Steady-state error of a type-1 system to a unit-ramp input
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A unity-feedback system has open-loop transfer function $$G(s)H(s) = \frac{100}{s\,(s+10)}.$$
The steady-state error for a unit-ramp input is ________.
Show the step-by-step derivation
Answer
0.1 unitless unitless
Step-by-step derivation
- Determine the system type, which is the number of poles at the origin in the open-loop transfer function. There is a single factor of $s$ in the denominator, so this is a type-1 system.
- A type-1 system tracks a ramp with finite error (a type-0 system would have infinite ramp error, a type-2 system zero). The relevant constant is the velocity error constant: $$K_v = \lim_{s \to 0} s\,G(s)H(s).$$
- Evaluate it: $$K_v = \lim_{s\to 0} s \cdot \frac{100}{s(s+10)} = \lim_{s \to 0}\frac{100}{s+10} = \frac{100}{10} = 10.$$
- The steady-state error for a unit ramp is the reciprocal: $$e_{ss} = \frac{1}{K_v} = \frac{1}{10} = \mathbf{0.1}.$$
- Where this comes from. With $E(s) = R(s)/(1+GH)$ and $R(s) = 1/s^2$, the final value theorem gives $$e_{ss} = \lim_{s\to 0}\frac{s \cdot 1/s^2}{1+GH} = \lim_{s\to 0}\frac{1}{s + sGH} = \frac{1}{K_v},$$ since $sGH \to K_v$ and the leading $s \to 0$.
The trap this question is built around
Using $K_p = \lim_{s\to 0} GH$, which is the position error constant for a step input. Here it is infinite, correctly telling you the step error is zero - but it says nothing about the ramp. Match the error constant to the input.