Question

A closed loop system is shown in the figure where $k > 0$ and $\alpha > 0$. The steady state error due to a ramp input \((R(s)=\alpha/s^{2})\) is _____________

• A. $\dfrac{2\alpha}{k}$
• B. $\dfrac{\alpha}{k}$
• C. $\dfrac{\alpha}{2k}$
• D. $\dfrac{\alpha}{4k}$

Hint:

For unity feedback: step $\Rightarrow e_{ss}=\dfrac{1}{K_p}$, ramp $\Rightarrow e_{ss}=\dfrac{\alpha}{K_v}$, parabolic $\Rightarrow e_{ss}=\dfrac{\beta}{K_a}$. Here $K_v=\lim_{s\to 0}sG(s)$.

Answer 1

The Correct Option is A

Unity negative feedback with forward path $G(s)=\dfrac{k}{s(s+2)}$.
For a ramp input $R(s)=\dfrac{\alpha}{s^{2}}$, the steady-state error is $e_{ss}=\dfrac{\alpha}{K_v}$ where $K_v=\displaystyle\lim_{s\to 0}sG(s)$.
Compute $K_v$: \; $sG(s)=\dfrac{ks}{s(s+2)}=\dfrac{k}{s+2}\xrightarrow[s\to 0]{}\dfrac{k}{2}$.
Therefore, $e_{ss}=\dfrac{\alpha}{K_v}=\dfrac{\alpha}{k/2}=\dfrac{2\alpha}{k}$.
\[ \boxed{\;e_{ss}=\dfrac{2\alpha}{k}\;} \]