Difference between revisions of "Talk:EGR 224/Spring 2009/Test 2"
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| − | 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? --[[User:Brs16|Brs16]] 18:28, 28 March 2009 (EDT) | + | *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? --[[User:Brs16|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: | |
| − | 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? --[[User:Brs16|Brs16]] 18:28, 28 March 2009 (EDT) | + | <center> |
| + | <math> | ||
| + | \begin{align} | ||
| + | \mathbb{H}&=\frac{a+jb}{c+jd}\\ | ||
| + | |\mathbb{H}|&=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}} | ||
| + | \end{align} | ||
| + | </math> | ||
| + | </center> | ||
| + | ::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.[[User:DukeEgr93|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? --[[User:Brs16|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. [[User:DukeEgr93|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, | ||
| + | <center> | ||
| + | <math> | ||
| + | \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} | ||
| + | </math> | ||
| + | </center> | ||
| + | ::where K is the gain at the high and low frequencies outside of the rejected range. [[User:DukeEgr93|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 averageing the logs and not the values. [[User:DukeEgr93|DukeEgr93]] 11:40, 29 March 2009 (EDT) | ||
Revision as of 15:40, 29 March 2009
- 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 averageing the logs and not the values. DukeEgr93 11:40, 29 March 2009 (EDT)
