
+	basicstyle=\ttfamily,
+	numberstyle=\ttfamily\small\color{gray},
+	keywordstyle=\color{blue},
+	commentstyle=\color{gray},
+	stringstyle=\color{red},
+	caption={Phasor plot code.},
+	captionpos=b
 ]{matlab/phasor.m}
 
 \lstinputlisting[
diff --git a/report.pdf b/report.pdf
index 45baed0..745dbb6 100644
--- a/report.pdf
+++ b/report.pdf