Question

If $g(t)=\dfrac{df(t)}{dt}$, and $F(s)=\dfrac{1+s}{s^{2}+12s+32}$ where $F(s)$ is the Laplace transform of $f(t)$, then the value of $g(t)$ at $t=0$ is _________.

• A. -11
• B. -5
• C. -17
• D. $\infty$

Hint:

Use the initial value theorem carefully: $\;f(0)=\lim\limits_{s\to\infty} sF(s)$ and $f'(0)=\lim\limits_{s\to\infty} s\left[sF(s)-f(0)\right]$.

Answer 1

The Correct Option is A

We are given that \[ F(s) = \frac{1+s}{s^{2}+12s+32}. \] We want the value of \[ g(0)=f'(0). \] Step 1: Use Laplace derivative property.
The Laplace transform of the derivative is \[ \mathcal{L}\{f'(t)\} = sF(s) - f(0). \] Thus, to find $f'(0)$, we use the initial value theorem: \[ f'(0) = \lim_{s\to\infty} s\left[sF(s) - f(0)\right]. \] Step 2: First compute $f(0)$ using initial value theorem.
Initial value theorem: \[ f(0)=\lim_{s\to\infty} sF(s)=\lim_{s\to\infty} s\cdot \frac{1+s}{s^{2}+12s+32}. \] Divide numerator and denominator by $s^{2}$: \[ f(0)=\lim_{s\to\infty} \frac{s(1+s)}{s^{2}(1+\frac{12}{s}+\frac{32}{s^{2}})} = \lim_{s\to\infty} \frac{1+\frac{1}{s}}{1+\frac{12}{s}+\frac{32}{s^{2}}}. \] As $s\to\infty$: \[ f(0)=1. \] Step 3: Now compute $f'(0)$.
\[ f'(0)=\lim_{s\to\infty} s\left[sF(s)-1\right]. \] Compute $sF(s)$: \[ sF(s)=\frac{s(1+s)}{s^{2}+12s+32}. \] Thus, \[ sF(s)-1 = \frac{s(1+s)}{s^{2}+12s+32} - 1. \] Combine over a common denominator: \[ = \frac{s(1+s)-(s^{2}+12s+32)}{s^{2}+12s+32} = \frac{s+s^{2}-s^{2}-12s-32}{s^{2}+12s+32}. \] Simplify numerator: \[ = \frac{-11s - 32}{s^{2}+12s+32}. \] Now multiply by $s$: \[ f'(0) = \lim_{s\to\infty} s\cdot \frac{-11s - 32}{s^{2}+12s+32} = \lim_{s\to\infty} \frac{-11s^{2}-32s}{s^{2}(1+\frac{12}{s}+\frac{32}{s^{2}})}. \] Divide numerator and denominator by $s^{2}$: \[ f'(0)=\lim_{s\to\infty} \frac{-11 - \frac{32}{s}}{1+\frac{12}{s}+\frac{32}{s^{2}}}. \] As $s\to\infty$: \[ f'(0) = -11. \] Therefore, \[ g(0)=f'(0)=-11. \]