Time response and steady-state error

For a unit step into $T(s)=K/(\tau s+1)$. At $t=\tau$ the output has made 63.2% of its final change, and its initial slope is $K/\tau$.

The exact settling values are $-\ln 0.02=3.91$ and $-\ln 0.05=3.00$ time constants. A first-order system without a zero has no peak time and no overshoot.

For $T(s)=1/(\tau s+1)$, which has unity DC gain. After the transient, the output runs parallel to the input but $\tau$ seconds behind it, so the steady-state error to a unit ramp is $\tau$. This is the same as a unity-feedback loop with $G(s)=1/(\tau s)$, whose $K_v=1/\tau$.

Unity negative feedback, with $K>0$ and $a>0$. Feedback moves the pole from $-a$ to $-(a+K)$, so the closed loop is faster than the plant. Its DC gain $K/(a+K)$ is less than 1. That gives a steady-state unit-step error of $a/(a+K)$, the same as $1/(1+K_p)$ with $K_p=K/a$.

Unity DC gain. If the numerator is some other constant, $\zeta$ and ${\omega }_n$ still come from the denominator. Only the final value changes.

Holds for $0<\zeta <1$. Here $\sigma =\zeta {\omega }_n$ is the decay rate, ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}$ is the damped frequency, and $\theta$ is the angle of the pole measured from the negative real axis.

For $0<\zeta <1$ in the standard form. The oscillation stays inside the envelope $1\pm e^{-\zeta {\omega }_{n}t}/\sqrt{1-{\zeta }^2}$. The settling-time formula is based on this envelope.

The derivative of the step response above, for $0<\zeta <1$. It starts at zero and decays at the same envelope rate, $\zeta {\omega }_n$.

For $\zeta >1$: two distinct real poles at $-p_1$ and $-p_2$, with $0<p_1<p_2$. Note that $p_1{p}_2={\omega }_n^2$ and $p_1+p_2=2\zeta {\omega }_n$. When $p_2$ is much larger than $p_1$, the $e^{-p_{2}t}$ term dies out quickly and $c(t)\approx 1-e^{-p_{1}t}$.

Unity negative feedback, with $K>0$ and $a>0$. The loop is critically damped at $K=a^2/4$ and underdamped for $K>a^2/4$. Raising $K$ raises ${\omega }_n$ and lowers $\zeta$. The real part of the poles, $\sigma =\zeta {\omega }_n=a/2$, stays fixed as long as $\zeta <1$.

Underdamped systems only. $\theta$ must be in radians.

Half of the damped period $T_d=2\pi /{\omega }_d$. It depends only on the imaginary part of the poles.

Valid for $0<\zeta <1$, and depends only on $\zeta$. For $\zeta \ge 1$ the step response has no overshoot.

Enter $M_p$ as a fraction (0.163, not 16.3). This is what you use in design questions that specify the overshoot.

Envelope approximations for $0<\zeta <1$. They give the time it takes $e^{-\zeta {\omega }_{n}t}$ to fall to about 2% ($e^{-4}=1.8\%$) or 5% ($e^{-3}=5.0\%$), and $1/(\zeta {\omega }_n)$ is the time constant of the envelope. They are not the exact time at which the response enters the band for good. For the worked example below, the 2% formula gives 1.33 s and the exact value is about 1.19 s. Use these formulas unless the question gives a different rule.

An empirical approximation for $0<\zeta <1$, within about 5% of the exact value over that range.

For $0<\zeta <1$. $x_k$ are overshoots on the same side of the final value, one damped period apart. More generally, the $k$-th extremum of the unit-step response (at $t=k\pi /{\omega }_d$) differs from the final value by $M_p^k$ times the final value, falling alternately above and below it.

The number of damped cycles completed within the settling time, using the approximation $t_s=4/(\zeta {\omega }_n)$. With the 5% rule, $t_s=3/(\zeta {\omega }_n)$, it is $3\sqrt{1-{\zeta }^2}/(2\pi \zeta )$.

The zero is at $s=-z$ with $z>0$, and the DC gain is unchanged. The derivative term is positive while the output rises, so the response rises faster and overshoots more (or starts to overshoot). The effect fades as $z$ grows.

Write the far pole in time-constant form $(1+s/p)$, then delete that factor. Here the pair $-1\pm j3$ dominates, the pole at $-10$ is 10 times further left, and $T(0)=1$ both before and after the reduction.

Valid only if every pole of $sF(s)$ lies in the open left half plane, that is, only if $f(t)$ actually settles. It gives wrong answers for unstable systems and for sinusoidal inputs (which put poles on the imaginary axis).

For a step of size $A$ into a stable closed loop $T(s)=C(s)/R(s)$. So for any loop structure, the step error $r-c$ settles at $A\left[1-T(0)\right]$.

The error here is $r-c$. The first form needs unity feedback. The second works for any loop structure, including non-unity feedback, as long as you know the closed-loop $T(s)=C(s)/R(s)$. Both need a stable closed loop. For a ramp or parabolic input the limit can come out infinite, which means the error grows without bound.

$N$ is the number of open-loop poles at the origin (integrators), counted after cancelling any factor of $s$ common to the numerator and denominator. $K$ here is the gain in time-constant form. Type belongs to the open-loop function, not to $T(s)$, and it has nothing to do with the order of the system. With unity feedback, the type of $G(s)$ decides which standard inputs give zero, finite or unbounded $r-c$ error. With non-unity feedback, the type of $G(s)H(s)$ describes the actuating error $R-HC$, not $r-c$.

The position, velocity and acceleration constants. For unity feedback, $H(s)=1$ and these constants give the $r-c$ error. With non-unity feedback, constants taken from $G(s)H(s)$ give the actuating error $R-HC$.

For unity feedback and a stable closed loop. When the input is a sum of these inputs, the errors add (linearity).

Turns a non-unity loop into a unity loop with the same $T(s)$. The type and error constants of $G_e(s)$ give the true $r-c$ error. $G_e(s)$ can have right-half-plane poles even when the closed loop is stable. That does not matter, because only closed-loop stability decides whether the final value theorem applies.

Unity negative feedback, with controller $G_c(s)$ followed by plant $G_p(s)$, and disturbance $D(s)$ added at the plant input (between $G_c$ and $G_p$). With $e=r-c$, $e_{ss,D}$ is the part of the steady-state error caused by the disturbance. Find it with $r=0$, then add it to the error caused by $r$. The closed loop must be stable. For a step disturbance of size $D_0$: if $G_c$ contains an integrator, $e_{ss,D}=0$. If only $G_p$ has an integrator and $G_c(0)$ is finite and nonzero, $e_{ss,D}=-D_0/G_c(0)$. If neither has one, $e_{ss,D}=-D_0{G}_p(0)/\left[1+G_c(0)G_p(0)\right]$.

For unity feedback, a stable closed loop, and $e=r-c$. The $k_i$ are the coefficients of the power series of $W(s)$ about $s=0$, which you find by long division in ascending powers of $s$. For an input that is a polynomial in $t$, the series stops after finitely many terms. Some texts write the series as $C_0+C_{1}s+C_2{s}^2/2!+\cdots$, so that $C_i=i!k_i$. The error comes out the same either way. Link to the static constants: for type 0, $k_0=1/(1+K_p)$; for type 1, $k_0=0$ and $k_1=1/K_v$; for type 2, $k_0=k_1=0$ and $k_2=1/K_a$. Example: $G(s)=10/(s(s+1))$ gives $W(s)=(s+s^2)/(10+s+s^2)=0.1s+0.09s^2-\cdots$. So for $r(t)=t^2/2$ the error is $e_{ss}(t)=0.1t+0.09$, which grows without bound, just as $K_a=0$ predicts.