Filter Linear Phase Property for non-integer time delays

[Master1022]

Homework Statement

What does it mean when the signal is 'reproduced at the output'?

Relevant Equations

Delay

Hi,

I just have a quick question regarding the linear phase property of filters. It might be easier to provide some context before getting to the question, but feel free to skip to the bottom.

Consider a system input as a discrete sequence obtained by sampling at t = 0, T, . . . , kT from an underlying continuous-time function cos(\omega t + \phi). Thus the input is:
$$x[k] = cos(\omega kT + \phi)$$
and the output of the system, with a linear phase response, is:
$$y[k]=Acos(\omega kT +\phi + \gamma)$$
where A is the gain of the system. If the phase response \gamma is linear, i.e. \gamma = −\omega k_0 T, then
$$ y[k] = A cos(\omega(k − k_0 ) T + \phi) $$
That is, the signal waveform has been preserved at the output, with only a delay in time and modification in amplitude. "Notice that when k_0 is not an integer, then the time delay is not integer multiple of the sampling period T . In this case, the input samples are not exactly reproduced at the output."

Question: Why is this final statement in quotation marks true?

I understand that if k_0 is not an integer, then the time delay is not an integer multiple of the sampling period, but I don't understand how this means that the input samples are not exactly reproduced at the output? I feel like this answer should be very simple, but for some reason, this statement doesn't make any sense to me.

Many thanks in advance for any help.

 

[DaveE]

Homework Statement:: What does it mean when the signal is 'reproduced at the output'?
Relevant Equations:: Delay

Hi,

I just have a quick question regarding the linear phase property of filters. It might be easier to provide some context before getting to the question, but feel free to skip to the bottom.

Consider a system input as a discrete sequence obtained by sampling at t = 0, T, . . . , kT from an underlying continuous-time function cos(\omega t + \phi). Thus the input is:
$$x[k] = cos(\omega kT + \phi)$$
and the output of the system, with a linear phase response, is:
$$y[k]=Acos(\omega kT +\phi + \gamma)$$
where A is the gain of the system. If the phase response \gamma is linear, i.e. \gamma = −\omega k_0 T, then
$$ y[k] = A cos(\omega(k − k_0 ) T + \phi) $$
That is, the signal waveform has been preserved at the output, with only a delay in time and modification in amplitude. "Notice that when k_0 is not an integer, then the time delay is not integer multiple of the sampling period T . In this case, the input samples are not exactly reproduced at the output."

Question: Why is this final statement in quotation marks true?

I understand that if k_0 is not an integer, then the time delay is not an integer multiple of the sampling period, but I don't understand how this means that the input samples are not exactly reproduced at the output? I feel like this answer should be very simple, but for some reason, this statement doesn't make any sense to me.

Many thanks in advance for any help.

Consider the input samples x[k] = cos(ωkT+Φ), let's assume that your lucky and one of the samples yields x[k] = -1 because ωkT+Φ = π for some value of k.

For the samples to be delayed without change there must be a k that yields y[k] = A cos(ω(k-k_o)T+Φ) = -1, so ω(k-k_o)T+Φ = π given that ωkT+Φ = π. Which means k_o must be an integer. If the delay isn't a multiple of the sample period you may miss and get the nearest sample to -1 at -0.99, for example. It will still look sinusoidal, but not with exactly the same sample values.

 

[Master1022]

Consider the input samples x[k] = cos(ωkT+Φ), let's assume that your lucky and one of the samples yields x[k] = -1 because ωkT+Φ = π for some value of k.

For the samples to be delayed without change there must be a k that yields y[k] = A cos(ω(k-k_o)T+Φ) = -1, so ω(k-k_o)T+Φ = π given that ωkT+Φ = π. Which means k_o must be an integer. If the delay isn't a multiple of the sample period you may miss and get the nearest sample to -1 at -0.99, for example. It will still look sinusoidal, but not with exactly the same sample values.

Many thanks. Okay, this is starting to make more sense. Why do we want to preserve the input signal though? If we have a linear phase response which can arise from an FIR filter, wouldn't the filter have some sort of objective (e.g. LPF, HPF, etc.) that would alter the signal?

Apologies if this is very obvious and I do appreciate you taking the time to help me out.

 

[DaveE]

Why do we want to preserve the input signal though?

Yea, I don't know. I seriously doubt that any practicing engineer will assume that these systems stay in phase like that. If they do, it would be a key part of the system, like feedback in a phase-locked loop or such. Delays in real systems aren't as stable as the mathematical equations that show up in homework problems. Nearly every sampled data system that I've heard of operates below the Nyquist limit and ignores most of the dynamics at the sample rate.

Having said that, jitter at the sampling rate is a huge deal that lots of people care about. There is some impressive engineering in making low jitter samplers. Sampling jitter appears as noise in the baseband. Also, if it's really bad it will add phase shifts to the feedback systems (like variable latency in real time operating systems, for example).

I'm sure there are others with more DSP experience that can comment.