Good Morning, Boys and Girls, (and those of you with the mental and emotional age of Boys and Girls; such age having been already demonstrated in your responses to Lesson 1. Perhaps there are those of you who, rather than coming up to University from high school, would have fared better by going back down to elementary school where you would find your equals in the playground?) In today's lecture, I want to aim right at the heart of the subject matter - that of the sampling of analogue signals. -----ooooo----- I have two things that I need to point out. Firstly, to highlight that sampling circuits are not LTI, (Linear, Time Invariant) and therefore are not analysable by our tools from Laplace. Certainly, we can describe the output signal that comes from our sampling circuits using Laplace, but we can neither describe, nor decompose, their operation using Laplace. The sampling circuit is, in fact, a SIGNAL GENERATOR that is certainly controlled by the incoming analogue waveform but it is definitely NOT an LTI filter! (A moment's thought will show the truth of this - if a sine-wave were to be presented to our sampling circuit, the output of the circuit would just not be represented by the convolution of that sine-wave with the output of the sampler) -----ooooo----- Secondly, I need to point out that the first-order hold presented as an attribute of the output of a DSP by many authors, is, in fact, an attribute of the input sampling. Samples are taken, and remain static until further samples are taken. -----ooooo----- .....I'm very sorry, children, I've been called away. I'll continue with this tomorrow!
A Sound Mathematical Basis For Sampling - Lesson 2
Started by ●November 13, 2004
Reply by ●November 14, 20042004-11-14
"Airy R. Bean" <me@privacy.net> wrote in message news:2vmf1bF2lqpf7U1@uni-berlin.de...> I have two things that I need to point out. Firstly, > to highlight that sampling circuits are not LTI, > (Linear, Time Invariant) and therefore are not analysable > by our tools from Laplace.I don't think anybody says that the sampling circuit is an LTI operation. If we were trying to treat the sampling as an LTI filter block then we would do convolution with some kind of frequency response. What we do is multiply by a train of impulses separated by 1/fs. In frequency this corresponds to convolution with a train of impulses separated by fs. This convolution results in periodicity of the input spectrum which ultimately gives rise to the Nyquist frequency such that the spectrums don't overlap. If you don't understand all of this, Mr. Bean, I can walk you through it slower. For example, a lot of people get lost at how the train of impulses in the time domain is also a train of impulses in the frequency domain. Brad
Reply by ●November 14, 20042004-11-14
Brad Griffis wrote:> "Airy R. Bean" <me@privacy.net> wrote in message > news:2vmf1bF2lqpf7U1@uni-berlin.de... > >>I have two things that I need to point out. Firstly, >>to highlight that sampling circuits are not LTI, >>(Linear, Time Invariant) and therefore are not analysable >>by our tools from Laplace. > > > I don't think anybody says that the sampling circuit is an LTI operation. > If we were trying to treat the sampling as an LTI filter block then we would > do convolution with some kind of frequency response. What we do is multiply > by a train of impulses separated by 1/fs. In frequency this corresponds to > convolution with a train of impulses separated by fs. This convolution > results in periodicity of the input spectrum which ultimately gives rise to > the Nyquist frequency such that the spectrums don't overlap. > > If you don't understand all of this, Mr. Bean, I can walk you through it > slower. For example, a lot of people get lost at how the train of impulses > in the time domain is also a train of impulses in the frequency domain. > > BradBrad, Bean is a clever troll who will lead you by the nose as far as you let him pull you. Any information he may get along the way is completely beside the point. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 16, 20042004-11-16
Amplitude modulation is not represented by convolution in the frequency domain, because the carriers are still present after multiplication/modulation. (In the act of convolving with a delta function, the delta disappears and the input signal re-appears faithfully reproduced in either time or frequency depending upon which domain you are working in, shifted by any delay factor) You are not the first to be confused by the multiplication of an input spectrum by the frequency response of a linear filter, as opposed to the multiplication/modulation of one signal by another. The former is a linear operation, the latter is not because more frequencies are present in the output than were there in the input. Now, who was it who said, "If you don't understand all of this, I can walk you through it slower." ????? "Brad Griffis" <bradgriffis@hotmail.com> wrote in message news:hQLld.21647$x62.8480@newssvr31.news.prodigy.com...> "Airy R. Bean" <me@privacy.net> wrote in message > news:2vmf1bF2lqpf7U1@uni-berlin.de... > > I have two things that I need to point out. Firstly, > > to highlight that sampling circuits are not LTI, > > (Linear, Time Invariant) and therefore are not analysable > > by our tools from Laplace. > I don't think anybody says that the sampling circuit is an LTI operation. > If we were trying to treat the sampling as an LTI filter block then wewould> do convolution with some kind of frequency response. What we do ismultiply> by a train of impulses separated by 1/fs. In frequency this correspondsto> convolution with a train of impulses separated by fs. This convolution > results in periodicity of the input spectrum which ultimately gives riseto> the Nyquist frequency such that the spectrums don't overlap. > If you don't understand all of this, Mr. Bean, I can walk you through it > slower. For example, a lot of people get lost at how the train ofimpulses> in the time domain is also a train of impulses in the frequency domain.
Reply by ●November 16, 20042004-11-16
Airy R. Bean wrote:> Amplitude modulation is not represented by > convolution in the frequency domain, because the > carriers are still present after multiplication/modulation. > > (In the act of convolving with a delta function, the > delta disappears and the input signal re-appears faithfully > reproduced in either time or frequency depending upon > which domain you are working in, shifted by any delay factor) > > You are not the first to be confused by the multiplication > of an input spectrum by the frequency response of a linear > filter, as opposed to the multiplication/modulation of one signal by > another. The former is a linear operation, the latter is not > because more frequencies are present in the output than > were there in the input. > > Now, who was it who said, "If you don't understand all > of this, I can walk you through it slower." ????? >Amplitude modulation is quite accurately represented by convolution in the frequency domain -- if you can't figure it out you could ask here and someone will explain. I think you need to revisit the definition of "linear system". Then you won't mistake a linear time-varying system for a nonlinear one. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●November 16, 20042004-11-16
Amplitude modulation by delta functions is not accurately represented by convolution in the frequency domain -- if you can't figure it out you could ask here and someone will explain. Convolution by a delta function in _ANY_ domain causes the other signal to be faithfully reproduced and the delta disappears. The delta does not disappear in amplitude modulation "Tim Wescott" <tim@wescottnospamdesign.com> wrote in message news:10pk73c3ulc62fa@corp.supernews.com...> Airy R. Bean wrote: > > > Amplitude modulation is not represented by > > convolution in the frequency domain, because the > > carriers are still present after multiplication/modulation. > > > > (In the act of convolving with a delta function, the > > delta disappears and the input signal re-appears faithfully > > reproduced in either time or frequency depending upon > > which domain you are working in, shifted by any delay factor) > > > > You are not the first to be confused by the multiplication > > of an input spectrum by the frequency response of a linear > > filter, as opposed to the multiplication/modulation of one signal by > > another. The former is a linear operation, the latter is not > > because more frequencies are present in the output than > > were there in the input. > > > > Now, who was it who said, "If you don't understand all > > of this, I can walk you through it slower." ????? > > > > Amplitude modulation is quite accurately represented by convolution in > the frequency domain -- if you can't figure it out you could ask here > and someone will explain. > > I think you need to revisit the definition of "linear system". Then you > won't mistake a linear time-varying system for a nonlinear one. > > -- > > Tim Wescott > Wescott Design Services > http://www.wescottdesign.com
Reply by ●November 16, 20042004-11-16
"Airy R. Bean" <me@privacy.net> wrote in message news:<2vtsvuF2pff1iU1@uni-berlin.de>...> You are not the first to be confused by the multiplication > of an input spectrum by the frequency response of a linear > filter, as opposed to the multiplication/modulation of one signal by > another. The former is a linear operation, the latter is not > because more frequencies are present in the output than > were there in the input.They are both linear. The latter is not time-invariant. For some reason this is a common mistake despite the fact that "linearity" has a pretty simple and straightforward definition. -Frederick Umminger
Reply by ●November 17, 20042004-11-17
"Brad Griffis" <bradgriffis@hotmail.com> wrote in message news:hQLld.21647$x62.8480@newssvr31.news.prodigy.com...> >> If you don't understand all of this, Mr. Bean, I can walk you through it > slower. For example, a lot of people get lost at how the train ofimpulses> in the time domain is also a train of impulses in the frequency domain. > > Brad > >That's right,scaled by 2*pi/T if my memory serves me well. Tom
Reply by ●November 17, 20042004-11-17
Oops - I seem to have made an (admittedly rare on my part) error. I was temporarily confused as to in which domain we were multiplying and in which domain we were convolving. Consider AM as cos(wct) * (1 + cos(wmt)) and all becomes clear. I was distracted by the rather childish and unnecessary ad hominem remarks by the two previous correspondents.. A good illustration of the use of emotive rather than intellectual commentaries by those adherents of the religion of sampling by delta functions, perhaps? "Airy R. Bean" <me@privacy.net> wrote in message news:2vuql7F2ph6tmU1@uni-berlin.de...> Amplitude modulation by delta functions is not accurately > represented by convolution in the frequency domain -- if > you can't figure it out you could ask here and someone will explain. > > Convolution by a delta function in _ANY_ domain causes the other > signal to be faithfully reproduced and the delta disappears. > > The delta does not disappear in amplitude modulation > > "Tim Wescott" <tim@wescottnospamdesign.com> wrote in message > news:10pk73c3ulc62fa@corp.supernews.com... > > Airy R. Bean wrote: > > > > > Amplitude modulation is not represented by > > > convolution in the frequency domain, because the > > > carriers are still present after multiplication/modulation. > > > > > > (In the act of convolving with a delta function, the > > > delta disappears and the input signal re-appears faithfully > > > reproduced in either time or frequency depending upon > > > which domain you are working in, shifted by any delay factor) > > > > > > You are not the first to be confused by the multiplication > > > of an input spectrum by the frequency response of a linear > > > filter, as opposed to the multiplication/modulation of one signal by > > > another. The former is a linear operation, the latter is not > > > because more frequencies are present in the output than > > > were there in the input. > > > > > > Now, who was it who said, "If you don't understand all > > > of this, I can walk you through it slower." ????? > > > > > > > Amplitude modulation is quite accurately represented by convolution in > > the frequency domain -- if you can't figure it out you could ask here > > and someone will explain. > > > > I think you need to revisit the definition of "linear system". Then you > > won't mistake a linear time-varying system for a nonlinear one. > > > > -- > > > > Tim Wescott > > Wescott Design Services > > http://www.wescottdesign.com > > >
Reply by ●November 17, 20042004-11-17
Why do you imply that it is time-variant? Surely the action of modulation does not change with time? If it _IS_ time-variant, then surely it is not describable with the signal processing theory that we all learn has to be LTI? "Frederick Umminger" <fumminger@umminger.com> wrote in message news:9b3f0b6d.0411160916.699465b8@posting.google.com...> "Airy R. Bean" <me@privacy.net> wrote in messagenews:<2vtsvuF2pff1iU1@uni-berlin.de>...> > You are not the first to be confused by the multiplication > > of an input spectrum by the frequency response of a linear > > filter, as opposed to the multiplication/modulation of one signal by > > another. The former is a linear operation, the latter is not > > because more frequencies are present in the output than > > were there in the input. > > They are both linear. The latter is not time-invariant. For some > reason this is a common mistake despite the fact that "linearity" has > a pretty simple and straightforward definition.
