Editing Step response (section)

====Analysis====
The two-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 + \tau_2 } {1 + \beta A_0} + (j \omega )^2 \frac { \tau_1 \tau_2} {1 + \beta A_0} }.  </math>

[[Image:Conjugate poles in s-plane.svg|thumbnail|250px|Figure 2: Conjugate pole locations for a two-pole feedback amplifier; Re(''s'') is the real axis and Im(''s'') is the imaginary axis.]]

The time dependence of the amplifier is easy to discover by switching variables to ''s'' = ''j''ω, whereupon the gain becomes:

:<math> A_{FB} = \frac {A_0} { \tau_1 \tau_2 } \; \cdot \; \frac {1} {s^2 +s \left( \frac {1} {\tau_1} + \frac {1} {\tau_2} \right) + \frac {1+ \beta A_0} {\tau_1 \tau_2}} </math>

The poles of this expression (that is, the zeros of the denominator) occur at:

:<math>2s = - \left( \frac {1} {\tau_1} + \frac {1} {\tau_2} \right)
 \pm \sqrt { \left( \frac {1} {\tau_1} - \frac {1} {\tau_2} \right) ^2 -\frac {4 \beta A_0 } {\tau_1 \tau_2 } },</math>

which shows for large enough values of ''βA''<sub>0</sub> the square root becomes the square root of a negative number, that is the square root becomes imaginary, and the pole positions are complex conjugate numbers, either ''s''<sub>+</sub> or ''s''<sub>−</sub>; see Figure 2:

:<math> s_{\pm} = -\rho \pm j \mu, </math>

with

:<math> \rho = \frac {1}{2} \left( \frac {1} {\tau_1} + \frac {1} {\tau_2} \right ), </math>

and
:<math> \mu = \frac {1} {2} \sqrt { \frac {4 \beta A_0} { \tau_1 \tau_2} - \left( \frac {1} {\tau_1} - \frac {1} {\tau_2} \right)^2 }. </math>
Using polar coordinates with the magnitude of the radius to the roots given by |''s''| (Figure 2):

:<math> | s | = |s_{ \pm } | =  \sqrt{ \rho^2 +\mu^2}, </math>

and the angular coordinate φ is given by:

: <math> \cos \phi = \frac { \rho} { | s | } , \sin \phi = \frac { \mu} { | s | }.</math>
Tables of [[Laplace transform]]s show that the time response of such a system is composed of combinations of the two functions:

:<math> e^{- \rho t} \sin ( \mu t) \quad\text{and} \quad  e^{- \rho t} \cos ( \mu t), </math>

which is to say, the solutions are damped oscillations in time. In particular, the unit step response of the system is:<ref name=Kuo>{{cite book |author=Benjamin C Kuo & Golnaraghi F|title=Automatic control systems|year= 2003 |pages=253|publisher=Wiley| edition=Eighth |location=New York|isbn=0-471-13476-7 |url=http://worldcat.org/isbn/0-471-13476-7}}</ref>
:<math>S(t) = \left(\frac {A_0} {1+ \beta A_0}\right)\left(1 -   e^{- \rho t} \  \frac {  \sin \left( \mu t + \phi \right)}{ \sin \phi}\right)\ , </math>

which simplifies to

:<math>S(t) = 1 -   e^{- \rho t} \  \frac {  \sin \left( \mu t + \phi \right)}{ \sin \phi}</math>

when ''A''<sub>0</sub> tends to infinity and the feedback factor ''β'' is one.

Notice that the damping of the response is set by ρ, that is, by the time constants of the open-loop amplifier. In contrast, the frequency of oscillation is set by μ, that is, by the feedback parameter through β''A''<sub>0</sub>. Because ρ is a sum of reciprocals of time constants, it is interesting to notice that ρ is dominated by the ''shorter'' of the two.