Editing Step response (section)

====Phase margin====
[[Image:Phase for Step Response.PNG|thumbnail|280px|Figure 5: Bode gain plot to find phase margin; scales are logarithmic, so labeled separations are multiplicative factors. For example, {{nowrap|1=''f''<sub>0 dB</sub> = ''βA''<sub>0</sub> × ''f''<sub>1</sub>.}}]]
Next, the choice of pole ratio ''τ''<sub>1</sub>/''τ''<sub>2</sub> is related to the phase margin of the feedback amplifier.<ref>The gain margin of the amplifier cannot be found using a two-pole model, because gain margin requires determination of the frequency ''f''<sub>180</sub> where the gain flips sign, and this never happens in a two-pole system. If we know ''f''<sub>180</sub> for the amplifier at hand, the gain margin can be found approximately, but ''f''<sub>180</sub> then depends on the third and higher pole positions, as does the gain margin, unlike the estimate of phase margin, which is a two-pole estimate.</ref>  The procedure outlined in the [[Bode plot#Examples using Bode plots|Bode plot]] article is followed. Figure 5 is the Bode gain plot for the two-pole amplifier in the range of frequencies up to the second pole position. The assumption behind Figure 5 is that the frequency ''f''<sub>0&nbsp;dB</sub> lies between the lowest pole at ''f''<sub>1</sub>&nbsp;=&nbsp;1/(2πτ<sub>1</sub>) and the second pole at ''f''<sub>2</sub>&nbsp;=&nbsp;1/(2πτ<sub>2</sub>). As indicated in Figure 5, this condition is satisfied for values of&nbsp;α&nbsp;≥&nbsp;1.

Using Figure 5 the frequency (denoted by ''f''<sub>0&nbsp;dB</sub>) is found where the loop gain β''A''<sub>0</sub> satisfies the unity gain or 0&nbsp;dB condition, as defined by:

:<math> | \beta A_\text{OL} ( f_\text{0 db} ) | = 1. </math>

The slope of the downward leg of the gain plot is (20&nbsp;dB/decade); for every factor of ten increase in frequency, the gain drops by the same factor:

:<math> f_\text{0 dB} = \beta A_0 f_1. </math>

The phase margin is the departure of the phase at ''f''<sub>0&nbsp;dB</sub> from −180°. Thus, the margin is:

:<math> \phi_m = 180 ^\circ - \arctan (f_\text{0 dB} /f_1) - \arctan ( f_\text{0 dB} /f_2). </math>

Because ''f''<sub>0&nbsp;dB</sub> / ''f''<sub>1</sub> =&nbsp;''βA''<sub>0</sub>&nbsp;≫&nbsp;1, the term in ''f''<sub>1</sub> is 90°. That makes the phase margin:

:<math>\begin{align}
\phi_m &= 90 ^\circ - \arctan ( f_\text{0 dB} /f_2) \\
&= 90 ^\circ -  \arctan \frac {\beta A_0 f_1} {\alpha \beta A_0 f_1 }  \\
&= 90 ^\circ - \arctan \frac {1} {\alpha } = \arctan \alpha \,.
\end{align}</math>

In particular, for case ''α'' = 1,  ''φ''<sub>m</sub> = 45°, and for ''α'' = 2, ''φ''<sub>m</sub> = 63.4°. Sansen<ref name=Sansen3>{{cite book 
|author=Willy M C Sansen
|title=§0526 p. 162
|date=2006-11-30
|publisher=Springer
|isbn=0-387-25746-2
|url=http://worldcat.org/isbn/0-387-25746-2}}</ref> recommends ''α'' = 3, ''φ''<sub>m</sub> = 71.6° as a "good safety position to start with".

If α is increased by shortening ''τ''<sub>2</sub>, the settling time ''t''<sub>S</sub> also is shortened. If ''α'' is increased by lengthening ''τ''<sub>1</sub>, the settling time ''t''<sub>S</sub> is little altered. More commonly, both ''τ''<sub>1</sub> ''and'' ''τ''<sub>2</sub> change, for example if the technique of [[pole splitting]] is used.

As an aside, for an amplifier with more than two poles, the diagram of Figure 5 still may be made to fit the Bode plots by making ''f''<sub>2</sub> a fitting parameter, referred to as an "equivalent second pole" position.<ref name=Palumbo>{{cite book 
|author=Gaetano Palumbo & Pennisi S
|title=Feedback amplifiers: theory and design
|year= 2002 
|pages=§ 4.4 pp. 97–98
|publisher=Kluwer Academic Press
|location=Boston/Dordrecht/London
|isbn=0-7923-7643-9 
|url=https://books.google.com/books?id=Xb0W1VsQFe0C&q=%22equivalent+two-pole+amplifier%22&pg=PA98 }}</ref>