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Finding phase angle of transfer function for hi and lo pass filters

Discussion in 'Electronics Homework Help' started by Jodles, Nov 19, 2010.

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  1. Jodles


    Nov 19, 2010

    I'm trying to derive the argument of H(jw) for hi and lo pass filters.

    So far I've gotten the following:
    for low pass:
    From the voltage dividor I've gotten: H(jw) = 1/ (1 + j(w/w_c)) (w_c = ang. cutoff freq.)
    Then I found the angle by converting it to polar form and I get:
    arg H(jw) = arctan (w / w_c)

    But to my knowledge, it should be H(jw) = - arctan(w/w_c). Where is this minus coming from?

    Same thing with hi pass. Following the same procedure as above, starting with H(jw) = 1 / (1-j( w_c / w ) ), I get:
    arg H(jw) = arctan (w_c / w)

    I also know I "should" be getting H(jw) 90 - arctan (w_c / w).

    I can see this by recognizing how the filters work and how the output voltage lags the input voltage. But I don't see how this happens mathematically?

    Any help would be much appreciated!
  2. barathbushan


    Sep 26, 2009
    step 1 ) factorise H(jw) = 1/ (1 + j(w/w_c))

    i mean multiply both numerator and denominator by 1 - j(w/w_c) , and then
    separate real and imaginary terms

    step 2) angle = tan (imaginary/real)

    step 3) its your homework
    Last edited: Nov 19, 2010
  3. Jodles


    Nov 19, 2010
    Thanks! I tried that earlier but ended up with a huge equation. Will try again!:)
  4. Jodles


    Nov 19, 2010
    Ok, I've now gotten the right answer for the lo pass. But for the hi pass filter, I get:
    1 / (1-j w_c/w) = (1+jw_c/w) / (1+w_c^2 /w^2), then separating these and taking the arctan of b/a, I get w_c /w as before. Where do I find the 90 - arctan w_c/w?
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