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Why and how does this active band pass filter change a triangle form to a sine form? I know that the band pass filter consists of a low pass and a high pass , so it cancels out some frequencies , but I don't understand graphically how this could happen. enter image description here

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  • \$\begingroup\$ Graphically? Are you trying to understand this looking at this fancy animation? \$\endgroup\$ – Eugene Sh. Nov 30 '17 at 16:42
  • \$\begingroup\$ First, if you picked the cut-off correctly, you don't need the high-pass part of the band-pass filter to convert triangle to sine. Second, the output will only be approximately a sine, and you can make the approximation better by using a higher-order LPF. \$\endgroup\$ – The Photon Nov 30 '17 at 17:01
  • \$\begingroup\$ @ThePhoton I am specifically asked to use this filter as is \$\endgroup\$ – maverick98 Nov 30 '17 at 17:04
  • \$\begingroup\$ @Maverick98, Okay, but you can get rid of the 10 uF cap on the input and understand why the simpler circuit converts triangle to sine. Then put it back in to satisfy whoever dictates that you must use it. \$\endgroup\$ – The Photon Nov 30 '17 at 17:09
  • \$\begingroup\$ @ThePhoton ok I'll try it :) \$\endgroup\$ – maverick98 Nov 30 '17 at 17:11
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Any repeating wave form - including a triangle wave - can be represented as a sum of sine waves. Look up "Fourier analysis" for details. These sine waves consist of a "fundamental", the lowest frequency, plus a series of "harmonics" at higher frequencies.

Applying a bandpass filter to the waveform just selects the frequency you want, and discards all the others. Pick the right filter and, in theory at least, you'd get a perfect sine wave out.

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