Editing Step response (section)

==Feedback amplifiers==
[[Image:Block Diagram for Feedback.svg|thumb|Figure 1: Ideal negative feedback model; open loop gain is ''A''<sub>OL</sub> and feedback factor is β.]]

This section describes the step response of a simple [[negative feedback amplifier]] shown in Figure 1. The feedback amplifier consists of a main '''open-loop''' amplifier of gain ''A''<sub>OL</sub> and a feedback loop governed by a '''feedback factor''' β. This feedback amplifier is analyzed to determine how its step response depends upon  the time constants governing the response of the main amplifier, and upon the amount of feedback used.

A negative-feedback amplifier has gain given by (see [[negative feedback amplifier]]):

:<math>A_{FB} = \frac {A_{OL}} {1+ \beta A_{OL}}, </math>

where ''A''<sub>OL</sub> = '''open-loop''' gain, ''A''<sub>FB</sub> = '''closed-loop''' gain (the gain with negative feedback present) and ''β'' = '''feedback factor'''.

===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.

===Two-pole amplifiers===
In the case that the open-loop gain has two poles (two [[time constant]]s, ''τ''<sub>1</sub>, ''τ''<sub>2</sub>), the step response is a bit more complicated. The open-loop gain is given by:

:<math>A_{OL} = \frac {A_0} {(1+j \omega \tau_1) (1 + j \omega \tau_2)}, </math>

with zero-frequency gain ''A''<sub>0</sub> and angular frequency ''ω'' = 2''πf''.

====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.

====Results====
[[Image:Step response for two-pole feedback amplifier.PNG|thumbnail|350px|Figure 3: Step-response of a linear two-pole feedback amplifier; time is in units of&nbsp;1/''ρ'', that is, in terms of the time constants of ''A''<sub>OL</sub>; curves are plotted for three values of ''mu''&nbsp;=&nbsp;''μ'', which is controlled by&nbsp;''β''.]]
Figure 3 shows the time response to a unit step input for three values of the parameter μ. It can be seen that the frequency of oscillation increases with μ, but the oscillations are contained between the two asymptotes set by the exponentials [&nbsp;1&nbsp;−&nbsp;exp(−''ρt'')&nbsp;] and [&nbsp;1&nbsp;+&nbsp;exp(−ρt)&nbsp;]. These asymptotes are determined by ρ and therefore by the time constants of the open-loop amplifier, independent of feedback.

The phenomenon of oscillation about the final value is called '''[[ringing (signal)|ringing]]'''. The '''[[overshoot (signal)|overshoot]]''' is the maximum swing above final value, and clearly increases with μ. Likewise, the '''undershoot''' is the minimum swing below final value, again increasing with μ. The '''[[settling time]]''' is the time for departures from final value to sink below some specified level, say 10% of final value.

The dependence of settling time upon μ is not obvious, and the approximation of a two-pole system probably is not accurate enough to make any real-world conclusions about feedback dependence of settling time. However, the asymptotes [&nbsp;1&nbsp;−&nbsp;exp(−''ρt'')&nbsp;] and [&nbsp;1&nbsp;+&nbsp;exp&nbsp;(−''ρt'')&nbsp;] clearly impact settling time, and they are controlled by the time constants of the open-loop amplifier, particularly the shorter of the two time constants. That suggests that a specification on settling time must be met by appropriate design of the open-loop amplifier.

The two major conclusions from this analysis are: 
#Feedback controls the amplitude of oscillation about final value for a given open-loop amplifier and given values of open-loop time constants,  τ<sub>1</sub> and τ<sub>2</sub>.
#The open-loop amplifier decides settling time. It sets the time scale of Figure 3, and the faster the open-loop amplifier, the faster this time scale.

As an aside, it may be noted that real-world departures from this linear two-pole model occur due to two major complications: first, real amplifiers have more than two poles, as well as zeros; and second, real amplifiers are nonlinear, so their step response changes with signal amplitude.
[[Image:Overshoot control.PNG|thumbnail|300px|Figure 4: Step response for three values of&nbsp;α. Top: α &nbsp;=&nbsp;4; Center: α&nbsp;=&nbsp;2; Bottom: α&nbsp;=&nbsp;0.5. As α is reduced the pole separation reduces, and the overshoot increases.]]

====Control of overshoot====
How overshoot may be controlled by appropriate parameter choices is discussed next.

Using the equations above, the amount of overshoot can be found by differentiating the step response and finding its maximum value. The result for maximum step response ''S''<sub>max</sub> is:<ref name=Kuo2>{{cite book |author=Benjamin C Kuo & Golnaraghi F| title=p. 259| year=2003| publisher=Wiley| isbn=0-471-13476-7 | url=http://worldcat.org/isbn/0-471-13476-7}}</ref>

:<math>S_\max= 1 + \exp \left( - \pi \frac { \rho }{ \mu } \right). </math>

The final value of the step response is 1, so the exponential is the actual overshoot itself. It is clear the overshoot is zero if ''μ'' = 0, which is the condition:

:<math> \frac {4 \beta A_0} { \tau_1 \tau_2} = \left( \frac {1} {\tau_1} - \frac {1} {\tau_2} \right)^2. </math>

This quadratic is solved for the ratio of time constants by setting ''x'' = (''τ''<sub>1</sub> / ''τ''<sub>2</sub>)<sup>1/2</sup> with the result

:<math>x = \sqrt{ \beta A_0 } + \sqrt { \beta A_0 +1 }. </math>

Because β ''A''<sub>0</sub> ≫ 1, the 1 in the square root can be dropped, and the result is

:<math> \frac { \tau_1} { \tau_2} = 4 \beta A_0. </math>

In words, the first time constant must be much larger than the second. To be more adventurous than a design allowing for no overshoot we can introduce a factor ''α'' in the above relation:

:<math> \frac { \tau_1} { \tau_2} = \alpha \beta A_0, </math>

and let α be set by the amount of overshoot that is acceptable.

Figure 4 illustrates the procedure. Comparing the top panel (α = 4) with the lower panel (α&nbsp;=&nbsp;0.5) shows lower values for α  increase the rate of response, but increase overshoot. The case α&nbsp;=&nbsp;2 (center panel) is the [[Butterworth filter#Maximal flatness|''maximally flat'']] design that shows no peaking in the [[Bode plot|Bode gain vs. frequency plot]]. That design has the [[rule of thumb]] built-in safety margin to deal with non-ideal realities like multiple poles (or zeros), nonlinearity (signal amplitude dependence) and manufacturing variations, any of which can lead to too much overshoot. The adjustment of the pole separation (that is, setting&nbsp;α) is the subject of [[frequency compensation]], and one such method is [[pole splitting]].

====Control of settling time ====
The amplitude of ringing in the step response in Figure 3 is governed by the damping factor exp(−''ρt''). That is, if we specify some acceptable step response deviation from final value, say Δ, that is:

:<math> S(t)  \le 1 + \Delta, </math>

this condition is satisfied  regardless of the value of β ''A''<sub>OL</sub> provided the time is longer than the settling time, say ''t''<sub>S</sub>, given by:<ref>This estimate is a bit conservative (long) because the factor 1 /sin(φ) in the overshoot contribution  to ''S'' (''t'') has been replaced by  1 /sin(''φ'') ≈ 1.</ref>

:<math> \Delta = e^{- \rho t_S }\text{ or }t_S = \frac { \ln \frac{1}{\Delta} } { \rho } = \tau_2 \frac {2 \ln \frac{1} { \Delta}  } { 1 + \frac { \tau_2 } { \tau_1} } \approx 2 \tau_2 \ln \frac{1} { \Delta}, </math>

where the  τ<sub>1</sub>&nbsp;≫&nbsp;τ<sub>2</sub> is applicable because of  the overshoot control condition, which makes ''τ''<sub>1</sub>&nbsp;=&nbsp;''αβA''<sub>OL</sub> τ<sub>2</sub>. Often the settling time condition is referred to by saying the settling period is inversely proportional to the unity gain bandwidth,  because 1/(2''π''&nbsp;''τ''<sub>2</sub>) is close to this bandwidth for an amplifier with typical [[Frequency compensation#Dominant-pole compensation|dominant pole compensation]]. However, this result is more precise than this [[rule of thumb]]. As an example of this formula, if {{nowrap|1=Δ = 1/e<sup>4</sup> = 1.8 %,}} the settling time condition is  ''t''<sub>S</sub>&nbsp;=&nbsp;8&nbsp;''τ''<sub>2</sub>.

In general, control of overshoot sets the time constant ratio, and settling time ''t''<sub>S</sub> sets&nbsp;τ<sub>2</sub>.<ref name=Johns>{{cite book 
|author=David A. Johns & Martin K W
|title=Analog integrated circuit design
|year= 1997
|pages=234–235
|publisher=Wiley
|location=New York
|isbn=0-471-14448-7
|url=http://worldcat.org/isbn/0-471-14448-7}}</ref><ref name=Sansen>{{cite book 
|author=Willy M C Sansen
|title=Analog design essentials
|page=§0528 p. 163
|year= 2006
|publisher=Springer
|location=Dordrecht, The Netherlands
|isbn=0-387-25746-2
|url=http://worldcat.org/isbn/0-387-25746-2}}</ref><ref>According to Johns and Martin, ''op. cit.'', settling time is significant in [[switched capacitor|switched-capacitor circuits]], for example, where an op amp settling time must be less than half a clock period for sufficiently rapid charge transfer.</ref>

==== System Identification using the Step Response: System with two real poles ====
[[File:PT2 System Step-Response Diagram with required Measurements (2018).png|thumb|340x340px|Step response of the system with <math>x(t)=1</math>. Measure the significant point <math>k</math>, <math>t_{25}</math>and <math>t_{75}</math>.]]
This method uses significant points of the step response. There is no need to guess tangents to the measured Signal. The equations are derived using numerical simulations, determining some significant ratios and fitting parameters of nonlinear equations. See also.<ref>{{Cite web|url=https://hackaday.io/page/4829-identification-of-a-damped-pt2-system|title=Identification of a damped PT2 system {{!}} Hackaday.io|website=hackaday.io|language=en|access-date=2018-08-06}}</ref>

Here the steps:

* Measure the system step-response <math>y(t)</math>of the system with an input step signal <math>x(t)</math>.
* Determine the time-spans <math>t_{25}</math>and <math>t_{75}</math>where the step response reaches 25% and 75% of the steady state output value.
* Determine the system steady-state gain <math>k=A_0</math>with <math>k=\lim_{t\to\infty} \dfrac{y(t)}{x(t)}</math>
* Calculate <math display="block">r=\dfrac{t_{25}}{t_{75}}</math> <math display="block">P=-18.56075\,r+\dfrac{0.57311}{r-0.20747}+4.16423</math> <math display="block">X=14.2797\,r^3-9.3891\,r^2+0.25437\,r+1.32148</math>
* Determine the two time constants <math display="block">\tau_2=T_2=\dfrac{t_{75}-t_{25}}{X\,(1+1/P)}</math> <math display="block">\tau_1=T_1=\dfrac{T_2}{P}</math>
* Calculate the transfer function of the identified system within the Laplace-domain <math display="block">G(s) = \dfrac{k}{(1+s\,T_1)\cdot(1+s\,T_2)}</math>

====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>