Transfer function of filter

  • Thread starter david90
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  • #1
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Main Question or Discussion Point

given

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?


Thanks:tongue:
 

Answers and Replies

  • #2
SGT
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.
 
  • #3
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how do you know poles/zeros cause attenuation/gain by a factor of 10?
 
  • #4
berkeman
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david90 said:
how do you know poles/zeros cause attenuation/gain by a factor of 10?
We learned it from the class that you are taking now.
 
  • #5
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I'm looking for a proof.:smile:
 
  • #6
SGT
david90 said:
how do you know poles/zeros cause attenuation/gain by a factor of 10?
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.
[tex]Gain_{dB}=10log_{10}\frac{A}{a}[/tex]
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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