Talk:EGR 224/Spring 2009/Final

General Questions

Post general questions or requests for clarification here.

• What is the significance of the quarter power frequency? Calculating it is easy enough, but why is this important for second order filters as opposed to the standard half power frequency? -Brs16 18:43, 20 April 2009 (EDT)
  • Assume you want a filter with a cutoff of \(\omega_{co}\). The typical first-order filter transfer function (i.e. 20 dB per decade slope after the cutoff) would be:\(\begin{align} \mathbb{H}_1(j\omega)&=\frac{K\omega_{co}}{j\omega+\omega_{co}} \end{align}\) which, at the cutoff, has a magnitude of \(\begin{align} |\mathbb{H}_1(j\omega_{co})|&=\left|\frac{K\omega_{co}}{j\omega_{co}+\omega_{co}}\right|=\frac{K\omega_{co}}{\sqrt{2\omega^2_{co}}}=\frac{K}{\sqrt{2}} \end{align}\) With that as a voltage ratio, you end up with half power.
    If you want a sharper filter, say 40 dB per decade, you might cascade two low-pass filters or design an LRC circuit that has a repeated root at the desired cutoff - that is:\(\begin{align} \mathbb{H}_2(j\omega)&=\frac{K\omega^2_{co}}{(j\omega+\omega_{co})^2} \end{align}\)which, at the cutoff, has a magnitude of\(\begin{align} |\mathbb{H}_2(j\omega_{co})|&=\left|\frac{K\omega^2_{co}}{(j\omega_{co}+\omega_{co})^2}\right|=\frac{K\omega_{co}}{2\omega^2_{co}}=\frac{K}{2}\end{align}\) thus yielding one-quarter power at \(\omega_{co}\). Strictly speaking, the half-power frequency for such a second-order repeated-root circuit would be:\(\begin{align} \left|\mathbb{H}_2(j\omega_{hp})\right|&=\left|\frac{K\omega^2_{co}}{(j\omega_{hp}+\omega_{co})^2}\right|=\frac{K}{\sqrt{2}}\\ \frac{K\omega^2_{co}}{\omega^2_{hp}+\omega^2_{co}}&=\frac{K}{\sqrt{2}}\\ \omega_{hp}&=\omega_{co}\sqrt{\sqrt{2}-1}\end{align}\) Note that there are myriad ways to build higher-order filters to get more sharply defined corners at particular locations - the repeated-term version above is just an entry level introduction to them. DukeEgr93 19:08, 20 April 2009 (EDT)

• With regard to second order passive filters, what is the difference between taking the voltage off the capacitor alone vs. taking it off the series combination of the resistor and capacitor? —Preceding unsigned comment added by Jlr37 (talk • contribs)
  • Mainly the transfer function; just the capacitor gives a "pure" lowpass while the resistor and cap together give a lowpass that, being made up of the lowpass and bandpass parts, has a half power frequency that is likely higher than that of just the lowpass part. You end up with something like:\(\begin{align}\mathbb{H}&=\mbox{constant}\frac{s+a}{(s+b)(s+c)}\end{align}\)so depending on the relative values of a, b, and c, the break points and directions can be different relative to each other. If a is smallest, you break up first, then down twice, so that's most likely a band pass filter. If a is between b and c, you break down, then up (to flat), then down again, which is probably lowpass. If a is largest, you break down, then down again, then up (this time to -20 dB/dec), so that is definitely a lowpass. DukeEgr93 22:06, 1 May 2009 (EDT)

• When graphing a Bode plot, how do you know where it starts? For example, on A&S page 626 the filter at the top starts at about 17 dB. How can you calculate this? —Preceding unsigned comment added by Brs16 (talk • contribs)
  • Two options - if the magnitude comes in "flat," take the limit of the magnitude as \(\omega\rightarrow 0\) and the flat part will be \(20\log_{10}\) of that amplitude. For others, if there is a passband, with the passband gain and use \(20\log_{10}\) on the flat part, then break from there. For particularly ugly ones - 14.4 on p. 626 comes to mind - they basically drew each individual straight-line approximation for all the terms and added them together - that works too. Key thing there is that a \(j\omega\) term on its own passes through 0 dB when \(\omega=1\) and that a constant term \(K\) shifts the entire graph up by \(20\log_{10}(K)\) DukeEgr93 00:04, 2 May 2009 (EDT)