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Rise time of RC Filter

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Amr Wael

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Hello ,
i am trying to design a low pass filter using the simple RC configuration , I have selected the values for the resistor and capacitor to be 10 nF and 390 Ohms for the cut off frequency to be around 40k.
My problem is that the rise time of a 20Khz signal entering this filter is making the digital pulse really bad. I know that the rise time and the cut off frequency are related through the following equation Tr = 0.35/fdB.
My question is there a method to decrease the rise time without affecting the cut off frequency ?
or should I use a different type of filter ?
This is a reference for the relationship between the rise time and the BW.
and attached is a photo of a simulation for the filter at the values mentioned before.
 

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Filtering a square wave with a low pass filter and requesting steep rise time is a contradiction in terms. Please explain what you want to achieve.
 

I want to filter any noise above my cut off frequency.
--- Updated ---

Basically I am filtering a digital signal which is I/O of a Mcu GPIO
 

Symmetrical 20 kHz square wave has odd harmonic components (1th, 3rd, 5th, 7th ...). Applying an ideal filter with 40 kHz cut-off only keeps the fundamental frequency, in other words a 20 kHz sine wave. Your sloppy RC filter passes a considerable amount of harmonics, so you see a slurred square wave instead of a sine.

If you still want a kind of square wave, you either need to use a max higher cut-off frequency or restore the pulse shape after the filter, e.g. by using a comparator.
 

    Amr Wael

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For single pole block 10-90 tr=2.2T and you cannot change it. If you want to keep cut off frequency the same and sharpening response edge, you need to use filter with complex poles. So, or active filter or RLC.
 

I want to filter any noise above my cut off frequency.
--- Updated Today at 11:11 AM ---

Basically I am filtering a digital signal which is I/O of a Mcu GPIO

So when data rate exceeds a given freq you want that data to be suppressed,
otherwise maintain the Tr, Tf inherent in the GPIO ?

Regards, Dana.
 

What's the minimum risetime you can tolerate with the 40kHz corner frequency?

How much filtering of the noise do you need?
No filter has infinite suppression of the noise.
 

For single pole block 10-90 tr=2.2T and you cannot change it. If you want to keep cut off frequency the same and sharpening response edge, you need to use filter with complex poles. So, or active filter or RLC.
Increasing the filter order with fixed cut-off frequency (- 3 dB) does not reduce the rise time, it rather increases it, depending on the filter type. This can be intuitively explained by the reduced harmonic content of the filtered signal.
 

Increasing the filter order with fixed cut-off frequency (- 3 dB) does not reduce the rise time, it rather increases it, depending on the filter type.
Yes, right. Mind shortcut, the peaking of impulse response is N*tau, so by increasing order to 2nd, time constant has to be reduced by factor of filter order of course. Thanks for correction.
 

Here's the LTspice simulation of a 1-pole passive RC filter, and a 3-pole active Sallen-Key filter (with basically Bessel response), both with an ≈1kHz, -3dB corner.
As can be seen, the 3-pole filter output (yellow trace) has more delay, but a less rounded rise and fall with a flatter top and bottom than the 1-pole (green trace), thus giving a more symmetrical and normal looking pulse output.

And, of course, the 3-pole filter will suppress any noise above the corner by a significantly greater amount.

1625190861562.png


1625190792954.png
 
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