Mixing
Laplace Transform of Derivative DeltaDirac(t-1)
Clash Royale CLAN TAG#URR8PPP
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0
down vote
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$mathscrLdelta'(t-1)=F(s) $
Im having a hard time thinking this.
My first thought was doing
$F(s) =e^-s s$
But when i do the inverse laplace transform of this i get
$f(t)= delta'(t-1) Heaviside(t-1)$
laplace-transform dirac-delta
asked Jul 25 at 16:45
German
91
 |Â
show 5 more comments
up vote
0
down vote
favorite
$mathscrLdelta'(t-1)=F(s) $
Im having a hard time thinking this.
My first thought was doing
$F(s) =e^-s s$
But when i do the inverse laplace transform of this i get
$f(t)= delta'(t-1) Heaviside(t-1)$
laplace-transform dirac-delta
asked Jul 25 at 16:45
German
91
How do you do the inverse Laplace transform?
– md2perpe
Jul 25 at 18:25
Wolfram can do it for me. (Funny story, when i try to do the laplace transform there, it cant calculate it)
– German
Jul 25 at 18:27
1
Strange result from WolframAlpha. The expression $delta'(t-1) , theta(t-1),$ where $theta$ is the Heaviside function, is not even well-defined as a distribution.
– md2perpe
Jul 25 at 18:33
I also think that $mathscrLdelta'(t-1)(s) = s e^-s.$
– md2perpe
Jul 25 at 18:34
A word of recommendation: Avoid using $*$ for multiplication. There is an operation called convolution which is written using an asterisk, and that is common when working with Laplace and Fourier transforms.
– md2perpe
Jul 25 at 18:38
 |Â
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$mathscrLdelta'(t-1)=F(s) $
Im having a hard time thinking this.
My first thought was doing
$F(s) =e^-s s$
But when i do the inverse laplace transform of this i get
$f(t)= delta'(t-1) Heaviside(t-1)$
laplace-transform dirac-delta
asked Jul 25 at 16:45
German
91
$mathscrLdelta'(t-1)=F(s) $
Im having a hard time thinking this.
My first thought was doing
$F(s) =e^-s s$
But when i do the inverse laplace transform of this i get
$f(t)= delta'(t-1) Heaviside(t-1)$
laplace-transform dirac-delta
asked Jul 25 at 16:45
German
91
edited Jul 25 at 18:39
asked Jul 25 at 16:45
German
91
asked Jul 25 at 16:45
German
91
asked Jul 25 at 16:45
German
91
91
How do you do the inverse Laplace transform?
– md2perpe
Jul 25 at 18:25
Wolfram can do it for me. (Funny story, when i try to do the laplace transform there, it cant calculate it)
– German
Jul 25 at 18:27
1
Strange result from WolframAlpha. The expression $delta'(t-1) , theta(t-1),$ where $theta$ is the Heaviside function, is not even well-defined as a distribution.
– md2perpe
Jul 25 at 18:33
I also think that $mathscrLdelta'(t-1)(s) = s e^-s.$
– md2perpe
Jul 25 at 18:34
A word of recommendation: Avoid using $*$ for multiplication. There is an operation called convolution which is written using an asterisk, and that is common when working with Laplace and Fourier transforms.
– md2perpe
Jul 25 at 18:38
 |Â
show 5 more comments
How do you do the inverse Laplace transform?
– md2perpe
Jul 25 at 18:25
Wolfram can do it for me. (Funny story, when i try to do the laplace transform there, it cant calculate it)
– German
Jul 25 at 18:27
1
Strange result from WolframAlpha. The expression $delta'(t-1) , theta(t-1),$ where $theta$ is the Heaviside function, is not even well-defined as a distribution.
– md2perpe
Jul 25 at 18:33
I also think that $mathscrLdelta'(t-1)(s) = s e^-s.$
– md2perpe
Jul 25 at 18:34
A word of recommendation: Avoid using $*$ for multiplication. There is an operation called convolution which is written using an asterisk, and that is common when working with Laplace and Fourier transforms.
– md2perpe
Jul 25 at 18:38
How do you do the inverse Laplace transform?
– md2perpe
Jul 25 at 18:25
How do you do the inverse Laplace transform?
– md2perpe
Jul 25 at 18:25
Wolfram can do it for me. (Funny story, when i try to do the laplace transform there, it cant calculate it)
– German
Jul 25 at 18:27
Wolfram can do it for me. (Funny story, when i try to do the laplace transform there, it cant calculate it)
– German
Jul 25 at 18:27
1
1
Strange result from WolframAlpha. The expression $delta'(t-1) , theta(t-1),$ where $theta$ is the Heaviside function, is not even well-defined as a distribution.
– md2perpe
Jul 25 at 18:33
Strange result from WolframAlpha. The expression $delta'(t-1) , theta(t-1),$ where $theta$ is the Heaviside function, is not even well-defined as a distribution.
– md2perpe
Jul 25 at 18:33
I also think that $mathscrLdelta'(t-1)(s) = s e^-s.$
– md2perpe
Jul 25 at 18:34
I also think that $mathscrLdelta'(t-1)(s) = s e^-s.$
– md2perpe
Jul 25 at 18:34
A word of recommendation: Avoid using $*$ for multiplication. There is an operation called convolution which is written using an asterisk, and that is common when working with Laplace and Fourier transforms.
– md2perpe
Jul 25 at 18:38
A word of recommendation: Avoid using $*$ for multiplication. There is an operation called convolution which is written using an asterisk, and that is common when working with Laplace and Fourier transforms.
– md2perpe
Jul 25 at 18:38
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
I also think that the Laplace transform of $delta'(t-1)$ is $s e^-s.$ Formally we have:
$$
mathscrLdelta'(t-1)
= int_0^infty delta'(t-1) , e^-st , dt \
= underbraceleft[ delta(t-1) , e^-st right]_0^infty_=0 - int_0^infty delta(t-1) , (-s) e^-st , dt \
= s int_0^infty delta(t-1) , e^-st , dt
= s e^-scdot 1 = s e^-s
$$
answered Jul 25 at 18:40
md2perpe
5,80511022
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I also think that the Laplace transform of $delta'(t-1)$ is $s e^-s.$ Formally we have:
$$
mathscrLdelta'(t-1)
= int_0^infty delta'(t-1) , e^-st , dt \
= underbraceleft[ delta(t-1) , e^-st right]_0^infty_=0 - int_0^infty delta(t-1) , (-s) e^-st , dt \
= s int_0^infty delta(t-1) , e^-st , dt
= s e^-scdot 1 = s e^-s
$$
answered Jul 25 at 18:40
md2perpe
5,80511022
add a comment |Â
up vote
1
down vote
I also think that the Laplace transform of $delta'(t-1)$ is $s e^-s.$ Formally we have:
$$
mathscrLdelta'(t-1)
= int_0^infty delta'(t-1) , e^-st , dt \
= underbraceleft[ delta(t-1) , e^-st right]_0^infty_=0 - int_0^infty delta(t-1) , (-s) e^-st , dt \
= s int_0^infty delta(t-1) , e^-st , dt
= s e^-scdot 1 = s e^-s
$$
answered Jul 25 at 18:40
md2perpe
5,80511022
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I also think that the Laplace transform of $delta'(t-1)$ is $s e^-s.$ Formally we have:
$$
mathscrLdelta'(t-1)
= int_0^infty delta'(t-1) , e^-st , dt \
= underbraceleft[ delta(t-1) , e^-st right]_0^infty_=0 - int_0^infty delta(t-1) , (-s) e^-st , dt \
= s int_0^infty delta(t-1) , e^-st , dt
= s e^-scdot 1 = s e^-s
$$
answered Jul 25 at 18:40
md2perpe
5,80511022
I also think that the Laplace transform of $delta'(t-1)$ is $s e^-s.$ Formally we have:
$$
mathscrLdelta'(t-1)
= int_0^infty delta'(t-1) , e^-st , dt \
= underbraceleft[ delta(t-1) , e^-st right]_0^infty_=0 - int_0^infty delta(t-1) , (-s) e^-st , dt \
= s int_0^infty delta(t-1) , e^-st , dt
= s e^-scdot 1 = s e^-s
$$
answered Jul 25 at 18:40
md2perpe
5,80511022
answered Jul 25 at 18:40
md2perpe
5,80511022
answered Jul 25 at 18:40
md2perpe
5,80511022
answered Jul 25 at 18:40
md2perpe
5,80511022
5,80511022
add a comment |Â
add a comment |Â
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How do you do the inverse Laplace transform?
– md2perpe
Jul 25 at 18:25
Wolfram can do it for me. (Funny story, when i try to do the laplace transform there, it cant calculate it)
– German
Jul 25 at 18:27
1
Strange result from WolframAlpha. The expression $delta'(t-1) , theta(t-1),$ where $theta$ is the Heaviside function, is not even well-defined as a distribution.
– md2perpe
Jul 25 at 18:33
I also think that $mathscrLdelta'(t-1)(s) = s e^-s.$
– md2perpe
Jul 25 at 18:34
A word of recommendation: Avoid using $*$ for multiplication. There is an operation called convolution which is written using an asterisk, and that is common when working with Laplace and Fourier transforms.
– md2perpe
Jul 25 at 18:38