Talk:BME 153/Spring 2009/Test 2

• Is Chapter 6.2 of the book going to be on the test? I know you mentioned Fourier Transforms but we never really did anything with it in the homework. Ibl 19:57, 20 March 2009 (EDT)IBL
  • The idea of a transfer function will certainly be on the test, but Fourier and inverse Fourier transforms themselves will not be. The parts of Chapter 6 that were in Homework 8 define the extents of what could be asked from that chapter. DukeEgr93 10:54, 21 March 2009 (EDT)
• What is the difference between linear and logarithmic center frequencies? How can you find one from the other if possible, and if not, how can the logarithmic center frequency be used to determine information about other circuit variables? My question essentially, is if you were to give us a problem in which you asked us to design a circuit given certain parameters, one of which was the logarithmic cutoff frequency, how would we use that piece of information? Egh4 15:49, 22 March 2009 (EDT)egh4
  • The linear center frequency for a bandpass/bandreject filter is the mean of the half-power frequencies; the logarithmic center frequency has a logarithm that is the average of the logarithms of the half-power frequencies. For the "typical" bandpass filter:

\( \begin{align} \mathbb{H}&=\frac{K2\zeta\omega_n(j\omega)}{(j\omega)^2+2\zeta\omega_n(j\omega)+\omega^2_n}\\ \mathbb{H}&=\frac{K}{1+jQ\left(\frac{\omega}{\omega_n}-\frac{\omega_n}{\omega}\right)} \end{align} \)

the logarithmic center frequency is the same as the resonant frequency, \(\omega_n\), and the linear center frequency can be calculated as

\( \begin{align} \omega_{lin ~ctr}&=\omega_n\frac{\sqrt{1+4Q^2}}{2Q} \end{align} \)

Note that as the quality increases, the bandwidth decreases and the linear and logarithmic centers get closer and closer. DukeEgr93 18:07, 22 March 2009 (EDT)

• On Test III, Fall 2001 of EE 61, you said in problem 5 that ix is 0 once the switch closes. Why is this? Ibl 18:36, 22 March 2009 (EDT)IBL
  • After t=0, the switch closes, and unfortunately, the switch is in parallel with R3. And ix is the current through R3, so when the switch closes, the voltage across the resistor must go to 0 (and so must the current). That is why the "solution" deals with the inductor instead - it is a more interesting problem. DukeEgr93 18:53, 22 March 2009 (EDT)

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In the interest of sleep, probably no more answers will be posted here before the test :) DukeEgr93 01:43, 23 March 2009 (EDT)