Step response: practice problems with step-by-step solutions
Free Step response practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: first-order systems, second-order parameters, physical systems and design.
Free practice problemEasy
An uncharged capacitor C = 47 µF is charged from a 5 V step source through a resistor R = 4.7 kΩ. Find the capacitor voltage vC at t = 220.9 ms.
A step is an input that jumps from 0 to a constant value at t = 0 and then stays there. Think of switching on a heater, or pressing the accelerator and holding it.
The step response is how the output follows. It cannot jump at once: it moves towards a new final value y(∞) and settles there.
Curve a rises smoothly and never overshoots. That is a first-order system, like an RC circuit or a heated tank.
Curve b overshoots, swings a few times and then settles. That is an underdamped second-order system, like a car suspension or a mass on a spring.
The final value comes from the DC gain G(0): a step of size U settles at y(∞) = G(0)·U. This is the final value theorem from Laplace transforms.
Self-check A system G(s) = 3/(2s + 1) receives a step of size 4. What is the final value y(∞)?
12
3
4
6
Answer: 12. G(0) = 3/(0 + 1) = 3, so y(∞) = 3·4 = 12. The value 4 forgets the gain, 3 forgets the step size, and 6 divides by the 2 in 2s, which vanishes at s = 0.
2First-order systems
A first-order system has the standard form below. K is the static gain and τ the time constant in seconds.
G(s) = K/(τs + 1) y(t) = K·U·(1 − e−t/τ)
If G(s) is written as b/(s + a), divide the top and bottom by a. Then K = b/a and τ = 1/a.
After 1τ the output has done 63% of its change, after 3τ 95%, after 5τ more than 99%.
The time to reach a fraction x of the final value: t = −τ·ln(1 − x). For 90%, t = τ·ln 10 = 2.3τ.
The 10–90% rise time is tr = τ·ln 9 ≈ 2.2τ.
For an RC circuit τ = R·C. Note that kΩ times µF gives milliseconds.
If you measure the response, you can find τ backwards. If the output reached a fraction x at time t, then τ = −t/ln(1 − x).
Worked example
G(s) = 10/(s + 5) receives a step of size 2 at t = 0. Find the final value and the time to reach 90% of it.
1
Standard form
Divide the top and bottom by 5:
G(s) = 2/(0.2s + 1) ⇒ K = 2, τ = 0.2 s
2
Final value
y(∞) = K·U = 2·2 = 4
3
Time to 90%
Solve 1 − e−t/τ = 0.9, so e−t/τ = 0.1:
t = −τ·ln(0.1) = 0.2·2.303 = 0.461 s
Answer
y(∞) = 4 and the output reaches 90% after t ≈ 0.461 s.
Self-check What is the time constant of G(s) = 10/(s + 5)?
0.2 s
10 s
2 s
5 s
Answer: 0.2 s. Written as 2/(0.2s + 1), the time constant is τ = 1/5 = 0.2 s. 5 s confuses τ with a, and 2 is the gain K, not a time.
3Second-order systems
A second-order system has a quadratic denominator. Compare it with the standard form:
G(s) = K/(s² + 2ζωns + ωn²)
The constant term is ωn², so the natural frequency is ωn = √(constant term), in rad/s.
The coefficient of s is 2ζωn, so the damping ratio is ζ = (coefficient of s)/(2ωn).
For 0 < ζ < 1 the response overshoots and oscillates. A smaller ζ means more overshoot.
For a unit step the final value is G(0) = K/ωn². Five numbers describe the shape of the response:
ωd is the damped frequency: how fast the response oscillates.
tp is the peak time, when the output is highest. %OS is the percent overshoot above the final value.
ts is the settling time, after which the output stays within 2% of the final value.
The two poles are s = σ ± jωd, with real part σ = −ζωn. Their distance from the origin is ωn, and cos θ = ζ.
Self-check A system has the denominator s² + 4s + 25. What is the damping ratio ζ?
0.08
0.4
0.8
2
Answer: 0.4. ωn = √25 = 5, and 2ζωn = 4 gives ζ = 4/(2·5) = 0.4. 0.8 forgets the 2, 0.08 divides by ωn² instead of ωn, and 2 is ζωn.
4The method step by step
Decide the order: a denominator of degree 1 is first order, degree 2 is second order.
First order: write G(s) as K/(τs + 1) and read K and τ.
Second order: compare the denominator with s² + 2ζωns + ωn². Find ωn first, then ζ.
Find the final value y(∞) = G(0)·U.
Apply the formula for what is asked: y(t), a time, tp, %OS, ts or σ.
Check the units: times in seconds, ω in rad/s, and the angle in e… and π formulas in radians.
5A worked example
This is a typical medium question. Try each step before you open it.
Worked example
G(s) = 16/(s² + 4s + 16) receives a unit step. Find ζ, ωn, the percent overshoot and the peak time.
1
Natural frequency
The constant term is ωn² = 16.
ωn = √16 = 4 rad/s
2
Damping ratio
The coefficient of s is 2ζωn = 4.
ζ = 4/(2·4) = 0.5
ζ is between 0 and 1, so the response overshoots.
3
Percent overshoot
ζπ/√(1 − ζ²) = 0.5·π/√0.75 = 1.814
%OS = 100·e−1.814 = 16.3%
4
Peak time
ωd = 4·√(1 − 0.25) = 3.464 rad/s
tp = π/3.464 = 0.907 s
5
Check the answer
The final value is G(0) = 16/16 = 1, so the peak is about 1.163. The figure in the previous section uses this same system.
The settling time 4/(ζωn) = 4/2 = 2 s is later than the peak, as it must be.
Answer
ωn = 4 rad/s, ζ = 0.5, %OS ≈ 16.3% and tp ≈ 0.907 s.
6Common mistakes
Treating the rise as linear. An exponential reaches 90% after 2.3τ, not after 0.9τ: use t = −τ·ln(1 − x).
Using ln(x) instead of ln(1 − x). The logarithm takes the part still missing.
Reading τ = a from b/(s + a). The time constant is 1/a.
Using ωn² as ωn. Take the square root of the constant term.
Forgetting the 2 in 2ζωn, or using ωn instead of ωd in tp = π/ωd.
Giving the overshoot as a fraction (0.163) when the question asks for percent (16.3%).
Self-check G(s) = 2/(s + 4) receives a step. How long does the output take to reach 90% of its final value?
0.026 s
0.225 s
0.576 s
9.21 s
Answer: 0.576 s. τ = 1/4 = 0.25 s and t = −0.25·ln(0.1) = 0.576 s. 0.225 s treats the rise as linear (0.9τ). 0.026 s uses ln(0.9) instead of ln(0.1). 9.21 s uses a = 4 instead of τ.
7How exam questions look
Easy: first-order systems. You find y(∞), the output at a given time, the time to reach a percentage, τ from a measured point, or the charging time of an RC circuit.
Medium: second-order parameters. From G(s) = K/(s² + a₁s + a₀) you find ζ, ωn, %OS, tp, ts or the real part σ of the poles.
Hard: physical systems and design. A mass–spring–damper or RLC circuit, or ζ and ωn worked back from a measured overshoot and peak time.
For a mass m, damper c and spring k, the equation m·x″ + c·x′ + k·x = F gives:
ωn = √(k/m) ζ = c/(2√(k·m)) x(∞) = F/k
To work back from an overshoot, solve the %OS formula for ζ:
A first-order system G(s) = 2/(2s + 1) receives a step input of size 2 at t = 0, starting from rest. Find the final value y(∞) of the output.
G(s) = 2/(2s + 1)U = 2
Find y(∞)
Show hints
Final value theorem: y(∞) = lim s·G(s)·U/s = G(0)·U.
G(0) is the DC gain.
G(0) = 2.
Show the answer and the step-by-step solution
Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).
DC gain
G(0) = 2
Final value
y(∞) = G(0)·U = 2·2 = 4
Answer
y(∞) = 4
Example 2Medium
A system has the transfer function G(s) = 213/(s² + 8s + 71) and receives a unit step input. Find the percent overshoot %OS of the step response.
G(s) = 213/(s² + 8s + 71)
Find %OS(%)
Show hints
Compare the denominator with s² + 2ζωₙs + ωₙ².
ωₙ = √71 = 8.426 rad/s, ζ = 8/(2ωₙ).
ζ = 0.4747.
Show the answer and the step-by-step solution
Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).
Compare with the standard form
The denominator must match s² + 2ζωₙs + ωₙ².
ωₙ² = 71 → ωₙ = 8.426 rad/s
2ζωₙ = 8 → ζ = 8/(2·8.426) = 0.4747
Result
%OS = 100·e−ζπ/√(1−ζ²) = 100·e−1.694 = 18.37 %
Answer
%OS = 18.37 %
Example 3Hard
A mass m = 2.5 kg hangs on a spring k = 800 N/m with a viscous damper c = 9 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 10 N is suddenly applied. Find the time tp at which the displacement is largest.
m = 2.5 kgk = 800 N/mc = 9 N·s/mF = 10 N
Find tp(s)
Show hints
Divide by m: x″ + (c/m)·x′ + (k/m)·x = F/m, so ωₙ = √(k/m) and ζ = c/(2√(km)).
ωₙ = 17.89 rad/s, ζ = 0.1006.
The final displacement is F/k = 12.5 mm.
Show the answer and the step-by-step solution
Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).
Natural frequency and damping ratio
ωₙ = √(k/m) = √(800/2.5) = 17.89 rad/s
ζ = c/(2√(km)) = 9/(2·√(800·2.5)) = 0.1006
Result
ωd = 17.8 rad/s
tp = π/ωd = 0.1765 s
Answer
tp = 0.1765 s
Questions and answers
What is Step response?
A step is an input that jumps from 0 to a constant value at t = 0 and then stays there. Think of switching on a heater, or pressing the accelerator and holding it.
The step response is how the output follows. It cannot jump at once: it moves towards a new final value y(∞) and settles there.
How do I solve Step response problems step by step?
Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).
What are common mistakes in Step response?
The output settles at K·U, not at U: the DC gain is G(0) = 2.
Express the overshoot in percent: multiply by 100.
The exponent is −ζπ/√(1 − ζ²).
What do the Step response exercises cover?
Easy: first-order systems
Medium: second-order parameters
Hard: physical systems and design
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.