What does it mean that spectra of a sampled signal is repeated at integer multiples of sampling frequency? I am struggling to visualise it.
Repeated Spectra
Started by ●December 10, 2014
Reply by ●December 10, 20142014-12-10
On Wed, 10 Dec 2014 07:31:58 -0800, muyihwah wrote:> What does it mean that spectra of a sampled signal is repeated at > integer multiples of sampling frequency? I am struggling to > visualise it.There are pictures in here, along with math to help you along: http://wescottdesign.com/articles/Sampling/sampling.pdf -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●December 10, 20142014-12-10
On Wednesday, December 10, 2014 7:00:58 PM UTC+1, Tim Wescott wrote:> On Wed, 10 Dec 2014 07:31:58 -0800, muyihwah wrote: > > > What does it mean that spectra of a sampled signal is repeated at > > integer multiples of sampling frequency? I am struggling to > > visualise it. > > There are pictures in here, along with math to help you along: > > http://wescottdesign.com/articles/Sampling/sampling.pdf > > -- > > Tim Wescott > Wescott Design Services > http://www.wescottdesign.comMany thanks for the link. First impression is good.
Reply by ●December 11, 20142014-12-11
On 12/11/2014 02:00 AM, Tim Wescott wrote:> On Wed, 10 Dec 2014 07:31:58 -0800, muyihwah wrote: > >> What does it mean that spectra of a sampled signal is repeated at >> integer multiples of sampling frequency? I am struggling to >> visualise it. > > There are pictures in here, along with math to help you along: > > http://wescottdesign.com/articles/Sampling/sampling.pdfWell done with page 23 :-) Steve
Reply by ●December 11, 20142014-12-11
muyihwah@gmail.com writes:> What does it mean that spectra of a sampled signal is repeated at > integer multiples of sampling frequency? I am struggling to > visualise it.Hi, While not as immediately useful as Tim's document and its figures, I always admire the beauty of the mathematics and often use it to explain to myself what's going on. To that end, here is a description of the math and the model at the high level: 1. Sampling is modelled by the multiplication of the signal s(t) by the infinite impulse train p(t). 2. A property of Fourier transforms states that multiplication of two signals in the time domain is equivalent to convolution of the the Fourier transform of those two signals the frequency domain. 3. The Fourier transform of the infinite impulse train, P(f), with impulses separated by Ts secons, is also an infinite impulse train, with pulses separated by Fs Hz, where Fs = 1 / Ts. 4. If you bandlimit the Fourier transform of the signal to Fs Hz (+/- Fs/2 Hz), you have a single image at baseband (0 Hz). 5. Convolution of the signal S(f) with the (frequency) infinite impulse train P(f) replicates the S(f) image at every Fs. This also shows: 1. If S(f) isn't bandlimited to Fs Hz, you get aliasing. 2. If you lowpass-filter the result S(f) <*> P(f) to Fs, you get the original spectrum back. 3. The ideal form of the filter in two, the so-called brickwall filter, is the sinc() function. I hope this helps. -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com
Reply by ●December 11, 20142014-12-11
Randy Yates <yates@digitalsignallabs.com> writes:> [...] > 4. If you bandlimit the Fourier transform of the signal to Fs Hz > (+/- Fs/2 Hz), you have a single image at baseband (0 Hz).Meh. I should have stated it this way: 4. If you bandlimit the signal to Fs Hz (+/- Fs/2 Hz), then the Fourier transform has a single image at baseband (0 Hz). -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com
Reply by ●December 11, 20142014-12-11
On 12/11/2014 11:47 AM, Randy Yates wrote:> Randy Yates <yates@digitalsignallabs.com> writes: >> [...] >> 4. If you bandlimit the Fourier transform of the signal to Fs Hz >> (+/- Fs/2 Hz), you have a single image at baseband (0 Hz). > > Meh. I should have stated it this way: > > 4. If you bandlimit the signal to Fs Hz (+/- Fs/2 Hz), then the > Fourier transform has a single image at baseband (0 Hz).I'm unclear on this clarification. Even if you bandlimit the signal, there are still all the aliases at N * Fs, no? I'm not sure what you mean by "single image". -- Rick
Reply by ●December 11, 20142014-12-11
rickman <gnuarm@gmail.com> writes:> On 12/11/2014 11:47 AM, Randy Yates wrote: >> Randy Yates <yates@digitalsignallabs.com> writes: >>> [...] >>> 4. If you bandlimit the Fourier transform of the signal to Fs Hz >>> (+/- Fs/2 Hz), you have a single image at baseband (0 Hz). >> >> Meh. I should have stated it this way: >> >> 4. If you bandlimit the signal to Fs Hz (+/- Fs/2 Hz), then the >> Fourier transform has a single image at baseband (0 Hz). > > I'm unclear on this clarification. Even if you bandlimit the signal, > there are still all the aliases at N * Fs, no? I'm not sure what you > mean by "single image".Rick, I meant the Fourier transform of the original analog signal, so no, that transform would not have energy outside +/- Fs/2 (presuming we have a baseband signal). After sampling, I would call the replicas at N*Fs "images." I wouldn't call them aliases. But I may not be using the term as others do. -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com
Reply by ●December 11, 20142014-12-11
On Thu, 11 Dec 2014 14:10:55 -0500, Randy Yates <yates@digitalsignallabs.com> wrote:>rickman <gnuarm@gmail.com> writes: > >> On 12/11/2014 11:47 AM, Randy Yates wrote: >>> Randy Yates <yates@digitalsignallabs.com> writes: >>>> [...] >>>> 4. If you bandlimit the Fourier transform of the signal to Fs Hz >>>> (+/- Fs/2 Hz), you have a single image at baseband (0 Hz). >>> >>> Meh. I should have stated it this way: >>> >>> 4. If you bandlimit the signal to Fs Hz (+/- Fs/2 Hz), then the >>> Fourier transform has a single image at baseband (0 Hz). >> >> I'm unclear on this clarification. Even if you bandlimit the signal, >> there are still all the aliases at N * Fs, no? I'm not sure what you >> mean by "single image". > >Rick, > >I meant the Fourier transform of the original analog signal, so no, >that transform would not have energy outside +/- Fs/2 (presuming >we have a baseband signal). > >After sampling, I would call the replicas at N*Fs "images." I wouldn't >call them aliases. But I may not be using the term as others do."Images" is consistent with how I've seen them referred to. "Aliased" energy is the energy that overlaps in the replicated spectra.>-- >Randy Yates >Digital Signal Labs >http://www.digitalsignallabs.comEric Jacobsen Anchor Hill Communications http://www.anchorhill.com
Reply by ●December 11, 20142014-12-11
On Thursday, December 11, 2014 8:11:00 PM UTC+1, Randy Yates wrote:> rickman <gnuarm@gmail.com> writes: > > > On 12/11/2014 11:47 AM, Randy Yates wrote: > >> Randy Yates <yates@digitalsignallabs.com> writes: > >>> [...] > >>> 4. If you bandlimit the Fourier transform of the signal to Fs Hz > >>> (+/- Fs/2 Hz), you have a single image at baseband (0 Hz). > >> > >> Meh. I should have stated it this way: > >> > >> 4. If you bandlimit the signal to Fs Hz (+/- Fs/2 Hz), then the > >> Fourier transform has a single image at baseband (0 Hz). > > > > I'm unclear on this clarification. Even if you bandlimit the signal, > > there are still all the aliases at N * Fs, no? I'm not sure what you > > mean by "single image". > > Rick, > > I meant the Fourier transform of the original analog signal, so no, > that transform would not have energy outside +/- Fs/2 (presuming > we have a baseband signal). > > After sampling, I would call the replicas at N*Fs "images." I wouldn't > call them aliases. But I may not be using the term as others do. > -- > Randy Yates > Digital Signal Labs > http://www.digitalsignallabs.comThanks a lot for the explanation. However I couldn't get from it in practical terms (i.e minus the Math) what it means that the spectra of the samples signal is repeated at integer multiples of the sampling frequencies
