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Subject: Re: Undersampling and its complement?
X-Newsreader: Microsoft Outlook Express 6.00.2600.0000
Date: Sat, 2 Nov 2002 22:25:35 -0800
NNTP-Posting-Date: Sat, 02 Nov 2002 22:22:57 MET
Organization: @Home Network
"Kevin Aylward" wrote in message
> "Jeroen" wrote in message
> > "James Meyer" wrote in message
> > news:email@example.com...
> > >
> > > I have to extract information from an 8 MHz signal. Nyquist says
> > > 16 MHz should be my minimum sampling frequency. However... if the
> > bandwidth of
> > > the signal I'm interested in is something like 10 KHz centered
> around 8
> > MHz,
> > > then I really only need to sample at 2 times 10 KHz, let the signal
> > "folded"
> > > or aliased down to 0 to 10 KHz and I wil not have lost any
> > That's
> > > undersampling.
> > >
> > > My question is.... is there anything similar regarding generating an
> > > MHz signal? Do I really need a 16 MHz or better DAC in order to
> > 8 MHz
> > > signals if the signal will not have a very wide bandwidth?
> > Yes you do.
> No you dont. The Nyquist Sampling theorem says that you have to sample
> at twice the *bandwidth* of the signal, not twice its frequency.
> >The bandwidth of the signal is 8Mhz, not the 10Khz you want
> No its not.
> > modulate it with. The 10kHz is modulation depth, not bandwidth.
> By assumption the poster said the BW was not large. The centre frequency
> is *not* the bandwidth of the signal. Depending on the modulation
> method, it might be +/- fmod, or approximately, say, 5 to 10 times the
> modulation frequency. For AM the BW is indeed 10Khz in this example.
But how do you create an 8MHz signal from an DAC if you feed only update it
I'm assuming the 8MHz a kind of a carrier signal and there are not special
filters after the DAC, just the standard LP (in this case 20KHz).
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