The maximum allowable value of \( n \) such that the output \( y(t) \), due to any step disturbance signal \( d(t) \), becomes zero at steady-state, is ________ (in integer).
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Apply the Final Value Theorem to the output response due to the disturbance input to determine the steady-state value. For the steady-state output to be zero, the limit as $s \to 0$ of $sY(s)$ must be zero.
Step 1: Determine the transfer function from the disturbance $D(s)$ to the output $Y(s)$.
$\frac{Y(s)}{D(s)} = \frac{s^2}{s^n (s^3 + s^2 + 1)} = \frac{1}{s^{n-2} (s^3 + s^2 + 1)}$ Step 2: Determine the Laplace transform of the step disturbance.
$D(s) = \frac{1}{s}$ Step 3: Find the Laplace transform of the output $Y(s)$ due to the step disturbance.
$Y(s) = \frac{1}{s^{n-2} (s^3 + s^2 + 1)} \cdot \frac{1}{s} = \frac{1}{s^{n-1} (s^3 + s^2 + 1)}$ Step 4: Apply the Final Value Theorem to find the steady-state output $y_{ss}$.
$y_{ss} = \lim_{s \to 0} s Y(s) = \lim_{s \to 0} s \cdot \frac{1}{s^{n-1} (s^3 + s^2 + 1)} = \lim_{s \to 0} \frac{1}{s^{n-2} (s^3 + s^2 + 1)}$ Step 5: Determine the condition for $y_{ss} = 0$.
For $y_{ss} = 0$, the power of $s$ in the denominator must be positive, i.e., $n - 2 > 0$, which means $n > 2$. The smallest integer value of $n$ satisfying this is $n = 3$.
Let's re-check the derivation of the transfer function.
$Y = \frac{1}{s+1} (U)$
$U = \frac{1}{s^n} D - \frac{1}{s^2} Y$
$Y(s+1) = \frac{1}{s^n} D - \frac{1}{s^2} Y$
$Y(s+1 + \frac{1}{s^2}) = \frac{1}{s^n} D$
$Y \frac{s^3 + s^2 + 1}{s^2} = \frac{1}{s^n} D$
$\frac{Y}{D} = \frac{s^2}{s^n (s^3 + s^2 + 1)} = \frac{1}{s^{n-2} (s^3 + s^2 + 1)}$
$Y(s) = \frac{1}{s^{n-1} (s^3 + s^2 + 1)}$
$y_{ss} = \lim_{s \to 0} \frac{1}{s^{n-2} (1)}$
For $y_{ss} = 0$, we need $n - 2 < 0$, so $n < 2$. The maximum integer value of $n$ satisfying this is $n = 1$.
Final Answer: The final answer is $\boxed{1}$
