Poles and Zeros Bode Plot

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TuAtAu

Hi Guys,

I am learning compensation and I want to know the poles and zeros from a circuit.
I learn from simple circuit 1st to make myself have a strong foundation.

But when I see this circuit and do AC analysis and Poles Zeroes Analysis.
It make myself confuse...

From my calculation, there should be a cascade from (RC) (CR) low/high pass filter..
Their poles and zeroes should be multiply together.
2 poles at -1000 and 1 zero at origin from calculation..

QUESTION: how come the simulator give me the unexpected answer??? Why give me 3 poles? but gain bode plot only -20dB/decade?
or my calculation is wrong?
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LvW

Hi,

poles are not at w=-1000 because both RC-CR sections are not decoupled. That means: Both 1st order transfer functions must not be calculated separately. Instead, find the pole distribution from the 2nd order denominator of the combined transfer function.
The slope of this function always is +- 20 dB/dec. (The high pass makes the positive and the lowpass the negative slope).

TuAtAu

TuAtAu

points: 2

TuAtAu

Hi LvW,

Thanks for point out the decoupling things for me to further research.

By the way, how about 3 poles from the picture? according to my understanding, each pole will give -20dB/decade, since there is a zero giving +20dB/decade from origin, 1st pole will make it 0dB/decade, 2nd pole will make it -20dB/decade and the 3rd pole should make it become -40dB/decade what.. how come the AC analysis still giving 20dB/decade?

LvW

By the way, how about 3 poles from the picture?

The shown circuit has only two poles.
For R1=R2=1k and C1=C2=1µF the poles are at: w1=-381.97 1/s and w2=2618 1/s .

I don`t know why you have found a third pole.
Regarding the slope of the transfer function:
* The zero makes +20 dB/dec
* The 1st pole reduces the slope to 0dB/dec
* The 2nd pole makes -20 dB/dec.

TuAtAu

points: 2