Thoughts on Filters and Bode Plots

[Omega Glory]

I'm a student focusing on digital design and don't have very much experience in the analog domain, but last night I was thinking about bode plots and realized that I had been viewing them incorrectly for quite some time.

If I have a simple first order RC low pass filter with a cut-off frequency of Fc= 10Hz, and I feed a 1v pp sine wave into it at 100Hz, then since the filter asymptotically has 20dB per decade attenuation, I ought to get a 0.1v pp sine wave on my output.

However, if I have another first order RC low pass filter with a cut-off frequency of Fc=100Hz, then in order to have my output attenuated to 0.1v pp like in the other example, my input frequency needs to be 1000Hz, as the rolloff is still 20db per decade.

So when looking at the bode plots of these two filters, one will see that they both have a constant rolloff of 20dB per decade, yet thinking in linear terms, the filter with the 10x higher cut-off frequency rolls off 10x slower. I somehow thought (through my carelessness) that the constant 20dB per decade roll-off meant their linear roll-off was identical. This probably also could have been avoided had I been thinking more carefully about the zero pole diagram, and the magnitude of the response along the imaginary j*omega axis.

This is an interesting point to think about, because it means that many bandpass filters which may appear to the naive to have symmetrical low frequency roll-off and high frequency roll-of, do not.

At first this realization troubled me quite a lot, but then I realized that at least for audio purposes, this might actually be a very nice property of filters, since the ear's determination of pitch is logarithmic (i.e. the musical note C5 is twice the frequency of C4).

This post isn't exactly a question, but I would be curious to know how this property of filters plays into designs. I can imagine that in RF designs this might be why signals are often brought down to be filtered at a lower intermediate frequency rather than a higher "intermediate" frequency. 

[Doctorandus_P]

You are making an error in your thinking process.

An RC filter is just a resistor and a capacitor.
For a resistor, impedance is the same, regardless of frequency.
For a capacitor impedance varies with frequency. When you double the frequency, the impedance halves.
A slope of 20dB/decade means that the amplitude changes with 20dB (which is a factor of 10 in "linear amplitude") when the frequency also changes with a factor of 10. It's the same as 6dB/octave. Double the frequency, and amplitude halves (or doubles, depending on filter type).

I don't know what your thoughts are about this linear frequency, but that just does not make sense when talking about filters.
From 10Hz to 100Hz is a factor 10, and from 100Hz to 1kHz is also a factor of 10, so the slopes and the filter characteristics are the same, they are just a bit shifted to the left or right. It really does not matter that 100-10 = 90Hz difference and 1k-100 = 900Hz difference. If you attempt to draw a bode diagram on a linear scale it just becomes a mess.

[Omega Glory]

Right, I agree with what you're saying. I think what I was trying to get at was that if I had a low pass filter and I wanted to attenuate my input signal down from 1v to 0.1 voltes, if my Fc is 10Hz, then I can pull that off with an additional 90Hz after my Fc. But if my Fc was 100Hz, and I wanted to attenuate 1v down to 0.1 voltes when my input frequency increases from 100Hz to 190Hz, then I couldn't do it with a first order filter, because my roll off is 20dB per decade and not 0.9v per 90Hz. So in order to realize that second filter, I would need a higher order filter. I should have understood this right off the bat, but somehow got these two attenuation slopes confused in my mind.

And then my thought on filtering after downconversion is that it might be possible to achieve one's goal with a lower order filter than if one were operating at a higher frequency.

[Benta]

I think the OP needs to read up on log-log scaled plots/graphs.

[Doctorandus_P]

when my input frequency increases from 100Hz to 90Hz, then I couldn't do it with a first order filter, because my roll off is 20db per decade and not 0.9v per 90Hz.

Meaning lost in typo errors.

and not 0.9v per 90Hz.

This does not work at all. If you have logarithmic axis on a bode plot, you can draw straight lines, but when you make the axis linear, the lines are not straight anymore.

[Omega Glory]

Typos fixed. And yes, what you've pointed out is exactly what I'm getting at. I'm not saying there's a problem with bode plots, but that I previously had precisely this misunderstanding you're pointing out. The fact that filter's don't work in that way has implications for downconversion followed by filtering (and other things, I assume).

[Omega Glory]

Here's a more flushed out example:

If I have some amplitude modulated signal with bandwidth W, centered around some high frequency wH, and I want to apply a low pass filter to this signal, then it is to my advantage to down convert the signal to be centered around some lower frequency wL, process it using a low order filter, and then re-center the signal around wH, because if I try and filter that signal where it currently is centered at wH, I will need a higher order filter to achieve the same attenuation goal.

[jonpaul]

A bode plot is not what you think, it shows the poles and zeros of a second oerder system to determine system stability and response, eg for control systems.


For filters you want a frequency/phase response over a range of frequencies.

Miltipoe filters can be of sharp cut like elliptical, linear phase like Bessel or easy construction like Butterworth.

To learn of filter design and specification read a good book on filter design, or see Don Lancasters Filter cookbook.

Jon