I have the input and output sinwave amplitudes as well as the cut off frequency how do I use that to calculate the sinewave frequency(Hz) for a low pass filter. I'm unsure of what I need to be a looking for to calculate it.

  • \$\begingroup\$ @Andy Aka I think he wants to calculate the frequency for which he knows the input/output amplitude ratio, and not the -3dB frequency, which he already knows.@Apple You need the transfer function, which is defined as H(s) = 1/(1+sT) for a first order low pass filter. \$\endgroup\$ – Bart Feb 21 '18 at 15:01
  • \$\begingroup\$ @Bart you may well be correct. \$\endgroup\$ – Andy aka Feb 21 '18 at 15:10
  • \$\begingroup\$ You will need the frequencies of the input and output signals at the known amplitudes. But that is not how it normally works. Analog filters go in steps of -6db/octave per stage. From that you have to determine how many stages you need to meet or exceed your requirements. \$\endgroup\$ – Oldfart Feb 21 '18 at 15:15
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    \$\begingroup\$ Dave dont be that way. From any filter design you can work out any unknown given all the other. DOH. If you cannot make a suggest to improve dont be that way (-1) If you dont understand how to design a filter or compute f from this understanding. Dont delete without warning... again./ \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 21 '18 at 18:12

LP Filters must be defined as follows

Pass f-3db.
Stop band f and attenuation -x dB.
This defines the breakpoint and order of filter which is -6dB per octave (2xf)

Eg if you have a 1st order LP filter with -6dB per octave given attenuation -X dB at some f1, and breakpoint fo Then what f1?

If attenuation X is 20 log fo/f1 then \$f1 = 10^{X/20} * fo\$. ( using X=+)


Assumptions: There are many traditional and fundamental ways to go about this in which you could use various filter equations and sampling techniques to fine tune a result that would give you the wide band operation of your filter. I propose the below solution with the assumption of a relatively ideal low pas filter and simple hand-calculated method given that your filter behaves as a simple low-pass filter. In reality you would use much more complicated filter assumptions to estimate the filter as mentioned in other comments.


We essentially need to solve the low pass filter response, as @Bart mentioned.

$$H(s) = {{1} \over 1 + (sT)} $$

Where s is a frequency Laplace variable and T is the time constant. In this example it is enough to just do magnitude of this equation. We can also replace s with jw in this case so that we can look into the specific frequency response.

$$|H(jw)| = {{1} \over \sqrt{1 + (wT)^2}} $$

Knowns: $$Ain, Ao, fc$$

Where Ao is the output amplitude you know, Ain is the input amplitude you know and fc is the cut off frequency.


First we find T using the cut off frequency. We know that at cut off the amplitude is -3db. To get dB from the |H(jw)| we must square it and take the log and then multiply by 10.

$$10\log{|H(jw)|^2} = 10 \log{ {{1} \over \sqrt{1 + (wT)^2}^2} \ }$$ $$-3 = {10 \log{1} - 10 \log{( 1 + (wT)^2 )} \ }$$ $$-3 = - 10*\log{(1 + (wT)^2)}$$ $$0.3 =\log{(1 + (wT)^2)}$$

$$1 + (wT)^2 = 2$$

$$T = \sqrt{{1} \over 2 \pi fc} $$

Now that you have T, simply plug it back into your magnitude equation and solve for fc.

$$|H(jw)| = {{1} \over \sqrt{1 + (2 \pi fc T)^2}} $$ $${{Ao} \over {Ain}} = {{1} \over \sqrt{1 + (2 \pi fc T)^2}} $$

I'm not going to go through this part with variables since actual numbers will make it a lot faster, but I hope you get the picture.


Take the example of a simple 1 kohm and 10 nF low pass filter. It has a cut-off frequency of 15.915 kHz. This is the frequency where the output voltage has dropped to 0.7071 of the input voltage. If you plotted a few points: -

  • 100 Hz, Vout/Vin = 0.99998
  • 1000 Hz, Vout/Vin = 0.998

You will notice that there is barely a change between 100 Hz and 1000 Hz so if you have a ratio that is approximately 1 then you cannot hope to pin-point the frequency with any degree of accuracy. However, from 1 kHz onwards it gets easier: -

  • 2000 Hz, Vout/Vin = 0.992 and possibly discernible from 1000 Hz
  • 5000 Hz, Vout/Vin = 0.954 and this should be easily pin-pointed to a fairly precise point in the spectrum.

However, knowing the cut-off point doesn't tell you anything about the filter order and it's shape. Here is a picture of a 2nd order filter with different damping ratios and notice how there can be two points on the response that have the same transfer function magnitude: -

enter image description here

So, if you have a slight peak in the response you cannot say for sure what the point on the spectrum is by knowing only the ratio of output to input levels. The problem compounds when taking higher order filter shapes into account: -

enter image description here

Above picture taken from wiki.


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