# Bode plot of a notch filter question

Discussion in 'Homework Help' started by bitrex, Dec 25, 2009.

1. ### bitrex Thread Starter Active Member

Dec 13, 2009
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4
So I have a notch filter consisting of a voltage divider - resistor in series, inductor and capacitor in shunt, output taken from the junction of the two legs. The transfer function of this filter with component values normalized to 1 is $\frac{s^2 +1}{s^2 +s +1}$. My question is, how do I calculate the response of the complex conjugate pair of zeros? It's 0 + i and 0 - i, so there doesn't seem to be any damping ratio, and so I don't know how to plot the response! Any advice would be appreciated.

2. ### Ron H AAC Fanatic!

Apr 14, 2005
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You need to post a schematic. What you described is not a notch filter, it's a bandpass filter. The transfer function also doesn't look right.

3. ### steveb Senior Member

Jul 3, 2008
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I agree with Ron that it's not clear how you arrived at the notch filter transfer function. The circuit description is not clear, and my interpretation of the description does not seem to indicate a notch filter. However, I'll ignore this and try to answer your question anyway.

The transfer function (whether correctly representing your circuit or not) does show a notch filter response. Your transfer function shows two zeros and two poles. The zeros are complex conjugates and they are located on the jw axis. The poles are complex conjugates in the left half plane, indicating a stable system.

With the zeros on the jw axis, you will get a transfer function of zero for a real frequency at the zero location (w=1 in this case). The explanation for this is simply that the bode plot is found by setting s=jw; hence, any zeros on the jw axis must become actual zero values in the transfer function (unless canceled by a pole).

For real frequencies far from the notch frequency, the poles and zeros behave as if they cancel each other's effects and you approach unity magnitude with zero phase. However, near the notch frequency, the poles and zeros do not cancel because although they appear to be at the same frequency, they do not exactly match in the complex frequency plane.

Last edited: Dec 28, 2009