Editing Step response (section)

===With one dominant pole===
In many cases, the forward amplifier can be sufficiently well modeled in terms of a single dominant pole of time constant τ, that it, as an open-loop gain given by:

:<math>A_{OL} = \frac {A_0} {1+j \omega \tau}, </math>

with zero-frequency gain ''A''<sub>0</sub> and angular frequency ω = 2π''f''.  This forward amplifier has unit step response

:<math>S_{OL}(t) = A_0(1 -   e^{-t / \tau})</math>,

an exponential approach from 0 toward the new equilibrium value of ''A''<sub>0</sub>.

The one-pole amplifier's transfer function leads to the closed-loop gain:

:<math>A_{FB} = \frac {A_0} {1+ \beta A_0} \; \cdot \; \  \frac {1} {1+j \omega \frac { \tau } {1 + \beta A_0} }.  </math>

This closed-loop gain is of the same form as the open-loop gain: a one-pole filter.  Its step response is of the same form: an exponential decay toward the new equilibrium value.  But the time constant of the closed-loop step function is ''τ'' / (1 + ''β'' ''A''<sub>0</sub>), so it is faster than the forward amplifier's response by a factor of 1 + ''β'' ''A''<sub>0</sub>:

:<math>S_{FB}(t) =  \frac {A_0} {1+ \beta A_0} \left(1 -   e^{-t (1 + \beta A_0)/ \tau}\right),</math>

As the feedback factor ''β'' is increased, the step response will get faster, until the original assumption of one dominant pole is no longer accurate.  If there is a second pole, then as the closed-loop time constant approaches the time constant of the second pole, a two-pole analysis is needed.