Inverse Laplace Transforms — Question 3

Question 3 – Solution

Strategy. Shift the numerator as well as the denominator; then distinguish phase, amplitude and exponential decay.

Step 1: Match the shifted pairs. The denominator is $(s+2)^2+9$, while $3s+1=3(s+2)-5$. Therefore $$F(s)=3\frac{s+2}{(s+2)^2+9}-\frac 53\frac 3{(s+2)^2+9}.$$

Step 2: Invert and diagnose the sign error. The matching time function is $$\boxed{f(t)=e^{-2t}[3\cos 3t-\tfrac 53\sin 3t].}$$ Its forward transform reproduces $3(s+2)-5=3s+1$. The proposed candidate instead has numerator $3(s+2)+1=3s+7$. Completing the square without adjusting the numerator loses a constant contribution.

Step 3: Recover phase and the first zero. In $R\cos(3t+\phi)=R\cos\phi\cos 3t-R\sin\phi\sin 3t$, compare coefficients: $$R=\frac{\sqrt{106}}3,\qquad \phi=\arctan(5/9).$$ Both sine and cosine of $\phi$ are positive, so the specified quadrant removes any phase ambiguity. Starting at $f(0)=3>0$, the first zero occurs when $3t+\phi=\pi/2$: $$\boxed{t_0=\frac{\pi/2-\arctan(5/9)}3 =\frac 13\arctan(9/5).}$$

Step 4: Separate envelope from oscillation. The bound $|f(t)|\le R e^{-2t}$ gives absolute convergence for $s>-2$. At $s=-2$, the weighted signal is a nonzero sinusoid, whose primitive oscillates without a limit. For $s<-2$, positive fixed-length subintervals centered at successive cosine peaks have integrals growing exponentially, contradicting the Cauchy criterion. Thus $\boxed{s>-2}$ is exact. The decay rate is two and the angular frequency is three; neither gives the phase by itself. The graph shows both signed envelopes and the first zero.