How Do Filter Transfer Functions Determine Attenuation and Gain?

[david90]

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

 

[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.

 

[david90]

how do you know poles/zeros cause attenuation/gain by a factor of 10?

 

[berkeman]

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.

 

[david90]

I'm looking for a proof.

 

[SGT]

how do you know poles/zeros cause attenuation/gain by a factor of 10?

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