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Transfer function of filter

  1. Sep 26, 2006 #1

    t(s)= [2(s^2+9.32)] / [s^4+1.322s^3+0.976s^2+.750s+1]

    how do you calculate the reate of attenuation increase in db per decade at high frequency?

    Determine the gain in db at dc. To do this, just replace s=0 and then do 10log(result) right?

    Also, at which freq is the attenation infinite? Is it when the denominator = 0?

  2. jcsd
  3. Sep 26, 2006 #2


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    Each zero increases the gain by a factor of 10 dB per decade. Each pole causes an attenuation of 10 dB per decade.
    Since you have 2 zeros and 4 poles, the attenuation at high frequency will be 20 dB per dacade.
  4. Sep 26, 2006 #3
    how do you know poles/zeros cause attenuation/gain by a factor of 10?
  5. Sep 26, 2006 #4


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    Staff: Mentor

    We learned it from the class that you are taking now.
  6. Sep 26, 2006 #5
    I'm looking for a proof.:smile:
  7. Sep 27, 2006 #6


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    Consider a system with a single pole:
    [tex]G(s) = \frac{A}{s+a}[/tex]
    the gain in low frequency is obtained making s = 0.
    The gain at a frequency [tex]\omega[/tex] is:
    [tex]Gain_{dB}=10log_{10}\frac{A}{|j\omega+a|} = 10log_{10}A - 10log_{10}|j\omega+a|[/tex]
    At a frequency [tex]\omega_1 >> a[/tex] the gain may be approximated by:
    [tex]Gain_{dB}=10log_{10}\frac{A}{|j\omega+a|} =
    10log_{10}A - 10log_{10}\omega_1[/tex]
    At a frequency [tex]\omega_2 = 10|omega_1[/tex] the gain will be:
    [tex]Gain_{dB}=10log_{10}\frac{A}{|j\omega+a|} = 10log_{10}A - 10log_{10}\10omega_1 = 10log_{10}\frac{A}{|j\omega+a|} = 10log_{10}A - 10log_{10}\omega_1 - 10log_{10}10 = 10log_{10}A - 10log_{10}\omega_1 - 10[/tex]
    So, you have a loss of 10dB in one decade.
    The reasoning for zeros is the inverse.
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