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There are a number of misconceptions here. bode plot db/Hz First, it's "dB", not "db". Second that's not a Bode plot. Those are dB/Log(Hz). In other words, a certain frequency ratio results in...
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#2: Post edited
by
Olin Lathrop
·
2022-07-06T20:00:26Z (over 2 years ago)
- There are a number of misconceptions here.
- <blockquote>bode plot db/Hz</blockquote>
- First, it's "dB", not "db".
- Second that's not a Bode plot. Those are dB/Log(Hz). In other words, a certain frequency <i>ratio</i> results in a fixed gain ratio as the slope approaches the asymptote. You were told in your previous question that dB/Hz doesn't make sense. If you're not going to listen answers, then there is no point in us writing them.
- <blockquote>is there a formula if we know the slope of the curve at the cutoff frequency of the 1st order filter to calculate the slope of the curve for a multistage filter</blockquote>
- For a first order low pass filter well past the rolloff frequency, gain is inversely proportional to frequency. For a second order, it's inversely proportional to the square of the frequency, and to the cube of the frequency for a third order, etc.
This is usually easier to express in dB. For first order, the voltage gain is -20 dB / Log(Hz), for a second order -40 dB / Log(Hz), etc.
- There are a number of misconceptions here.
- <blockquote>bode plot db/Hz</blockquote>
- First, it's "dB", not "db".
- Second that's not a Bode plot. Those are dB/Log(Hz). In other words, a certain frequency <i>ratio</i> results in a fixed gain ratio as the slope approaches the asymptote. You were told in your previous question that dB/Hz doesn't make sense. If you're not going to listen answers, then there is no point in us writing them.
- <blockquote>is there a formula if we know the slope of the curve at the cutoff frequency of the 1st order filter to calculate the slope of the curve for a multistage filter</blockquote>
- For a first order low pass filter well past the rolloff frequency, gain is inversely proportional to frequency. For a second order, it's inversely proportional to the square of the frequency, and to the cube of the frequency for a third order, etc.
- This is usually easier to express in dB. For first order, the voltage gain is -20 dB / Log(Hz), for a second order -40 dB / Log(Hz), etc.
- <hr>
- <blockquote>Isn't the gain inversely proportional to frequency from the rolloff frequency of a filter? Why "well past"?</blockquote>
- Because it approaches "gain inversely proportional to frequency" gradually. A single pole low pass filter isn't totally flat up to the rolloff frequency, then suddenly falls off proportional to frequency above that. There isn't a hard corner in the Bode plot. It's smoothed out. That's why I don't like the term "cutoff" frequency, since it incorrectly implies a hard transition.
- If there was a hard transition, the gain at the rolloff frequency would still be 1 (0 dB). Instead, the gain has already started to fall off, and is actually -3 dB at the rolloff frequency. The further from the transition, the closer the gain becomes to the hard transition asymptotes.
- Put another way, the real gain of the filter is off by 3 dB from the asymptotes at the rolloff frequency. Either side of that frequency, the gain approaches the asymptotes more closely. Therefore "well past" the rolloff frequency, the gain falls off inversely proportional with frequency. "Well below" the rolloff frequency, the gain is 1.
#1: Initial revision
by
Olin Lathrop
·
2022-07-06T19:08:51Z (over 2 years ago)
There are a number of misconceptions here. <blockquote>bode plot db/Hz</blockquote> First, it's "dB", not "db". Second that's not a Bode plot. Those are dB/Log(Hz). In other words, a certain frequency <i>ratio</i> results in a fixed gain ratio as the slope approaches the asymptote. You were told in your previous question that dB/Hz doesn't make sense. If you're not going to listen answers, then there is no point in us writing them. <blockquote>is there a formula if we know the slope of the curve at the cutoff frequency of the 1st order filter to calculate the slope of the curve for a multistage filter</blockquote> For a first order low pass filter well past the rolloff frequency, gain is inversely proportional to frequency. For a second order, it's inversely proportional to the square of the frequency, and to the cube of the frequency for a third order, etc. This is usually easier to express in dB. For first order, the voltage gain is -20 dB / Log(Hz), for a second order -40 dB / Log(Hz), etc.