waah

Final.tex

@@ -72,6 +72,9 @@ TODO Lead into circuit analysis
 
 TODO Describe the circuit
 
+A design for a first-order inverting active high-pass filter with
+amplification is depicted below.
+
 \begin{figure}[h]
   \caption{A first-order high-pass filter}
   \begin{circuitikz}[american voltages]
@@ -92,13 +95,31 @@ TODO Describe the circuit
 \end{circuitikz}
 \end{figure}
 
-A more accurate representation of the op-amp foregoes the assumption,
-oft made regarding op-amps wired in a feedback configuration, that
-$v_n = v_p$. Avoiding this assumption requires a more precise
-description of the properties of the op-amp itself, describing its
-input resistance $R_i$, output resistance $R_o$, and the relationship
-between the input and output - the open-loop gain $A$. Figure
-\ref{img:opamp_internal} depicts such a model.
+Ordinarily, the analysis of this filter is easy. With an ideal op-amp
+configured in negative feedback, $v_n = v_p$: with awareness of the
+properties of an ideal op-amp, this is easy to see by inspection, as
+when $v_p > v_n$, $v_{out}$ is driven as high as possible, and
+vice-versa when $v_p < v_n$.
+
+At this point, it would be easy to derive the relationship between
+$v_{out}$ and $v_{in}$: $v_n = v_p = \SI{0}{V}$; since all current
+which flows from $v_{in}$ into $v_n$ then flows from $v_n$ into
+$v_{out}$, the network of $R_1$ and $C_1$ into $R_2$ forms a voltage
+divider, and so
+
+\begin{equation}
+  \frac{v_{out}}{v_{in}} = - \frac{R_2}{R_1+Z_{C_1}}
+\end{equation}
+
+To generate a cutoff frequency of $\omega_c = \SI{100}{kHz}$,
+
+A more accurate representation of the op-amp foregoes the
+assumption, oft made regarding op-amps wired in a feedback
+configuration, that $v_n = v_p$. Avoiding this assumption requires a
+more precise description of the properties of the op-amp itself,
+describing its input resistance $R_i$, output resistance $R_o$, and
+the relationship between the input and output - the open-loop gain
+$A$. Figure \ref{img:opamp_internal} depicts such a model.
 
 \begin{figure}[h]
   \caption{A more accurate depiction of an op-amp}
@@ -107,6 +128,42 @@ between the input and output - the open-loop gain $A$. Figure
   \includegraphics[width=0.4\textwidth]{opamp_internal}
 \end{figure}
 
+TODO Add citation for this image
+
+An ideal op-amp operates as though $R_i \to \infty$, $R_o \to 0$, and
+$A \to \infty$: it is these three properties which in the ideal case
+allow $v_n$ to be considered equal to $v_p$ in the closed-loop
+feedback configuration.
+
+% TODO: Cite Nilsson file:///home/mkennedy/Sync/Books/James W. Nilsson/ELECTRIC CIRCUITS,12e (49)/ELECTRIC CIRCUITS,12e - James W. Nilsson.pdf#page=194
+
+% > Note that Eq. 5.22 reduces to Eq. 5.4 as R 0,o → R ,i → ∞ and A ∞
+
+This paper focuses only on perturbing $A$ in our model; the
+assumptions that $R_o = 0$ and $R_i \to \infty$ remain in place.
+
+\subsection{Notes on Terms related to Open-Loop Gain}
+
+Manufacturers document the value of $A$ primarily as a function of
+frequency in an attribute called the ``open-loop gain''.
+
+In some cases, manufacturers give more precise details about the
+conditions under which the op-amp amplifies higher and lower
+frequencies than would be specified under generic ``open-loop gain'':
+for example, the LM741 datasheet documents the \textit{Open-Loop
+  Large-Signal Differential Voltage Amplification} as a function of
+frequency, as seen in Figure \ref{img:lm741_oclsg}. This differs from
+``open-loop gain'' in that the operating output is known to be a
+meaningful fraction of the supply voltage; here, it is
+$V_o = \SI{10}{V}$ for $V_{CC} = \pm \SI{15}{V}$.
+
+\begin{figure}[h]
+  \caption{The large-signal open-loop gain of the LM741}
+  \label{img:lm741_oclsg}
+  \centering
+  \includegraphics[width=0.4\textwidth]{lm741_oclsg}
+\end{figure}
+
 \section{Ease of Use}
 
 \subsection{Maintaining the Integrity of the Specifications}