Step response

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

${\displaystyle {\boldsymbol {x}}|_{t}=\Phi _{\{H(t)\}}\left(t,{{\boldsymbol {x}}_{0}}\right).}$

It is the evolution function when the control inputs (or source term, or forcing inputs) are Heaviside functions: the notation emphasizes this concept showing H(t) as a subscript.

For a linear time-invariant (LTI) black box, let ${\displaystyle {\mathfrak {S}}\equiv S}$ for notational convenience: the step response can be obtained by convolution of the Heaviside step function control and the impulse response h(t) of the system itself

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

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 or time-variant system.[1]

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 gain A₀ 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 of A₀.

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 is τ / (1 + β A₀), so it is faster than the forward amplifier's response by a factor of 1 + β A₀:

${\displaystyle S_{FB}(t)={\frac {A_{0}}{1+\beta A_{0}}}\left(1-e^{-t(1+\beta A_{0})/\tau }\right),}$

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.

In the case that the open-loop gain has two poles (two time 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 gain A₀ and angular frequency ω = 2πf.

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

${\displaystyle A_{FB}={\frac {A_{0}}{1+\beta A_{0}}}\;\cdot \;\ {\frac {1}{1+j\omega {\frac {\tau _{1}+\tau _{2}}{1+\beta A_{0}}}+(j\omega )^{2}{\frac {\tau _{1}\tau _{2}}{1+\beta A_{0}}}}}.}$

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 = 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 of βA₀ 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₊ or s₋; 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:

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

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

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

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

${\displaystyle S(t)=\left({\frac {A_{0}}{1+\beta A_{0}}}\right)\left(1-e^{-\rho t}\ {\frac {\sin \left(\mu t+\phi \right)}{\sin \phi }}\right)\ ,}$

which simplifies to

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

when A₀ 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 βA₀. 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.

Figure 3: Step-response of a linear two-pole feedback amplifier; time is in units of 1/ρ, that is, in terms of the time constants of A_OL; curves are plotted for three values of mu = μ, which is controlled by 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 [ 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 called ringing. The 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 [ 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.

Figure 4: Step response for three values of α. Top: α  = 4; Center: α = 2; Bottom: α = 0.5. As α is reduced the pole separation reduces, and the overshoot increases.

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_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 if μ = 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 setting x = (τ₁ / τ₂)^1/2 with the result

${\displaystyle x={\sqrt {\beta A_{0}}}+{\sqrt {\beta A_{0}+1}}.}$

Because β A₀ ≫ 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 factor α 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 the maximally flat design that shows no peaking in the 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 α) is the subject of frequency compensation, and one such method is pole 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:

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

this condition is satisfied regardless of the value of β A_OL provided the time is longer than the settling time, say t_S, given by:[4]

${\displaystyle \Delta =e^{-\rho t_{S}}{\text{ or }}t_{S}={\frac {\ln {\frac {1}{\Delta }}}{\rho }}=\tau _{2}{\frac {2\ln {\frac {1}{\Delta }}}{1+{\frac {\tau _{2}}{\tau _{1}}}}}\approx 2\tau _{2}\ln {\frac {1}{\Delta }},}$

where the τ₁ ≫ τ₂ is applicable because of the overshoot control condition, which makes τ₁ = 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 typical dominant pole compensation. However, this result is more precise than this rule of thumb. As an example of this formula, if Δ = 1/e⁴ = 1.8 %, the settling time condition is t_S = 8 τ₂.

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

Step response of the system with ${\displaystyle x(t)=1}$. Measure the significant point ${\displaystyle k}$, ${\displaystyle t_{25}}$and ${\displaystyle t_{75}}$.

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 ${\displaystyle k=\lim _{t\to \infty }{\dfrac {y(t)}{x(t)}}}$
• Calculate
  $${\displaystyle r={\dfrac {t_{25}}{t_{75}}}}$$

  
  $${\displaystyle P=-18.56075\,r+{\dfrac {0.57311}{r-0.20747}}+4.16423}$$

  
  $${\displaystyle X=14.2797\,r^{3}-9.3891\,r^{2}+0.25437\,r+1.32148}$$

  
• Determine the two time constants
  $${\displaystyle \tau _{2}=T_{2}={\dfrac {t_{75}-t_{25}}{X\,(1+1/P)}}}$$

  
  $${\displaystyle \tau _{1}=T_{1}={\dfrac {T_{2}}{P}}}$$

  
• Calculate the transfer function of the identified system within the Laplace-domain
  $${\displaystyle G(s)={\dfrac {k}{(1+s\,T_{1})\cdot (1+s\,T_{2})}}}$$

  

Figure 5: Bode gain plot to find phase margin; scales are logarithmic, so labeled separations are multiplicative factors. For example, f_{0 dB} = βA₀ × f₁.

Next, the choice of pole ratio τ₁/τ₂ is related to the phase margin of the feedback amplifier.[9] The procedure outlined in the 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_0 dB lies between the lowest pole at f₁ = 1/(2πτ₁) and the second pole at f₂ = 1/(2πτ₂). As indicated in Figure 5, this condition is satisfied for values of α ≥ 1.

Using Figure 5 the frequency (denoted by f_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 at f_0 dB from −180°. Thus, the margin is:

${\displaystyle \phi _{m}=180^{\circ }-\arctan(f_{\text{0 dB}}/f_{1})-\arctan(f_{\text{0 dB}}/f_{2}).}$

Because f_0 dB / f₁ = βA₀ ≫ 1, the term in f₁ is 90°. That makes the phase margin:

${\displaystyle {\begin{aligned}\phi _{m}&=90^{\circ }-\arctan(f_{\text{0 dB}}/f_{2})\\&=90^{\circ }-\arctan {\frac {\beta A_{0}f_{1}}{\alpha \beta A_{0}f_{1}}}\\&=90^{\circ }-\arctan {\frac {1}{\alpha }}=\arctan \alpha \,.\end{aligned}}}$

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

If α is increased by shortening τ₂, the settling time t_S also is shortened. If α is increased by lengthening τ₁, the settling time t_S is little altered. More commonly, both τ₁ and τ₂ 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 f₂ a fitting parameter, referred to as an "equivalent second pole" position.[11]