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Transfer function

M

Michael

Jan 1, 1970
0
Polymath said:
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!

2. 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.

3. 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.

2. 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.)

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

4. 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.

5. 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!
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
www.mickpc.bigpondhosting.com
 
F

Fred Marshall

Jan 1, 1970
0
Michael said:
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

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
 
R

Reg Edwards

Jan 1, 1970
0
Fred, you seem to know a bit about it.

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

It appears all over the shop.
 
Z

ZZZZPK

Jan 1, 1970
0
: 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.
 
J

Jerry Avins

Jan 1, 1970
0
Reg said:
Fred, you seem to know a bit about it.

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

It appears all over the shop.

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

Jerry
 
F

Fred Marshall

Jan 1, 1970
0
Reg Edwards said:
Fred, you seem to know a bit about it.

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

It appears all over the shop.

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
 
R

Reg Edwards

Jan 1, 1970
0
Fred, you seem to know a bit about it.
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,

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?"
 
R

Reg Edwards

Jan 1, 1970
0
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.
 
J

Jerry Avins

Jan 1, 1970
0
On 8/28/05 7:06 PM, in article [email protected], "Jerry Avins"
...

...

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.

'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
 
F

Fred Marshall

Jan 1, 1970
0
Polymath said:
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.

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

Fred
 
J

Jerry Avins

Jan 1, 1970
0
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.

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
 
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