Table Of Laplace Transforms: Question 4-10.4 & Solution

Question 4

Use causal one-sided Laplace transforms. Write $H_a(t)=0$ for $t<a$ and $1$ for $t\ge a$ . Ordinary functions are zero for $t<0$ . Justify the table entries and operational rules you use; give exact expressions.

An unknown function has transform $F(s)=\frac{As+B}{s^2+4},\qquad s>0,$ where $A,B$ are real. Measurements give $f(0)=1$ and $f'(0)=-2$ .

Tasks

Invert the general transform by matching numerators to table entries, then recover $A,B$ from the data.
For the recovered function, find its amplitude, all positive zeros, and the conserved quantity $f'(t)^2+4f(t)^2$ .
Classify all real pairs $(A,B)$ for which the inverse is nonnegative for every $t\ge 0$ . Prove the classification.
A student argues that $\lim_{s\to 0^+}sF(s)=0$ proves $\lim_{t\to\infty}f(t)=0$ . Refute the conclusion with two sequences of times and identify the missing hypothesis in that use of the final-value theorem.

Original worksheet page 1: question and worked solution for 4-10-004

Show solutionHide solution+

Question 4 – Solution

Strategy. Matching a denominator is insufficient: the numerator determines the phase, initial state and sign changes.

Step 1: Match and use the data. The table gives $f=A\cos 2t+(B/2)\sin 2t$ , so $f(0)=A$ and $f'(0)=B$ . Therefore $A=1$ , $B=-2$ and $\boxed{f(t)=\cos 2t-\sin 2t=\sqrt 2\cos(2t+\pi/4).}$

Step 2: Determine the motion. The amplitude is $\sqrt 2$ . The positive zeros are $t=\pi/8+k\pi/2$ , $k=0,1,2,\ldots$ . Direct differentiation gives $f''+4f=0$ , so $(f'^2+4f^2)'=2f'(f''+4f)=0$ . Its initial value is $(-2)^2+4(1)^2=\boxed{8}$ .

Step 3: Prove the positivity classification. If $(A,B)\ne(0,0)$ , the amplitude $\sqrt{A^2+B^2/4}$ is positive. The sinusoid attains both that amplitude and its negative at arbitrarily large nonnegative times. It cannot be nonnegative on the whole half-line. Only $\boxed{A=B=0}$ works. An alternative proof uses zero integral over one period: a continuous nonnegative function with zero integral must vanish.

Step 4: Check the claimed final value. Although $sF(s)\to 0$ as $s\to 0^+$ , the selected solution obeys $f(n\pi)=1$ and $f(n\pi+\pi/2)=-1$ . Hence it has no time limit. The usual rational final-value theorem requires the poles of $sF(s)$ to lie strictly in the left half-plane after cancellation. Here the uncanceled poles are $s=\pm 2i$ , supporting persistent oscillations. An algebraic limit of the transform does not by itself establish a limit of the function.

Original worksheet page 2: question and worked solution for 4-10-004

Table Of Laplace Transforms — Question 4

Question 4

Question 4 – Solution