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)