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Can feedback delay change correction timing even when the physical task stays the same? Susilaradeya and colleagues tested this in a human tracking experiment. Participants pressed with both index fingers against fixed force sensors; the two forces controlled a screen cursor along two diagonal axes. A target moved in slow circular motion, usually at $0.2$ rotations per second. The researchers sampled the forces at $50$ samples per second and inserted an additional delay of $0$, $100$, $200$, $300$, or $400$ milliseconds before displaying the cursor. Each trial lasted $20$ seconds; the $70$ trials included $14$ per delay, in pseudorandom order. ^[ref]

The readout was fluctuation in cursor angular velocity, analyzed by Fourier transforms over approximately $10$ seconds from the middle of each trial. The principal peak was around $2$ hertz without added delay, around $1.4$ hertz with $100$ milliseconds, and around $1$ hertz with $200$ milliseconds. Thus changing observation timing changed the timing of corrective motion. ^[ref] This is an empirical motivation for taking stale observations seriously. Human feedback control includes prediction and other dynamics; these observations do not measure $k$ in the ideal integrator model or test its $k\tau =\pi /2$ boundary.

A feedback delay is a time interval between an event and its returning influence. Multiple biochemical steps can introduce an effective lag; a fixed delay is an idealization, not a universal description of signal transduction. To isolate timing, let $x$ be a signed displacement from an equilibrium point and use normalized dimensionless variables:The local removal is $-x$ and the returning influence is negative at every delay. At $d=0$ the ordinary differential equation reduces to the decay already illustrated under signaling feedback. For $d>0$ a delay differential equation requires a history on $[-d,0]$, not just an initial condition at one instant. Negative values of $x$ are displacements, not negative chemical concentrations.

An exponential function $x=e^{\lambda t}$ gives the characteristic equation of a delay differential equationPutting $\lambda =i\omega$ and equating real and imaginary parts gives $\cos (\omega d)=-1/2$ and $\omega =2\sin (\omega d)$. The first crossing with $\omega >0$ therefore occurs atFor this particular mathematical model, the equilibrium point is asymptotically stable for $0\le d<d_c$. No nonnegative real delayed feedback characteristic root can satisfy the equation; the first loss of asymptotic stability is the indicated pair of imaginary delayed feedback characteristic roots. Differentiating the characteristic equation of a delay differential equation along a root yieldsThus the pair moves into growth as the delay increases through $d_c$. The neutral mode at $d_c$ is a periodic function, but its amplitude depends on history: this linear mathematical model does not select an attracting limit cycle.

The figure plots $x(t)=e^{at}\cos (bt)$ using a delayed feedback characteristic root $\lambda =a+ib$ and that same formula on the history interval. At $d=0.7$ the displayed mode decays; at $d=1.5$ it grows. These are exact modes up to numerical root accuracy, not step responses with identical histories. Molecular bounds and a changed return law would be needed to interpret what happens after an unstable biological response becomes large.

For the constant-coefficient model $x^{\prime }(t)=-kx(t-\tau )$ with $k>0$ and $\tau \ge 0$, the zero equilibrium point is asymptotically stable if and only if $k\tau <\pi /2$. For positive $\tau$, perturbations are measured by the supremum norm of the initial history of delayed feedback; in the stable regime, every continuous history produces a solution tending to zero. At equality, nondecaying sinusoids prevent asymptotic stability. For $k\tau >\pi /2$, growing oscillatory solutions exist. These statements concern this ideal linear time-invariant system, rather than an arbitrary controlled plant. The corresponding linear equation and its threshold are developed in Cain and Reynolds, section 6.4, pages 174–177.

Substitute the exponential function $x(t)=e^{st}$, allowing a complex number $s$. The characteristic equation isFor positive feedback delay this is not a characteristic polynomial: it has infinitely many roots. A root's real part determines growth or decay of its exponential mode, and its imaginary part determines the angular frequency. To find the first boundary, set $s=i\omega$ with $\omega >0$ and use Euler's formula:The smallest positive crossing is $\omega =k$ and $\omega \tau =\pi /2$. Thus at ${\tau }_c=\pi /(2k)$ the complex conjugate roots are $s=\pm ik$. In fact, with the matching initial history of delayed feedback,solves the delay differential equation exactly: a quarter-cycle feedback delay turns the past displacement into precisely the current derivative. Its cycle lasts $2\pi /k=4{\tau }_c$. The amplitude is arbitrary; this linear model does not select a finite-amplitude oscillation.

The roots cross toward growth as feedback delay increases. Differentiating the characteristic equation at fixed $k$ givesThis calculation identifies the direction of the first destabilization. The full threshold also requires ruling out earlier unstable roots: if $s=\alpha +i\beta$ with $\alpha \ge 0$ and $\beta >0$, the characteristic equation implies $\cos (\beta \tau )\le 0$ and $0<\beta \le k$, hence $k\tau \ge \beta \tau \ge \pi /2$. A real nonnegative root is impossible because $s+k{e}^{-s\tau }>0$ there. This supplies the geometric reason the threshold is a quarter-cycle rather than an arbitrary fitting constant.

The tradeoff is concrete. Without feedback delay, increasing $k$ shortens the exponential decay time constant $1/k$. With known $\tau >0$, the same proportional control must satisfy $k<\pi /(2\tau )$. For $\tau =0.2\mathrm{s}$, the threshold is $k\approx 7.854{\mathrm{s}}^{-1}$: $k=5{\mathrm{s}}^{-1}$ is stable and $k=8{\mathrm{s}}^{-1}$ is unstable. A design close to the boundary leaves little room for additional feedback delay; reducing $k$ or reducing measurement latency increases that room. The zero-delay time constant is only a comparison scale, not the decay time of the delayed response. Frank discusses the gain–delay compromise in Control Theory Tutorial, section 13.4 and figure 13.3.

A feedback delay means a controller reacts to an earlier plant output. An accurate measurement can still be stale: when the plant output has already crossed the reference signal, a controller using the earlier value may continue pushing it away from the target. In a pure constant-delay model, $y_m(t)=y(t-\tau )$ with $\tau \ge 0$. This isolates timing from measurement noise; it does not describe every sampling or communication mechanism. ^[ref]

For the ideal scalar controlled plant $\dot{x}=u$, a zero reference signal, no control disturbance, and proportional control with constant feedback gain $k>0$, the equations areWith $\tau =0$, this is an ordinary differential equation with exponential decay $x(t)=x(0)e^{-kt}$. With $\tau >0$, it is a delay differential equation, requiring an initial history of delayed feedback. The delay stability of proportional integrator control depends on the product $k\tau$: even this simple controlled plant loses asymptotic stability at $k\tau =\pi /2$.

For positive feedback delay $\tau$, an initial condition at a single instant is insufficient. Prescribe a continuous function $\varphi$ throughout $[-\tau ,0]$: for $0\le t\le \tau$, the derivative depends on $x(t-\tau )=\varphi (t-\tau )$. A right derivative at $t=0$ is understood; the earlier history need not itself obey the subsequent control law. Once this first interval is computed, it supplies the delayed values for the next interval. This successive integration is the method of steps for a delay differential equation. ^[ref]

For the constant initial history of delayed feedback $\varphi (t)=1$, direct integration givesFor $k=1$, $\tau =1.2$, the plant output crosses zero at $t=1$, while its measured output remains positive until $t=2.2$. At $t=1.4$, the delayed command therefore increases the magnitude of the actual deviation. This short exact calculation explains the first diagram without relying on a numerical approximation.

The measured output is the information about the plant output available to the controller. It may differ from the actual plant output because of measurement noise or feedback delay. In the ideal examples $y_m=y=x$, so the controller observes the actual current state exactly.

Proportional control chooses a control input proportional to the current tracking error. For the ideal controlled plant and constant reference signal $r$, its ordinary differential equation is $\dot{x}=k(r-x)+w$. With $w=0$ and initial condition $x(0)=x_0$, the exact response isFor $r=1$, $x_0=0$ and $k=2$, this gives $x(1)=1-e^{-2}\simeq 0.864665$ and $x(2)=1-e^{-4}\simeq 0.981684$. To verify the formula, substitute it into the ordinary differential equation and check the initial condition; its decaying exponential function explains the approach to the goal.

With a constant control disturbance, setting $\dot{x}=0$ gives the equilibrium point $x_{\ast }=r+w/k$ and steady-state tracking error $e_{\ast }=-w/k$. For $w=0.2$, doubling the feedback gain from $k=2$ to $k=4$ reduces the offset from $0.1$ to $0.05$. This improvement is a statement about the ideal model: it neglects feedback delay and measurement noise, so it does not establish that arbitrarily large feedback gains are safe. ^[ref]

In the first diagram's switched control disturbance experiment, $k=2$ and $w=0.2$ from $t=2$ give, for $t\ge 2$,Thus $x(4)\simeq 1.097833$ and $x(8)\simeq 1.099999$. At the equilibrium point, the controller supplies $u=-0.2$ to cancel the control disturbance. Proportional control needs a nonzero tracking error to produce that persistent correction.

The steady-state tracking error iswhen this finite limit exists. A finite steady-state tracking error does not require the reference signal, the plant output, or the control input to become constant: all can grow while their difference settles. If the tracking error grows without bound or keeps oscillating, report that behavior rather than assigning a finite steady-state tracking error. Eventual accuracy and asymptotic stability are different questions.

For the first-order low-pass filter $\tau \dot{y}+y=Ku+d$ introduced in reference signal, assume exact measurement, constant control disturbance $d$, and no actuator saturation or feedback delay. Under proportional control $u=k_p(r-y)$ with $k_p>0$, a constant reference signal $r=R$ givesThe decaying homogeneous solution has rate $(1+K{k}_p)/\tau >0$, so this equilibrium point is approached from any finite initial condition. A nonzero tracking error is needed to maintain the control input that balances the model's tendency to relax toward $y=d$. This differs from the accumulating controlled plant $\dot{x}=u+w$, for which proportional control already tracks a constant reference signal exactly when $w=0$. The presence or absence of a step offset depends on both the controlled plant and the controller.

For a ramp reference signal $r=vt$, the same proportional control givesIts long-time particular solution grows with slope $v/(1+K{k}_p)$ when $v\ne 0$. Thus the controlled plant and homogeneous closed-loop response are stable, but the tracking error has no finite steady-state tracking error. A changing goal requires a different performance test from a fixed destination.

Now use proportional-integral control $u=k_{p}e+k_{i}z$, $\dot{z}=e$, with $k_i>0$. For the same controlled plant, differentiate the error equation on intervals where the inputs are smooth:The characteristic polynomial $\tau {\lambda }^2+(1+K{k}_p)\lambda +K{k}_i$ has both roots of a polynomial in the open left half-plane: real roots have negative sum and positive product, and a nonreal complex conjugate pair has negative real part. This proves decay of the homogeneous error for this model. It does not prove stability of an arbitrary controlled plant controlled by positive gains. At a reference jump, use the original ordinary differential equations to set the new initial conditions, rather than treating the reference as smooth at that jump.

With a constant reference signal and constant control disturbance, proportional-integral control therefore gives $e_{\mathrm{ss}}=0$. At its equilibrium point, $y_{\ast }=R$, $u_{\ast }=(R-d)/K$, and $z_{\ast }=(R-d)/(K{k}_i)$. The stored integral maintains the required command while the current tracking error vanishes. For a ramp reference signal, however, the forcing above is $v$, andThe tracking error settles even though the accumulated integral and the required control input grow. This finite offset is the error needed to keep increasing that command. For $r=a{t}^2/2$ with $a\ne 0$, the forcing is $\tau a+at$ and even this proportional-integral control leaves a tracking error growing with slope $a/(K{k}_i)$. Integral action is therefore a precise mechanism for selected input classes, not a promise of perfect tracking for every reference signal. ^[ref]

Here are completely explicit synthetic examples with $y(0)=z(0)=0$, $d=0$, and no sensor bias. Choose $\tau =2\mathrm{s}$, $K=1{}^{\circ }\mathrm{C}$ per percentage point, $k_p=1$ percentage point per degree, and $k_i=0.5$ percentage points per degree-second. Thus $K{k}_p=1$ is dimensionless, and $K{k}_i=0.5{\mathrm{s}}^{-1}$. In the following formulas, $t$ denotes the numerical time in seconds and $y,e,r$ denote numerical temperature rises in degrees. For the unit-step reference signal $r=1$, proportional control and proportional-integral control respectively giveFor the ramp reference signal $r=0.2t$, the corresponding solutions areThe control inputs here are numerical heater-command percentage points. Substitution into $2\dot{y}+y=u$ and into the controller equations checks these solutions without invoking an unverified limiting rule. The proportional control step offset is $0.5$ degrees; the proportional-integral control step offset is zero. The proportional control ramp error grows; the proportional-integral control ramp offset is $0.4$ degrees, exactly $v/(K{k}_i)$.

The conclusions have specific limits. With constant sensor bias $b$, stable proportional-integral control for a feasible constant reference signal drives the measured tracking error to zero but leaves the physical discrepancy ${\epsilon }_{\mathrm{ss}}=b$. Under proportional control, the measured offset instead becomes $(R-b-d)/(1+K{k}_p)$. If actuator saturation prevents the required control input, zero step offset is not achievable by these formulas. A changing control disturbance contributes $-\dot{d}$ to the error equation, while feedback delay can change stability; neither is covered by the constant-disturbance result. These restrictions are what make steady-state tracking error useful: it identifies the input, measurement convention, and model for which an accuracy claim is actually justified.