Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] Integrator - Pole-at-Zero

Status
Not open for further replies.

rightmederek

Junior Member level 1
Junior Member level 1
Joined
Sep 10, 2013
Messages
16
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Visit site
Activity points
100
Hi,

I have a question about integrator.

Since the pole of an ideal integrator is at 0, in a bode plot the y-axis of the pole should be at infinite left, log0 = -∞, isn't it?

Thanks,
DD
 

Hi,
I have a question about integrator.
Since the pole of an ideal integrator is at 0, in a bode plot the y-axis of the pole should be at infinite left, log0 = -∞, isn't it?

Thanks,
DD
I suppose, you mean that the transfer function 1/sT approaches ∞ for s=0, correct?
Yes - that`s correct from the mathematical point of you.
However, in practice there is no IDEAL integrator. You always can realize a first-order lowpass only [A/(1+sT)] which can be used as an intergator - with suffficient accuracy - for frequencies well above the corner frequency (pole) wc=1/T.
For this reason, we try to keep this lowpass pole as low as possible.
 

Yes, the gain goes towards infinity as the input frequency goes towards zero. That's why a real integrator will drift towards saturation of the amp at the voltage rails, even with no input, just due to the small input offset dc voltages in the real amp.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top