hi,

I would like some clarification on Smith's description
on p. 544. The material is discussed between p. 537 - p. 544
Specifically, the regions of convergence, or value boundedness, or
no-convergence in the s-place.

Q:
On the surface, it looks fine. However, the details of the specific
descriptions of the pole/zero a-e fig 30-4 appear to me confusing
(as far as I can understand the material ). It appears the LHS
and RHS descriptions are reversed again based on the text.
For example, Smith says there is not a stable system in the RHS !!
Perhaps someone could clarify the poles/zero a-e description ??

background:
I see he uses a sigma that is substituted
into exp(sigma) rather than exp(- (sigma) ).
This results in the exponential growth on the LHS of the
complex chart. Another EE. circuits text shows the exponential
growth on the RHS. Of course the e-decay is on the other side.

I can accept this.

However, the example description on p.544 which uses fig 30-4 is
confusing. The book error list shows nothing on this.

This book is extememely helpful to my learning DSP,
and has inspired me to revisit a quite old EE circuit text
by Huelsman, which again, as far as my understanding, appears
at odds. I'd prefer the clarification to be directly on the Smith
content.

thanks,


Walt.....

Hello Walt,
I'll bet someone here can help you.
But we'll need to know:

* Which Smith book (exact title & publisher) you're reading.
* Which Edition of that book.
* Which printing number.
* Exactly which lines, on which pages, did you find the
statement that "there is not a stable system in the RHS"?
I, for one, don't know what those words mean.

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.

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 ??

On Nov 5, 2:43?pm, "waltech" <waltechm... at (no spam) yahoo.com> wrote:

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.

brent <buleg... at (no spam) columbus.rr.com> wrote:
On Nov 5, 2:43?pm, "waltech" <waltechm... at (no spam) 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

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

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 ..