Question

A linear time invariant (LTI) system has a transfer function \[ G(s)=\frac{10(s+1)}{s(s^2+2s+5)} \] The system is placed in unity negative feedback configuration.
For a unit ramp reference, the steady state error is ________. (rounded off to two decimal places)

Hint:

For a Type 1 system, the steady state error to a ramp input equals \(1/K_v\), where \(K_v=\lim_{s\to0}sG(s)\).

Answer 1

Correct Answer: 0.5

Step 1: Write the error transfer function.
For unity negative feedback, the error signal in the Laplace domain is $E(s)=\dfrac{R(s)}{1+G(s)}$, where $R(s)$ is the reference input.

Step 2: Insert the ramp input.
A unit ramp has $R(s)=\dfrac{1}{s^2}$, so
\[ E(s)=\frac{1}{s^2\left(1+G(s)\right)} \]

Step 3: Apply the final value theorem.
Since the closed loop system is stable, the steady state error is
\[ e_{ss}=\lim_{s\to0} sE(s)=\lim_{s\to0}\frac{s}{s^2\left(1+G(s)\right)}=\lim_{s\to0}\frac{1}{s\left(1+G(s)\right)} \]

Step 4: Expand and take the limit as $s\to0$.
\[ s\left(1+G(s)\right) = s + sG(s) \] As $s\to0$, the first term $s\to0$. The second term is exactly the velocity constant,
\[ \lim_{s\to0} sG(s) = \lim_{s\to0}\frac{10(s+1)}{s^2+2s+5}=\frac{10}{5}=2 \] So
\[ \lim_{s\to0} s\left(1+G(s)\right) = 0+2 = 2 \]

Step 5: Read off the steady state error.
\[ e_{ss}=\frac{1}{2}=0.5 \]
\[ \boxed{0.5} \]