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I have the function $$ F(s)=\frac{1}{1+s^2}\frac{1}{1+4s^2} $$ and I would like to know if exists a non-decreasing function $f(t)$ such that $F(s)$ is the two-sided Laplace transform of $f(t)$.

Of course $F(s)$ is the two-sided Laplace transform of $f(t)=u(t)[2\sin(t/2)-\sin(t)]/3$ with $u(t)$ Heaviside function, but is not non-decreasing. Somebody knows how I can find out if such a $f(t)$ exists?

Thank you

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Consider the decomposition

$$\frac{1}{1+s^2}\frac{1}{1+4s^2}=\frac{A}{1+s^2}+\frac{B}{1+4s^2},$$

with $A$ and $B$ coefficients to determine.

Then

$$\frac{A}{1+s^2}+\frac{B}{1+4s^2}=\frac{(A+B)+s^2(4A+B)}{(1+s^2)(1+4s^2)} $$

implies $A+B=1$ and $4A+B=0$, or $A=-\frac{1}{3}$ and $ B=\frac{4}{3}$. In summary

$$F(s)=\frac{1}{1+s^2}\frac{1}{1+4s^2}=-\frac{1}{3}\frac{1}{1+s^2}+\frac{4}{3}\frac{1}{1+4s^2},$$

i.e.

$$F(s)=-\frac{1}{3}\mathcal L(\sin(t))+\frac{4}{3}\frac{1}{2}\mathcal L(\sin(2t))= -\frac{1}{3}\mathcal L(\sin(t))+\frac{2}{3}\mathcal L(\sin(2t)),$$

as $\mathcal L(\sin(2t))=\frac{\frac{1}{2}}{s^2+\frac{1}{4}}=2\frac{1}{4s^2+1}$.

EDIT: The OP asks for the 2-sided Laplace transform.

Using the above computations, and remembering that the bilateral Laplace transform $\mathcal B(f)$ of a function $f$ is defined as the improper integral $$\mathcal B(f)(s)=\int_{-\infty}^{\infty}e^{-st}f(t)dt, $$

then the causal functions $f$ and $g$ s.t.

$$f(t)=-\frac{1}{3}\sin(t),$$

$$g(t)=\frac{2}{3}\sin(2t),$$

for $t\geq 0$ and $f(t)=g(t)=0$ for $t<0$ satisfy $\mathcal B(f+g)=F(s)$.

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  • $\begingroup$ I'm sorry, but your answer is not useful..I already did this calculation as I wrote in my question. Moreover, yours is wrong because I'm interested in two-sided Laplace transform and because you change the values of A and B during your calculation. $\endgroup$ – alemou Jun 27 '13 at 14:00
  • $\begingroup$ I did not understand your question, then. I had a switch between the coeffs. but I promptly corrected it, as you can check. I try to work on the bilateral version, then. $\endgroup$ – Avitus Jun 27 '13 at 14:14
  • $\begingroup$ I'm wondering if there exists another function, instead of the one that we wrote, that is non-decreasing and such that its two-sided Laplace transform is $F(s)$. $\endgroup$ – alemou Jun 27 '13 at 14:19
  • $\begingroup$ Why don't you consider the causal functions $f(t)=\sin(t), t>0$ $f(t)=0, t\leq 0$ and similarly for $g(t)=\sin(2t)$? Up to normalization, their bilateral Laplace transform is the $F(s)$ you need. The "trick" is to consider causal functions, however. Have you particular applications in mind? $\endgroup$ – Avitus Jun 27 '13 at 14:29
  • $\begingroup$ If with casual function you mean heaviside function, it is what I did in my question. Yes I would like to use this result to prove that $F(y)=\frac{1}{1-y^2}\frac{1}{1-4y^2}$, real, is such that $\sum_{i,j=1}^{n}\xi_i\bar{\xi}_jF(y_i+y_j)\geq 0$, for every $n\in\mathbb{N}$, $\xi_i \in \mathbb{C}$ and $y_i \in (1,+\infty)$ $\endgroup$ – alemou Jun 27 '13 at 14:39

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