3
\$\begingroup\$

A filter I have is showing to have a positive phase shift of ~30° at 1 kHz:

enter image description here enter image description here

Also, it seems that the output signal is actually advanced by 30°. enter image description here

But we all know that this is impossible because of causality. So what's really happening here, and how much degrees of phase shift should I consider?

\$\endgroup\$
9
  • 1
    \$\begingroup\$ It is V(vout), I'm editing this right away \$\endgroup\$
    – c.leblanc
    May 9 at 9:22
  • 1
    \$\begingroup\$ Note that Vin is a periodic signal. \$\endgroup\$
    – uriyabsc
    May 9 at 9:28
  • 1
    \$\begingroup\$ The current in a capacitor when connected to a sinewave voltage source leads by 90 degrees yet, do we worry about causality? \$\endgroup\$
    – Andy aka
    May 9 at 9:32
  • 1
    \$\begingroup\$ Trust the simulator and don't worry about causality (a red-herring). \$\endgroup\$
    – Andy aka
    May 9 at 9:38
  • 1
    \$\begingroup\$ I still don't know what's the phase shift I should consider to adjust my second signal on it with the all pass filter of my real circuit (and it's not precise enough to measure it manually there because of the shape of the signals) \$\endgroup\$
    – c.leblanc
    May 9 at 9:41

2 Answers 2

4
\$\begingroup\$

But we all know that this is impossible because of causality.

If I had a simple high-pass filter like this: -

enter image description here

I can do a spectral plot and see this: -

enter image description here

And, that spectral plot tells me that the output at 1 kHz is leading the input by about 30°. Yet, if I do a transient response for a newly applied sinewave of amplitude 0.5 volts p-p, I see this: -

enter image description here

The output doesn't begin before the input but, after a cycle it does end-up leading the input by 30°. However, causality isn't broken.

So what's really happening here, and how much degrees of phase shift should I consider?

It settles down to leading by approximately 30° but, in the wider scheme of things and, what you are trying to ultimately achieve, I can't answer because I don't know anything about these things.

\$\endgroup\$
4
  • \$\begingroup\$ Thank you. Here is what I'm trying to achieve: I have 2 signals, on one of them 2 filters are applied : 1) the filter I have posted with 30° phase shift and 2) a low pass filter with -110° phase shift. I would like to adjust the second signal on it, so can I just give it a -80° phase shift and expect it to work properly ? \$\endgroup\$
    – c.leblanc
    May 9 at 9:57
  • \$\begingroup\$ (1) I don't think this circuit relates to your posted question (other than you had it in mind when you asked a fairly simple question about causality). (2) I try to encourage folk to draw a circuit. I mean, even a simple RC filter (2 components with an input and an output) are drawn as schematics because, words can be ambiguous and don't impart anything like the information that a circuit diagram does. \$\endgroup\$
    – Andy aka
    May 9 at 10:00
  • \$\begingroup\$ It is actually working (the way I described just before) on a simple example. The phase correction is not correct in the real circuit though, so it must be another problem causing this, I will seek for other reasons. Thank you for the clarification. \$\endgroup\$
    – c.leblanc
    May 9 at 10:03
  • 1
    \$\begingroup\$ Maybe ask a new question but, show the circuit @c.leblanc \$\endgroup\$
    – Andy aka
    May 9 at 10:04
4
\$\begingroup\$

The 30 degrees phase lead doesn't violate the causality. That phase lead is true only for continuous sinewaves - signals which have never started, they only have existed since t = minus eternity. In practice you must start the input signal. If you record the result to a storage oscilloscope or calculate it with math or time domain simulator you'll see that the lead is at first zero, but approaches gradually 30 degrees.

OOPS, forget this. The answer is already given and written there better.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.