EGR 119/Spring 2009/Test 2/Post Test

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Problem I: Filter Design

$ω$ needs to be in rad/s; remember, $ω = 2π f$ !
For first-order low-pass filters, the general form is

$\begin{align} \mathbb{G}&=\frac{Ka}{j\omega+a} \end{align}$

where K is the gain and a is the cutoff (half-power) frequency.

An easy first-order low-pass filter is either an RC circuit, reading the voltage across the C; or an LR circuit, reading the voltage across the R. These yield, respectively,

$\begin{align} \mathbb{G}_1&=\frac{1/RC}{j\omega+1/RC} & \mathbb{G}_2&=\frac{R/L}{j\omega+R/L} \end{align}$

For second-order low-pass filters, a general form is

$\begin{align} \mathbb{G}&=\frac{Ka^2}{(j\omega+a)^2}=\frac{Ka^2}{(j\omega)^2+2aj\omega+a^2} \end{align}$

where K is the gain and a is the cutoff frequency. For this particular combination, the natural frequency is a and the damping is 1 (i.e. critically damped response). However, for a double corner like this, there is a 6 dB shift from the maximum rather than a 3 dB shift; a is thus more like a "quarter power" frequency. This was taken care of by the wording of the problem.

An easy second-order low-pass filter is an LRC circuit, reading the voltage across the C. This yields:

$\begin{align} \mathbb{G}&=\frac{1/LC}{(j\omega)^2+j\omega(R/L)+(1/LC)} \end{align}$

Problem II: Switched Circuit

Main issue here was solving for variables just after the switch. Technique is to determine the continuity variables (capacitor voltages and inductor currents) before the switch, redraw the circuit as it appears after the switch, and put in the continuity variables, From there, KVL and KCL should be used to get all other variables.

Problem III: RL Circuit

The first transfer function was in the form

$\begin{align} \mathbb{G}_a&=\frac{j\omega L+R_2}{j\omega L + R_1 + R_2} \end{align}$

There are several methods to determine what kind of filter this is. One is to consider the topology of the circuit - the inductor is going to get a larger impedance as the frequency increases, so a greater percentage of the current will flow through R1 - this denotes a high-pass filter. Mathematically, you can determine the magnitude and note that

$\begin{align} \lim_{\omega\rightarrow 0}|\mathbb{G}_a|&=\frac{R_2}{R_1 + R_2} \\ \lim_{\omega\rightarrow \infty}|\mathbb{G}_a|&=1 \end{align}$

which, for a first-order circuit is sufficient to determine that it is a high-pass. Finally, you can look at the corner frequencies for a Bode approximation - there is a break down at R2/L and a break back up at (R1+R2)/L, which is a higher frequency.

To find the differential equation using frequency techniques, get $I x (j ω)$ and then cross-multiply to get rid of any denominators. Next, recall that $j ω$ is the frequency-space representation of a time derivative.
When solving a circuit that has multiple source frequencies, each frequency must be considered independently and solved completely in time. You cannot add phasors representing multiple frequencies together.

Problem IV: Operational Amplifiers

For the design, the requirement for no power from the source means the source must be buffered.
For the design, the requirement that the ideal op-amp assumptions apply means that resistor values must be selected that do not interfere with the large (2 M $Ω$ ) input impedance or the small (75 $Ω$ ) output impedance. Using k $Ω$ order resistors works great.
For the third part, note that the inverting input is directly connected to the output; this means that location cannot be used for a KCL equation unless your goal is to determine the unknown current coming out of the op amp.

EGR 119/Spring 2009/Test 2/Post Test

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Contents

Problem I: Filter Design

Problem II: Switched Circuit

Problem III: RL Circuit

Problem IV: Operational Amplifiers

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