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Transfer function

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

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  1. Michael

    Michael Guest

    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. Michael

    Michael Guest

    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. 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
     
  4. Reg Edwards

    Reg Edwards Guest

    Fred, you seem to know a bit about it.

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

    It appears all over the shop.
     
  5. ZZZZPK

    ZZZZPK Guest

    : 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 Avins

    Jerry Avins Guest

    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. 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
     
  8. Michael

    Michael Guest

  9. Reg Edwards

    Reg Edwards Guest

    Fred, you seem to know a bit about it.
    ===================================
    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?"
     
  10. Reg Edwards

    Reg Edwards Guest

    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 Avins

    Jerry Avins Guest

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

    Fred
     
  13. Jerry Avins

    Jerry Avins Guest

    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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