Final value of a signal from its Laplace transform
BlinkNBuild practice problem · GATE standard, authored and verified in-house
The Laplace transform of a signal $y(t)$ is $$Y(s) = \frac{10}{s\,(s+2)(s+5)}.$$ The value of $\displaystyle\lim_{t \to \infty} y(t)$ is ________.
Show the step-by-step derivation
Answer
1 unitless unitless
Step-by-step derivation
- Before applying the final value theorem, check that it is valid. The theorem requires every pole of $sY(s)$ to lie strictly in the left half plane.
- Form $sY(s)$: $$sY(s) = \frac{10}{(s+2)(s+5)}.$$ Its poles are at $s = -2$ and $s = -5$, both in the open left half plane. The theorem applies. ✓
- Apply the final value theorem: $$\lim_{t \to \infty} y(t) = \lim_{s \to 0} s\,Y(s) = \lim_{s \to 0} \frac{10}{(s+2)(s+5)}.$$
- Substitute $s = 0$ directly, since the expression is continuous there: $$\frac{10}{(0+2)(0+5)} = \frac{10}{10} = \mathbf{1}.$$
- Cross-check by partial fractions. Writing $Y(s) = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+5}$ gives $A = \frac{10}{(2)(5)} = 1$. The other two terms are decaying exponentials $Be^{-2t}$ and $Ce^{-5t}$, which vanish as $t \to \infty$, leaving the constant $A = 1$. ✓
The trap this question is built around
Applying the theorem without checking the poles. For $Y(s) = \frac{1}{s^2+1}$, blindly computing $\lim_{s\to 0} sY(s)$ gives $0$ - but $y(t) = \sin t$ has no final value at all. The check is not a formality.