Transfer function

There seems to be some confusion that is driving your questions - you really ought to complain to your lecturers that they are ineffective and incompetent if they have left you in such confusion.

Their job is to train you, and to ensure that you become qualified even if you didn't pay attention at some point!

1. 1/(1+jwrc) _IS_ the transfer funcion!

1. The transfer function tells you how individual frequencies (assuming a linear system) pass through your system. As such, the transfer function is a plot of gain against frequency.

2. You use the Laplace transform to give you the transfer function, but what you transform is the reaction of your system to being given a "short sharp shock" (Thank-you G&S)

That answers what you were actually asking, but consider the following....

1. You hit a bell with a drumstick, and deliver a short, sharp shock (known as an "Impulse" in the trade) and the bell responds with a waveform (ringing) that you could plot in time.

1. If you applied the Laplace Transform to this waveform, you will get the frequencies that made it up. (Yes, OK, to the purists, we should really be discussing the Fourier Transform, but the questioner is unlikely to be able to understand yet the "windowing" effect of the factor "e^(-ct)" on the frequency response.)

2. The Laplace Transform takes a time-based "Twang" and converts it into a frequency spectrum.

1. You could find the frequency response of your bell by a different method, and that is to whistle into it and determine at what frequencies it resounds (not necessarily at resonance) back at you. By plotting the amplitude of its response at each frequency, you should be able to duplicate the Laplace transform.

2. ANother way of achieving (4) is to hit it with all possible frequencies AT THE SAME TIME. Difficult, but once you have a mathematical model you can excite that model with Dirac's Unit Impulse which contains all possible cosine frequencies (including DC). They all add up in phase at time 0, to give a height of infinity and an area of unity, and cancel out everywhere else.

HTH!

Michael wrote:

Sorry for the lack of understanding, I am studying via correspondence and am covering material that has not been taught to us yet, so tell me what is the difference between 1/(1+jwc) and something like 1/(s+1) wich I can put into my graphing calculator and do a bode plot? Just trying to get a handle on a few things! Mick Carey

Then you are to be congratulated and encouraged - too many students these days are taught, very few are inspired as you are!

"s" and "jw" are the same thing, for the moment.

"jw" comes from the Fourier Transform.

"s" comes from the Laplace Transform.

s is a combination of sigma + jw. (Sorry, no Greek keyboard for the sigma!)

The sigma is introduced as a decaying exponential (which I discussed as e^(-ct) in the last article to ensure that the transform from Fourier will converge, so giving us the Laplace Transform.

The value of sigma is never defined! It has to be big enough to ensure convergence!

The sigma is a parameter that can mostly be ignored, unless you are into regions of convergence, which you won't be for

99% of the transforms that you come up against.

Michael wrote:

Perhaps that's why the standards of the GC(S)E have been lowered to produce the incredible pass rates published yesterday!

Now even the Ploddity can get the 5 'O' Levels passes that they were > "Polymath" wrote:

Mick,

Perhaps this will help:

Systems that are described by ordinary, linear, differential equations with constant coefficients can be analyzed using the Laplace Transform with the variable "s" representing the differential operator d/dt. I'm supposing that you're getting exposed to this now or you wouldn't have asked the question.

If you have a transfer function like the one you posed that is 1/(ks + 1) ... where I've added the constant "k" ... then a very typical thing to want to do is to evaluate the transfer function for its frequency response. You know that s=sigma + jw. The "w" part is frequency and I will discuss "sigma" later, below. For now, the steady state frequency response of the system is determined by evaluating the transfer function along the jw axis in the s plane. So:

1/(ks + 1) for s=jw becomes 1/(1 + jwk) In your case, k=c so you have 1/(1 + sc) and for the special case of the jw axis where we are only interested in values of s=jw, you evaluate 1/(1 + jwc). You vary "w" to determine the steady state frequency response of the system. "c" is a constant.

I think that answers your question. Perhaps I should add:

1/(ks + 1) is the "Transfer Function" 1/(jwk +1) is the transfer function *evaluated on the jw axis* which, when evaluated for all values of "w" is the "Steady State Frequency Response" The former provides information across the entire s-plane. The latter provides information only along a single line in that plane. The Transfer Function allows you to analyze transient response and stability. The Frequency Response is obviously more limited.

Now, what about sigma? Well, we're working in a complex plane which is just another way to say we're working in 2 dimensions. It doesn't really matter which is Real and which is Imaginary as long as we have a consistent system - and here we do have a consistent system.

Imagine a family of all sinusoids of all possible frequencies that decay or grow exponentially at all possible rates. Each point on the s-plane represents one of this infinite family of sinusoids.

- those in the left hand plane decay exponentially

- those in the right hand plane grow exponentially

- those on the jw axis are steady, neither growing nor decaying.

So, we have the two dimensions "sigma" and "w". "w" is the frequency of each sinusoid "sigma" is the exponential decaying or growth term

As in: [e^(sigma*t)]*sin(wt)

When sigma=0 we have sin(wt) with no decay or growth. When sigma is negative, the exponential term decays. When sigma is positive, the exponential term grows.

For system frequency response analysis, we find the steady sinusoids to be very interesting because they correspond to the steady state frequency response. Thus we use s=jw where sigma=0 and we see that we are evaluating the function along the jw axis only.

For system stability analysis, we find the location of poles and zeros of the transfer function to be interesting because they represent the transient response of the system - its "natural frequencies" including how they decay or grow in response to external excitation or a set of initial conditions of the system ... getting back to those differential equations, eh?

Of course, we don't really need to separate the two but it's a handy way to think about it.

If a system has no input at all but does have some nonzero initial state (capacitors charged, masses moving, etc.) then the system response will be some linear combination of the family of sinusoids represented by the poles of the transfer function. So, if any of the poles are in the right half plane the resulting response to the initial conditions can grow without bound and the sytem is unstable - because the pole is at a point in the plane that looks like [e^5t]*sin(12t) which *grows* exponentially and which is the definition of "unstable".

I hope this helps.

Fred

Fred, you seem to know a bit about it.

If s = jw, wot does sqrt(s) mean?

It appears all over the shop.

: Their job is to train you, and to ensure that you : become qualified even if you didn't pay attention at some : point!

that *was* their job.

now its a job of produce a batch of people whose marks fit the bell curve.

The same s? I guess not. Anyhow, s = sigma + j*omega. Sometimes we set sigma = zero, but don't let that confuse you.

Jerry

Reg,

I didn't say that s=jw exactly. I said that s = sigma + jw and we often *set* sigma = 0 or s = jw for purposes of evaluation of transfer functions on the jw axis only - that is, for steady state sinusoidal inputs.

what "shop"? sqrt(s) in what context please?

Fred

Without going into detail, sqrt(x) shows up in the solution of diffusion like phenomena such as heat conduction or twisted pair telephone lines where the inductance is low.

Bill

How do I graph 1/(1+jwc) on a program that take two polynomials as its arguments eg

steps do I have to go about to do this? Cheers Mick Carey

=================================== Fred,

Sqrt(s) appears in contexts such as diffusion of heat and in propagation along real transmission lines. See also tables of Laplace Transforms.

s and 1/s are operators. But what is Sqrt(s)? It defies the imagination.

Perhaps only Oliver Heaviside, 1850-1925, knew the answer. The university professors who ridiculed Heaviside certainly didn't. Not even when use of sqrt(s) gave the correct answers.

Self-educated Heaviside retaliated by asking "Shall I refuse to eat my dinner because I do not fully understand the process of digestion?"

That's not the case.

Whether you refer to "s" or to "jw", the roots of the various polynomials will be of the form x+jy whether you choose to evaluate them as functions of "jw" or as functions of "s".

These roots will not, except for a very few simple cases, be on the jw axis, but lie in the whole of the complex plane.

The sigma to which yo hav referred is merely the agent of convergence in evaluating the Laplace Transform.

Fred Marshall wrote:

Expand 1/Functions(s+a) binomially into an infinite series and then perform the indicated integrations to obtain time series.

Functions of time, eg., waveforms, are just as interesting and useful as functions of frequency. People who are restricted to thinking in terms only of CW, reflections and SWR are handicapped.

...

's', or 'x'? In any case sqrt(a + jb) isn't arcane knowledge. In polar coordinates, sqrt(1 + j) is 1 at an angle of +/- pi/4.

Jerry

I should have used sqrt(s). The letter, however, is unimportant. Some people, such as Bode, used p as the letter for the transform variable. That is merely a distinction without a difference.

In the end, the inverse transform has to be determined by integration, usually contour integration. The sqrt indicates the existence of branch point in the transform function. That affects how the integration contour is selected.

Bill

I don't disagree but I think you missed the point....

Fred

It isn't necessary to construe 's' as an operator. Ir 'j', for that matter. It's convenient, but if the construction of the square root is over the top for someone, just construe it as a variable.

Jerry

Hi,

When I read our teams response, it seems like everybody shown thier thought. That we call team work (sorry if I make a mistaken). Why do not you referred to prof. to clarifiy your problems. I thinks it is the best way.

Tks

magic