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\title{\ifnum\TeacherEdition=1 TEACHER VERSION \else STUDENT VERSION \fi\\MYSTERY CIRCUIT}







\markboth{Mystery Circuit}{Mystery Circuit}


\author{Brian Winkel,
Director SIMIODE\\
Chardon OH 44024 USA\\
BrianWinkel@SIMIODE.org}

\makeStitlePDFLaTex


\begin{abstract}
We require students to build a system of differential equations which model a circuit. We give each student a unique input voltage frequency parameter and ask for system response (gain) to that input as a voltage over one of the resistors in the second loop of the circuit. Students collect the data, plot their results, and discover a cascaded high-low filter circuit is the object of their study. 
\end{abstract}






\section*{SCENARIO DESCRIPTION}

We present a two loop circuit (see Figure~\ref{fig:4-23-CircuitFig2}) for analysis.  Using different frequency voltages we drive the circuit and asking you and your classmates to report back {\it gain\/} (defined below) from the circuit, wherein one can plot the gain versus input voltage frequency data from the entire class to analyze this circuit.  Some basic electrical circuit notions are offered from a textbook source we have used \cite[pp.~118-119]{Kreyszig1999}.  Of course, there are other sources, e.g., electrical engineering and circuits texts, physics texts, and web-based material.



\subsection*{Introductory Notions for Circuit Study}
Consider  a typical one-loop RLC circuit (see Figure~\ref{fig:4-23-CircuitFig1}). $R$ is resistance in ohms, $L$ is inductance in henrys, and $C$ is capacitance in farads, or more usually microfarads. There is an electromotive force, $E(t) = E_0\sin(\omega t)$, in volts driving the circuit. This voltage produces an alternating  current, $I(t)$, in amperes at time $t$, in the circuit as a measure of the flow of electrons in the particular branch of the circuit. 

{\it Kirchhoff's Voltage Law} says that  in the circuit at all times  $t$  the voltage source, $E(t) = E_0\sin(\omega t)$, equals  the sum of the three voltage drops -- over $R$, $L$, and $C$, i.e. $E = E_R + E_L + E_C$ (see Figure~\ref{fig:4-23-CircuitFig1}).   The voltage drop over each of the devices, resistor $R$, inductor $L$, and capacitor $C$ is given by the following formulae, respectively:
\vspace*{.3 true cm}

\centerline{(R) \hspace*{.2 true cm} $E_R = R I$, \hspace*{.7 true cm} 
(L)\hspace*{.7 true cm} $E_L = L I'$,   \hspace*{.2 true cm} and  \hspace*{.2 true cm}
(C) \hspace*{.2true cm} $E_C = \frac{1}{C}\int I(t)\, dt$.}







\begin{figure}[h]
\begin{center}
\includegraphics[width=10.6cm]{4-23-CircuitFig1.eps}



\vspace*{-7.3 true cm}
\parbox{12.2 true cm}{\caption{\label{fig:4-23-CircuitFig1} Single loop RLC circuit.}}
\end{center}

\end{figure}


Consider the following   differential equation in $I(t)$, the current in the one loop circuit of Figure~\ref{fig:4-23-CircuitFig1}, using Kirchhoff's Voltage Law as an elementary example  of Figure~\ref{fig:4-23-CircuitFig2}. Sum the voltages across the resistor, the inductor, and the capacitor and set this sum equal to the voltage from our source or driver, $E(t)$:

\begin{equation}\label{eq:initial}
L I'(t) + R I(t) + \frac{1}{C}\int I(t)\, dt = E_0\sin(\omega t) \,\, . 
\end{equation}
Convert this into a second order differential equation in $I(t)$ by differentiating both sides with respect to $t$:

\begin{equation}\label{eq:convert}
L I''(t) + R I'(t) + \frac{1}{C}I(t) = E_0\omega \cos(\omega t) \,\, .
\end{equation} Note that prior to the voltage source being applied there is nothing happening in the circuit, i.e.~there is no voltage or force to move the electrons in the circuit. Hence, the reasonable initial conditions are $I(0) = 0$ and $I'(0) = 0$.

Consider the electrical circuit in Figure~\ref{fig:4-23-CircuitFig2} and we give each student a unique parameter - the frequency of the driver voltage, i.e. $\omega$. Each student should take a different input frequency, $\omega$, aqnd  model the circuit, solve for the gain associated with the specific assigned parameter, and report back this gain value. The data is to be collected in class where it will be  plotted  to see what  can inferred from such a display.  Describe what   this circuit really does. The circuit  offered is shown in Figure~\ref{fig:4-23-CircuitFig2}.

\begin{figure}[h]
\begin{center}
\includegraphics[width=10.6cm]{4-23-CircuitFig2.eps}



\vspace*{-9.5 true cm}
\parbox{12.2 true cm}{\caption{\label{fig:4-23-CircuitFig2} Two loop RLC circuit.}}
\end{center}
  \end{figure}

The values for this circuit are $C_1 = 2.5 \times 10^{-6}$ F (farads), $C_2 = 1.0 \times 10^{-6}$ F, $R_1 = 200\, \Omega$ (ohms), and $R_{\rm load} = 1000\, \Omega$. Use  $E(t)  =  \sin( \omega t)$  (volts) with $\omega = 100$ (radians/sec) as a trial run.
 
\subsubsection*{Activity}


\begin{itemize}
\item [1)] 
Build equations to describe this two  loop circuit using {\it Kirchhoff's Voltage Law\/} as above and  {\it Kirchhoff's Current Law} which says that the sum of the currents at any node in the circuit must be 0, i.e.~what goes into the node has to go out without loss or gain.
\end{itemize}

We identify variables:
\begin{itemize}
\item [] $x(t)$ is the current from Point 1 to Point 2, 
\item [] $y(t)$ is the current from Point 2 to Point 5,  and
\item [] $z(t)$ is the current from Point 2 to Point 3 to Point 4 to Point 5.
\end{itemize}

One way to measure the behavior of this circuit is to compare the amplitude of the ``output'' voltage $z(t) R_{\rm load}$ (recall voltage is current times resistance!) with that of the input voltage $E(t)$.

From Kirchhoff's two laws one can find the relationship between the source voltage, $E(t)$, in Loop I and the sum of the voltages across the devices in Loop I,


\begin{equation}\label{eq:LoopI}
\sin(100 t) = x(t) R_1 + \frac{1}{C_1}\int y(t) \, dt 
\end{equation}
and in Loop II using the resultant voltage across $C_1$ as a source voltage and adding the voltages across $C_2$ and $R_{\rm load}$,
\begin{equation}\label{eq:cap}
\frac{1}{C_1} \int y(t)\, dt = \frac{1}{C_2} \int z(t)\, dt + z(t) R_{\rm load} \,\, . 
\end{equation}
Finally,  use Kirchhoff's Current Law at node 2 to relate the three currents, $x(t)$,  $y(t)$,  and  $z(t)$.

Imposing initial conditions that all currents and all change (derivative) in currents are 0, we have in (\ref{eq:initial}), (\ref{eq:convert}), and (\ref{eq:LoopI}) what we need to solve for each of the three currents, $x(t)$, $y(t)$, and $z(t)$. In particular, we can solve for $z(t)$ and determine the amplitude of $z(t) R_{\rm load}$, the output voltage,  to compare to the amplitude of $E(t) = \sin(100 t)$, the input voltage as a ratio. This ratio of the amplitude of the output voltage to the amplitude of the input or source voltage is called {\it gain\/} and we seek the gain for various input voltage frequencies $\omega$ (frequency here is in radians/second) with $\omega = 100$  being the trial run frequency. We recall that the input or source voltage frequency is exactly the same as the output voltage and so in studying gain we are merely comparing amplitudes of the same frequency voltages.


\subsection*{Assignment of Input Voltage Frequencies}

We now assign a span of   frequencies $\omega$, say from $\omega = 100$ to 10,000, for an input voltages $E(t) = \sin(\omega t)$ and ask you and your classmates to compute a gain  over the resistance $R_{\rm Load}$ for the specific assigned frequency in our circuit of Figure~\ref{fig:4-23-CircuitFig2}. 


\begin{itemize}
\item [2)]Use your assigned frequency, $\omega$, and determine your corresponding gain, i.e. the ratio of amplitude of the output voltage to the amplitude of the input or source voltage 


When the class has returned with their circuit gains for their respective input frequencies we  will plot gain vs.~input frequency and see what such a plot shows us when we consider  input voltage frequencies $\omega$ in the range $(0, 10,000]$.


\item [3)] For further study, consider the same two loop circuit in Figure~\ref{fig:4-23-CircuitFig2} and determine the sensitivity of the Gain vs. Voltage input frequency to changes in one or more of the parameters $C_1$, $C_2$, $R_1$, or $R_{\rm Load}$. Namely, if we change one of these parameters how does such change effect the circuit's performance, its gain attributes?
\end{itemize}


\teacher{
\section*{NOTES FOR TEACHERS}


\subsection*{Review of Problem Assignment}

In an engineering mathematics course we provide sufficient conceptual framework (Conservation of Charge and Kirchhoff's Circuit Laws) and background in differential equations theory and solutions using Mathematica and Laplace Transforms to enable students to build a system of differential equations for a given two loop circuit and solve the system for various values of  parameters of the input voltage to the system. 

We assign each student a different input voltage frequency in the range $(0, 10,000]$ and ask them to compute the gain over $R_{\rm Load}$ in the second loop.  

When the class has returned with their circuit gains for their respective input frequencies we  will plot gain vs.~input frequency and see what such a plot shows us when we consider  input voltage frequencies $\omega$ in the range $(0, 10000]$.

We should get something like Figure~\ref{fig:4-23-GainDataFigure} from their results. Of course, any errors will stand out when plotted. 

\begin{figure}[h]
\begin{center}
\includegraphics[width=10.6cm]{4-23-GainDataFigure.eps}



\vspace*{-9.1 true cm}
\parbox{8.2 true cm}{\caption{\label{fig:4-23-GainDataFigure} Typical plot of  student results for gain vs. input voltage frequency for two loop RLC circuit of Figure~\ref{fig:4-23-CircuitFig2}.}}
\end{center}

\end{figure}


A paper describing the approach and use of this material was originally published in \cite{MyersWinkel2007}


\subsection*{Analysis}

In our notes to the teacher, below and in the accompanying Mathematica notebook 4-23-Mma-T-MysteryCircuit-TeacherVersion.nb (along with a pdf version of the notebook) we offer a further enhancement of the modeling perspective by interpreting the system with Laplace Transforms and the transfer function for the system. This path will permit us to compute the gain in an electrical circuit by easy use of the transfer function of the circuit. In this way the students see the meaning of the transfer function and the power of mathematics and technology in their study of a system of differential equations to model a circuit. 

What follows is essential the Laplace analysis from \cite[pp.~18-24]{MyersWinkel2007} which can also be seen in the accompanying Mathematica notebook 4-23-Mma-T-MysteryCircuit-TeacherVersion.nb (along with a pdf version of the notebook).




\subsection*{Laplace Transform Solution Strategy}


Let us demonstrate the analysis through  which we guide our students for $w = 100$ to determine the gain of the circuit. We then leave them to analyze the gain from their own input voltage frequency for homework. Of course, we do not have to use Laplace Transforms to compute specific gain for a given input voltage frequency. We can just solve the differential equation by non-Laplace methods and proceed. However, if it is appropriate for the class to apply Laplace Transforms then this is a good place to do so. Moreover, the analysis offered in the accompanying Mathematica notebook is best done with general Laplace Transform methods.
 
We use Laplace Transforms in Mathematica ({\tt Laplace[f[t],t,s]}  means transform $f(t)$ from the time domain, $t$, to the frequency domain, $s$) to solve this system of differential equations, but there are a number of other tools to use. For  after differentiating  (\ref{eq:LoopI}) and (\ref{eq:cap}) and using  (\ref{eq:xyz}) one can obtain a system of two first order differential equations, say in $x(t)$ and $z(t)$ ($y(t)$ being eliminated by  (\ref{eq:xyz}).  However, we will take the reader on the Laplace Transform route as there are conceptual ``gains" to be made with this approach and we investigate them in the last section of this paper.


We asked the students to use Kirchhoff's Current Law at node 2 to relate the three currents, $x(t)$,  $y(t)$,  and  $z(t)$ and they should obtain (\ref{eq:xyz}):

\begin{equation}\label{eq:xyz}
x(t)+y(t)= z(t)\,\, .
\end{equation}
Here are the Laplace transforms of Equations (\ref{eq:LoopI}), (\ref{eq:cap}), and (\ref{eq:xyz}) respectively, where $$F(s) = \int_0^{\infty} e^{-s t} f(t)\, dt$$ is the Laplace transform of the function $f(t)$:

$$\frac{100}{10000 + s^2} = 200 X(s)+\frac{400000 Y(s)}{s}\,\, , \eqno(3') $$
$$\frac{400000 Y(s)}{s}=\frac{1000000
   Z(s)}{s}+1000 Z(s)\,\, , \quad{\rm and} \eqno(4') $$
$$X(s) = Y(s) + Z(s)\,\, . \eqno(5') $$

\noindent Using Mathematica's powerful {\tt Solve} command we obtain:

$$Z(s) = \frac{200
   s}{(s+756.6)(s+2643.4)
   (s^2+10000
)} \, .$$
The {\tt Apart} command permits students to see the transforms of the corresponding transient state and steady state portions of the solutions:


$$Z(s) = \underbrace{-\frac{0.00013769}{ s+756.6}+\frac{0.000040046
 }{s+2643.4}}_{{\rm transient \,\, solution}} + \underbrace{\frac{0.0000976519s+0.0016684}{s^2+10000}}_{\rm steady \,\, state\,\, solution}\,\, ,$$
the transient and steady-state  Laplace Transform pieces being identified either by familiarity with the Inverse Laplace Transforms of $\frac{1}{s+a}$, $\frac{1}{s^2+a^2}$, and $\frac{s}{s^2+a^2}$ or by taking the Inverse Laplace Transform and making the transient and steady state identifications in hindsight.
\noindent Applying the {\tt InverseLaplace} command we obtain $z(t)$, the current over the resistor $R_{\rm load}$.
\begin{eqnarray*}
 z(t) &=& \underbrace{0.000040046 e^{-2643.4t} - 0.00013769 e^{-756.602 t}}_{{\rm transient \,\, solution}}\\
 && \quad + \underbrace{0.000097652 \cos(100 t) + 0.000016684\sin(100 t)}_{\rm steady \,\, state\,\, solution}\,\, . 
\end{eqnarray*}

Thus, to obtain the amplitude of the steady state output voltage, for the contribution of the transient disappears very rapidly, we use trigonometric identities to combine the sine and cosine terms into one phase-shifted sine term with  the amplitude of our current $z(t)$ passing through resister $R_{\rm load}$:
$${\rm amplitude} = \sqrt{0.000097652^2 + 0.000016684^2} = 0.0000990669 . $$
This means the  voltage across  $R_{\rm load}$ has amplitude, 
$${\rm amplitude}\cdot R_{\rm load} = 0.0000990669 \cdot 1000 = 0.0990669  \, \, .$$
Now the gain for our chosen $\omega = 100$  is the ratio of  amplitude of the output voltage over the resistor $R_{\rm load}$, which is 0.0990669, divided by the amplitude of the input voltage $E(t) = \sin(\omega t)$ which  is 1:
$${\rm gain} \,\,\, = \frac{0.0990669 }{1} = 0.0990669  \,\, .$$ 

\subsection*{Student Activity}

Upon return to class the next day we ask for a report from each student, i.e.~``What was your assigned frequency, $\omega$, and what gain did you obtain?" We actually sketch these on the board with a rough set of axes, before inputting them as data pairs into Mathematica to get a {\tt ListPlot} (see Figure \ref{fig:4-23-GainDataFigure}). Incidentally, errors in student work are glaring when each individual student's data point  is plotted in this manner!

\begin{figure}[h]
\begin{center}
\includegraphics[width=10.6cm]{4-23-GainDataFigure.eps}



\vspace*{-9.1 true cm}
\parbox{8.2 true cm}{\caption{\label{fig:4-23-GainDataFigure} Typical plot of  student results for gain vs. input voltage frequency for two loop RLC circuit of Figure~\ref{fig:4-23-CircuitFig2}.}}
\end{center}

\end{figure}
Once we all see this plot we  know something about what this circuit does. It appears to have low gain for both very low frequency input voltages and high frequency input voltages. Almost always there is some student who says this is the characteristic of a filter, i.e.~it takes all the mid frequency input voltages and keeps their gains high while dropping the gains for very low and high input voltage frequencies. Indeed, during class with their analysis on their own laptops brought to class we ask the students to change their $\omega$ to either a very high or very low frequency and compute the new gain. From their new, added results we see that this puts data at the left hand side of the plot in Figure~\ref{fig:4-23-GainDataFigure} for low frequency with very low gain or on the right hand side of the plot in  Figure~\ref{fig:4-23-GainDataFigure} for high frequency with very low gain, confirming their interpretation of the plot.

Another activity to follow this one is to ask the students to consider several variations on the source voltage, perhaps a sum of different frequencies as one source voltages, e.g., 
$$E(t) = \sin(200 t) + \sin(2000 t) + \sin(20000 t) \,\,, $$
to see that the respective gains for the components of the source voltage are $1.92\times 10^{-4}$ for $\omega = 200$, $5.64\times 10^{-4}$ for $\omega = 2000$, and $0.99\times 10^{-4}$  for $\omega = 20000$. Clearly the low and high frequency source components have lower gain than the mid frequency component!

\section*{TRANSFER FUNCTIONS IN LAPLACE TRANSFORMS}

One of our objectives in our engineering mathematics course is to use the vocabulary of the prospective engineering disciplines our students will encounter when possible. The notion of the transfer function, indeed, of living in the frequency domain of the Laplace Transform, is something we discuss when we do second order, linear, constant coefficient, ordinary differential equation work earlier in the course. Now it pays off as we try to analyze what is going on in this circuit with  attention to the transfer function of the system of differential equations.



Basically, given a differential equation, in  $y(t)$, with driver, $f(t)$,  and zero initial conditions,

\begin{equation}\label{eq:basic}
a \,y''(t) + b \,y'(t) + c\,y(t) = f(t), \quad y(0) = 0, \quad y'(0) = 0 \,\, ,
\end{equation}
the transfer function is the ratio of the Laplace Transform of the solution, $Y(s)$ in our notation, to the Laplace Transform of the driver, here $F(s)$. We show how this all fits together with  (\ref{eq:basic}). Taking the Laplace Transform of both sides and incorporating the initial conditions give us:

$$(a s^2 + b s + c) Y(s) = F(s) \,\, , $$ which, when we solve for $Y(s)$, en route to obtaining the solution $y(t)$, by applying the inverse Laplace Transform to $Y(s)$, gives

\begin{equation}\label{eq:Y}
Y(s) = \frac{1}{a s^2 + b s + c} F(s) \,\, .
\end{equation}
 Now in  (\ref{eq:Y}), $T(s) = \frac{1}{a s^2 + b s + c}$ is called the {\it transfer function} so we would have, in simplest terms, a relationship between the Laplace Transforms of the input or driver,  $F(s)$, and the output or solution, $Y(s)$. We simply multiply the Laplace Transform of the input by the transfer function to represent the behavior of the system on the input to produce an output Laplace Transform. How elegant!  Here is how it looks symbolically:
$$ Y(s) = T(s) F(s)\,\, . $$
Or
$$T(s) = \frac{Y(s)}{F(s)} \,\, . $$
This is almost like our notion of gain, ratio of ``out" to ``in," but not quite.

In our circuit's system of differential equations we see that $T(s) = Z(s)/U(s)$, i.e.~the system Laplace Transform or transfer function $T(s)$  IS the output Transform, $Z(s)$, divided by the input Laplace Transform, $U(s)$. Here we use $U(s)$ to denote the Laplace Transform of $E(t)$, the source voltage. If we multiply the output transform by a resistance, $R_{\rm Load}$, we get the Laplace Transform of the output voltage using Ohm's Law, $E = I*R$. 

Now if we take the amplitude of each of these, i.e.~$U(s)$, $T(s)$, and  $Z(s)$, we see that the amplitude of $T(s)$ is (amplitude of $Z(s) R_{\rm load}$)/(amplitude of $U(s)$) and using $i\omega$ for $s$ will permit us to understand the amplitude (or gain) ratio at a particular frequency $\omega$ while plotting $\vert T(i\omega)\vert$ permits us to see the gain for all values of $\omega$. 

Let us explain the previous paragraph in a more general setting and  show the power of the transfer function, by examining a second order, constant coefficient, linear, ordinary differential equation, $a y''(t) + b y'(t) +  c y(t) = f(t)$ with driver $f(t) = C \sin(\omega t)$ and initial conditions $y(0) = 0$ and $y'(0) = 0$. The solution of this differential equation will have a transient portion, which in our case will go to 0 quickly, leaving only the steady state portion of the solution for $y(t)$, namely $A(w) \cos(\omega t) + B(w) \sin(\omega t)$, where the coefficients of $\cos(\omega t)$ and $\sin(\omega t)$ ($A(w)$ and $B(w)$, respectively) are real  functions of $\omega$. Now $y(t)$ and $A(w) \cos(\omega t) + B(w) \sin(\omega t)$ differ, after a very short time, by a negligible transient solution, thus we say $y(t)\sim A(w) \cos(\omega t) + B(w) \sin(\omega t)$. So let us compute the Laplace Transform of all the terms in the steady state solution only and examine the transfer function as it applies to the input voltage only, i.e.:

$$F(s) = \frac{C \omega }{s^2 + \omega^2} \quad {\rm and} \quad 
Y(s) = \frac{A(w) s}{s^2 +\omega^2} + \frac{B(w) \omega}{s^2 +\omega^2}\, .$$ 
The transfer function is then
$$T(s) = \frac{Y(s)}{F(s)} =\displaystyle{\frac{ \frac{A(w) s}{s^2 +\omega^2} + \frac{B(w) \omega}{s^2 +\omega^2}}{\frac{C \omega}{s^2 + \omega^2}}} = \frac{A(w) s + B(w) \omega}{C \omega}\, . $$


\noindent We  substitute $s = i \omega$ into $T(s)$ and see what happens:
$$T(i \omega) =  \frac{A(w) i \omega + B(w) \omega}{C  \omega}=\frac{A(w) i + B(w) }{C  }$$ and we obtain
$$\vert T(i \omega) \vert = \frac{\sqrt{A(w)^2 + B(w)^2}}{C} \,   \eqno(8) $$
since the terms $A(\omega)$, $B(\omega)$, and $C$ are real. We see that $\vert T(i \omega) \vert$ gives the amplitude of the steady state output  function $y(t) = A(w) \cos(\omega t) + B(w) \sin(\omega t)$ divided by the amplitude of the input function $f(t) = C \sin(\omega t)$. This is simply the gain.  This is an ``Aha!" moment, for we see that by taking the absolute value (magnitude) of the transfer function, $T(s)$, with  $i \omega$ substituted  for $s$ we easily produce the gain {\it for any\/} $\omega$.




Mathematica can show, in the case of the circuit, that the transfer function $T(s)$ which is the Laplace Transform of the output voltage  across $R_{\rm load}$, i.e.~$Z(s)\cdot R_{\rm load}$, divided by the Laplace Transform of the input function $u(t) = \sin(\omega t)$, i.e.~$U(s) = \frac{\omega}{s^2 +\omega^2}$, is simply
$$T(s) = \frac{2000s}{2\times  10^6 + 3400 s + s^2}\, . $$
When we substitute $s = i \omega$ and take the absolute value of $T(i \omega)$ we obtain

\begin{equation}\label{eq:gainomega}
{\rm gain}(\omega)=\left\vert  \frac{2000 \omega }{2\times  10^6 + 3400 i\omega - \omega^2}\right\vert\, . 
\end{equation}

We offer a plot of gain($\omega$) vs. $\omega$ in Figure~\ref{fig:4-23-GainPlotFigure} and finally,  we offer a plot of our student computed data over the theoretical gain (see Figure~\ref{fig:4-23-GainDataAndPlotFigure}). 


\begin{figure}[h]
\begin{center}
\includegraphics[width=10.6cm]{4-23-GainPlotFigure.eps}



\vspace*{-9.1 true cm}
\parbox{10.2 true cm}{\caption{\label{fig:4-23-GainPlotFigure} Plot of gain$(\omega)$ for our two loop  RLC circuit of Figure~\ref{fig:4-23-CircuitFig2}.}}
\end{center}

\end{figure}





\begin{figure}[h!]
\begin{center}
\includegraphics[width=10.6cm]{4-23-GainDataAndPlotFigure.eps}



\vspace*{-9.1 true cm}
\parbox{8.2 true cm}{\caption{\label{fig:4-23-GainDataAndPlotFigure} Plot of gain$(\omega)$ and student data for our two loop  RLC circuit of Figure~\ref{fig:4-23-CircuitFig2}.}}
\end{center}

\end{figure}


A picture is worth a thousand words and so the students see the value of the transfer function in easily computing the gain  for our circuit over a range of input voltage frequencies, $\omega$. They can also see qualitatively that ${\rm gain}(\omega)$ gets small as $\omega$ increases, for the denominator in the fraction of (\ref{eq:gainomega}) has a square term in $\omega$ while the numerator is first order in $\omega$. In the case that $\omega$ gets small they can see that the $\omega^2$ term in the denominator gets very small while the numerator goes to 0  and the denominator goes to $2 \times 10^6$, hence  ${\rm gain}(\omega)$  goes to 0, this time as $\omega$ gets small. Thus the gain on either end, $\omega$ small or big, is small and so this circuit keeps amplitude of midrange frequencies high while lowering, considerably, the amplitude of the lower and higher range frequencies - it is a filter, for low and high input voltage frequencies have low gain while the midrange input voltage frequencies have higher gain.
}


\begin{thebibliography}{99}
\bibitem{Kreyszig1999} Kreyszig, E. 1999. {\it Advanced Engineering Mathematics, Eighth Edition\/}. New York:  John Wiley and Sons, Inc. 

\teacher{
\bibitem{MyersWinkel2007} Myers, J. and B. Winkel. 2007. Mystery Circuit in Mathematics Class. {\it The UMAP Journal}. 28(1): 15-25.
}
\end{thebibliography}






\end{document}
