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: Identify the system type.
The transfer function \(G(s)=\dfrac{10(s+1)}{s(s^2+2s+5)}\) has exactly one pole at the origin (\(s=0\)) in the denominator. A system with one pole at the origin in the open loop transfer function is called a Type 1 system.

Step 2: Recall the steady state error rule for a ramp input.
For a unity feedback system, the steady state error to a unit ramp input is
\[ e_{ss}=\frac{1}{K_v} \] where \(K_v\) is the velocity error constant, defined as
\[ K_v=\lim_{s\to0} sG(s) \] For a Type 1 system \(K_v\) is finite and non zero, so \(e_{ss}\) comes out finite, which is exactly what should happen here.

Step 3: Compute the velocity error constant.
\[ K_v=\lim_{s\to0} s\cdot\frac{10(s+1)}{s(s^2+2s+5)}=\lim_{s\to0}\frac{10(s+1)}{s^2+2s+5} \] Substituting \(s=0\):
\[ K_v=\frac{10(0+1)}{0+0+5}=\frac{10}{5}=2 \]

Step 4: Compute the steady state error.
\[ e_{ss}=\frac{1}{K_v}=\frac{1}{2}=0.5 \]

Final Answer:
The steady state error for a unit ramp input is \[ \boxed{0.5} \]