Looking at the phase Bode plot for the output of a filter as shown below:
1-) Taking the point -460° as an example, can we say that this is equivalent to +100°?
2-) Does that mean the output signal is leading the input signal by 100°?
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4\$\begingroup\$ -460 degrees is equivalent to -100 degrees, not +100 degrees. \$\endgroup\$– Math Keeps Me BusyCommented Dec 3, 2020 at 2:06
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1\$\begingroup\$ Generally speaking, you can add or subtract multiples of 360 deg without affecting the meaning. In the case of -460 deg, adding 1x+360 gives -100 deg. \$\endgroup\$– AJNCommented Dec 3, 2020 at 2:47
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\$\begingroup\$ It depends on what information you're seeking from the phase angle. If it's determining the phase margin, for example, then the absolute phase angle is important. If you're determining gain, it's not. \$\endgroup\$– ChuCommented Dec 3, 2020 at 9:24
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1 Answer
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At 0.5 Hz, the output signal is lagging the input signal by 460 degrees and this tells you a lot about what is going on in terms of overall signal delay: -
If you viewed the waveforms associated with the construction of this bode plot, it would look like (on the face of it) that the output is lagging the input by 100 degrees. In other words a lag of 460 degrees appears to be a lag of 100 degrees. And, that may be acceptable to you but, it disguises the fact that the system or filter producing this lag is actually introducing a delay equivalent to 460 degrees at 0.5 Hz.
Given that 0.5 Hz has a period of 2 seconds, the overall input to output delay is in fact 2.556 seconds. And, this may be more important to know than assuming the delay is only 0.556 seconds (100 degree lag) in terms of how stable the system or filter might be.
Taking the point -460° as an example, can we say that this is equivalent to +100°?
No, it is equivalent to -100° but, it's only equivalent if you are ignoring that the overall delay is equivalent to a phase lag of 460°.
