Thestep response of a system in a given initial state consists of the time evolution of its outputs when its control inputs areHeaviside step functions. Inelectronic engineering andcontrol theory, step response is the time behaviour of the outputs of a generalsystem when its inputs change from zero to one in a very short time. The concept can be extended to the abstract mathematical notion of adynamical system using anevolution parameter.

From a practical standpoint, knowing how the system responds to a sudden input is important because large and possibly fast deviations from the long term steady state may have extreme effects on the component itself and on other portions of the overall system dependent on this component. In addition, the overall system cannot act until the component's output settles down to some vicinity of its final state, delaying the overall system response. Formally, knowing the step response of a dynamical system gives information on thestability of such a system, and on its ability to reach one stationary state when starting from another.

This section provides a formal mathematical definition of step response in terms of the abstract mathematical concept of adynamical system${\displaystyle \mathfrak{S}}$: all notations and assumptions required for the following description are listed here.

• ${\displaystyle t\in T}$ is theevolution parameter of the system, called "time" for the sake of simplicity,
• ${\displaystyle \mathit{x}{|}_t\in M}$ is thestate of the system at time${\displaystyle t}$, called "output" for the sake of simplicity,
• ${\displaystyle \mathrm{\Phi }:T\times M\to M}$ is the dynamical systemevolution function,
• ${\displaystyle \mathrm{\Phi }(0,\mathit{x})={\mathit{x}}_0\in M}$ is the dynamical systeminitial state,
• ${\displaystyle H(t)}$ is theHeaviside step function

For a general dynamical system, the step response is defined as follows:

${\displaystyle \mathit{x}{|}_t={\mathrm{\Phi }}_{\{H(t)\}}\left(t,{\mathit{x}}_0\right).}$

It is theevolution function when the control inputs (orsource term, orforcing inputs) are Heaviside functions: the notation emphasizes this concept showingH(t) as a subscript.

For alineartime-invariant (LTI) black box, let${\displaystyle \mathfrak{S}\equiv S}$ for notational convenience: the step response can be obtained byconvolution of theHeaviside step function control and theimpulse responseh(t) of the system itself

${\displaystyle a(t)=(h\ast H)(t)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}h(\tau )H(t-\tau )d\tau ={\int }_{-\mathrm{\infty }}^{t}h(\tau )d\tau .}$

which for an LTI system is equivalent to just integrating the latter. Conversely, for an LTI system, the derivative of the step response yields the impulse response:

${\displaystyle h(t)=\frac{d}{dt}a(t).}$

However, these simple relations are not true for a non-linear ortime-variant system.^[1]

Instead of frequency response, system performance may be specified in terms of parameters describing time-dependence of response. The step response can be described by the following quantities related to itstime behavior,

• overshoot
• rise time
• settling time
• ringing

In the case oflinear dynamic systems, much can be inferred about the system from these characteristics.Below the step response of a simple two-pole amplifier is presented, and some of these terms are illustrated.

In LTI systems, the function that has the steepest slew rate that doesn't create overshoot or ringing is the Gaussian function. This is because it is the only function whose Fourier transform has the same shape.

This section describes the step response of a simplenegative feedback amplifier shown in Figure 1. The feedback amplifier consists of a mainopen-loop amplifier of gainA_OL and a feedback loop governed by afeedback 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 (seenegative feedback amplifier):

${\displaystyle A_{FB}=\frac{A_{OL}}{1+\beta A_{OL}},}$

whereA_OL =open-loop gain,A_FB =closed-loop gain (the gain with negative feedback present) andβ =feedback factor.

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:

${\displaystyle A_{OL}=\frac{A_0}{1+j\omega \tau },}$

with zero-frequency gainA₀ and angular frequency ω = 2πf. This forward amplifier has unit step response

${\displaystyle S_{OL}(t)=A_0(1-e^{-t/\tau })}$,

an exponential approach from 0 toward the new equilibrium value ofA₀.

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

${\displaystyle A_{FB}=\frac{A_0}{1+\beta A_0}\cdot \frac{1}{1+j\omega \frac{\tau }{1+\beta A_0}}.}$

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 isτ / (1 +βA₀), so it is faster than the forward amplifier's response by a factor of 1 +βA₀:

${\displaystyle S_{FB}(t)=\frac{A_0}{1+\beta A_0}\left(1-e^{-t(1+\beta A_0)/\tau }\right),}$

As the feedback factorβ 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.

In the case that the open-loop gain has two poles (twotime constants,τ₁,τ₂), the step response is a bit more complicated. The open-loop gain is given by:

${\displaystyle A_{OL}=\frac{A_0}{(1+j\omega {\tau }_1)(1+j\omega {\tau }_2)},}$

with zero-frequency gainA₀ and angular frequencyω = 2πf.

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

${\displaystyle A_{FB}=\frac{A_0}{1+\beta A_0}\cdot \frac{1}{1+j\omega \frac{{\tau }_1+{\tau }_2}{1+\beta A_0}+(j\omega {)}^2\frac{{\tau }_1{\tau }_2}{1+\beta A_0}}.}$

The time dependence of the amplifier is easy to discover by switching variables tos =jω, whereupon the gain becomes:

${\displaystyle A_{FB}=\frac{A_0}{{\tau }_1{\tau }_2}\cdot \frac{1}{s^2+s\left(\frac{1}{{\tau }_1}+\frac{1}{{\tau }_2}\right)+\frac{1+\beta A_0}{{\tau }_1{\tau }_2}}}$

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

${\displaystyle 2s=-\left(\frac{1}{{\tau }_1}+\frac{1}{{\tau }_2}\right)\pm \sqrt{{\left(\frac{1}{{\tau }_1}-\frac{1}{{\tau }_2}\right)}^2-\frac{4\beta A_0}{{\tau }_1{\tau }_2}},}$

which shows for large enough values ofβA₀ 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, eithers₊ ors₋; see Figure 2:

${\displaystyle s_{\pm }=-\rho \pm j\mu ,}$

with

${\displaystyle \rho =\frac{1}{2}\left(\frac{1}{{\tau }_1}+\frac{1}{{\tau }_2}\right),}$

and

${\displaystyle \mu =\frac{1}{2}\sqrt{\frac{4\beta A_0}{{\tau }_1{\tau }_2}-{\left(\frac{1}{{\tau }_1}-\frac{1}{{\tau }_2}\right)}^2}.}$

Using polar coordinates with the magnitude of the radius to the roots given by |s| (Figure 2):

${\displaystyle |s|=|s_{\pm }|=\sqrt{{\rho }^2+{\mu }^2},}$

and the angular coordinate φ is given by:

${\displaystyle \cos \varphi =\frac{\rho }{|s|},\sin \varphi =\frac{\mu }{|s|}.}$

Tables ofLaplace transforms show that the time response of such a system is composed of combinations of the two functions:

${\displaystyle e^{-\rho t}\sin (\mu t)\text{and}{e}^{-\rho t}\cos (\mu t),}$

which is to say, the solutions are damped oscillations in time. In particular, the unit step response of the system is:^[2]

${\displaystyle S(t)=\left(\frac{A_0}{1+\beta A_0}\right)\left(1-e^{-\rho t}\frac{\sin \left(\mu t+\varphi \right)}{\sin \varphi }\right),}$

which simplifies to

${\displaystyle S(t)=1-e^{-\rho t}\frac{\sin \left(\mu t+\varphi \right)}{\sin \varphi }}$

whenA₀ tends to infinity and the feedback factorβ 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 βA₀. Because ρ is a sum of reciprocals of time constants, it is interesting to notice that ρ is dominated by theshorter of the two.

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 [ 1 − exp(−ρt) ] and [ 1 + exp(−ρt) ]. 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 calledringing. Theovershoot is the maximum swing above final value, and clearly increases with μ. Likewise, theundershoot is the minimum swing below final value, again increasing with μ. Thesettling 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 [ 1 − exp(−ρt) ] and [ 1 + exp (−ρt) ] 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:

1. Feedback controls the amplitude of oscillation about final value for a given open-loop amplifier and given values of open-loop time constants, τ₁ and τ₂.
2. 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.

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 responseS_max is:^[3]

${\displaystyle S_{max}=1+\exp \left(-\pi \frac{\rho }{\mu }\right).}$

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

${\displaystyle \frac{4\beta A_0}{{\tau }_1{\tau }_2}={\left(\frac{1}{{\tau }_1}-\frac{1}{{\tau }_2}\right)}^2.}$

This quadratic is solved for the ratio of time constants by settingx = (τ₁ /τ₂)^1/2 with the result

${\displaystyle x=\sqrt{\beta A_0}+\sqrt{\beta A_0+1}.}$

Because βA₀ ≫ 1, the 1 in the square root can be dropped, and the result is

${\displaystyle \frac{{\tau }_1}{{\tau }_2}=4\beta A_0.}$

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 factorα in the above relation:

${\displaystyle \frac{{\tau }_1}{{\tau }_2}=\alpha \beta A_0,}$

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 = 0.5) shows lower values for α increase the rate of response, but increase overshoot. The case α = 2 (center panel) is themaximally flat design that shows no peaking in theBode gain vs. frequency plot. That design has therule 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 α) is the subject offrequency compensation, and one such method ispole splitting.

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:

${\displaystyle S(t)\le 1+\mathrm{\Delta },}$

this condition is satisfied regardless of the value of βA_OL provided the time is longer than the settling time, sayt_S, given by:^[4]

${\displaystyle \mathrm{\Delta }=e^{-\rho t_S}\text{ or }{t}_S=\frac{\ln \frac{1}{\mathrm{\Delta }}}{\rho }={\tau }_2\frac{2\ln \frac{1}{\mathrm{\Delta }}}{1+\frac{{\tau }_2}{{\tau }_1}}\approx 2{\tau }_2\ln \frac{1}{\mathrm{\Delta }},}$

where the τ₁ ≫ τ₂ is applicable because of the overshoot control condition, which makesτ₁ = αβA_OL τ₂. Often the settling time condition is referred to by saying the settling period is inversely proportional to the unity gain bandwidth, because 1/(2π τ₂) is close to this bandwidth for an amplifier with typicaldominant pole compensation. However, this result is more precise than thisrule of thumb. As an example of this formula, ifΔ = 1/e⁴ = 1.8 %, the settling time condition ist_S = 8 τ₂.

In general, control of overshoot sets the time constant ratio, and settling timet_S sets τ₂.^[5][6][7]

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.^[8]

Here the steps:

• Measure the system step-response${\displaystyle y(t)}$of the system with an input step signal${\displaystyle x(t)}$.
• Determine the time-spans${\displaystyle t_{25}}$and${\displaystyle t_{75}}$where the step response reaches 25% and 75% of the steady state output value.
• Determine the system steady-state gain${\displaystyle k=A_0}$with${\displaystyle k=\underset{t\to \mathrm{\infty }}{lim}{\displaystyle \frac{y(t)}{x(t)}}}$
• Calculate$${\displaystyle r={\displaystyle \frac{t_{25}}{t_{75}}}}$$$${\displaystyle P=-18.56075r+{\displaystyle \frac{0.57311}{r-0.20747}}+4.16423}$$$${\displaystyle X=14.2797r^3-9.3891r^2+0.25437r+1.32148}$$
• Determine the two time constants$${\displaystyle {\tau }_2=T_2={\displaystyle \frac{t_{75}-t_{25}}{X(1+1/P)}}}$$$${\displaystyle {\tau }_1=T_1={\displaystyle \frac{T_2}{P}}}$$
• Calculate the transfer function of the identified system within the Laplace-domain$${\displaystyle G(s)={\displaystyle \frac{k}{(1+s{T}_1)\cdot (1+s{T}_2)}}}$$

Next, the choice of pole ratioτ₁/τ₂ is related to the phase margin of the feedback amplifier.^[9] The procedure outlined in theBode 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 frequencyf_0 dB lies between the lowest pole atf₁ = 1/(2πτ₁) and the second pole atf₂ = 1/(2πτ₂). As indicated in Figure 5, this condition is satisfied for values of α ≥ 1.

Using Figure 5 the frequency (denoted byf_0 dB) is found where the loop gain βA₀ satisfies the unity gain or 0 dB condition, as defined by:

${\displaystyle |\beta A_{\text{OL}}(f_{\text{0 db}})|=1.}$

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

${\displaystyle f_{\text{0 dB}}=\beta A_0{f}_1.}$

The phase margin is the departure of the phase atf_0 dB from −180°. Thus, the margin is:

${\displaystyle {\varphi }_m={180}^{\circ }-\arctan (f_{\text{0 dB}}/f_1)-\arctan (f_{\text{0 dB}}/f_2).}$

Becausef_0 dB /f₁ = βA₀ ≫ 1, the term inf₁ is 90°. That makes the phase margin:

In particular, for caseα = 1,φ_m = 45°, and forα = 2,φ_m = 63.4°. Sansen^[10] recommendsα = 3,φ_m = 71.6° as a "good safety position to start with".

If α is increased by shorteningτ₂, the settling timet_S also is shortened. Ifα is increased by lengtheningτ₁, the settling timet_S is little altered. More commonly, bothτ₁andτ₂ change, for example if the technique ofpole 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 makingf₂ a fitting parameter, referred to as an "equivalent second pole" position.^[11]

• Impulse response
• Overshoot (signal)
• Pole splitting
• Rise time
• Settling time
• Time constant

• Kuo power point slides; Chapter 7 especially

• Analog circuits
• Electronic design
• Dynamical systems
• Classical control theory
• Signal processing
• Electronic amplifiers
• Transient response characteristics

• Articles with short description
• Short description is different from Wikidata