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
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 idea behind this question
The final value theorem finds $\lim_{t\to\infty} y(t)$ from $\lim_{s\to 0} sY(s)$ without an inverse transform. It is valid only if $y(t)$ actually settles, which means every pole of $sY(s)$ must be in the left half of the $s$-plane. A single pole at the origin in $Y(s)$ is allowed, because the factor $s$ cancels it.
Try a variation
What is the final value of $y(t)$ if $Y(s) = \dfrac{6}{s(s+1)(s+3)}$?
Show the answer
Answer: $2$
All poles of $sY(s)$ are in the left half-plane, and $\lim_{s\to0} sY(s) = 6/3 = 2$.
Other mistakes to avoid
- Forgetting to multiply by $s$ before taking the limit.
- Using the theorem when $sY(s)$ has poles on the imaginary axis, where $y(t)$ keeps oscillating.