suppose I have a stable analogue system (no feedback - open loop) with two complex poles only. If I take the frequency response it will have a small overshoot in the frequency domain but the gain does not go to infinity because the pole(s) are not on the jw axis. Is it possible to excite the same system with a complex signal sigma+jw so that you excit the stable pole. ie normally frequency response is carried out along the jw axis only, can it be shifted left?
Injecting a complex frequency
Started by ●October 10, 2016
Reply by ●October 10, 20162016-10-10
gyansorova@gmail.com wrote:> suppose I have a stable analogue system (no feedback - open loop) with two complex poles only. If I take the frequency response it will have a small overshoot in the frequency domain but the gain does not go to infinity because the pole(s) are not on the jw axis. > > Is it possible to excite the same system with a complex signal sigma+jw so that you excit the stable pole. ie normally frequency response is carried out along the jw axis only, can it be shifted left?Your initial response is stable, meaning that it gives bounded output for a bounded input. If you want to excite that complex pole pair, I'm pretty sure that means providing an input that is growing exponentially without limit. -- Rob Gaddi, Highland Technology -- www.highlandtechnology.com Email address domain is currently out of order. See above to fix.
Reply by ●October 10, 20162016-10-10
On Monday, October 10, 2016 at 5:42:27 PM UTC-4, Rob Gaddi wrote:> gyansorova@gmail.com wrote: > > > suppose I have a stable analogue system (no feedback - open loop) with two complex poles only. If I take the frequency response it will have a small overshoot in the frequency domain but the gain does not go to infinity because the pole(s) are not on the jw axis. > > > > Is it possible to excite the same system with a complex signal sigma+jw so that you excit the stable pole. ie normally frequency response is carried out along the jw axis only, can it be shifted left? > > Your initial response is stable, meaning that it gives bounded output > for a bounded input. If you want to excite that complex pole pair, I'm > pretty sure that means providing an input that is growing exponentially > without limit.i guess it's a semantic issue, but i don't see why an exponentially increasing "input" is what is needed to "excite the ... system". and i think i understand what you mean by "excit[ing a] complex pole pair", and an **impulse** will do that. it's because the impulse contains all frequency components and is not missing any. i think you can continuously "excite [a] complex pair" with a single sinusoid of constant amplitude at the resonant frequency of the complex pole. and that complex pole will remain excited without decay. r b-j
Reply by ●October 11, 20162016-10-11
Rob Gaddi <rgaddi@highlandtechnology.invalid> writes:> gyansorova@gmail.com wrote: > >> suppose I have a stable analogue system (no feedback - open loop) >> with two complex poles only. If I take the frequency response it >> will have a small overshoot in the frequency domain but the gain >> does not go to infinity because the pole(s) are not on the jw axis. >> >> Is it possible to excite the same system with a complex signal sigma+jw so that you excit the stable pole. ie normally frequency response is carried out along the jw axis only, can it be shifted left? > > Your initial response is stable, meaning that it gives bounded output > for a bounded input. If you want to excite that complex pole pair, I'm > pretty sure that means providing an input that is growing exponentially > without limit.Rob, At first I thought I agreed with you, but after thinking this through I believe I don't agree. Because complex exponentials are eigenvectors to LTI systems, an input e^(s*t) produces an output H(s) * e^(s*t) (cite{signalsandsystems} p. 167). If the system is causal and stable, then a complex-conjugate pole (e.g.) will be in the left half of the s-plane. Thus sigma_0 < 0 in s_0 = sigma_0 + j*omega_0, and e^(s*t) = e^(sigma*t) * e^(j*omega*t), which is a _decaying_ exponential. But since s_0 is a pole in H(s), H(s) will blow up and the composite e^(s*t) * H(s) blows up. --Randy @BOOK{signalsandsystems, title = "{Signals and Systems}", author = "{Alan~V.~Oppenheim, Alan~S.~Willsky, with Ian~T.~Young}", publisher = "Prentice Hall", year = "1983"} -- Randy Yates, DSP/Embedded Firmware Developer Digital Signal Labs http://www.digitalsignallabs.com
Reply by ●October 11, 20162016-10-11
On Mon, 10 Oct 2016 13:42:41 -0700, gyansorova wrote:> suppose I have a stable analogue system (no feedback - open loop) with > two complex poles only. If I take the frequency response it will have a > small overshoot in the frequency domain but the gain does not go to > infinity because the pole(s) are not on the jw axis. > > Is it possible to excite the same system with a complex signal sigma+jw > so that you excit the stable pole. ie normally frequency response is > carried out along the jw axis only, can it be shifted left?I suppose it could -- I'm not sure what you'd learn that'd be useful, though. -- www.wescottdesign.com
Reply by ●October 12, 20162016-10-12
> suppose I have a stable analogue system (no feedback - open loop) with two complex poles only. If I take the frequency response it will have a small overshoot in the frequency domain but the gain does not go to infinity because the pole(s) are not on the jw axis. > > Is it possible to excite the same system with a complex signal sigma+jw so that you excit the stable pole. ie normally frequency response is carried out along the jw axis only, can it be shifted left?It depends what exactly you mean by "excite" -- as others have noted, for a stable system a bounded input will produce a bounded output. But the poles still do have significance. Normally if you feed an complex exponential (decaying or growing or neither, doesn't matter) to an LTI system, your output will converge toward some multiple of the same exponential. This fails to be true at the poles -- if your input is e^st and s is a pole, your output will grow as t*e^st. So the thing that's special about poles, regardless of stability, is that the output is not bounded by any multiple of the input. -Ethan
Reply by ●October 12, 20162016-10-12
On Thursday, October 13, 2016 at 4:43:56 AM UTC+13, et...@polyspectral.com wrote:> > suppose I have a stable analogue system (no feedback - open loop) with two complex poles only. If I take the frequency response it will have a small overshoot in the frequency domain but the gain does not go to infinity because the pole(s) are not on the jw axis. > > > > Is it possible to excite the same system with a complex signal sigma+jw so that you excit the stable pole. ie normally frequency response is carried out along the jw axis only, can it be shifted left? > > It depends what exactly you mean by "excite" -- as others have noted, for a stable system a bounded input will produce a bounded output. But the poles still do have significance. > > Normally if you feed an complex exponential (decaying or growing or neither, doesn't matter) to an LTI system, your output will converge toward some multiple of the same exponential. This fails to be true at the poles -- if your input is e^st and s is a pole, your output will grow as t*e^st. So the thing that's special about poles, regardless of stability, is that the output is not bounded by any multiple of the input. > > -EthanBut what would such a waveform look like? The system is BIBO stable I cannot see how you could inject a sine wave of any form that would give an unstable output except at the poles - so how would you illustrate this?
Reply by ●October 12, 20162016-10-12
On Wed, 12 Oct 2016 10:56:02 -0700, gyansorova wrote:> On Thursday, October 13, 2016 at 4:43:56 AM UTC+13, > et...@polyspectral.com wrote: >> > suppose I have a stable analogue system (no feedback - open loop) >> > with two complex poles only. If I take the frequency response it will >> > have a small overshoot in the frequency domain but the gain does not >> > go to infinity because the pole(s) are not on the jw axis. >> > >> > Is it possible to excite the same system with a complex signal >> > sigma+jw so that you excit the stable pole. ie normally frequency >> > response is carried out along the jw axis only, can it be shifted >> > left? >> >> It depends what exactly you mean by "excite" -- as others have noted, >> for a stable system a bounded input will produce a bounded output. But >> the poles still do have significance. >> >> Normally if you feed an complex exponential (decaying or growing or >> neither, doesn't matter) to an LTI system, your output will converge >> toward some multiple of the same exponential. This fails to be true at >> the poles -- if your input is e^st and s is a pole, your output will >> grow as t*e^st. So the thing that's special about poles, regardless of >> stability, is that the output is not bounded by any multiple of the >> input. >> >> -Ethan > > But what would such a waveform look like? The system is BIBO stable I > cannot see how you could inject a sine wave of any form that would give > an unstable output except at the poles - so how would you illustrate > this?If the system is BIBO stable then, by definition, you couldn't even inject a signal "at the poles" and have it grow without bound unless the signal you inject is unbounded. -- www.wescottdesign.com
Reply by ●October 12, 20162016-10-12
On Thursday, October 13, 2016 at 9:13:21 AM UTC+13, Tim Wescott wrote:> On Wed, 12 Oct 2016 10:56:02 -0700, gyansorova wrote: > > > On Thursday, October 13, 2016 at 4:43:56 AM UTC+13, > > et...@polyspectral.com wrote: > >> > suppose I have a stable analogue system (no feedback - open loop) > >> > with two complex poles only. If I take the frequency response it will > >> > have a small overshoot in the frequency domain but the gain does not > >> > go to infinity because the pole(s) are not on the jw axis. > >> > > >> > Is it possible to excite the same system with a complex signal > >> > sigma+jw so that you excit the stable pole. ie normally frequency > >> > response is carried out along the jw axis only, can it be shifted > >> > left? > >> > >> It depends what exactly you mean by "excite" -- as others have noted, > >> for a stable system a bounded input will produce a bounded output. But > >> the poles still do have significance. > >> > >> Normally if you feed an complex exponential (decaying or growing or > >> neither, doesn't matter) to an LTI system, your output will converge > >> toward some multiple of the same exponential. This fails to be true at > >> the poles -- if your input is e^st and s is a pole, your output will > >> grow as t*e^st. So the thing that's special about poles, regardless of > >> stability, is that the output is not bounded by any multiple of the > >> input. > >> > >> -Ethan > > > > But what would such a waveform look like? The system is BIBO stable I > > cannot see how you could inject a sine wave of any form that would give > > an unstable output except at the poles - so how would you illustrate > > this? > > If the system is BIBO stable then, by definition, you couldn't even > inject a signal "at the poles" and have it grow without bound unless the > signal you inject is unbounded. > > -- > www.wescottdesign.comThat's what my reasoning is, but by definition the system is unstable at the poles since it's magnitude (in 3D goes to infinity). Therefore it should be possible to inject a complex sine wave at that real and imaginary frequency.
Reply by ●October 13, 20162016-10-13
On 13.10.16 00:41, gyansorova@gmail.com wrote:> On Thursday, October 13, 2016 at 9:13:21 AM UTC+13, Tim Wescott wrote: >> On Wed, 12 Oct 2016 10:56:02 -0700, gyansorova wrote: >> >>> On Thursday, October 13, 2016 at 4:43:56 AM UTC+13, >>> et...@polyspectral.com wrote: >>>>> suppose I have a stable analogue system (no feedback - open loop) >>>>> with two complex poles only. If I take the frequency response it will >>>>> have a small overshoot in the frequency domain but the gain does not >>>>> go to infinity because the pole(s) are not on the jw axis. >>>>> >>>>> Is it possible to excite the same system with a complex signal >>>>> sigma+jw so that you excit the stable pole. ie normally frequency >>>>> response is carried out along the jw axis only, can it be shifted >>>>> left? >>>> >>>> It depends what exactly you mean by "excite" -- as others have noted, >>>> for a stable system a bounded input will produce a bounded output. But >>>> the poles still do have significance. >>>> >>>> Normally if you feed an complex exponential (decaying or growing or >>>> neither, doesn't matter) to an LTI system, your output will converge >>>> toward some multiple of the same exponential. This fails to be true at >>>> the poles -- if your input is e^st and s is a pole, your output will >>>> grow as t*e^st. So the thing that's special about poles, regardless of >>>> stability, is that the output is not bounded by any multiple of the >>>> input. >>>> >>>> -Ethan >>> >>> But what would such a waveform look like? The system is BIBO stable I >>> cannot see how you could inject a sine wave of any form that would give >>> an unstable output except at the poles - so how would you illustrate >>> this? >> >> If the system is BIBO stable then, by definition, you couldn't even >> inject a signal "at the poles" and have it grow without bound unless the >> signal you inject is unbounded. >> >> -- >> www.wescottdesign.com > > That's what my reasoning is, but by definition the system is unstable at the poles since it's magnitude (in 3D goes to infinity). Therefore it should be possible to inject a complex sine wave at that real and imaginary frequency.The imaginary component of such a signal is an exponential function which, in the real world, grows unbounded. -- -TV
