The step, ramp and parabolic test input signals can respectively be expressed as
\(R(s)=\dfrac{A}{s};\ R(s)=\dfrac{A}{s^{2}};\ R(s)=\dfrac{2A}{s^{3}}\)
Each test input is the integral of the one before it, and integration divides by s — so the transforms form the chain A/s, A/s2, 2A/s3, option 1.
| Input | r(t), t ≥ 0 | R(s) |
|---|---|---|
| Impulse | \(A\delta(t)\) | \(A\) |
| Step | \(A\) | \(\dfrac{A}{s}\) |
| Ramp | \(At\) | \(\dfrac{A}{s^{2}}\) |
| Parabolic | \(\dfrac{At^{2}}{2}\) | \(\dfrac{A}{s^{3}}\) |
The factor of 2 explains itself once the standard definition is used. The parabolic input is conventionally defined as \(At^{2}/2\), precisely so that its transform is the clean \(A/s^{3}\). If instead the input is written as plain \(At^{2}\), then since \(\mathcal{L}\left\{t^{n}\right\}=n!/s^{n+1}\) the transform carries the factorial:
\(\mathcal{L}\left\{At^{2}\right\}=\dfrac{2A}{s^{3}}\)
which is the form option 1 uses. Either convention is defensible, and the 2A in the numerator is the marker that this question uses the second.
Why the other options fail immediately. Options 2 and 4 begin with \(R(s)=A\), which is the impulse, not the step — the whole list is shifted by one place. Option 3 multiplies by powers of s instead of dividing, which corresponds to repeated differentiation: it describes the impulse and its derivatives, signals of increasing violence rather than the increasingly gentle test inputs intended.
Why these three inputs are the standard set. They test progressively harder tracking tasks — a sudden change of set point, a constant rate of change, a constant acceleration — and each is matched to one error constant through the final-value theorem:
\(e_{ss}=\lim_{s\to0}\dfrac{sR(s)}{1+G(s)H(s)}\)
giving \(K_{p}\) for the step, \(K_{v}\) for the ramp and \(K_{a}\) for the parabola. A type-1 system tracks a step with no error but a ramp with a constant one; each extra integrator in the forward path moves the system one rung up this ladder.
Hence, the correct set is A/s, A/s2 and 2A/s3.
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