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Transfer function to phase transfere

  1. Oct 26, 2014 #1


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    This is problem is a part of a bigger problem wich I solved and came to a formula which is correct.

    This is the equation for one transfer function. The next thing I would like to find out here is when arg(H(jw)) is -90 or 90 degrees but I get stuck.

    Transfer function:
    ## H(jw) = \frac{R}{R(1-w^2LC)+jwL} ##

    This is how I proceeded till I got stuck.

    ## H(jw) = \frac{R}{R(1-w^2LC)+jwL} \\ H(jw) = \frac{R}{\sqrt{(R(1-w^2LC))^2+(wL)^2}e^{jarctan( \frac{wL}{R(1-w^2LC)})}} \\ H(jw) = \frac{ R }{ \sqrt{ (R(1-w^2LC))^2+(wL)^2} } e^{-jarctan( \frac{wL}{R(1-w^2LC)})} \\ ##

    Then we want to know where the argument is -90 or 90 degrees.

    ## 90=-jarctan( \frac{wL}{R(1-w^2LC)}) ##

    Here is the step where I get stuck. Could you please help me out?
    Thanks in advance!
  2. jcsd
  3. Oct 26, 2014 #2
    Where does the "j" on the right hand side of the last line come from?
  4. Oct 26, 2014 #3


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    Thank you for your reply!

    From j here:
    ##e^{ -jarctan( \frac{wL}{R(1-w^2LC)})} ##

    But I guess it shouldnt be there. Since I an looking for the phase.

    What about ## 90=-arctan( \frac{wL}{R(1-w^2LC)}) ## on the last line then :) ?

    EDIT: the problem I have is that 90=-arctan(x) has no solutions :,(. But the solution in my book is ## w= \frac{1}{\sqrt{LC}} ##. When I try to insert the solution from the book inside the equation I have gives something that is not defined:
    ## 90=-arctan( \frac{\frac{1}{\sqrt{LC}}L}{R(1-(\frac{1}{\sqrt{LC}})^2LC)}) \\ 90= -arctan( \frac{\frac{1}{\sqrt{LC}}L}{R(1-(\frac{1}{1}))}) \\ 90= -arctan( \frac{\frac{1}{\sqrt{LC}}L}{0})##
    Last edited: Oct 26, 2014
  5. Oct 26, 2014 #4
    What does the tangent function look like, in particular, what is the value of tan(90 deg) and tan(-90 deg)?

    The mathematically correct way would be to calculate the solution for +/-(90 + epsilon), and then take the limit epsilon-> 0.
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