I am trying to make a fourth order passive RC low pass filter. I simulate it in Multisim but find that the filter creates THD. And the higher the order of the filter, the larger the THD. I can`t figure out why. Here is the schematic:


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ How are you measuring THD? Yes, the filter will change the waveform, since it affects different frequencies differently, but this is not usually classified as "distortion", a term normally used only for nonlinear effects. \$\endgroup\$ – Dave Tweed Apr 9 '14 at 13:21
  • \$\begingroup\$ oh, I found that maybe it is due to the mistakes of the Multisim or my setting. I use the "Multisim" to simulate it before. \$\endgroup\$ – billyzhao Apr 9 '14 at 13:53
  • \$\begingroup\$ But one more question, could the filter cause distortion if I just give if a 1kHz ac voltage source? \$\endgroup\$ – billyzhao Apr 9 '14 at 13:54
  • \$\begingroup\$ No. The output will also be a sinewave. The amplitude and phase of the output will be different from the input, but neither of these effects is considered "distortion". \$\endgroup\$ – Dave Tweed Apr 9 '14 at 13:59
  • \$\begingroup\$ Thank you,Dave. Do you know how to select the value of R and C?Would it be suitable to select the same value just like my schematic? \$\endgroup\$ – billyzhao Apr 9 '14 at 14:24

I'll paraphrase what Dave Tweed says.

If you input a sinewave then there is no misshaping of the signal just an attenuation that is dependent on how high the input frequency is. If you input a square wave, the output will be misshaped because the harmonics (that comprise the square wave) will be attenuated at significantly different levels and you'll likely start to see an output that more resembles a sine wave or triangular wave.

Here's what a 1kHz square wave looks like when it is filtered with the same circuit as the OP has in his question: -

enter image description here

The input is 1Vp-p centred on 3V dc and as you can see the square wave is nearly triangular.

  • 4
    \$\begingroup\$ But this doesn't mean that new harmonics have been introduced, does it? Visually, the output is different from the input, but the square wave is the sum of sine waves of different frequencies and amplitudes. A linear filter will reduce the amplitude at some frequencies, but it should never increase the amplitude of one of those sine waves or introduce new sine wave frequencies, should it? \$\endgroup\$ – Joe Hass Apr 9 '14 at 13:58
  • \$\begingroup\$ This is still linear. If \$e_1,...,e_n...\$ is a basis of some (infinite) dimensional vector space and \$a_1,....,a_n, ..\$ are some fixed coefficients then \$L(\sum_n c_n e_n) = \sum_n a_n c_n e_n\$ is linear. \$\endgroup\$ – SomeEE Apr 9 '14 at 15:04
  • \$\begingroup\$ @JoeHass New harmonics have not been introduced. \$\endgroup\$ – Andy aka Apr 9 '14 at 15:07
  • \$\begingroup\$ So, following along with Dave Tweed, what you are describing is not considered distortion. The linear filter described by the OP should not, therefore, cause increased THD. That's the point...the effect you show visually is not considered distortion. I'm just trying to make a connection between your answer and the original question. \$\endgroup\$ – Joe Hass Apr 9 '14 at 15:21
  • \$\begingroup\$ @joehass I consider it as distortion sure BUT the phrase "total harmonic distortion" is generally reserved when testing audio amps with sinwaves. \$\endgroup\$ – Andy aka Apr 9 '14 at 16:21

To answer the question in your comment: in order to determine what resistor values you need, you actually need to perform the math to obtain the transfer function of your circuit. The most I've seen books go up to for RC ladder filters is second-order. Beyond that you'll spend a lot of time working through the math to obtain the transfer function. One of my classmates had to work out the math for a 6th order RC ladder and it took him more than a full days work to obtain the transfer function so that he could determine what values of R and C should be used. The reason the higher order filters take so much math, time and effort is due to the fact that each stage affects the previous stages. That's worked around using active filters where input impedance to each stage is high and output impedance is low.

To get you started in the maths of non-active filters here's a starting point: http://www.johnhearfield.com/RC/RC4.htm


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