Question

A unit step input is applied to the negative feedback system whose closed loop transfer function is $\frac{63}{S^2 + 4.8S + 9}$, then the steady state value of the output is}

• A. 7
• B. 1
• C. 63
• D. 9

Hint:

For a unit step input, the steady-state output of a stable system is simply the DC gain of the transfer function.
To find the DC gain, just substitute $s = 0$ directly into the transfer function: $T(0) = \frac{63}{9} = 7$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem requires finding the steady-state value of the output, $y(\infty)$, for a given closed-loop transfer function $T(s)$ when subjected to a unit step input.

Step 2: Key Formula or Approach:
The Final Value Theorem (FVT) is used to find the steady-state value of a system:
\[ y(\infty) = \lim_{s \to 0} s Y(s) \]
where $Y(s) = T(s) X(s)$, and $X(s)$ is the Laplace transform of the input.
For a unit step input:
\[ X(s) = \frac{1}{s} \]

Step 3: Detailed Explanation:

• Write the expression for the output $Y(s)$:
\[ Y(s) = \frac{63}{s^2 + 4.8s + 9} \cdot \frac{1}{s} \]

• Apply the Final Value Theorem:
\[ y(\infty) = \lim_{s \to 0} s \left( \frac{63}{s^2 + 4.8s + 9} \cdot \frac{1}{s} \right) \]
\[ y(\infty) = \lim_{s \to 0} \frac{63}{s^2 + 4.8s + 9} \]

• Substituting $s = 0$:
\[ y(\infty) = \frac{63}{0 + 0 + 9} = \frac{63}{9} = 7 \]

Step 4: Final Answer:
The steady-state value of the output is $7$, which corresponds to Option (A).