Placement of pole at origin for buckboost feedback compensation?

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treez

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Dr Ridley, in his book, "power supply design", volume 1 , control, states on page 73 that regarding using a type 2 compensator for a buckboost, "the first pole of the compensator is placed at the origin to form an integrator"

However, in the attached type 2 compensator, the first pole is f = 1/[2.pi.R1(C1+C2)]......and this cannot be placed at the origin (f=0) unless either R1 or (c1+c2) is made infinite, which is nonsensical.

Do you know what is meant ?
 

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When you have a 1/s term in your transfer function, it is clearly a pole at the origin. However, how can you put it in your bode plot when you have a frequency scale of powers of 10. You can put this pole in your bode diagram at location frequency f=1 at 0 dB. Below 1, the value 1/s is increasing to the left 20dB/dec. For f higher than 1, the bode plot is decreasing steadily -20dB/dec. It is similar to when you have a C/s term which is a pole at the origin where C is a constant. You can draw it on your bode diagram as a slant line with slope -20dB/dec passing through the point f=C at 0dB.

In your case, C = 1/[2.pi.R1(C1+C2)].
 
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The circuit has two poles, one of which is at the origin. A brief glance should be sufficient to show this (infinite gain at DC).

And R1 does not affect any pole or zero frequencies (just the K).
 
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