1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Refreshing stuff - transfer function

  1. Sep 10, 2012 #1
    I'm trying to determine the transfer function to a simple circuit, see pdf. Is the simplest way to describe H(jω) by finding Y(jω) and U(jω) in H = Y/U. Further using the impedances for each component. If so, is each function Y(jω) and U(jω) determine by 'walking' through each impedance path giving something like this.

    [itex]Y = \frac{1}{\frac{1}{Z_L} + \frac{1}{Z_C} }[/itex] and [itex]U = R +\frac{1}{\frac{1}{Z_L} + \frac{1}{Z_C} }[/itex]

    [itex]Y(jω) = \frac{1}{\frac{1}{jωL} + \frac{1}{1/jωC} }[/itex] and [itex]U(jω) = R +\frac{1}{\frac{1}{jωL} + \frac{1}{1/jωC} }[/itex]

    And next step simplify H(jω) to get the transfer function?

    I'd really appreciate some help or pointers on this. Is this how it's done or..?

    Attached Files:

  2. jcsd
  3. Sep 10, 2012 #2

    Assuming your Y and U functions are voltages, then your Y function is incorrect for starters. You have equated your Y and U voltages to resistances, which is an incorrect approach (it doesnt make sense to have volts on the left side and ohms on the right side). You got steps 1 and 2 mixed up:

    Step 1: Define a complex impedance Z, which is equal to the C and L impedance in parallel. Basically lump your C and L into an equivalent single component. You attempted to do this, but you mistakenly equated the impedance to your voltage function Y.

    Step 2: Define your output voltage Y as a function of your input voltage U. The C and L in parallel, Z, acts as a voltage divider with R on U(jw): Z/(Z+R). this should be straight forward: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html

    Step 3: Solve the transfer function from the above equation by dividing out U from both sides.
    Last edited: Sep 10, 2012
  4. Sep 10, 2012 #3

    What is u(t), current or voltage? What is y(t), output admittance? The transfer function is determined by the components, not by the input voltage/current.

  5. Sep 10, 2012 #4
    Sorry! For violating Ohms law. :shy:

    I didn't post any result in my last post, I get: [itex]H(jω) = \frac{ {jωL} }{{(jω)^2 CLR + R+jωL}}[/itex]

    If I like to perform a Laplace transformation to get into the time domain, H(s) to h(t) to examining a pulse. How do I go from H(jω) to H(s)?
  6. Sep 10, 2012 #5
    You are already there when you define s as s = jw. There is actually a real part too (s = jω + σ), but the sigma is ignored.
  7. Sep 10, 2012 #6
    That transfer function result is the same that I got :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook