Mixing
laplace transform of impulse function
$begingroup$
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
asked Aug 12 '17 at 6:44
TapasXTapasX
192
$endgroup$
add a comment |
$begingroup$
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
asked Aug 12 '17 at 6:44
TapasXTapasX
192
$endgroup$
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
add a comment |
$begingroup$
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
asked Aug 12 '17 at 6:44
TapasXTapasX
192
$endgroup$
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
asked Aug 12 '17 at 6:44
TapasXTapasX
192
asked Aug 12 '17 at 6:44
TapasXTapasX
192
asked Aug 12 '17 at 6:44
TapasXTapasX
192
asked Aug 12 '17 at 6:44
TapasXTapasX
192
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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Hint. Note that by the sampling property of the delta function (which is actually a distribution)
$$int_{-infty}^{+infty}f(t)delta(t-t_0)dt=f(t_0).$$
answered Aug 12 '17 at 7:06
Robert ZRobert Z
95.1k1063134
$endgroup$
$begingroup$
@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
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Hint. Note that by the sampling property of the delta function (which is actually a distribution)
$$int_{-infty}^{+infty}f(t)delta(t-t_0)dt=f(t_0).$$
answered Aug 12 '17 at 7:06
Robert ZRobert Z
95.1k1063134
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
Hint. Note that by the sampling property of the delta function (which is actually a distribution)
$$int_{-infty}^{+infty}f(t)delta(t-t_0)dt=f(t_0).$$
answered Aug 12 '17 at 7:06
Robert ZRobert Z
95.1k1063134
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
Hint. Note that by the sampling property of the delta function (which is actually a distribution)
$$int_{-infty}^{+infty}f(t)delta(t-t_0)dt=f(t_0).$$
answered Aug 12 '17 at 7:06
Robert ZRobert Z
95.1k1063134
$endgroup$
Hint. Note that by the sampling property of the delta function (which is actually a distribution)
$$int_{-infty}^{+infty}f(t)delta(t-t_0)dt=f(t_0).$$
answered Aug 12 '17 at 7:06
Robert ZRobert Z
95.1k1063134
edited Aug 12 '17 at 7:12
answered Aug 12 '17 at 7:06
Robert ZRobert Z
95.1k1063134
answered Aug 12 '17 at 7:06
Robert ZRobert Z
95.1k1063134
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
add a comment |
$begingroup$
@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
add a comment |
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$begingroup$
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