Hi,
The main mathematical tool engineers use in control is Laplace
transforms which enable one to take an ordinary differential
equation(s) and express it in the form of a transfer function.
For example, given
x_dot_dot + 3 x_dot + 2x = u
we can create the transfer function
x/u = 1/((s+1)(s + 2)) = h(s)
The poles are -1 and -2, and there are no zeros.
Now, if I where to construct a Bode or Nyquist plot I would set s = jw
and evaluate the magnitude and phase of h(s = jw). My question is, why
can we do this? Where does this idea of setting s=jw come from?
Shouldn't we be setting s = sigma + jw because s is a complex variable
with a real and imaginary part?
If anyone has a decent explanation I would appreciate. Or even better,
if someone has a few links I could look at that would be good too.
JRF

When doing a Bode or Nyquist plot we are looking at the steady state
forced response and this is equivalent to evaluating the frequency
response while moving up along the imaginary jw axis.
fred

Hi Jerry,
You use Bode and Nyquist plots to analyze the asymptotic response of
an LTI system when applying a sinusoidal input.
When analyzing only asymptotic response we can reduce the frequency
response to being only along the imaginary axis because the real
values of s only contributes the an exponentially decaying function of
time, e.g. e^(-a*t). Basically we disregard this part of the response.
I hope this answered your question.
/Kasper

...
It wasn't my question. Your answer was more complete than mine. Mr.
Forbes asked a question about a point that eluded him. I had hoped that
setting him on the right track would encourage him to think the rest
through for himself. Sometimes, a more complete understanding develops
that way.
Jerry

Hi,
Thanks for all the responses. I am aware that both a Nyquist plot and
Bode plot provide insight into the response of a system after the
transient effects have died out. More to the point, I am curious as to
where this magical complex analysis using this magical s = a + j w
comes from.
Let me elaborate using my original example, just modified slightly.
x_dot_dot + 1 x_dot + 2x = u
Back in my differential equations class we first solved equations such
as the one above as follows:
Assume the steady state response is x = A e^(L t), then x_dot = A L
e^(L t) and x_dot_dot = A L^2 e^(L t). We will sub these in:
A L^2 e^(L t) + 1 A L e^(L t) + 2 A e^(L t) = u
We will first solve the homogeneous solutions, so set u = 0 for the
time being.
A L^2 e^(L t) + 1 A L e^(L t) + 2 A e^(L t) = 0
( L^2 + 1 L + 2 ) A e^(L t) = 0
(L + 1/2 + (7/4)^0.5j)(L + 1/2 - (7/4)^0.5j) = 0
Thus the eigenvalues of the above equation are L = -1/2 +/- (7/4)^0.5j
= a +/- j w . Then, using our assumed solution:
x = A e^ (a t +/- j w t)
x = e^(a t) (A e^( j w t) + B e^( j w t ))
Next you can solve for the particular solution if u is known and
augment it with the homogeneous solution. If u is not known the
convolution integral must be employed.
SO, when we look at the asymptotic response we are assuming the e^(a
t) effect has died out, and we are just looking at A e^( j w t) + B
e^( j w t ) system. Okay, I buy that. To me however, there is a
suspicious similarity to what I define as L (i.e. L = a +/- j w ) and
the Laplace variable s.
I am looking for the connection between what I have been taught in
controls classes (using Laplace transforms and the like) and what I
have learned in DE classes. I all ready know the answer is in both
classes we are being taught the same thing, just a different way (i.e
many ways of of skinning a cat), but still, I want to know the
connection or the thrust to always use the Laplace domain. I am not
looking for an answer like "replace differential equations in the time
domain, with an equivalent algebraic representation in the Laplace
domain" either. I think one of the reasons we use Laplace stuff it
enables us to better understand signals in the frequency domain, but
evidently I don't know if this is true, of if it is true why we do it.
I find while I am using these tools (Laplace transforms, bode plots,
Nyquist diagrams etc.) I am "faking it", or "going through the
motions". I am executing what I have been taught over and over, but I
feel I don't understand why we are doing things the way we do them. I
no longer want to "fake it", but rather really understand the history
and background as to why we use all this complex math.
JRF

The Laplace transformation is an extension of the D operator and the
less rigorous Heavyside transformation. They are all ways to mechanize
the solution of homogeneous linear differential equations, which are
summed up as passing from the "frequency domain" to the "time domain".
The Fourier transform is the other side of that same coin.
General solutions involve complex frequency, to express the decaying
behavior along with the cylic behavior of systems whose action isn't
entirely forced. When there is no unforced behavior, the real part --
the sigma part -- vanishes, ans only the imaginary part (i.e. real
frequencies) remain. Setting sigma to zero in s = sigma + jw leaves one
with only jw.
Jerry

James Forbes wrote in
news: snipped-for-privacy@22g2000hsm.googlegroups.com:
This is the way things are commonly done, certainly. You should pay a
modicum of attention to the ROC of the Laplace Xform, though.

You are off to an excellent start. I used to teach college EE
courses, and so few students ever made an attempt to connect theory
with reality. Just the fact that you are trying to take ownership of
your knowledge will lead you to an answer. I wish I could give you a
good answer for your question, but I am sadly out of practice. I will
give it some thought.
Andy

Or perhaps a fortuitous similarity? If the Laplace transform does nothing
but let you replace something that must be proved each time to something
that can be used with confidence, then it's a valuable thing.
Replacing the differential equations with algebraic ones _is_ a valuable
thing! If there's any deeper meaning to extract the s = jw stuff, it's
that s = jw (or z = e^jw) implies that you're dealing with a sinusoidal
signal. You can couple this with the fact that tracing the value of a
function in the s (or z) domain along _any_ closed boundary (IIRC, perhaps
along a closed boundary that encompasses all of the poles) gives you _all_
of the information about that signal. Choosing a closed boundary that is
not only an easy signal to generate in the time domain, but is the
stability boundary of the particular domain, is particularly valuable.
With the Laplace transform (or z transform) you can introduce the concept
of a transfer function. Since a transfer function always stays the same
regardless of the input, it is an invariant property of the system itself,
so you can use the Laplace (or z) transform directly to analyze your
_system_; you can't really do this as easily with differential equations.
Also, if you have a system that is difficult to describe mathematically or
that contains continuous states like pure delays or (horrors) pure delays
with leakage, you can still use Bode, Nyquist and Nichols diagrams to
analyze the system, even when you cannot write a closed-form transfer
function of it.
I try to explain this in my book (see the link below), in a way that keeps
it rooted in the physical world. I'm not sure that the explanation is
worth running out and buying the book, but if you're close to a university
it may be worth while to see if they have a copy (or put them up to buying
a copy -- please!).

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