Hello i have just been introduced to low pass filters and am unsure of 1 thing. I dont understand why having a larger minimum attenuation in the stopband is better? What does it do? I understand that moving the cutoff frequency closer to the stop band frequency is better as then more frequencies are retained. I dont understand how having say a minimum stop band attentuation of 10db is worse than 80db? enter image description here

I thought aslong as the filter removes the frequencies that arent wanted its fine. However someone told me that its better to have a larger minimum attenuation for the stopband. Basically how does the gain(db) (yaxis) affect the filter?

  • \$\begingroup\$ Note that saying "the more stopband, the better" is just one case, because that will make the transition band narrower, which comes with its own prices to pay. Filter design is really about goals: what do you want it to do, time or frequency? Remove frequencies, filter pulses, have flat group delay, etc? If it's the first, then go on and make the stopband -120dB, it won't hurt, but if it's the second (for example), then not only the type of filter will be different, but you'll also want to stay away from narrow transition widths. \$\endgroup\$ – a concerned citizen May 7 '17 at 6:12
  • \$\begingroup\$ How does the stopband affect it though. I dont get how it makes the transition band narrower? Since a larger db doesnt that increase the size of the stopband? \$\endgroup\$ – Student May 7 '17 at 6:19
  • \$\begingroup\$ Please note that a lowpass cannot "remove" unwanted frequencies. It only can attenuate these frequencies - and a better damping factor is an indication of a "better" filter. \$\endgroup\$ – LvW May 7 '17 at 9:04
  • \$\begingroup\$ A larger stopband attenuation isn't necessarily better - you might want to control amplitudes to specific levels as per tone controls on a stereo or a graphic equalizer. \$\endgroup\$ – Andy aka May 7 '17 at 9:15

Here is example of an active waveform-inverting low-pass-filter with a stop-band problem; notice the input pulse appears (initially) as part of the output waveform, not what we expect from a low-pass-filter. The initial pulse-leakage into Output Wave is 240 milliVolts.

enter image description here

Here is the circuit: the OpAmp is slow, only 1MHz UGBW by default (we'll later return to this menu and edit UGBW to 10MHz), and the OpAmp has a high Rout of 1,000 Ohms (typical values are 100 ohms). Notice at lower left, the C1 is not enabled. enter image description here

What is going on? At the higher frequencies, the opAmp is unable to control Vout, because charge passing through the "Cf" path is stronger than the opamp.

What is cure? (1) Turn on (in the simulator) the very first capacitor, the C1 at junction of the first 2 resistors. (2) Use a faster opamp.

Here is the benefit of C1 in place. The initial pulse-leakage is 13 milliVolts.

enter image description here

What if we use a faster OpAmp (almost guaranteed to consume more power)? Here is 10x faster OpAmp, with C1 turned off. enter image description here

Here is that same fast OpAmp, with C1 on, and Rout reduced to 100 Ohms; notice the pulse-leakage is "invisible": enter image description here

Here is the frequency response for Fast OpAmp, C1 enabled, Rout of 100 Ohms; see how the BODE rolloff is very very close to perfect 1_pole rolloff (that is, straight line fit to 6dB/octave once past 1Mhz; there is a tiny rise at 100MHz, but only a couple db, quite unlike the 20dB rise in the prior BODE): enter image description here

We encountered this "noise reduction" challenge some years back, for a client's touchscreen product. The column-drivers were magnetically coupling into the top 1/4 of the Y-axis pickup loops. The signal conditioning already had a good GND plane and bypass caps. Turns out the cure was to split the Rin into 2 resistors and add that "T" cap (C1) to keep the very fast transient energy off the Virtual Gnd node of the OpAmp LPF. The solution was so useful, we included it as 1 of the 13 Examples in the tool.


I thought aslong as the filter removes the frequencies that arent wanted its fine.

The "frequencies that aren't wanted" are the stop band.

Higher attenuation in the stop band means you did a better job removing the frequencies that aren't wanted.

If the stop band attenuation is 10 dB, it means 10% of signal power in the stop band is let through. If it's 20 dB only 1% of the signal power in the stop band is let through.

Note that practically, the response of the filter won't continue to drop by 20 dB/decade forever. At some point, some parasitic element will disturb the behavior and introduce a zero. Depending how well the filter is designed, at some point the response might even start to increase again at high frequencies.

  • \$\begingroup\$ thanks, how did you calculate the percentages or where they just as a guide? \$\endgroup\$ – Student May 7 '17 at 5:52
  • \$\begingroup\$ @user6186979, from the definition of decibels. \$\endgroup\$ – The Photon May 7 '17 at 5:53
  • \$\begingroup\$ -For a standard passive low pass filter is it possible to set the minimum attenuation for the stopband? I undersatnd you can with butterworths etc. Although i have to filter a PWM signal and to do this i was just going to use a standard low pass filter. The problem i dont get is that how do you calculate the stop band attenuation? \$\endgroup\$ – Student May 7 '17 at 6:00
  • 1
    \$\begingroup\$ Not sure what you mean by "standard" low pass filter. If you mean a one-pole RC filter, then you've only got one free parameter, so you can't set cut-off frequency separate from stop-band behavior. But if you want to do real filter design, you start with requirements (like stop-band attenuation) and use that do decide which filter type, what filter order, etc. \$\endgroup\$ – The Photon May 7 '17 at 14:50

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