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A Basic Aliasing Question

Started by August 23, 2004
Here's something I don't remember about aliasing - someone please
verify whether or not this is correct. Assume the sample rate is
2000 Hz for the sake of illustration.

Conventional wisdom tells us that a 1200 Hz signal will look
like an 800 Hz signal due to aliasing. I say it may 
look the same on a spectrum analyzer (i.e., its magnitude
spectrum may look the same), but it may actually be different
since the negative and positive frequency components of such
a wave are swapped. Hence if the original wave was a cosine
wave, the aliased wave will still be cosine (i.e., will look
identical to the original) since the positive and negative
frequency components are identical (cos x = (e^(i*x) + e^(-i*x))/2).
However, if the wave is any other type, it will look different. 

Said perhaps more simply, xa(t) = -x(t). Since a cosine is 
even-symmetric, xa(t) = -x(t) = x(t). Not true for a sine
wave or sinusoids of other phases.

Trivial, but something I'd never thought of before. Or
am I wrong? 
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124
Randy Yates <randy.yates@sonyericsson.com> writes:
> [...] > Said perhaps more simply, xa(t) = -x(t). Since a cosine is > even-symmetric, xa(t) = -x(t) = x(t).
Woops. That should be xa(t) = x(-t), and xa(t) = x(-t) = x(t). -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124
Randy Yates wrote:

 > Here's something I don't remember about aliasing - someone please
 > verify whether or not this is correct. Assume the sample rate is
 > 2000 Hz for the sake of illustration.
 >
 > Conventional wisdom tells us that a 1200 Hz signal will look
 > like an 800 Hz signal due to aliasing. I say it may
 > look the same on a spectrum analyzer (i.e., its magnitude
 > spectrum may look the same), but it may actually be different
 > since the negative and positive frequency components of such
 > a wave are swapped. Hence if the original wave was a cosine
 > wave, the aliased wave will still be cosine (i.e., will look
 > identical to the original) since the positive and negative
 > frequency components are identical (cos x = (e^(i*x) + e^(-i*x))/2).
 > However, if the wave is any other type, it will look different.
 >
 > Said perhaps more simply, xa(t) = -x(t). Since a cosine is
 > even-symmetric, xa(t) = -x(t) = x(t). Not true for a sine
 > wave or sinusoids of other phases.
 >
 > Trivial, but something I'd never thought of before. Or
 > am I wrong?

> Randy Yates <randy.yates@sonyericsson.com> writes: > >>[...] >>Said perhaps more simply, xa(t) = -x(t). Since a cosine is >>even-symmetric, xa(t) = -x(t) = x(t). > > > Woops. That should be xa(t) = x(-t), and xa(t) = x(-t) = x(t).
First, lets explicitly assume that there's a low-pass reconstruction filter. Without that assumption, the samples could validly represent, say, 2800 Hz. I won't do the math. It's a chore to draw pictures like the ones in books that illustrate aliasing and I couldn't represent one with ASCII art anyway, so I'll ask you to imagine one with me. So we have this nice 1200 Hz solid (analog) cosine wave with dots on it that correspond to 2000 Hz sample points. Through the points, one can imagine an 800 Hz cosine drawn. (Let's show it as a light dashed line.) Have I adequately illustrated your hypothesized case? Good! The solid curve can represent a cosine or a sine, depending on where the origin is located. Shifting the phase of the sampling clock -- the dots common to the solid and dashed curves -- can change the relative positions of those curves.* Every positive analog peak can be the origin of the sampled cosine. The 800 Hz alias will be a cosine only with respect to special ones. The issue that probably puzzled you boils down to upper vs. lower sidebands. Having said that, I don't imagine it's a puzzle any more. Thanks for this opportunity to clarify my thoughts. Jerry ________________________________________ * Discussions like this illustrate the impropriety of comparing the phases of signals at different frequencies with explicitly defining what that means in each instance. -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Hi Jerry,

See comments below.

Jerry Avins <jya@ieee.org> writes:

> Randy Yates wrote: > > > Here's something I don't remember about aliasing - someone please > > verify whether or not this is correct. Assume the sample rate is > > 2000 Hz for the sake of illustration. > > > > Conventional wisdom tells us that a 1200 Hz signal will look > > like an 800 Hz signal due to aliasing. I say it may > > look the same on a spectrum analyzer (i.e., its magnitude > > spectrum may look the same), but it may actually be different > > since the negative and positive frequency components of such > > a wave are swapped. Hence if the original wave was a cosine > > wave, the aliased wave will still be cosine (i.e., will look > > identical to the original) since the positive and negative > > frequency components are identical (cos x = (e^(i*x) + e^(-i*x))/2). > > However, if the wave is any other type, it will look different. > > > > Said perhaps more simply, xa(t) = -x(t). Since a cosine is > > even-symmetric, xa(t) = -x(t) = x(t). Not true for a sine > > wave or sinusoids of other phases. > > > > Trivial, but something I'd never thought of before. Or > > am I wrong? > > > Randy Yates <randy.yates@sonyericsson.com> writes: > > > > >>[...] > >> Said perhaps more simply, xa(t) = -x(t). Since a cosine is > >> even-symmetric, xa(t) = -x(t) = x(t). > > > Woops. That should be xa(t) = x(-t), and xa(t) = x(-t) = x(t). > > > First, lets explicitly assume that there's a low-pass reconstruction > filter. Without that assumption, the samples could validly represent, > say, 2800 Hz. > > I won't do the math. It's a chore to draw pictures like the ones in > books that illustrate aliasing and I couldn't represent one with ASCII > art anyway, so I'll ask you to imagine one with me. > > So we have this nice 1200 Hz solid (analog) cosine wave with dots on it > that correspond to 2000 Hz sample points. Through the points, one can > imagine an 800 Hz cosine drawn. (Let's show it as a light dashed line.) > Have I adequately illustrated your hypothesized case? Good! > > The solid curve can represent a cosine or a sine, depending on where the > origin is located. Shifting the phase of the sampling clock -- the dots > common to the solid and dashed curves -- can change the relative > positions of those curves.* Every positive analog peak can be the origin > of the sampled cosine. The 800 Hz alias will be a cosine only with > respect to special ones.
That's true too. I was assuming that the sample clock is synchronized so that a sample is taken at continuous time t = 0.
> The issue that probably puzzled you boils down to upper vs. lower > sidebands.
Yeah, I thought that's what I said, if by sidebands you mean the "positive" and "negative" frequency components of the sinusoid. I don't know, though - trying to be careful here with words.
> Having said that, I don't imagine it's a puzzle any more.
Does the problem you see go away with the sample clock synchronized to t = 0? If so, then we're not seeing the same thing. -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124
"Randy Yates" <randy.yates@sonyericsson.com> wrote in message
news:xxpsmaermef.fsf@usrts005.corpusers.net...
> Here's something I don't remember about aliasing - someone please > verify whether or not this is correct. Assume the sample rate is > 2000 Hz for the sake of illustration. > > Conventional wisdom tells us that a 1200 Hz signal will look > like an 800 Hz signal due to aliasing. I say it may > look the same on a spectrum analyzer (i.e., its magnitude > spectrum may look the same), but it may actually be different > since the negative and positive frequency components of such > a wave are swapped. Hence if the original wave was a cosine > wave, the aliased wave will still be cosine (i.e., will look > identical to the original) since the positive and negative > frequency components are identical (cos x = (e^(i*x) + e^(-i*x))/2). > However, if the wave is any other type, it will look different. > > Said perhaps more simply, xa(t) = -x(t). Since a cosine is > even-symmetric, xa(t) = -x(t) = x(t). Not true for a sine > wave or sinusoids of other phases. > > Trivial, but something I'd never thought of before. Or > am I wrong?
I see nothing wrong here. If you extend your example to more than just tones, say a signal with some BW that is not symmetric around it's center, then your point becomes even more important. For designs where undersampling is used, it now becomes important to pick the right alias (not all aliases have the correct spectral orientation). If you don't have the flexibility to pick the right alias, then you have to adjust for that in the final digital quadrature mixing stage (which isn't too bad, since it's just a sign inversion on the complex exponential). I'm probably just restating what you already knew.... Cheers Bhaskar
> -- > Randy Yates > Sony Ericsson Mobile Communications > Research Triangle Park, NC, USA > randy.yates@sonyericsson.com, 919-472-1124
Randy Yates wrote:

   ...

> Does the problem you see go away with the sample clock synchronized > to t = 0? If so, then we're not seeing the same thing.
No, but agreeing to sample always at t=0 will simplify what I want to say. When sampling a unity-amplitude cosine, the first sample will be +1 and the next (if it isn't also +1) must be less. It's clear that the alias is a cosine too. (My daughter brought home straight-A report cards all through high school. I used to tell her, "This is terrible: the only way you can go is down.") When sampling a sine, the first sample is zero. The sign of the second sample depends in a simple way on the ratio of the analog signal to the sampling frequency. If the second sample falls in the first (or any odd) half cycle of the signal, the alias will be a positive sine. If it falls in the second (or any even) half cycle, the alias will be a negative sine. Simple. Pat. Right? Please excuse the unnecessary stuff I wrote before. The problem hadn't settled in yet. (I was always naive yesterday.) Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Bhaskar Thiagarajan wrote:
>[SNIP] For designs where undersampling > is used, it now becomes important to pick the right alias...
Does this mean that in some cases you can sample at less than Nyquist criterion? [ If true, just point out some good keywords for Google search. ]
"Bhaskar Thiagarajan" <bhaskart@my-deja.com> writes:

> "Randy Yates" <randy.yates@sonyericsson.com> wrote in message > news:xxpsmaermef.fsf@usrts005.corpusers.net... > > Here's something I don't remember about aliasing - someone please > > verify whether or not this is correct. Assume the sample rate is > > 2000 Hz for the sake of illustration. > > > > Conventional wisdom tells us that a 1200 Hz signal will look > > like an 800 Hz signal due to aliasing. I say it may > > look the same on a spectrum analyzer (i.e., its magnitude > > spectrum may look the same), but it may actually be different > > since the negative and positive frequency components of such > > a wave are swapped. Hence if the original wave was a cosine > > wave, the aliased wave will still be cosine (i.e., will look > > identical to the original) since the positive and negative > > frequency components are identical (cos x = (e^(i*x) + e^(-i*x))/2). > > However, if the wave is any other type, it will look different. > > > > Said perhaps more simply, xa(t) = -x(t). Since a cosine is > > even-symmetric, xa(t) = -x(t) = x(t). Not true for a sine > > wave or sinusoids of other phases. > > > > Trivial, but something I'd never thought of before. Or > > am I wrong? > > I see nothing wrong here. If you extend your example to more than just > tones, say a signal with some BW that is not symmetric around it's center, > then your point becomes even more important. For designs where undersampling > is used, it now becomes important to pick the right alias (not all aliases > have the correct spectral orientation).
Right. That's exactly my point (I think?!). The aliases of a frequency f at k*Fs - f have their spectra reversed; the ones at k*Fs + f do not. Both alias to the same frequency.
> If you don't have the flexibility to > pick the right alias, then you have to adjust for that in the final digital > quadrature mixing stage (which isn't too bad, since it's just a sign > inversion on the complex exponential).
I think I follow that.
> I'm probably just restating what you already knew....
The term "know" can be slippery! ... Thanks for the verification, Bhaskar! -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124
Richard Owlett <rowlett@atlascomm.net> writes:

> Bhaskar Thiagarajan wrote: > >[SNIP] For designs where undersampling > > is used, it now becomes important to pick the right alias... > > Does this mean that in some cases you can sample at less than Nyquist > criterion?
Yes.
> [ If true, just point out some good keywords for Google search. ]
Bandpass sampling. It's also in Rick's book. An easy way to think about it is like this: You know that an analog signal that is to be sampled is normally bandlimited between 0 and +Fs/2. The process of sampling replicates this frequency spectrum every Fs cycles. This can be generalized to sampling between k*Fs and k*Fs + Fs/2, where k is an integer. You still have a bandwidth of Fs/2, and the duplicates that come from sampling still won't overlap. -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124
"Randy Yates" <randy.yates@sonyericsson.com> wrote in message
news:xxp3c2dsoig.fsf@usrts005.corpusers.net...
> Richard Owlett <rowlett@atlascomm.net> writes: > > > Bhaskar Thiagarajan wrote: > > >[SNIP] For designs where undersampling > > > is used, it now becomes important to pick the right alias... > > > > Does this mean that in some cases you can sample at less than Nyquist > > criterion? > > Yes.
I guess that depends on how you define the "Nyquist criterion". If you define it to require sampling at more than twice the highest frequency component in the original signal, then the answer is yes. But if you define the Nyquist criterion to require sampling at more than twice the bandwidth of the original signal, than no. I'm not sure what the 'official' definition is, but here is some relevant info from http://www.wordiq.com/definition/Nyquist_theorem: [It has to be noted that even if the concept of "twice the highest frequency" is the more commonly used idea, it is not absolute. In fact the theorem stands for "twice the bandwidth", which is totally different. Bandwidth is related with the range between the first frequency and the last frequency that represent the signal. Bandwidth and highest frequency are identical only in baseband signals, that is, those that go very nearly down to DC. This concept led to what is called undersampling, that is very used in software-defined radio.]
> > [ If true, just point out some good keywords for Google search. ] > > Bandpass sampling. It's also in Rick's book.
See also the link above.