s = jw

Briefly, you don't claim that s is everywhere the same as jw, you evaluate f(s) along the line where sigma is zero. Can you take it from there?

Jerry

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

Jerry knows the answer, he didn't ask the question!

You are right, sorry about that

...

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

Not to worry! :-)

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

You know Euler's identity

e^jw = cos w + j sin w

and have calculated some Fourier Series (sine + cosine as well as e^jw),

and have calculated some LaPlace transforms?

Working out a bunch of problems in those areas (preferably out of a calculus text) may make you more comfortable using the frequency domain shortcuts.

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!).