Mixing
convergence- Laplace Transform
$begingroup$
The bilateral Laplace transform is defined as,
$$X(s) = int_{-infty}^{infty}x(t)e^{-st}dt~~,s = sigma +jomega$$
where both $sigma$ and $omega$ are real. Then,
$$X(s) = X(sigma+jomega) = int_{-infty}^{infty}x(t)e^{-sigma t}e^{-jomega t}dt leqint_{-infty}^{infty}|x(t)e^{-sigma t}e^{-jomega t}|dt = int_{-infty}^{infty}|x(t)|e^{-sigma t}dt$$
So we can deduce that,
$$int_{-infty}^{infty}|x(t)|e^{-sigma t}dt < infty Rightarrow X(s)~~~text{converges if}~~ Re{s} = sigma$$
What about the reverse? That is, if $X(s)$ is convergent then $x(t)exp(-Re{s}t)$ is absolutely integrable.
Can you please provide either a proof or a counter example. The book Signals and Systems by Oppenheim and Wilsky uses that as a fact but I am not convinced.
convergence laplace-transform
asked Jan 4 at 18:27
Bora DoğanBora Doğan
113
$endgroup$
add a comment |
$begingroup$
The bilateral Laplace transform is defined as,
$$X(s) = int_{-infty}^{infty}x(t)e^{-st}dt~~,s = sigma +jomega$$
where both $sigma$ and $omega$ are real. Then,
$$X(s) = X(sigma+jomega) = int_{-infty}^{infty}x(t)e^{-sigma t}e^{-jomega t}dt leqint_{-infty}^{infty}|x(t)e^{-sigma t}e^{-jomega t}|dt = int_{-infty}^{infty}|x(t)|e^{-sigma t}dt$$
So we can deduce that,
$$int_{-infty}^{infty}|x(t)|e^{-sigma t}dt < infty Rightarrow X(s)~~~text{converges if}~~ Re{s} = sigma$$
What about the reverse? That is, if $X(s)$ is convergent then $x(t)exp(-Re{s}t)$ is absolutely integrable.
Can you please provide either a proof or a counter example. The book Signals and Systems by Oppenheim and Wilsky uses that as a fact but I am not convinced.
convergence laplace-transform
asked Jan 4 at 18:27
Bora DoğanBora Doğan
113
$endgroup$
add a comment |
$begingroup$
The bilateral Laplace transform is defined as,
$$X(s) = int_{-infty}^{infty}x(t)e^{-st}dt~~,s = sigma +jomega$$
where both $sigma$ and $omega$ are real. Then,
$$X(s) = X(sigma+jomega) = int_{-infty}^{infty}x(t)e^{-sigma t}e^{-jomega t}dt leqint_{-infty}^{infty}|x(t)e^{-sigma t}e^{-jomega t}|dt = int_{-infty}^{infty}|x(t)|e^{-sigma t}dt$$
So we can deduce that,
$$int_{-infty}^{infty}|x(t)|e^{-sigma t}dt < infty Rightarrow X(s)~~~text{converges if}~~ Re{s} = sigma$$
What about the reverse? That is, if $X(s)$ is convergent then $x(t)exp(-Re{s}t)$ is absolutely integrable.
Can you please provide either a proof or a counter example. The book Signals and Systems by Oppenheim and Wilsky uses that as a fact but I am not convinced.
convergence laplace-transform
asked Jan 4 at 18:27
Bora DoğanBora Doğan
113
$endgroup$
The bilateral Laplace transform is defined as,
$$X(s) = int_{-infty}^{infty}x(t)e^{-st}dt~~,s = sigma +jomega$$
where both $sigma$ and $omega$ are real. Then,
$$X(s) = X(sigma+jomega) = int_{-infty}^{infty}x(t)e^{-sigma t}e^{-jomega t}dt leqint_{-infty}^{infty}|x(t)e^{-sigma t}e^{-jomega t}|dt = int_{-infty}^{infty}|x(t)|e^{-sigma t}dt$$
So we can deduce that,
$$int_{-infty}^{infty}|x(t)|e^{-sigma t}dt < infty Rightarrow X(s)~~~text{converges if}~~ Re{s} = sigma$$
What about the reverse? That is, if $X(s)$ is convergent then $x(t)exp(-Re{s}t)$ is absolutely integrable.
Can you please provide either a proof or a counter example. The book Signals and Systems by Oppenheim and Wilsky uses that as a fact but I am not convinced.
convergence laplace-transform
convergence laplace-transform
asked Jan 4 at 18:27
Bora DoğanBora Doğan
113
asked Jan 4 at 18:27
Bora DoğanBora Doğan
113
edited Jan 5 at 13:13
Bora Doğan
asked Jan 4 at 18:27
Bora DoğanBora Doğan
113
asked Jan 4 at 18:27
Bora DoğanBora Doğan
113
asked Jan 4 at 18:27
Bora DoğanBora Doğan
113
113
add a comment |
add a comment |
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