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Answer by periblepsis for Theoretically, if we send a square wave, of frequency equal to the pole frequency, through an RC lowpass filter is the output shifted by -45 degrees?

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  • Wrong: "with harmonics at integer multiples". The square wave is every odd integer multiple, starting at 1. Not at all integer multiples.
  • Right: "main component would be phase shifted by -45 degrees". An RC lowpass set with its \$-3.0103\:\text{dB}\$ at the \$\times 1\$ frequency of the square wave will be \$-45^\circ\$.
  • Roughly right: "lesser harmonics would be phased shifted by differing amounts but these higher frequency components would be attenuated more heavily from the output signal". At \$\times 3\$ it will be \$-10\:\text{dB}\$ and \$\approx -71.6^\circ\$. At \$\times 5\$ it will be \$-14.15\:\text{dB}\$ and \$\approx -78.7^\circ\$.
  • Wrong: "the output approximately be another square wave phased shifted by -45 degrees". With the RC is set for \$-3.0103\:\text{dB}\$ at the primary square wave frequency, you'd expect to see a mostly RC-decay curve during each half-cycle. (Peak to peak would be about 90%, at a guess.)

Suppose the square wave is \$1\:\text{kHz}\$ and that you are using \$R=1.59154943\:\text{k}\Omega\$ and \$C=100\:\text{nF}\$. The RC time constant is \$\tau=159.154943\:\mu\text{s}\$ and the period between half-cycles of the square wave is \$500\:\mu\text{s}\$, or \$\pi\times\tau\$ put another way.

Here's a nice table that google threw at me:

enter image description here

At \$3\times\tau\$ you'd expect about 95% of the signal. \$\pi\gt 3\$, so better than that in this case.

In short, this should look "mostly" like an RC decay curve, going up 95% and down 95%, or at least 90% peak to peak, I think.


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