thanks Rick and Jerry for the reply. The book is the first printing of the Smith book. Will furnish details in a moment: I'm still searching for the words to describe what I'm after. In particular, in the several days since the orig. post, I've learned that the words "convergence" and unbounded" are steps toward discovery of what and how to determine stability. I'm clearly interested in stability. The fig's 30-3 and fig. 30-6, have allowed me to visualize what happens if a pole or zero is moved about. My "familiarity" has improved somewhat by the study of poles/zeros using my old standard circuit book. (Lots of reading, but more to go ). Again, I don't get how to determine "stability". I fail to understand how the display of the pole on the LHS ( left hand side ) but often seems to be called LHP ( left hand plane ) .. seems to be in the "normal" place when the decay and exp growth waveforms both in my text and the Smith book are on opposite sides. Surely if the decay and e growth sides are reversed, the *pole* should also have switched sides. Fig 30-9 again has the poles on the LS. And agrees with my old circuit book. But, again, the decay in the circuit book is on the left side. The statement quoted is on p. 544. First partial paragraph: " In Fact, poles in the right-half of the s-place show that the system is unstable. (i.e. an impulse response that increases with time." On that page, descriptions of cases a-e follow from the diagram P. 543 fig 30-5. Here besides stabilty, I'm trying to determine the relationship of the * increasing/decreasing amplitudes vs the integral *. One example ( but not the only ) is that of case "a" fig 30-5. where both the probe and impulse response decrease. To me with what I know, this would be a zero since the amplitude "goes toward zero". (Should I say in the limit to infinity ?) but the integral has some finite number. Clearly I don't get it !! thanks, Walt......

# re: s-plane; Laplace; Smith book

Started by ●November 5, 2009

Reply by ●November 5, 20092009-11-05

waltech wrote:> thanks Rick and Jerry for the reply. > > The book is the first printing of the Smith book. > Will furnish details in a moment: > > I'm still searching for the words to describe what I'm > after. In particular, in the several days since the orig. post, > I've learned that the words "convergence" and unbounded" are > steps toward discovery of what and how to determine stability. > I'm clearly interested in stability. > > The fig's 30-3 and fig. 30-6, have allowed me to > visualize what happens if a pole or zero is moved about. > My "familiarity" has improved somewhat by the study of poles/zeros > using my old standard circuit book. (Lots of reading, but more > to go ). Again, I don't get how to determine "stability". > > I fail to understand how the display of the pole on the LHS > ( left hand side ) but often seems to be called LHP ( left hand plane )left _half_ plane> .. seems to be in the "normal" place when the decay and exp > growth waveforms both in my text and the Smith book are on opposite > sides. Surely if the decay and e growth sides are reversed, the > *pole* should also have switched sides. > Fig 30-9 again has the poles on the LS. And agrees with my > old circuit book. But, again, the decay in the circuit book is > on the left side. > > > The statement quoted is on p. 544. First partial paragraph: > " In Fact, poles in the right-half of the s-place show that the system > is unstable. (i.e. an impulse response that increases with time." > > On that page, descriptions of cases a-e follow from the diagram > P. 543 fig 30-5. Here besides stabilty, I'm trying to determine > the relationship of the > * increasing/decreasing amplitudes vs the integral *. > > One example ( but not the only ) is that of case "a" fig 30-5. > where both the probe and impulse response decrease. To me with what > I know, this would be a zero since the amplitude "goes toward > zero". (Should I say in the limit to infinity ?) but the integral > has some finite number. Clearly I don't get it !!The s plane allows one to plot a point as (σ,jω), just as the x-y plane deals with (x,y). If that's garbled for you, read the first pair as (sigma,j_omega). Note that poles with non-zero complex parts come in conjugate pairs. The response of a pair of poles (a,jb) and (a,-jb) is exp(a+jb)*t + exp(a-jb)*t = exp(at)*[exp(jbt)+exp(-jbt)]. Apply exp(jx)=cos(x)+jsin(x) to get the response as exp(at)*cos(bt). Negative values of σ are to the left of the jω axis, in other words, in the left half of the σ,jω plane. In the left half plane, exp(σt) decays because σ is negative. In the right half plane, exp(σt) grows without bound as time progresses. Clear as mud, no? Let me know what still bugs you. Jerry P.S. View as UTF-8 Unicode. -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Reply by ●November 5, 20092009-11-05

AH ! this list is bottom-post. see below.>waltech wrote: >> thanks Rick and Jerry for the reply. >> >> The book is the first printing of the Smith book. >> Will furnish details in a moment: >> >> I'm still searching for the words to describe what I'm >> after. In particular, in the several days since the orig. post, >> I've learned that the words "convergence" and unbounded" are >> steps toward discovery of what and how to determine stability. >> I'm clearly interested in stability. >> >> The fig's 30-3 and fig. 30-6, have allowed me to >> visualize what happens if a pole or zero is moved about. >> My "familiarity" has improved somewhat by the study of poles/zeros >> using my old standard circuit book. (Lots of reading, but more >> to go ). Again, I don't get how to determine "stability". >> >> I fail to understand how the display of the pole on the LHS >> ( left hand side ) but often seems to be called LHP ( left hand plane)> >left _half_ plane > >> .. seems to be in the "normal" place when the decay and exp >> growth waveforms both in my text and the Smith book are on opposite >> sides. Surely if the decay and e growth sides are reversed, the >> *pole* should also have switched sides. >> Fig 30-9 again has the poles on the LS. And agrees with my >> old circuit book. But, again, the decay in the circuit book is >> on the left side. >> >> >> The statement quoted is on p. 544. First partial paragraph: >> " In Fact, poles in the right-half of the s-place show that the system >> is unstable. (i.e. an impulse response that increases with time." >> >> On that page, descriptions of cases a-e follow from the diagram >> P. 543 fig 30-5. Here besides stabilty, I'm trying to determine >> the relationship of the >> * increasing/decreasing amplitudes vs the integral *. >> >> One example ( but not the only ) is that of case "a" fig 30-5. >> where both the probe and impulse response decrease. To me with what >> I know, this would be a zero since the amplitude "goes toward >> zero". (Should I say in the limit to infinity ?) but the integral >> has some finite number. Clearly I don't get it !! > >The s plane allows one to plot a point as (σ,jω), just as the x-y plane> deals with (x,y). If that's garbled for you, read the first pair as >(sigma,j_omega). Note that poles with non-zero complex parts come in >conjugate pairs. The response of a pair of poles (a,jb) and (a,-jb) is >exp(a+jb)*t + exp(a-jb)*t = exp(at)*[exp(jbt)+exp(-jbt)]. Apply >exp(jx)=cos(x)+jsin(x) to get the response as exp(at)*cos(bt). > >Negative values of σ are to the left of the jω axis, in other words, in>the left half of the σ,jω plane. In the left half plane, exp(σt)decays>because σ is negative. In the right half plane, exp(σt) grows without >bound as time progresses. > >Clear as mud, no? Let me know what still bugs you. > >Jerry >I quite agree with your statements. I am quite familiar with sigma + j-omega plane as tool for graphing: F(S) = (s-z1)(s-z2)(s-z3).../ (s-p1)(s-p2)(s-p3) ... either as network response or description from impulse response. In particular, the figures I mentioned were viewed with idea of using factors of form (s minus delta-s) where neighborhoods plotted describe surrounding areas to the poles etc... For example, it's clear that if the pole is close to the j-omega axis, then small movements of freq. result in large phase changes. I believe my question is fairly clear: I don't understand how to use the tools to determine stability. And how unboundedness is different than infinite boundedness. For example, fig 30-5 shows a too-fast increase of probe, in section "e". This diagram is interesting in that it shows only 5 examples, one is too fast increasing sinusoidal in the LHP. Your statement is different than what the book states. I wondered if there was a real error, or difference in interpretation. In trying to learn, I accepted it's a difference in substitution for either sigma, or (-sigma). I checked at least one other website which shows the exp growth in the LHP as comparison. But the pole diagrams all seem to agree with each other !! It may very well be my misunderstanding the probing waveforms themselves rather than the s-plane characteristics. ( fig 30-5 ). Now, while here, let me add one other point of confusion: the Fourier transform places the freq response on the j-omega line. In Laplace, the circuit response is on the j-omega axis, but poles and zeros can be any part of the space. This assumes, like Fourier, that sigma is zero. In a real circuit, with a sigma, why isn't the frequency response on some vertical line sigma = some value, rather than sigma = 0, since a real circuit has a sigma ?? By the way, the circuit text I'm also using for reference is quite a large text, whose copyright is 1972. It's been a long while, as a computer ( not dsp ) designer, since I wanted to know this material. The Smith book became known to me about a year ago, and motivated me to begin to learn more in this area. Yes, I purchased a hard copy. thanks, Walt ......

Reply by ●November 5, 20092009-11-05

On Nov 5, 2:43�pm, "waltech" <waltechm...@yahoo.com> wrote: I will try to answer some of your question, but will not get to the specifics of the diagrams:> > Now, while here, let me add one other point of confusion: > the Fourier transform places the freq response on the j-omega > line. �In Laplace, the circuit response is on the j-omega axis, > but poles and zeros can be any part of the space. �This assumes, > like Fourier, that sigma is zero. > > In a real circuit, with a sigma, why isn't the frequency response > on some vertical line �sigma = some value, rather than sigma = 0, > since a real circuit has a sigma ?? >The laplace transform is providing a means to look at how your system would respond if you excited it with every possible combination of exponentially increasing and exponentially decreasing sinusoidal input wave forms. It additionally provides a look into how your system would respond if you excited it with simple eponentially increasing and exponentially decreasing (non-sinusoidal) signals. In other words, If you have a system the laplace diagram gives insight into what happens when the system is excited with this vast array of possible input signals. Now, in a heck of a lot of circuit analysis, we are interested in the simple frequency response of the system. That is, what happens to the system when it is excited with simple sinusoidal waves that do not increase or decrease in amplitude. That particular condition is simply the jw axis of the laplace diagram. Now, the laplace diagram show stability issues (You will only have stablity issues when there is gain being implemented in the system) by looking at how the system responds to exponentially decaying sinusoids (that is the left hand plane). If you can put in an exponentially decreasing sinusoid into a system and the system somehow turns that signal into an exponentially increasing sinusoid, then you have a problem. On the other hand, if you can put an exponentially increasing sinusoid into a system and the system turns it into an exponentially decreasing sinusoid at the output (I use output in a general sense) then you are not going to have a stability problem. All real world inputs are assumed to be either flat sinusoidal signals or exponentially decreasing signals. You can't really excite a system with an exponentially increasing signal for very long. So , with laplace you look at how your system reacts with every possible exponentially increasing or decreasing signal to dtermine stability. If the pole ( an integration that indicates that the output goes to infinity - which means the output turned into an exponentially increasing sinusoid) is in the RHP then you can put "normal" signals in and have instability. If the pole is in the LHP, then you need a supernatural signal to cause this pole. Since this "supernatural " signal is assumed to not exist, then you are OK. Done :-)

Reply by ●November 5, 20092009-11-05

brent <bulegoge@columbus.rr.com> wrote:> On Nov 5, 2:43?pm, "waltech" <waltechm...@yahoo.com> wrote:(snip)> The laplace transform is providing a means to look at how your system > would respond if you excited it with every possible combination of > exponentially increasing and exponentially decreasing sinusoidal input > wave forms. It additionally provides a look into how your system > would respond if you excited it with simple eponentially increasing > and exponentially decreasing (non-sinusoidal) signals. In other > words, If you have a system the laplace diagram gives insight into > what happens when the system is excited with this vast array of > possible input signals.Especially for systems described by nice simple differential equations...> Now, in a heck of a lot of circuit analysis, we are interested in the > simple frequency response of the system. That is, what happens to the > system when it is excited with simple sinusoidal waves that do not > increase or decrease in amplitude. That particular condition is > simply the jw axis of the laplace diagram.Or can be handled more easily with the Fourier transform.> Now, the laplace diagram show stability issues (You will only have > stablity issues when there is gain being implemented in the system) by > looking at how the system responds to exponentially decaying sinusoids > (that is the left hand plane). If you can put in an exponentially > decreasing sinusoid into a system and the system somehow turns that > signal into an exponentially increasing sinusoid, then you have a > problem.Note that an exponentially increasing or decreasing sinusoid is an exponential of a complex value. The imaginary part supplies the sinusoid, the real part the exponential increase or decrease. They do this even if you in the end take the real part of the exponential as a physical quanitity.> On the other hand, if you can put an exponentially > increasing sinusoid into a system and the system turns it into an > exponentially decreasing sinusoid at the output (I use output in a > general sense) then you are not going to have a stability problem.> All real world inputs are assumed to be either flat sinusoidal signals > or exponentially decreasing signals. You can't really excite a system > with an exponentially increasing signal for very long.Well, there are many important systems that operate in the exponentially increasing region. There must be a non-linearity to limit the growth, though. Oscillators in electronics and lasers in optics for example.> So , with laplace you look at how your system reacts with every > possible exponentially increasing or decreasing signal to dtermine > stability. If the pole ( an integration that indicates that the > output goes to infinity - which means the output turned into an > exponentially increasing sinusoid) is in the RHP then you can put > "normal" signals in and have instability. If the pole is in the LHP, > then you need a supernatural signal to cause this pole. Since this > "supernatural " signal is assumed to not exist, then you are OK.-- glen

Reply by ●November 5, 20092009-11-05

On Nov 5, 9:38�pm, glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote:> brent <buleg...@columbus.rr.com> wrote: > > On Nov 5, 2:43?pm, "waltech" <waltechm...@yahoo.com> wrote: > > (snip) > > > The laplace transform is providing a means to look at how your system > > would respond if you excited it with every possible combination of > > exponentially increasing and exponentially decreasing sinusoidal input > > wave forms. �It additionally provides a look into how your system > > would respond if you excited it with simple eponentially increasing > > and exponentially decreasing (non-sinusoidal) signals. �In other > > words, If you have a system the laplace diagram gives insight into > > what happens when the system is excited with this vast array of > > possible input signals. > > Especially for systems described by nice simple differential > equations... > > > Now, in a heck of a lot of circuit analysis, we are interested in the > > simple frequency response of the system. �That is, what happens to the > > system when it is excited with simple sinusoidal waves that do not > > increase or decrease in amplitude. �That particular condition is > > simply the jw axis of the laplace diagram. > > Or can be handled more easily with the Fourier transform. >To clarify: the jw axis of the Laplace Tranform is the Fourier transform.> > Now, the laplace diagram show stability issues (You will only have > > stablity issues when there is gain being implemented in the system) by > > looking at how the system responds to exponentially decaying sinusoids > > (that is the left hand plane). �If you can put in an exponentially > > decreasing sinusoid into a system and the system somehow turns that > > signal into an exponentially increasing sinusoid, then you have a > > problem. � > > Note that an exponentially increasing or decreasing sinusoid > is an exponential of a complex value. �The imaginary part supplies > the sinusoid, the real part the exponential increase or decrease. > They do this even if you in the end take the real part of the > exponential as a physical quanitity. � > > > On the other hand, if you can put an exponentially > > increasing sinusoid into a system and the system turns it into an > > exponentially decreasing sinusoid at the output (I use output in a > > general sense) then you are not going to have a stability problem. > > All real world inputs are assumed to be either flat sinusoidal signals > > or exponentially decreasing signals. �You can't really excite a system > > with an exponentially increasing signal for very long. > > Well, there are many important systems that operate in the > exponentially increasing region.But not for very long :-). Just from the time you turn them on until the signal has grown enough to put the system into compression an * exactly* make the steady state input a flat sinusoidal signal. >�There must be a non-linearity Actually, I would not really consider it a non-linearity, but rather an automatic gain adjustment phenomenon. The oscillator can still be modeled as a linear system after the gain has perfectly adjusted (through compression) such that the over all system has unity gain and "perfect" 360 degree phase feedback.> to limit the growth, though. �Oscillators in electronics and > lasers in optics for example. � > > > So , with laplace you look at how your system reacts with every > > possible exponentially increasing or decreasing signal to dtermine > > stability. �If the pole ( an integration that indicates that the > > output goes to infinity - which means the output turned into an > > exponentially increasing sinusoid) is in the RHP then you can put > > "normal" signals in and have instability. �If the pole is in the LHP, > > then you need a supernatural signal to cause this pole. �Since this > > "supernatural " signal is assumed to not exist, then you are OK. > > -- glen

Reply by ●November 11, 20092009-11-11

I would like to thanks you all for the most enlightening answer. Brents was quite comprehensive. In studying my circuit book, to get more familiar with s-plane, it appears some simple filter with biquad equation such as Butterworth with N-poles and no zeros is a "root-locus". Meaning, the poles reside on a circle about the (0,0) axis. Suppose the input frequency is moved up and down, it rides this circle ( ?). But it's clear that if the circuit has gain ( 0-3), poles also move. For all of this, I ask ONLY regarding LHP ! It appears the root locus is for a stable sine wave ( no decay). (please clarify if wrong) Q1. If the input signal rides this circle, is there some input signal fed into the fixed-component circuit such that the charactoristics fall off this circle and fall elsewhere into the s-plane? For example, some decaying sine wave with sigma other than what is on the circle ? Q2. Is the circle my ONLY interest for all input signals since the circuit is fixed with some RLC and maybe opamp ? Another way to ask is, would any DSP processing be necessary over the rest of the domain for analysis? Q3. If I collect raw data from a scientific instrument which performs some sort of scan, and repeats with maybe some small change in parameter for each scan, is the LaPlace useful here? I realize I can take the data and perform a Fourier and examine the frequency domain. thanks, Walt ..>On Nov 5, 9:38=A0pm, glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote: >> brent <buleg...@columbus.rr.com> wrote: >> > On Nov 5, 2:43?pm, "waltech" <waltechm...@yahoo.com> wrote: >> >> (snip) >> >> > The laplace transform is providing a means to look at how yoursystem>> > would respond if you excited it with every possible combination of >> > exponentially increasing and exponentially decreasing sinusoidalinput>> > wave forms. =A0It additionally provides a look into how your system >> > would respond if you excited it with simple eponentially increasing >> > and exponentially decreasing (non-sinusoidal) signals. =A0In other >> > words, If you have a system the laplace diagram gives insight into >> > what happens when the system is excited with this vast array of >> > possible input signals. >> >> Especially for systems described by nice simple differential >> equations... >> >> > Now, in a heck of a lot of circuit analysis, we are interested inthe>> > simple frequency response of the system. =A0That is, what happens toth=>e >> > system when it is excited with simple sinusoidal waves that do not >> > increase or decrease in amplitude. =A0That particular condition is >> > simply the jw axis of the laplace diagram. >> >> Or can be handled more easily with the Fourier transform. >> >To clarify: the jw axis of the Laplace Tranform is the Fourier >transform. > > >> > Now, the laplace diagram show stability issues (You will only have >> > stablity issues when there is gain being implemented in the system)by>> > looking at how the system responds to exponentially decayingsinusoids>> > (that is the left hand plane). =A0If you can put in an exponentially >> > decreasing sinusoid into a system and the system somehow turns that >> > signal into an exponentially increasing sinusoid, then you have a >> > problem. =A0 >> >> Note that an exponentially increasing or decreasing sinusoid >> is an exponential of a complex value. =A0The imaginary part supplies >> the sinusoid, the real part the exponential increase or decrease. >> They do this even if you in the end take the real part of the >> exponential as a physical quanitity. =A0 >> >> > On the other hand, if you can put an exponentially >> > increasing sinusoid into a system and the system turns it into an >> > exponentially decreasing sinusoid at the output (I use output in a >> > general sense) then you are not going to have a stability problem. >> > All real world inputs are assumed to be either flat sinusoidalsignals>> > or exponentially decreasing signals. =A0You can't really excite asyste=>m >> > with an exponentially increasing signal for very long. >> >> Well, there are many important systems that operate in the >> exponentially increasing region. > >But not for very long :-). Just from the time you turn them on until >the signal has grown enough to put the system into compression an * >exactly* make the steady state input a flat sinusoidal signal. > > >=A0There must be a non-linearity > >Actually, I would not really consider it a non-linearity, but rather >an automatic gain adjustment phenomenon. The oscillator can still be >modeled as a linear system after the gain has perfectly adjusted >(through compression) such that the over all system has unity gain and >"perfect" 360 degree phase feedback. > > >> to limit the growth, though. =A0Oscillators in electronics and >> lasers in optics for example. =A0 >> >> > So , with laplace you look at how your system reacts with every >> > possible exponentially increasing or decreasing signal to dtermine >> > stability. =A0If the pole ( an integration that indicates that the >> > output goes to infinity - which means the output turned into an >> > exponentially increasing sinusoid) is in the RHP then you can put >> > "normal" signals in and have instability. =A0If the pole is in theLHP,>> > then you need a supernatural signal to cause this pole. =A0Sincethis>> > "supernatural " signal is assumed to not exist, then you are OK. >> >> -- glen > >

Reply by ●November 11, 20092009-11-11

waltech wrote:> I would like to thanks you all for the most > enlightening answer. Brents was quite comprehensive. > > In studying my circuit book, to get more familiar with > s-plane, it appears some simple filter with biquad equation > such as Butterworth with N-poles and no zeros is a "root-locus". > Meaning, the poles reside on a circle about the (0,0) axis. > Suppose the input frequency is moved up and down, it rides this > circle ( ?). But it's clear that if the circuit has gain ( 0-3), > poles also move. For all of this, I ask ONLY regarding LHP ! > > It appears the root locus is for a stable sine wave ( no decay). > (please clarify if wrong) > > Q1. If the input signal rides this circle, is there some > input signal fed into the fixed-component circuit such that > the charactoristics fall off this circle and fall elsewhere > into the s-plane? For example, some decaying sine wave with > sigma other than what is on the circle ? > > Q2. Is the circle my ONLY interest for all input signals since the > circuit is fixed with some RLC and maybe opamp ? > > Another way to ask is, would any DSP processing be necessary > over the rest of the domain for analysis? > > Q3. If I collect raw data from a scientific instrument which > performs some sort of scan, and repeats with maybe some small > change in parameter for each scan, > is the LaPlace useful here? I realize I can take the data and perform > a Fourier and examine the frequency domain. > > thanks, > Walt .."Root locus" is a technique usually used for designing servos and examining their stability. See http://www.google.com/search?q=root+locus and ask more questions if it matters. I suspect that where you write /root locus/, you mean something else. I won't guess what that it. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●December 12, 20092009-12-12

hi Brent, Glen and Jerry, and others, I've been giving this some thought. I plan to review the Laplace math. Before I do so, I believe I am better able to present my problem. In the RHP, + sigma, the network function is exponentally increasing. In a vertical line, if the line is moved closed to the jw axis, the increase is "slower". (I believe it's correct the say that the s-plane is a represenation of a "network function", as well as an "impulse response". I just want to let you know where I am in this thinking. now, assume there are no poles or zeros in the RHP. Are the following true ?? They all refer to the RHP. If I have as input a steady state sine wave, with response in the RHP, it will always increase (i.e. in every RHP position) ? If I have a decreasing sine wave as input, there is a place where it must increase (further to the right ) ? and to the left of the above, a place where a decreasing sine must decrease and ? somewhere in the middle where the input decay exp's cancel with the network resulting in a stable sine wave ? If the input signal is merely an impulse that results in a response in the RHP ( is this possible? ), then that impulse response MUST decay, since it is "less" of a signal than a steady state sinusoid? If there are poles in the RHP, then instability is guaranteed ? or is there a situation where the RHP may have a response greater than a pole in the RHP? -- I hope I have made the idea of my Q clear; if the language is not precise enough, please correct. thanks for your assistance, Walt.....>waltech wrote: >> I would like to thanks you all for the most >> enlightening answer. Brents was quite comprehensive. >> >> In studying my circuit book, to get more familiar with >> s-plane, it appears some simple filter with biquad equation >> such as Butterworth with N-poles and no zeros is a "root-locus". >> Meaning, the poles reside on a circle about the (0,0) axis. >> Suppose the input frequency is moved up and down, it rides this >> circle ( ?). But it's clear that if the circuit has gain ( 0-3), >> poles also move. For all of this, I ask ONLY regarding LHP ! >> >> It appears the root locus is for a stable sine wave ( no decay). >> (please clarify if wrong) >> >> Q1. If the input signal rides this circle, is there some >> input signal fed into the fixed-component circuit such that >> the charactoristics fall off this circle and fall elsewhere >> into the s-plane? For example, some decaying sine wave with >> sigma other than what is on the circle ? >> >> Q2. Is the circle my ONLY interest for all input signals since the >> circuit is fixed with some RLC and maybe opamp ? >> >> Another way to ask is, would any DSP processing be necessary >> over the rest of the domain for analysis? >> >> Q3. If I collect raw data from a scientific instrument which >> performs some sort of scan, and repeats with maybe some small >> change in parameter for each scan, >> is the LaPlace useful here? I realize I can take the data and perform >> a Fourier and examine the frequency domain. >> >> thanks, >> Walt .. > >"Root locus" is a technique usually used for designing servos and >examining their stability. See http://www.google.com/search?q=root+locus>and ask more questions if it matters. I suspect that where you write >/root locus/, you mean something else. I won't guess what that it. > >Jerry >-- >Engineering is the art of making what you want from things you can get. >����������������������������������������������������������������������� >