Read a step response
A single target change reveals speed, overshoot, settling and residual error.
Read overshoot and settling from a trace using a stated target and tolerance band.
Block diagrams show how signals are connected. A step response shows what that closed loop does after a sudden target change.
Review the previous lesson: Follow the signal, then reduce the blocksFirst, picture it
A fast response is not automatically good. A controller may reach the target quickly, overshoot it, oscillate, and use far more actuator effort than a slightly slower design.
Build the idea step by step
A step input changes suddenly and then stays fixed. Use its response to answer three questions. How long does the output take to rise through a stated range, often 10% to 90% of its final change? How far does it pass the target? When does it enter and stay inside an allowed error band? These are rise time, overshoot, and settling time. State the range, band, and observation time. If a recorded run ends before settling, report that it has not settled within the recording.
Read the equations
For a nonzero step target and suitable starting point, overshoot is the peak amount above target as a percentage of target size.
What the symbols mean
- nonzero step target, in output units
- highest output reached, in output units
- percentage overshoot, in percent
The example starts at zero and steps to a positive target. Other initial values and negative steps need a definition based on the step direction and size. A short recording can measure a peak, but cannot establish long-term stability.
Read a speed step
A speed loop starts at zero and aims for 10 radians per second. Its highest measured speed is 12.
The peak is 2 above target, or 20% overshoot.
For a 2% settling band, output must enter and stay between 9.8 and 10.2. If it stays there from 5 seconds onward, report a 5 second settling time and the 2% band.
More detail and model assumptions
Read one step response in stages
At the step, the reference changes first; the measured output cannot usually jump with it. Rise time describes the early climb, overshoot describes a peak beyond the final target, and settling time asks when the response enters a stated band and stays there. Final error compares the response near the end with the target. These are different questions about the same curve.
A run that crosses the target once has not necessarily settled. If it leaves the tolerance band again, the clock must continue. If a short recorded run ends while still moving, report ‘not settled’ rather than inventing a settling time.
Metrics need conditions
Overshoot as a percentage needs a meaningful nonzero target change. Rise time and the usual percent overshoot are step-tracking metrics; they do not automatically fit cart-pole recovery from an initial tilt or a disturbance pulse. In those runs, inspect peak angle, rail travel and recovery instead.
Compare two runs only after fixing the target, initial state, load, noise seed and duration. A faster response produced by a larger applied voltage is a design tradeoff, so read control effort and saturation beside speed.
Build a second-order system you can see
A horizontal mass is attached to a spring and viscous damper. Choose displacement x positive to the right. A positive external force F pushes right; the spring pulls with −kx and the damper resists velocity with −c x′. Newton's law gives m x″ = F − c x′ − kx. The acceleration needs both position and velocity, so the state is [x, x′], not position alone.
With zero initial state the force-to-position transfer function is G(s)=1/(ms²+cs+k). Dividing its denominator by m gives s²+2ζωn s+ωn², where ωn=√(k/m) and ζ=c/(2√(km)). The physical numerator remains 1/m; the standard unit-gain step model has numerator ωn² instead. Here the static force-to-position gain is 1/k m/N.
Separate release motion from a force response
In the free-response preset, F=0 but x(0)=0.25 m. The spring already stores energy, so the mass moves without an applied force. This is the zero-input response. In the step preset, the initial position and velocity are zero and a constant force starts at t=0: this is the zero-state response. A nonzero initial state plus a force produces their sum in this linear model.
Energy connects the equations to the animation. E=½m(x′)²+½kx² is kinetic plus spring energy. Its derivative is F x′−c(x′)². A damper with c≥0 removes energy; it does not store it. With F=0 and c=0, energy is constant even though position repeatedly crosses zero. Crossing the equilibrium is not the same as coming to rest.
More damping is not an unlimited speed improvement
For m=1 kg and k=16 N/m, ωn=4 rad/s and critical damping is ccrit=8 N·s/m. Compare c=1.6, 8 and 16 with the same 2 N step and zero initial state. All positively damped cases approach x∞=F/k=0.125 m. Damping changes the transient, not this static balance.
Underdamping has a decaying oscillation. Critical damping separates complex poles from real poles. In the overdamped case the slower real pole can approach zero as c increases, giving a long tail. Critical damping gives the fastest monotonic zero-state step response within this fixed-m, fixed-k family; it is not a universal optimum for every objective or initial condition.
Three frequencies, three different questions
ωn is the undamped natural frequency. For ζ<1, free oscillation has frequency ωd=ωn√(1−ζ²). A damped force-to-displacement gain has a resonance maximum at ωr=ωn√(1−2ζ²) only for 0<ζ<1/√2. With ζ=0.2, these are 4, about 3.92 and about 3.84 rad/s. Divide rad/s by 2π to get Hz; entering 4 Hz instead of 4 rad/s is a different experiment.
The sinusoidal steady-state amplitude is |F0|/√((k−mω²)²+(cω)²) when the decaying transient has become negligible. The first visible peak is not that steady amplitude. With zero damping the free part never decays; at ω=ωn a nonzero sine force grows the oscillation without a bounded steady state. The frequency panel is a plant analysis, not a feedback-loop margin calculation.
Connect the example to the lab
Compare two motor runs with identical target and load. One may settle earlier but require a larger voltage spike. Inspect both speed and control-input traces.
Compare physical second-order responsesMake a prediction before moving a control.
- PREDICTWill the same 2 N step finish at a different position when only damping increases?
- CHANGE ONE THINGOpen One force, different transients. Pin c=1.6, then change only c to 8 and 16 N·s/m.
- OBSERVECompare the equilibrium line, overshoot and the late tail. Check that initial state, mass, stiffness and force are identical.
- EXPLAINUse kx∞=F for the common equilibrium and the poles of ms²+cs+k for the different paths. Explain why a short window may show Not settled.
A metric without its definition and conditions can mislead. Saturation may also make the textbook linear step-response formulas inapplicable.
Check your understanding+
Why should a settling-time report state its tolerance band?
A 2% band and a 5% band can yield different times for the same trace.Why the next lesson follows
A nice-looking finite trace does not prove the loop is stable or that its final error is zero. We check those separately.