First-Order Step Response Lab

adjust gain, step size, and time constant to see the exponential response evolve

Worked example — First-Order Step Response Lab

Output y(t)	0.6321
Steady-state value	1.0000
Progress to steady state	63.21%
Elapsed, in time constants	1.000 τ
Rise time, 10% → 90%	2.197
Settling time, to within 2%	3.912
Your values, substituted	0.6321
final value K·A	0

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — First-Order Step Response Lab

A first-order system is one whose rate of change is proportional to how far it still has to go. Push it with a sudden constant input — a step — and it does not jump; it closes the remaining gap at a steady fractional rate, tracing an exponential approach to a new resting value. The time constant τ is how long that gap takes to shrink to 1/e of its original size.

What each symbol means

K: the gain: how much output the system eventually gives per unit of input.
A: the size of the step applied at t = 0.
τ: the time constant, in the same units as t. It is the only symbol here that affects the timing.
K·A: the final value the output settles at, once the exponential has run out.

Where the formula comes from

Start from the physical premise rather than the formula. A first-order system changes at a rate proportional to its remaining distance from the target: the further it has to go, the faster it moves. Written down, dy/dt = (K·A − y)/τ, which rearranges to τ·dy/dt + y = K·A.
Separate the variables and integrate: dy/(K·A − y) = dt/τ gives −ln(K·A − y) = t/τ + c.
Apply the starting condition. The system is at rest before the step, so y(0) = 0, which fixes the constant and leaves K·A − y = K·A·e^(−t/τ). Read that line on its own: the gap decays exponentially, and it is the gap — not the output — that is the simple thing.
So y(t) = K·A(1 − e^(−t/τ)). The fraction still to go is e^(−t/τ), which depends on t and τ only through their ratio. That single fact produces the whole milestone table: at one τ the gap is 1/e, so 63.21% is done; at three τ, 95.02%; at five τ, 99.33%. Nothing in those numbers can be changed by the gain or the step size.

How to read what you see

The green curve is y(t) and the grey dashed line is the final value K·A it never quite reaches. The vertical gridlines mark 1τ to 5τ, so you can read the milestone percentages straight off the curve. The orange dashed line is step 4 made visible: it leaves the origin at the initial slope K·A/τ and meets the grey line at exactly one time constant, with a dot on the crossing. The blue band is ±2% of the final value — the usual engineering definition of “arrived” — and the vertical marker is the instant in the Time box. The table below highlights whichever τ row you are currently standing on.

Assumes: One storage element, a system at rest before the step, and a step that arrives instantly. It also assumes linearity: double the step and every point on the curve doubles, which is exactly why the milestone percentages are fixed. Nothing here models saturation, so a gain of 10 and a step of 10 will happily predict an output of 100 that a real actuator could not deliver.
Breaks when: A first-order response is monotonic — it can never overshoot its final value, and it can never oscillate. So if a real system rings, or crosses its target before settling back, it is not first order, and no value of τ will fit it. That takes at least two storage elements, which is where the second-order behaviour of a damped oscillator comes from. Dead time breaks it differently: a transport delay shifts the entire curve to the right without changing its shape, so it cannot be absorbed into τ either.

The time constant belongs to the system; the step size does not 🖖

Change the input and the whole curve scales, but the clock does not move. With τ = 3 the response has covered 63.21% of its journey after three seconds — and it still reads 63.21% whether the step is 1 or 5, and whether the gain is 1 or 4. Only τ touches the timing: raise it to 10 and those same three seconds buy just 25.92%. This is why a time constant is quoted as a property of a motor, a thermometer or an RC circuit without anyone mentioning what was done to it. The gain and the step decide where the output ends up; τ alone decides how long it takes to get there.

The first instant already knows where it will finish 🖖

The response leaves the origin with slope K·A/τ. Extend that straight line — the dashed orange one above — and it crosses the final value at exactly t = τ. Always: any gain, any step size, any time constant, and negative steps too. This is how a time constant is read off an oscilloscope trace with a ruler and no arithmetic. Draw the tangent at the start, see where it meets the final level, read off the time. The curve itself is nowhere near finished at that moment — it has covered 63.21% — because it has been slowing down since the first instant, and the tangent is the only part of the motion that never slowed.

A thermometer, a motor and a capacitor are the same equation 🖖

The axes on this page carry no units, and that is deliberate. τ·dy/dt + y = K·A describes anything with one place to put something and one route to lose it, and in every case τ is the size of the store divided by how fast the store drains. For a capacitor through a resistor that is R·C. For a thermometer in moving air it is the thermal capacity over the heat-transfer coefficient times area. For a motor coming up to speed it is inertia over damping. For a well-stirred tank it is volume over flow rate. Four different sciences, one curve, one number — which is why control engineering can say “a first-order lag” without ever saying what the thing is made of.

Problem solved in full

Output that has reached 63.21% one time constant into a step input 5 steps
A first-order system is one time constant into a step input, and the output has reached 63.21%. Show that 63.21% — and the rise time, and the settling time — are not properties of this system at all.
1. The governing equation says the output chases the input at a rate proportional to how far behind it is. Solving it gives the exponential approach, with the gain and the step size appearing only as a multiplier out front.
  
  $τ \frac{d y}{d t} + y = K A, y (0) = 0 ⟹ y (t) = K A (1 - e^{- t / τ})$
2. So at t = τ the fraction reached is 1 − e⁻¹. Gain and step cancel: the 63.21% is a fact about e, not about this system.
  
  $t = τ : \frac{y}{K A} = 1 - e^{- 1} = 0.6321 = 63.21%$
3. The same cancellation gives the rise time. Invert the response for the instant at any fraction, then subtract the 10% and 90% times. The difference is τ ln 9, and ln 9 = 2.197 is a pure number.
  
  $t_{p} = - τ ln (1 - p) : t_{r} = t_{90} - t_{10} = τ ln 9 = 2.197 τ$
4. Settling to within 2% asks when the remaining exponential falls to 0.02, which is τ ln 50 = 3.912τ. Again no gain, no step size.
  
  $1 - e^{- t / τ} = 0.98 ⟹ t_{s} = τ ln 50 = 3.912 τ$
5. Which means the whole thing runs backwards. Measure a rise time on real hardware and τ = tᵣ/2.197 — one stopwatch reading characterises the dynamics completely.
  
  $2 τ : 86.47%, 3 τ : 95.02%, 5 τ : 99.33% τ = \frac{t _{r}}{2.197}$
Answer

The tool prints 0.6321, 63.21%, a rise time of 2.197 and a settling time of 3.912. Every one is τ multiplied by a constant that came out of e, and not one of them depends on the gain or the step. Click the fast and slow presets in turn and you can watch the split: the output, the steady-state value and the 63.21% do not move at all, while every time on the panel multiplies by exactly ten — 2.197 becomes 21.972 and 3.912 becomes 39.120. There is only one first-order step response in the world, and τ merely decides how fast you watch it. Two time constants reaches 86.47%, three reaches 95.02%, five reaches 99.33% — which is where engineering's "five time constants and it's settled" comes from.

$y = K A (1 - e^{- t / τ}) : 0.6321, ln 9, ln 50 are constants, not settings$

Learning path

Resistance, then reactance

Leads to RLC resonance

References (1)

The standard treatment of the first-order lag, including the rise and settling times this page reports and why they are pure multiples of τ: K. J. Åström & R. M. Murray, Feedback Systems: An Introduction for Scientists and Engineers. Princeton University Press, 2008.

By the 42Math team · Updated 2026-05-21

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

fast response - τ = 1, read at t = 1 → 63.21% of the way to a steady 1
slow response - τ = 10, read at t = 10 → 63.21% again: ten times slower, identical progress, because the clock runs in units of τ
four time constants - τ = 1 read at t = 4 → 98.17%, which is the usual definition of “finished”
double gain - K = 2 moves the destination to 2, and t = 5 against τ = 3 reaches 1.6222 — 81.11% of the way
negative step - a step of −1 drives the output down to −0.6321 at t = τ, the same 63.21% of an inverted journey