Sampling question

The sampling theorem tells us that we must sample at least twice the
bandwidth of a signal. When the signal is centered around dc to some
max freq say Fmax than we can say Fs>=2Fmax. However, when the signal
is centred around some other higher frequency (say Fc), how is
bandwidth defined?

is it from Fc (the center freq) to the max value or from Fc-minfreq
to Fc + maxfreq ? If you apply teh same rules it should be Fc to Fc +
maxfreq only if you consider it to be a shifter version of one
centered around dc?

eg suppose we have a signal centered on 1GHz that goes up to 1GHz +
5kHz and down to 1GHz-5kHz. Is the bandwidth 5kHz or 10 kHz and
therefore what do I sample at?

kronecker@yahoo.co.uk writes:
> [...]
> However, when the signal is centred around some other higher frequency
> (say Fc), how is bandwidth defined?

It is b2 - b1, where S(f) = 0 when f > b2 and when f < b1.
--
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%%% 919-577-9882 % that's the way it goes..."
%%%% <yates@ieee.org> % Getting To The Point', *Balance of Power*, ELO
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On Sep 7, 10:28 pm, kronec...@yahoo.co.uk wrote:
> The sampling theorem tells us that we must sample at least twice the
> bandwidth of a signal.

actually, i think it's *more* than twice. we cannot, from one number,
represent *both* the phase and the magnitude of the sinusoidal
component at the Nyquist frequency.

> When the signal is centered around dc to some
> max freq say Fmax than we can say Fs>=2Fmax.

no, just Fs > 2*Fmax .

> However, when the signal
> is centred around some other higher frequency (say Fc), how is
> bandwidth defined?
>
> is it from Fc (the center freq) to the max value or from Fc-minfreq
> to Fc + maxfreq ? If you apply teh same rules it should be Fc to Fc +
> maxfreq only if you consider it to be a shifter version of one
> centered around dc?

it's a different definition of bandwidth. remember, a real signal
that is centered at Fc has a mirror component at -Fc. so comparing
bandwidths centered only at +Fc to ones centered at DC, you should
only compare the centered at Fc to one centered at maxfreq/2 with
width on both sides also maxfreq/2. they would both have bandwidth of
maxfreq, if defined consistently.

> eg suppose we have a signal centered on 1GHz that goes up to 1GHz +
> 5kHz and down to 1GHz-5kHz. Is the bandwidth 5kHz or 10 kHz and
> therefore what do I sample at?

if there was no inherent symmetry between the upper half and lower
half, it's 10 kHz. if there is a mirror symmetry of the upper and
lower half, it's 5 kHz.

r b-j

On Sep 8, 3:27 pm, robert bristow-johnson <r...@audioimagination.com>
wrote:
> On Sep 7, 10:28 pm, kronec...@yahoo.co.uk wrote:
>
> > The sampling theorem tells us that we must sample at least twice the
> > bandwidth of a signal.
>
> actually, i think it's *more* than twice. we cannot, from one number,
> represent *both* the phase and the magnitude of the sinusoidal
> component at the Nyquist frequency.
>
> > When the signal is centered around dc to some
> > max freq say Fmax than we can say Fs>=2Fmax.
>
> no, just Fs > 2*Fmax .
>
> > However, when the signal
> > is centred around some other higher frequency (say Fc), how is
> > bandwidth defined?
>
> > is it from Fc (the center freq) to the max value or from Fc-minfreq
> > to Fc + maxfreq ? If you apply teh same rules it should be Fc to Fc +
> > maxfreq only if you consider it to be a shifter version of one
> > centered around dc?
>
> it's a different definition of bandwidth. remember, a real signal
> that is centered at Fc has a mirror component at -Fc. so comparing
> bandwidths centered only at +Fc to ones centered at DC, you should
> only compare the centered at Fc to one centered at maxfreq/2 with
> width on both sides also maxfreq/2. they would both have bandwidth of
> maxfreq, if defined consistently.
>
> > eg suppose we have a signal centered on 1GHz that goes up to 1GHz +
> > 5kHz and down to 1GHz-5kHz. Is the bandwidth 5kHz or 10 kHz and
> > therefore what do I sample at?
>
> if there was no inherent symmetry between the upper half and lower
> half, it's 10 kHz. if there is a mirror symmetry of the upper and
> lower half, it's 5 kHz.
>
> r b-j

So if I had a complex signal with a bandwidth from maxfreq to minfreq
(ie no mirror image frequencies) I would need to sample at (for the
same example)
greater than 2X 10kHz since 10Khz is the bandwidth?

On Sun, 07 Sep 2008 19:28:58 -0700, kronecker wrote:

> The sampling theorem tells us that we must sample at least twice the
> bandwidth of a signal. When the signal is centered around dc to some max
> freq say Fmax than we can say Fs>=2Fmax. However, when the signal is
> centred around some other higher frequency (say Fc), how is bandwidth
> defined?
>
> is it from Fc (the center freq) to the max value or from Fc-minfreq to
> Fc + maxfreq ? If you apply teh same rules it should be Fc to Fc +
> maxfreq only if you consider it to be a shifter version of one centered
> around dc?
>
> eg suppose we have a signal centered on 1GHz that goes up to 1GHz + 5kHz
> and down to 1GHz-5kHz. Is the bandwidth 5kHz or 10 kHz and therefore
> what do I sample at?

See if this helps, I have a section on bandpass sampling in it: http://
http://www.wescottdesign.com/article.../sampling.html.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

On Sep 8, 7:28*am, kronec...@yahoo.co.uk wrote:
> The sampling theorem tells us that we must sample at least twice the
> bandwidth of a signal. When the signal is centered around dc to some
> max freq say Fmax than we can say Fs>=2Fmax. However, when the signal
> is centred around some other higher frequency (say Fc), how is
> bandwidth defined?
>
> is it from Fc (the center freq) *to the max value or from Fc-minfreq
> to Fc + maxfreq ? If you apply teh same rules it should be Fc to Fc +
> maxfreq only if you consider it to be a shifter version of one
> centered around dc?
>
> eg suppose we have a signal centered on 1GHz that goes up to 1GHz +
> 5kHz and down to 1GHz-5kHz. Is the bandwidth 5kHz or 10 kHz and
> therefore what do I sample at?

Bandwidth is the range overwhich signal exists.
For the dc centered one it starts from 0 to maximum frequency.
For Fc centered signal we have to consider the frequency ranges two
sides of the center frequency.
Suppose your signal is centered but it is one sided then it is like dc
centered and bandwidth is (Fc+fmax- Fc).So your signal is two sided
you have to consider both sides for bandwidth calculation (Fc+famx-Fc
+fcmin).
For your example the bandwidth is 10khz
so your sampling frequency should be >=2*10khz

On Sep 8, 12:19*am, kronec...@yahoo.co.uk wrote:
>
> So if I had a complex signal with a bandwidth from maxfreq to minfreq
> (ie no mirror image frequencies)

if there is *no* mirror image frequencies (at the negative
frequencies), then your complex signal is a special kind of complex
signal that we call the "analytic signal" and the imaginary part is
the Hilbert transform of the real part and thus doesn't offer new
information.

> I would need to sample at (for the
> same example)
> greater than 2X 10kHz since 10Khz is the bandwidth?

assuming that, with some analog process, you heterodyne this 10 kHz
wide complex signal down with the left edge at DC. sampling will
cause the spectrum to be duplicated, shifted by multiples of your
sampling rate (a milli-smidgen more than 10 kHz), overlapped and
added. you don't want non-zero parts to overlap and add. once two
numbers are added and you know only the sum, it's hard to split it
back up and recover the knowledge of what the two numbers were.

r b-j

>The sampling theorem tells us that we must sample at least twice the
>bandwidth of a signal.

In this statement, "bandwidth" means the highest frequency in the signal
measured from DC.


> suppose we have a signal centered on 1GHz that goes up to 1GHz +
>5kHz and down to 1GHz-5kHz. ... what do I sample at?

The highest frequency in this signal is 1 GHz + 5 kHz. To comply with the
sampling theorem you must sample at twice this rate.


[I should mention that there is a technique where you can sample such a
narrow-band signal at a lower rate, and use the aliasing to downconvert the
signal to baseband in the process. But I doubt this is the question you
have.]

Regards,
Steve

kronecker@yahoo.co.uk wrote:
> On Sep 8, 3:27 pm, robert bristow-johnson <r...@audioimagination.com>
> wrote:
>> On Sep 7, 10:28 pm, kronec...@yahoo.co.uk wrote:
>>
>>> The sampling theorem tells us that we must sample at least twice the
>>> bandwidth of a signal.
>> actually, i think it's *more* than twice. we cannot, from one number,
>> represent *both* the phase and the magnitude of the sinusoidal
>> component at the Nyquist frequency.
>>
>>> When the signal is centered around dc to some
>>> max freq say Fmax than we can say Fs>=2Fmax.
>> no, just Fs > 2*Fmax .
>>
>>> However, when the signal
>>> is centred around some other higher frequency (say Fc), how is
>>> bandwidth defined?
>>> is it from Fc (the center freq) to the max value or from Fc-minfreq
>>> to Fc + maxfreq ? If you apply teh same rules it should be Fc to Fc +
>>> maxfreq only if you consider it to be a shifter version of one
>>> centered around dc?
>> it's a different definition of bandwidth. remember, a real signal
>> that is centered at Fc has a mirror component at -Fc. so comparing
>> bandwidths centered only at +Fc to ones centered at DC, you should
>> only compare the centered at Fc to one centered at maxfreq/2 with
>> width on both sides also maxfreq/2. they would both have bandwidth of
>> maxfreq, if defined consistently.
>>
>>> eg suppose we have a signal centered on 1GHz that goes up to 1GHz +
>>> 5kHz and down to 1GHz-5kHz. Is the bandwidth 5kHz or 10 kHz and
>>> therefore what do I sample at?
>> if there was no inherent symmetry between the upper half and lower
>> half, it's 10 kHz. if there is a mirror symmetry of the upper and
>> lower half, it's 5 kHz.
>>
>> r b-j
>
> So if I had a complex signal with a bandwidth from maxfreq to minfreq
> (ie no mirror image frequencies) I would need to sample at (for the
> same example)
> greater than 2X 10kHz since 10Khz is the bandwidth?

No. 10 KHz would be enough for _each_ of the I and Q parts.

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

On Sep 9, 4:33 am, "SteveSmith" <Steve.Smi...@SpectrumSDI.com> wrote:
> >The sampling theorem tells us that we must sample at least twice the
> >bandwidth of a signal.
>
> In this statement, "bandwidth" means the highest frequency in the signal
> measured from DC.
>
> > suppose we have a signal centered on 1GHz that goes up to 1GHz +
> >5kHz and down to 1GHz-5kHz. ... what do I sample at?
>
> The highest frequency in this signal is 1 GHz + 5 kHz. To comply with the
> sampling theorem you must sample at twice this rate.
>
> [I should mention that there is a technique where you can sample such a
> narrow-band signal at a lower rate, and use the aliasing to downconvert the
> signal to baseband in the process. But I doubt this is the question you
> have.]
>
> Regards,
> Steve

Actually it was the question since I don't classify sampling at (more
than) twice the bandwidth of a bandpass signal as sub-sampling. This
is Nyquists theorem and not sampling at twie the highest frequency.