"Eduard Kais" <who_me@hotmail.com> wrote in message news:62pp8rF24la75U1@mid.individual.net...> "Randy Yates" <yates@ieee.org> wrote in message > news:m3prugacgc.fsf@ieee.org... >> "Eduard Kais" <who_me@hotmail.com> writes: >> >>> "Jerry Avins" <jya@ieee.org> wrote in message >>> news:q-2dnXhg36t_f1vanZ2dnUVZ_qjinZ2d@rcn.net... >>>> Eduard Kais wrote: >>>>> "Andre" <alodwig@gmx.de> wrote in message >>>>> news:fq694f$i7a$02$1@news.t-online.com... >>>>>> Yes. >>>>>> >>>>>> Eduard Kais wrote: >>>>>>> Hello >>>>>>> >>>>>>> If I have a 1MHz signal, and I sample at 100kHz, then I get a DC >>>>>>> output. True? >>>>>>> >>>>>>> If I sample a 1,000,001Hz signal at 100kHz, what do I get as an >>>>>>> output? >>>>>>> Is it 1Hz? >>>>> >>>>> I guess my next question(s) is(are): >>>>> If I sample at 200kHz, is the output still 1Hz? >>>>> At what point does the output changed from being 1Hz? >>>> >>>> When you sample at 101 KHz, e.g. You have a specific idea in mind. When >>>> you figure out how to ask it unambiguously, you will already have the >>>> answer. >>> >>> No doubt you're right. >>> A dentist by trade, I'm going to have to do some more reading up on math >>> and >>> DSP. >>> >>> The situation is a signal with a frequency range from 19.5 to 20.5MHz, >>> and >>> the sampler can sample up to 2MHz. >>> I'd like to see what the sampled signal looks like at different sample >>> rates. >>> Is there a particular sample rate that will provide a sampled signal >>> that >>> has a frequency range 0-1MHz? (bandwidth?) >>> Or what specific searches can I do to find the explanations myself? >>> Sorry if I haven't got the terminology correct. >> >> Hi Eduard, >> >> I suggest you Google on the terms "bandpass sampling" and/or >> "undersampling." >> For example, here's one of the top hits for the latter: >> >> >> http://www.analog.com/UploadedFiles/Associated_Docs/3689418379346Section5.pdf >> >> You can also see this described in Rick Lyons' book: >> >> @BOOK{lyonsnew, >> title = "{Understanding Digital Signal Processing}", >> edition = "Second", >> author = "{Richard~G.~Lyons}", >> publisher = "Prentice Hall", >> year = "2004"} > > Thanks Jerry, Randy. > I know what to look for now. And will probably be back after I've digested > it all.Ed, Just to be a bit pedantic but also to keep it simple, you will get sum and difference frequency components. The sampling waveform can be viewed as having an infinite set of harmonics so you'd have 100,000, 200,000,300,000 ...... on up and zero and -100,000, -200,000, ..... which convolves with the signal being sampled containing components at -1,000,001 and +1,000,001. -1,000,001 + 1,000,000 is -1Hz where 1,000,000 is 10x100,000, the 10th harmonic of the sampling waveform. +1,000,001 - 1,000,000 is +1Hz and, together there is your 1Hz signal you mention. And, then because you essentially get discrete time with this sampling waveform (we have not introduced A/D yet here), the spectrum will be periodic at 100,000Hz intervals. One can well calculate a rather exhaustive set of sums and differences up to some reasonable limitng frequency on a continuous scale from -inf to +inf and then, because it's periodic, "wrap" or roll it onto itself in a circular fashion. Then all the components end up in that one period from zero to 100,000. Many of them will be on top of one another which just confirms the periodicity of the result. An approach that I like to use to keep things fairly straight in my head is to make a set of "cartoons". Time Domain on the Left <> Frequency Domain on the Right Plot time waveform #1 <> Plot spectrum #1 | | Multiply (or convolve) Convolve (or multiply) | | The time result The frequency result then the next operation then the next operation Of course, these are always "real" functions so things like special phase relationships don't get handled. Just something to keep in mind. As above, if one multiplies two waveforms in time then their respective Fourier Transforms get convolved and vice versa. Generally I plot these from -in to +inf in concept so that the periods lay out side by side for periodic functions. When you do the frequency convolution for your case, it all comes out pretty clear what's happening. When you slide 1Hz, there's an overlap and thus an "output" from the convolution. The next nonzero output comes when you slide 99,998 Hz more or a total of 99,999. The next nonzero output comes when you slide to 100,001 .... So, there are "doublets" 100,000 apart that are +/-1Hz from the harmonics of the sampling frequency ... in the cartoon. When we recognize that this is a periodic frequency domain function then we interpret the nonzero results as being at 1Hz and at 99,999Hz or at 1Hz and -1Hz equivalently for a periodic spectrum. If you use this graphic method it can help with insight and often lead you to the right math or pose math questions like "what happened to the component that should have been there?" when there's cancellation due to phase. Fred
Simple sampling question
Started by ●February 28, 2008
Reply by ●February 29, 20082008-02-29
