[Geezer]

since you can't sample the step with perfect resolution.

Ah, right! Doesn't that boil down to giving it a certain amount of slope that doesn't really exist, in which case a step function would really be a ramp?

[JP]

since you can't sample the step with perfect resolution.

Ah, right! Doesn't that boil down to giving it a certain amount of slope that doesn't really exist, in which case a step function would really be a ramp?

Something like that, I think.  When you actually sample the function, you have to do a discrete Fourier transform, which isn't quite the same as the continuous integral.  If the function has sharp features that you can't resolve perfectly with your sampling, it throws off the result a little bit.

[Geezer]

Something like that, I think.  When you actually sample the function, you have to do a discrete Fourier transform, which isn't quite the same as the continuous integral.  If the function has sharp features that you can't resolve perfectly with your sampling, it throws off the result a little bit.

Right, but what about a "vertical" section of the signal? If it really is vertical, there is no phase angle between the start and end of the vertical section.

[JP]

Something like that, I think.  When you actually sample the function, you have to do a discrete Fourier transform, which isn't quite the same as the continuous integral.  If the function has sharp features that you can't resolve perfectly with your sampling, it throws off the result a little bit.

Right, but what about a "vertical" section of the signal? If it really is vertical, there is no phase angle between the start and end of the vertical section.

The way I'm picturing this, the signal is sin(t) to the left of the discontinuity and sin(t+phi) to the right.  The phase of the sine shifts by phi at the discontinuity.

[Geezer]

The way I'm picturing this, the signal is sin(t) to the left of the discontinuity and sin(t+phi) to the right.  The phase of the sine shifts by phi at the discontinuity.

Ewe! You mean a DC section. I'm not sure that constitutes a phase shift exactly. I think you have to instantaneously "jump" from one sine wave to another sine wave of the same frequency but different phase, and that results in an "infinite" slope rather than no slope.

[JP]

Maybe we're at odds with each other on terminology.  Usually when I see DC, I think of a constant added to the signal.  I would tell you that sin(t)+C has a DC component of C.

When I think of phase of a sine function, I think of whatever's in the parentheses.  For sin(t), the phase is t, which increases linearly and continuously with time.  If it jumps from sin(t) to sin(t+phi) suddenly, that's a discontinuous change in phase, which I would probably call a phase jump or phase discontinuity, although I assumed (perhaps wrongly) that's what we meant by the term phase shift.

At any rate, I think I get your point.  Your sampling density can only put a lower limit on the slope.  If you know it was sin(t) on one side and sin(t+phi) on the other, you don't know if it took the entire interval to climb or if it jumped in a fraction of the interval.  Your discrete Fourier transform will look identical whether it did either of those, you you've lost information.  (I believe the usual discrete Fourier transform assumes linear changes in phase across intervals.)

[Geezer]

I'm pretty sure an instantaneous phase change constitutes a vertical "jump". If the signal had a horizontal section and you analysed only that section, there would be no frequency components at all, whereas the vertical jump tends towards infinite frequencies.

EDIT: Of course, if you do the "jump" at precisely the right time, the signal would just have a sharp bend in it.

[JP]

I'm pretty sure an instantaneous phase change constitutes a vertical "jump". If the signal had a horizontal section and you analysed only that section, there would be no frequency components at all, whereas the vertical jump tends towards infinite frequencies.

EDIT: Of course, if you do the "jump" at precisely the right time, the signal would just have a sharp bend in it.

Yep.  I agree.  The DC term I was talking about is a flat, horizontal component of the signal, or a flat, horizontal component added to a changing signal, which just shifts it's amplitude globally by a certain amount.  A discontinuous jump in the signal would have infinite slope. 

But if you sampled the signal, you'd only be able to say that it jumped between samples by a certain amount.  You couldn't tell if that was an instantaneous jump with infinite slope, or a steep but linear rise.  For that reason, sampling and doing a discrete version of the Fourier transform can lose information if you aren't careful about sampling finely enough.  (And you can never sample vertical segments densely enough.) 

By the way, one of the basic rules of determining sampling density is to use Nyquist sampling:
http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem ,
which essentially says that you have to sample your signal more finely if you want to accurately recover high frequency components.

[Geezer]

Great! Well, I think this cow has been more than adequately flogged.

[JP]

Great! Well, I think this cow has been more than adequately flogged.

We went from cow farts to Fourier transforms in just 6 pages of posts!