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This lecture shows the frequency response of source follower.

As you can see, the transfer function has one pole, one zero and they are close to each other.

So from the phase plot, we can see that there is a phase bump here. I have heard that the phase bump causes signal distortion.

However, I can't find the explanation anywhere now.

If the pole and zero are far from each other, the output signal is also shifted in phase.

Could anyone explain that? Thank you.

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It really depends on what you mean by the term "distortion".

Yes, the bump in the phase response will cause the output waveform to be different from the input waveform. But the non-flat frequency response has that kind of effect, too. But both of these effects are linear and can easily be compensated for, restoring the original waveform.

A more useful definition of "distortion" is any nonlinear effect that cannot readily be compensated for, such a clipping, crossover distortion, slew rate limiting, etc. The effects you're talking about do not fall within this definition.

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  • \$\begingroup\$ Thank you. With "distortion" I thought that it is phase shift that can not be compensated for. I saw it somewhere but I can't find it now. With "both of these effects" do you mean that one with phase bump and the one when pole and zero are far from each other? \$\endgroup\$
    – emnha
    Commented Oct 19, 2016 at 5:29
  • \$\begingroup\$ No, I'm talking about the phase "bump" and the fact that amplitudes to the right of the bump are lower than to the left. How far apart the pole and zero are merely affects the magnitude of these two effects. \$\endgroup\$
    – Dave Tweed
    Commented Oct 19, 2016 at 11:28
  • \$\begingroup\$ Sorry I still don't understand what you mean by this "the bump in the phase response will cause the output waveform to be different from the input waveform". Output and input waveforms are always different at frequencies around pole and zero regardless they are close or not. \$\endgroup\$
    – emnha
    Commented Oct 19, 2016 at 18:11
  • \$\begingroup\$ It seems that we are talking past each other without reaching a mutual understanding. Perhaps you should more precisely define what you meant by "distortion" in your original question. \$\endgroup\$
    – Dave Tweed
    Commented Oct 19, 2016 at 18:27
  • \$\begingroup\$ With distortion, I think we have the same definition - output waveform is different from input waveform. However, it may be too general. I this case I mean the output waveform is of the same phase with input waveform. The magnitude is assumed to be the same. So if pole and zero are far from each other, the output and input waveforms are also different in phase. Not Not sure if that is also what you meant by distortion. \$\endgroup\$
    – emnha
    Commented Oct 20, 2016 at 3:00

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