Time response formulas

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 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.

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.

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).

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, $e_{ss,D}=-D_0{G}_p(0)/\left[1+G_c(0)G_p(0)\right]$.