Talk:EGR 224/Spring 2009/Test 2

• How exactly does one take the magnitude of a transfer function? My understanding was to take the square root of the sum of the squares of the components of the numerator over the same for the denominator. Why/how is this done different is A&W equation 14.10.3? I see that they use j^2=-1, but I don't get the same answer if I leave it a j^2. Also what is the reason for dropping the j's when taking a magnitude? --Brs16 18:28, 28 March 2009 (EDT)
  • The magnitude of the transfer function is as you describe; you basically need to get it in the form:

\( \begin{align} \mathbb{H}&=\frac{a+jb}{c+jd}\\ |\mathbb{H}|&=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}} \end{align} \)

where above, a and c are the real parts and b and d are the imaginary parts. You are dropping the j because you are taking a magnitude - basically, using pythagorean theorem in the complex plane.DukeEgr93 11:40, 29 March 2009 (EDT)

• Generating a Bode plot for functions that are already factored or easily factored is straightforward. All the examples in the book or that we did in class are like this so the poles and zeros and easily found. How do you approach generating a bode plot for a transfer function that cannot easily be factored (like a typical bandpass)? Are these considered "too tricky?" Do "multiple zero/pole systems" mean transfer functions that are already factored? --Brs16 18:28, 28 March 2009 (EDT)
  • For the test, a problem like that will either already be factored or be easily factored. I will not give a case with complex poles. DukeEgr93 11:40, 29 March 2009 (EDT)
• Via e-mail: Is finding the gain on a band-reject filter the same as on a bandpass filter?
  • For a prototypical band-reject,

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

where K is the gain at the high and low frequencies outside of the rejected range. DukeEgr93 11:40, 29 March 2009 (EDT)

• Via e-mail: Are the log center frequency and linear center frequencies just the middle of the bandwidth on their respective plots?
  • Pretty much. Just remember that to get the center from a Bode plot, you will be averaging the logs and not the values. DukeEgr93 11:40, 29 March 2009 (EDT)
• How do you convert between the simplified filter transfer function and the pole zero factored form? also how do you find the gain?
  • Easiest way is to assume that \(j\omega\) is your variable and factor it. With respect to the gain, if the transfer function can be put into any of the prototypical forms:

\( \begin{align} \begin{array}{cc} \mbox{1st order low-pass} & \mathbb{G}(j\omega)=\frac{Ka}{j\omega+a}\\ \mbox{1st order high-pass} & \mathbb{G}(j\omega)=\frac{Kj\omega}{j\omega+a}\\ \mbox{2nd order low-pass} & \mathbb{G}(j\omega)=\frac{K\omega_n^2}{(j\omega)^2+2\zeta\omega_n j\omega+\omega_n^2}\\ \mbox{2nd order band-pass} & \mathbb{G}(j\omega)=\frac{K2\zeta\omega_n j\omega}{(j\omega)^2+2\zeta\omega_n j\omega+\omega_n^2}\\ \mbox{2nd order band-reject} & \mathbb{G}(j\omega)=\frac{K((j\omega)^2+\omega_n^2)}{(j\omega)^2+2\zeta\omega_n j\omega+\omega_n^2}\\ \mbox{2nd order high-pass} & \mathbb{G}(j\omega)=\frac{K(j\omega)^2}{(j\omega)^2+2\zeta\omega_n j\omega+\omega_n^2}\\ \end{array} \end{align} \)

then the maximum gain is *usually* \(K\) except sometimes, in the second order cases, if the damping ratio is small enough the actual maximum gain for the high and low-pass filters may be much higher. DukeEgr93 23:19, 29 March 2009 (EDT)