laplace transform of impulse function












0












$begingroup$


The laplace tranform of the following function impulse function is



$$int_{0}^{infty}delta(t).e^{-st},dt$$
$$=int_{0}^{infty}delta(t),dt=1$$
(area under unit impulse is always 1)
$$=1.int_{0}^{infty}e^{-st},dt$$
$$=1.frac{1}{S}$$
$$=frac{1}{S}$$
but the correct answer is 1, I don't know why.






calculus







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You can't integrate a factor only and take it out from the integral.
    $endgroup$
    – Sergio Enrique Yarza Acuña
    Aug 12 '17 at 6:46










  • $begingroup$
    Also $int_0^infty delta(t) e^{-st}dt$ is not a very good notation. You should say instead that $int_{-infty}^infty delta(t) e^{-st}dt = e^{-st}|_{s=0} = 1$
    $endgroup$
    – reuns
    Aug 12 '17 at 7:06
















0












$begingroup$


The laplace tranform of the following function impulse function is



$$int_{0}^{infty}delta(t).e^{-st},dt$$
$$=int_{0}^{infty}delta(t),dt=1$$
(area under unit impulse is always 1)
$$=1.int_{0}^{infty}e^{-st},dt$$
$$=1.frac{1}{S}$$
$$=frac{1}{S}$$
but the correct answer is 1, I don't know why.






calculus







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You can't integrate a factor only and take it out from the integral.
    $endgroup$
    – Sergio Enrique Yarza Acuña
    Aug 12 '17 at 6:46










  • $begingroup$
    Also $int_0^infty delta(t) e^{-st}dt$ is not a very good notation. You should say instead that $int_{-infty}^infty delta(t) e^{-st}dt = e^{-st}|_{s=0} = 1$
    $endgroup$
    – reuns
    Aug 12 '17 at 7:06














0












0








0





$begingroup$


The laplace tranform of the following function impulse function is



$$int_{0}^{infty}delta(t).e^{-st},dt$$
$$=int_{0}^{infty}delta(t),dt=1$$
(area under unit impulse is always 1)
$$=1.int_{0}^{infty}e^{-st},dt$$
$$=1.frac{1}{S}$$
$$=frac{1}{S}$$
but the correct answer is 1, I don't know why.






calculus







share|cite|improve this question









$endgroup$




The laplace tranform of the following function impulse function is



$$int_{0}^{infty}delta(t).e^{-st},dt$$
$$=int_{0}^{infty}delta(t),dt=1$$
(area under unit impulse is always 1)
$$=1.int_{0}^{infty}e^{-st},dt$$
$$=1.frac{1}{S}$$
$$=frac{1}{S}$$
but the correct answer is 1, I don't know why.






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 12 '17 at 6:44









TapasXTapasX

192




192








  • 2




    $begingroup$
    You can't integrate a factor only and take it out from the integral.
    $endgroup$
    – Sergio Enrique Yarza Acuña
    Aug 12 '17 at 6:46










  • $begingroup$
    Also $int_0^infty delta(t) e^{-st}dt$ is not a very good notation. You should say instead that $int_{-infty}^infty delta(t) e^{-st}dt = e^{-st}|_{s=0} = 1$
    $endgroup$
    – reuns
    Aug 12 '17 at 7:06














  • 2




    $begingroup$
    You can't integrate a factor only and take it out from the integral.
    $endgroup$
    – Sergio Enrique Yarza Acuña
    Aug 12 '17 at 6:46










  • $begingroup$
    Also $int_0^infty delta(t) e^{-st}dt$ is not a very good notation. You should say instead that $int_{-infty}^infty delta(t) e^{-st}dt = e^{-st}|_{s=0} = 1$
    $endgroup$
    – reuns
    Aug 12 '17 at 7:06








2




2




$begingroup$
You can't integrate a factor only and take it out from the integral.
$endgroup$
– Sergio Enrique Yarza Acuña
Aug 12 '17 at 6:46




$begingroup$
You can't integrate a factor only and take it out from the integral.
$endgroup$
– Sergio Enrique Yarza Acuña
Aug 12 '17 at 6:46












$begingroup$
Also $int_0^infty delta(t) e^{-st}dt$ is not a very good notation. You should say instead that $int_{-infty}^infty delta(t) e^{-st}dt = e^{-st}|_{s=0} = 1$
$endgroup$
– reuns
Aug 12 '17 at 7:06




$begingroup$
Also $int_0^infty delta(t) e^{-st}dt$ is not a very good notation. You should say instead that $int_{-infty}^infty delta(t) e^{-st}dt = e^{-st}|_{s=0} = 1$
$endgroup$
– reuns
Aug 12 '17 at 7:06










1 Answer
1






active

oldest

votes


















0












$begingroup$

Hint. Note that by the sampling property of the delta function (which is actually a distribution)
$$int_{-infty}^{+infty}f(t)delta(t-t_0)dt=f(t_0).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @reuns Yes, it is better to recall that.
    $endgroup$
    – Robert Z
    Aug 12 '17 at 7:13










  • $begingroup$
    To make it clear, you can define it as $int_{-infty}^infty delta(t) f(t)dt overset{def}= lim_{epsilon to 0^+} int_{-infty}^infty frac{1_{|t| < epsilon}}{2 epsilon} f(t)dt$
    $endgroup$
    – reuns
    Aug 12 '17 at 7:22











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Hint. Note that by the sampling property of the delta function (which is actually a distribution)
$$int_{-infty}^{+infty}f(t)delta(t-t_0)dt=f(t_0).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @reuns Yes, it is better to recall that.
    $endgroup$
    – Robert Z
    Aug 12 '17 at 7:13










  • $begingroup$
    To make it clear, you can define it as $int_{-infty}^infty delta(t) f(t)dt overset{def}= lim_{epsilon to 0^+} int_{-infty}^infty frac{1_{|t| < epsilon}}{2 epsilon} f(t)dt$
    $endgroup$
    – reuns
    Aug 12 '17 at 7:22
















0












$begingroup$

Hint. Note that by the sampling property of the delta function (which is actually a distribution)
$$int_{-infty}^{+infty}f(t)delta(t-t_0)dt=f(t_0).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @reuns Yes, it is better to recall that.
    $endgroup$
    – Robert Z
    Aug 12 '17 at 7:13










  • $begingroup$
    To make it clear, you can define it as $int_{-infty}^infty delta(t) f(t)dt overset{def}= lim_{epsilon to 0^+} int_{-infty}^infty frac{1_{|t| < epsilon}}{2 epsilon} f(t)dt$
    $endgroup$
    – reuns
    Aug 12 '17 at 7:22














0












0








0





$begingroup$

Hint. Note that by the sampling property of the delta function (which is actually a distribution)
$$int_{-infty}^{+infty}f(t)delta(t-t_0)dt=f(t_0).$$






share|cite|improve this answer











$endgroup$



Hint. Note that by the sampling property of the delta function (which is actually a distribution)
$$int_{-infty}^{+infty}f(t)delta(t-t_0)dt=f(t_0).$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 12 '17 at 7:12

























answered Aug 12 '17 at 7:06









Robert ZRobert Z

95.1k1063134




95.1k1063134












  • $begingroup$
    @reuns Yes, it is better to recall that.
    $endgroup$
    – Robert Z
    Aug 12 '17 at 7:13










  • $begingroup$
    To make it clear, you can define it as $int_{-infty}^infty delta(t) f(t)dt overset{def}= lim_{epsilon to 0^+} int_{-infty}^infty frac{1_{|t| < epsilon}}{2 epsilon} f(t)dt$
    $endgroup$
    – reuns
    Aug 12 '17 at 7:22


















  • $begingroup$
    @reuns Yes, it is better to recall that.
    $endgroup$
    – Robert Z
    Aug 12 '17 at 7:13










  • $begingroup$
    To make it clear, you can define it as $int_{-infty}^infty delta(t) f(t)dt overset{def}= lim_{epsilon to 0^+} int_{-infty}^infty frac{1_{|t| < epsilon}}{2 epsilon} f(t)dt$
    $endgroup$
    – reuns
    Aug 12 '17 at 7:22
















$begingroup$
@reuns Yes, it is better to recall that.
$endgroup$
– Robert Z
Aug 12 '17 at 7:13




$begingroup$
@reuns Yes, it is better to recall that.
$endgroup$
– Robert Z
Aug 12 '17 at 7:13












$begingroup$
To make it clear, you can define it as $int_{-infty}^infty delta(t) f(t)dt overset{def}= lim_{epsilon to 0^+} int_{-infty}^infty frac{1_{|t| < epsilon}}{2 epsilon} f(t)dt$
$endgroup$
– reuns
Aug 12 '17 at 7:22




$begingroup$
To make it clear, you can define it as $int_{-infty}^infty delta(t) f(t)dt overset{def}= lim_{epsilon to 0^+} int_{-infty}^infty frac{1_{|t| < epsilon}}{2 epsilon} f(t)dt$
$endgroup$
– reuns
Aug 12 '17 at 7:22


















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