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This is a fundamental question about the idea of filters. As an example lets assume we have a passive LP filter with 1uF cap and a 100 Ohm resistor. Now lets assume our input signal is composed of addition of a 1V 4Hz signal and a 1V 400Hz.

Now if I plot the Bode diagram, the filter amplitude response is unity both at 400Hz and 4Hz.

But there will be a 15° phase difference between these signals at the output. We know that adding two sines with same amplitude but with different phases will yield a sine with less than twice the one of the two amplitude.

So even a filter is extremely flat in passband, the phase response seems will corrupt the output signal shape in time domain. I guess this will have no effect in frequency domain.

So what is that phenomenon called? And seems no filter has flat phase response in passband. Can we say this is important only if we are dealing the shape of output signal in time domain? Is there a rule of thumb for the maximum phase shift of a filter in passband?

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  • \$\begingroup\$ What filter - single-pole? what corner frequency? if you have a simple passive R/C filter, the phase is already noticeably shifted at even a roughly tenth of the corner frequency... \$\endgroup\$ – aschipfl Jun 6 '19 at 18:11
  • \$\begingroup\$ Im asking the consequences of large phase shifts in a flat passband. In my example I described a single pole LP filter with values. But it is just an example. When we should be concerned about the phase shifts in passband and how much phase shift is fine ect. \$\endgroup\$ – pnatk Jun 6 '19 at 18:18
  • \$\begingroup\$ The F3dB corner is 1,600Hz (10,000 rad/sec / 2*pi). At that frequency, expect 45 degree phaseshift. If you drop 2ocaves below 1,600Hz to 400Hz, I'd expect about 45/2*2 or 11 degrees. And for 4Hz, expect about 4/1600 or 1/400thdegree. \$\endgroup\$ – analogsystemsrf Jun 7 '19 at 2:02
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We know that adding two sines with same amplitude but with different phases will yield a sine with less than twice the one of the two amplitude.

You forgot to say "at the same frequency". Since your two components are not at the same frequency, this doesn't necessarily apply.

A delay that causes 15 degree change in the phase of the 400 Hz signal is only a 0.0015 degree change in the phase of the 4 Hz signal.

The overall amplitude of your signal won't be noticeably changed.

So even a filter is extremely flat in passband, the phase response seems will corrupt the output signal shape in time domain.

Yes, the output shape will be slightly changed.

You could plot the signals in a SPICE simulator, or Matlab or whatever to see the effect. Given the 2 decades difference in frequency between your signal components, you'll have trouble seeing the change if you don't know what to look for.

I guess this will have no effect in frequency domain.

If you consider the phase of the signal as part of its frequency domain representation then of course it will have an effect. It'll actually be much easier to see on the phase-vs-frequency plot than on a time domain plot.

And seems no filter has flat phase response in passband. Can we say this is important only if we are dealing the shape of output signal in time domain? Is there a rule of thumb for the maximum phase shift of a filter in passband?

Whether it's important and how much shift you can tolerate depends on the requirements for the application where you're using the filter.

For what it's worth, here's the plot. The upper plot shows a 400 Hz signal summed with a 4 Hz signal, and the lower plot shows the same thing, with the 400 Hz signal shifted by 15 degrees:

enter image description here

Good luck seeing the difference without zooming in to very carefully check the phase of the 400 Hz component at some distinguishable position relative to the 4 Hz component.

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  • \$\begingroup\$ Would an ideal LP filter phase freq plot be flat? I thought flat i.e same for all frequencies but here it is not i.stack.imgur.com/qsTeg.png \$\endgroup\$ – pnatk Jun 6 '19 at 18:27
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    \$\begingroup\$ @panicattack, That's an ideal filter defined to have a certain phase response (given by \$H(f)=\exp(-j2\pi f t_0)\$). You could define some other "ideal filter" to have an entirely different phase response, depending what you want to be ideal about your filter. \$\endgroup\$ – The Photon Jun 6 '19 at 18:30

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