transfer functions

Discussion in 'Math' started by suzuki, Apr 1, 2012.

  1. suzuki

    Thread Starter Member

    Aug 10, 2011
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    0
    I think this should be the correct place to post this.

    I am wondering about deriving transfer functions. Let's say typically, we have a basic circuit with a source and two series resistors. We can do a voltage division to find the relationship between Vo/Vin.

    Would this be the same as calculating the output voltage, and then dividing both sides by Vin? For example, lets say I know the output voltage is equal to v(t) = 5cos(wt). Does that mean Vo/Vin = 5cos(wt) / Vin? Would this outcome be the same as the scenario as described above?

    Hope that makes sense.

    tia
     
  2. panic mode

    Senior Member

    Oct 10, 2011
    1,321
    304
    if you know exact function (and phase) yes, you just divide the two.
     
    suzuki likes this.
  3. suzuki

    Thread Starter Member

    Aug 10, 2011
    119
    0
    hi,

    thank you for confirming that. it seems obvious, but i just wasn't sure as that is not the normal "approach".

    I have another question, which i thought would just add here instead of creating a new thread. Here goes...

    Usually, when we have a transfer function, it is a function of 's'. i.e. H(s) = s^2 / s+1

    we also know that s = jw.

    Something strange I have recently come across, which i have not really seen before is the following

    H(s) = 1/w *(s^2 / s+1).

    i can't quite seem to wrap my head around what seems to be pretty trivial problem. according to the equation, we want the equation to be a function of 's', so 1/w should be constant. But as stated before, s= jw, and it is the 'w' value that is varied in the frequency response for example. Does this imply that 1/w should also be changing along with the value of s?

    thanks again for all replies.
     
  4. panic mode

    Senior Member

    Oct 10, 2011
    1,321
    304
    can't tell without seeing entire question but

    if you know that s=jω

    then ω=s/j or ω=-js
     
  5. suzuki

    Thread Starter Member

    Aug 10, 2011
    119
    0
    hmm that is a good point.

    so as an example, if we had

    H(s) = w*(s/(s+1))

    we could actually rewrite this as

    H(s) = -js*(s/(s+1))

    which gives an equation as a function of 's'?
     
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