# Transfer function

Discussion in 'Electrical Engineering' started by Michael, Aug 28, 2005.

1. ### MichaelGuest

What is the transfer function for a low pass filter 1/(1+jwrc)?
Is the transfer function derived from the laplace transform of the gain?
Thankyou
Mick Carey
www.mickpc.bigpondhosting.com

2. ### MichaelGuest

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

3. ### Fred MarshallGuest

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.

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.

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

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

4. ### Reg EdwardsGuest

Fred, you seem to know a bit about it.

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

It appears all over the shop.

5. ### ZZZZPKGuest

: 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.

6. ### Jerry AvinsGuest

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

Jerry

7. ### Fred MarshallGuest

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"?

Fred

9. ### Reg EdwardsGuest

Fred, you seem to know a bit about it.
===================================
Fred,

Sqrt(s) appears in contexts such as diffusion of heat and in
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?"

10. ### Reg EdwardsGuest

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.

11. ### Jerry AvinsGuest

'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

12. ### Fred MarshallGuest

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

Fred

13. ### Jerry AvinsGuest

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