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Phase relationship at high frequencies

  1. May 7, 2017 #1
    1. The problem statement, all variables and given/known data
    Given the transfer function:
    upload_2017-5-7_10-37-25.png

    Find the phase relationship between the input and output voltage at high frequencies

    2. Relevant equations
    NA

    3. The attempt at a solution
    My approach was to find the phase of the transfer function, which I got to be:
    upload_2017-5-7_10-38-45.png
    I assumed that "very high frequencies" meant w = ∞, plugging it in, I get the answer to be 0°.

    However, the answer from my teacher was that it was -180, because the transfer function had 2 poles, each at -90. While I can understand this, I wanted to know why my method was wrong.
     
  2. jcsd
  3. May 7, 2017 #2

    scottdave

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    Homework Helper
    Gold Member

    Try substituting j*w {j is imaginary, and w is omega} for s and then see what the angle becomes. So j is equivalent to +90 degrees. 1/j = -j is equivalent to -90 degrees. So what does j^2 do? (or 1/j^2) What angle would that approach? This is how Bode plots (phase) can be made, if you are familiar with them.
     
  4. May 8, 2017 #3

    LvW

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    I rather think, the questioner has already replaced s by jw while deriving the formula as given in 3) .
    And - yes - the formula is correct. Hence, there is nothing left than to evaluate the formula for very large frequencies (considering that the arctan function is not unambiguous).
     
  5. May 8, 2017 #4
    Yes; this is what i was hoping to do. However, plugging w as infinity does not get the -180, so I am confused.
     
  6. May 8, 2017 #5

    gneill

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    Staff: Mentor

    Keep the x and y components separate rather than forming the ratio y/x for the arctan function. That means retaining any signs associated with them rather than letting them get entangled in the ratio. Then evaluate by inspection how those components will trend as ω gets larger, and where the vector (x,y) formed from them is headed.
     
  7. May 8, 2017 #6
    I think this works well. I got the awnser I was looking for. Thank You!
     
  8. May 8, 2017 #7
    Just another question. Can higher frequencies be assumed that frequencies higher than the cutoff frequency?
     
  9. May 8, 2017 #8

    gneill

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    Staff: Mentor

    Yes.
     
  10. May 8, 2017 #9
    Thank you !!!
     
  11. May 8, 2017 #10

    scottdave

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    That is a good question, because it never will actually get to -180 degrees. It approaches that at very high frequencies (much higher than the cutoff). Wolfram Alpha has a nice tool (search for Bode Plot) and you can type in the transfer function and see both the gain plots and phase plots.
     
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