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For $\ddot{\theta }+2{\omega }_0\dot{\theta }+{\omega }_0^2\theta =F$ switched on from rest, $\theta (t)=(F/{\omega }_0^2)[1-(1+{\omega }_{0}t)e^{-{\omega }_{0}t}]$. critical damping is the fastest nonoscillatory step response at fixed natural frequency within the ordinary second-order model.

The step response describes a sudden persistent input change. Suppose the input becomes a constant $V_s$ at $t=0$, and the initial capacitor voltage is $v_0$. Finite electric current leaves that voltage continuous. Subtract $V_s$ from the RC low-pass filter equation: the remaining error satisfies $\tau de/dt+e=0$. Separating variables givesFor an initially uncharged capacitor and a $0$ to $3\mathrm{V}$ step in the constructed example,At one time constant, $t=103.4\mu \mathrm{s}$, the voltage is $1.8964\mathrm{V}$, about $63.2\%$ of its final value. The voltage error then continues its exponential decay. This is different from a pure delay: the output starts changing immediately, yet takes time to approach the target.

Define settling time relative to the total voltage change: require $|v_{\mathrm{out}}-V_s|\le ϵ|v_0-V_s|$ thereafter, with $0<ϵ<1$. The exponential decay is monotone, soFor $1\%$ settling, $t_{0.01}=4.6052\tau \approx 476.17\mu \mathrm{s}$; the $10\%$ to $90\%$ rise takes $\tau \ln 9\approx 227.19\mu \mathrm{s}$. Reducing the cutoff frequency by doubling $C$ doubles both times. This reciprocal relation is the useful connection between signal bandwidth and response speed: stronger suppression of rapid noise comes with slower tracking of wanted changes.

The zero-state response is the output produced by an applied input when the system starts with zero internal state. For a linear time-invariant system it separates the effect of the input from the zero-input response caused by an initially stored state. A zero initial output alone need not mean zero internal state.

For a causal linear time-invariant system with impulse response $h$, the zero-state response is the convolutionAn instantaneous feedthrough term, if present, is represented by a Dirac delta term in $h$. Equivalently, for constant matrices in $\dot{x}=Ax+Bu$, $y=Cx+Du$ with $x(0)=0$,For nonzero $x(0)=x_0$, linearity adds the zero-input response $C{e}^{At}{x}_0$. Here the matrix exponential propagates the state left by the earlier input.

For example, $\tau \dot{y}+y=u$ with $\tau >0$, $y(0)=0$ and a constant input $u(t)=A$ for $t\ge 0$ has the step response $y_{\mathrm{zs}}=A(1-e^{-t/\tau })$. At one time constant it has reached about $63.2\%$ of its final value. If the initial value instead equals $y_0$, the total output is $A(1-e^{-t/\tau })+y_0{e}^{-t/\tau }$. A charged capacitor in an RC circuit provides a direct physical example: discharge it before repeating the same input step to isolate the zero-state response, and compare with a run from a known initial electric potential difference.^[ref]

With compatible transforms, the zero-state input-output relation is $Y=HU$, where $H$ is the transfer function. For a nonzero initial state, a unilateral Laplace transform includes additional initial-condition terms. Those terms cannot be absorbed into an input-independent transfer function multiplying $U$.