On Wednesday, October 19, 2016 at 12:00:51 PM UTC+13, et...@polyspectral.com wrote:
> > Taking an easier example, a first order system 1/(s+2) has a pole at s=-2.
> > No imaginary part. How to excite the pole at s=-2? There is a 3D graph which has magnitude in dB and sigma jw (s plane) as the other two dimensions. When you take frequency response you go along the imaginary axis in 2D on the s plane (missing the pole of course).
> 
> If we feed any complex exponential ( x(t)=e^jwt, t>0 ) to this system, we'll see that the output y(t) tends to a multiple of the input, after the transient response has worn off. Said differently, the limit of y(t)/x(t) as t->+inf is well defined... and in fact that limit gives the value of the frequency response, H(jw)=1/(jw+2).
> 
> This isn't only true on the imaginary axis. For any x(t)=e^st where Re(s)>-2 we'll get the same result, that the limit of y(t)/x(t) is H(s). But since there's a pole at s=-2, we'll find if we approach s=-2 from the right, this limit gets arbitrarily large.
> 
> So, approaching the pole, the ratio y(t)/x(t) becomes unbounded. It's not a very dramatic effect since both x and y are exponentially decreasing, but it's still a resonance of sorts.
> 
> -Ethan

I agree but how would you perform such an experiment with for example a simple RC network.

> Taking an easier example, a first order system 1/(s+2) has a pole at s=-2.
> No imaginary part. How to excite the pole at s=-2? There is a 3D graph which has magnitude in dB and sigma jw (s plane) as the other two dimensions. When you take frequency response you go along the imaginary axis in 2D on the s plane (missing the pole of course).

If we feed any complex exponential ( x(t)=e^jwt, t>0 ) to this system, we'll see that the output y(t) tends to a multiple of the input, after the transient response has worn off. Said differently, the limit of y(t)/x(t) as t->+inf is well defined... and in fact that limit gives the value of the frequency response, H(jw)=1/(jw+2).

This isn't only true on the imaginary axis. For any x(t)=e^st where Re(s)>-2 we'll get the same result, that the limit of y(t)/x(t) is H(s). But since there's a pole at s=-2, we'll find if we approach s=-2 from the right, this limit gets arbitrarily large.

So, approaching the pole, the ratio y(t)/x(t) becomes unbounded. It's not a very dramatic effect since both x and y are exponentially decreasing, but it's still a resonance of sorts.

-Ethan

On Thursday, October 13, 2016 at 9:08:00 PM UTC+13, Tauno Voipio wrote:
> 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

That bit is confusing me because if the pole is at say -1+j that is not unbounded.

Taking an easier example, a first order system 1/(s+2) has a pole at s=-2.
No imaginary part. How to excite the pole at s=-2? There is a 3D graph which has magnitude in dB and sigma jw (s plane) as the other two dimensions. When you take frequency response you go along the imaginary axis in 2D on the s plane (missing the pole of course).

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

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.

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

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?

> 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

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

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