Aliasing Phase Shift

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How do you determine the phase shift of an aliased signal?
 

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  • #2
Baluncore
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You want to measure the phase relative to what ?
An alias appears at the wrong frequency because the frequency has been reflected or folded one or more times from the ends of the 2π wide spectrum.
 
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  • #3
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You want to measure the phase relative to what ?
An alias appears at the wrong frequency because the frequency has been reflected or folded one or more times from the ends of the 2π wide spectrum.
My apologies, I think was referring to the perceived signal. Would there be a phase shift between that and the signal being sample?
 
  • #4
sophiecentaur
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My apologies, I think was referring to the perceived signal. Would there be a phase shift between that and the signal being sample?

You can only really talk about phase between signals if they are the same frequency. If they are different frequencies then the phase will be changing by (2π times the frequency difference), every second.
Tell us the context of your question and we may find out what and how the actual misconception is arising.
 
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  • #5
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You can only really talk about phase between signals if they are the same frequency. If they are different frequencies then the phase will be changing by (2π times the frequency difference), every second.
Tell us the context of your question and we may find out what and how the actual misconception is arising.
I guess I am a little confused here. I am studying for my FE Exam. I am stuck on problem 8.1c and d. I get that the highest frequency being sample is 1500 and the sampling frequency of 2000 is not high enough. It needs to be at least 3000. That I am good with. It's the recovered signal that I am stuck. I think they are asking for the reconstructed signal. It wouldn't have a frequency of 1500 but something much lower at 500. I am lost on the phase shift here.
 

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  • #6
sophiecentaur
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I am lost on the phase shift here.
If you alter the phase of the sampled signal then the phase of the reconstructed signal will be affected but relative to what?
I think what they mean is to compare the changes in phase of input signal and output signals relative 'before and after' the phase is altered. You take the reference phases of the input and output signals as zero and then change the phase of the input signal. Advancing the phase of the input signal can either advance or retard the phase of the output product(s). Imagine locking a 500kHz oscillator to the alias - as reference) and see what happens when you nudge the input signal phase.

But you don't have to bother with phases if you think in terms of a small frequency shift in the input frequency and what will happen to the alias. Input f goes up towards the sampling frequency and lower alias f goes down.
 
  • #7
Baluncore
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The answer to 9.1D does not require consideration of phase, it only requires computation of the alias frequency. But frequency and the phase can be solved most easily by considering that when you sample a signal you are performing a frequency mixing operation, and you can treat the sampling frequency, Fs, as a harmonic-rich local oscillator.

Frequencies f, in the range 0 to Fs/2 are analysed normally as being at 0 to Fs/2. Frequencies in the range Fs/2 to Fs map to an alias in the reversed range Fs/2 to 0. The alias frequency will be Falias = Fs – f, but the sampled signal will be the phase conjugate, which is effectively backwards in time, so can be seen as having negative phase.

The general equation of a harmonic mixer is; Falias = f – n·Fs;
https://en.wikipedia.org/wiki/Harmonic_mixer
https://en.wikipedia.org/wiki/Undersampling
 
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  • #8
sophiecentaur
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The answer to 9.1D does not require consideration of phase, it only requires computation of the alias frequency.
Starting with phase is certainly the hard way round, imo but, as frequency is the time derivative of phase, it should be possible to go at it either way. The 'trig identities' allow you to find the passage of a phase change θ on the way through. I guess the examiners had their own priorities about this.
 
  • #9
Baluncore
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Starting with phase is certainly the hard way round, imo but, as frequency is the time derivative of phase, it should be possible to go at it either way.
Either way, except that there is a constant of integration to be found.
 
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  • #10
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The answer to 9.1D does not require consideration of phase, it only requires computation of the alias frequency. But frequency and the phase can be solved most easily by considering that when you sample a signal you are performing a frequency mixing operation, and you can treat the sampling frequency, Fs, as a harmonic-rich local oscillator.

Frequencies f, in the range 0 to Fs/2 are analysed normally as being at 0 to Fs/2. Frequencies in the range Fs/2 to Fs map to an alias in the reversed range Fs/2 to 0. The alias frequency will be Falias = Fs – f, but the sampled signal will be the phase conjugate, which is effectively backwards in time, so can be seen as having negative phase.

The general equation of a harmonic mixer is; Falias = f – n·Fs;
https://en.wikipedia.org/wiki/Harmonic_mixer
https://en.wikipedia.org/wiki/Undersampling
Phase conjugate? Negative phase? Wow, okay, that does seem to make sense. The problem wasn’t asking me for a specific angle but whether it was positive or negative as compared to the frequency being sampled. Thanks!
 
  • #11
sophiecentaur
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Either way, except that there is a constant of integration to be found.
Or just ignored on many occasions.
 

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