Laplace transform and delay
Clash Royale CLAN TAG#URR8PPP
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Hi guys if i have to transform $e^t-1u(t-1)$ , why the result is $frace^-s(s-1)$ and not $frace^-(s-1)(s-1)$ if the $e^t-1$ multiply all the "function" ?
complex-analysis laplace-transform
add a comment |Â
up vote
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down vote
favorite
Hi guys if i have to transform $e^t-1u(t-1)$ , why the result is $frace^-s(s-1)$ and not $frace^-(s-1)(s-1)$ if the $e^t-1$ multiply all the "function" ?
complex-analysis laplace-transform
What is $u$ here?
â Kavi Rama Murthy
Aug 29 at 10:04
is the step function
â xmaionx
Aug 29 at 10:25
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Hi guys if i have to transform $e^t-1u(t-1)$ , why the result is $frace^-s(s-1)$ and not $frace^-(s-1)(s-1)$ if the $e^t-1$ multiply all the "function" ?
complex-analysis laplace-transform
Hi guys if i have to transform $e^t-1u(t-1)$ , why the result is $frace^-s(s-1)$ and not $frace^-(s-1)(s-1)$ if the $e^t-1$ multiply all the "function" ?
complex-analysis laplace-transform
asked Aug 29 at 9:53
xmaionx
44
44
What is $u$ here?
â Kavi Rama Murthy
Aug 29 at 10:04
is the step function
â xmaionx
Aug 29 at 10:25
add a comment |Â
What is $u$ here?
â Kavi Rama Murthy
Aug 29 at 10:04
is the step function
â xmaionx
Aug 29 at 10:25
What is $u$ here?
â Kavi Rama Murthy
Aug 29 at 10:04
What is $u$ here?
â Kavi Rama Murthy
Aug 29 at 10:04
is the step function
â xmaionx
Aug 29 at 10:25
is the step function
â xmaionx
Aug 29 at 10:25
add a comment |Â
1 Answer
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$$cal Lleft(u(t-c)f(t-c)right)=e^-cscal L(f)$$
now let $c=1$:
$$cal Lleft(u(t-1)e^t-1right)=e^-sdfrac1s-1$$
so after i have multiplied by $e^-cs$ i only have to make the transorm as it was f(t) and not f(t-c) ? so in my example i have to make the transform of $e^tu(t)$ as if this was my transform and after all multiply by $e^-1s$ ?
â xmaionx
Aug 29 at 10:31
first statement is 'yes'. I don't understand the second. sorry :)
â Nosrati
Aug 29 at 10:35
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$$cal Lleft(u(t-c)f(t-c)right)=e^-cscal L(f)$$
now let $c=1$:
$$cal Lleft(u(t-1)e^t-1right)=e^-sdfrac1s-1$$
so after i have multiplied by $e^-cs$ i only have to make the transorm as it was f(t) and not f(t-c) ? so in my example i have to make the transform of $e^tu(t)$ as if this was my transform and after all multiply by $e^-1s$ ?
â xmaionx
Aug 29 at 10:31
first statement is 'yes'. I don't understand the second. sorry :)
â Nosrati
Aug 29 at 10:35
add a comment |Â
up vote
0
down vote
accepted
$$cal Lleft(u(t-c)f(t-c)right)=e^-cscal L(f)$$
now let $c=1$:
$$cal Lleft(u(t-1)e^t-1right)=e^-sdfrac1s-1$$
so after i have multiplied by $e^-cs$ i only have to make the transorm as it was f(t) and not f(t-c) ? so in my example i have to make the transform of $e^tu(t)$ as if this was my transform and after all multiply by $e^-1s$ ?
â xmaionx
Aug 29 at 10:31
first statement is 'yes'. I don't understand the second. sorry :)
â Nosrati
Aug 29 at 10:35
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$$cal Lleft(u(t-c)f(t-c)right)=e^-cscal L(f)$$
now let $c=1$:
$$cal Lleft(u(t-1)e^t-1right)=e^-sdfrac1s-1$$
$$cal Lleft(u(t-c)f(t-c)right)=e^-cscal L(f)$$
now let $c=1$:
$$cal Lleft(u(t-1)e^t-1right)=e^-sdfrac1s-1$$
answered Aug 29 at 10:09
Nosrati
22k51747
22k51747
so after i have multiplied by $e^-cs$ i only have to make the transorm as it was f(t) and not f(t-c) ? so in my example i have to make the transform of $e^tu(t)$ as if this was my transform and after all multiply by $e^-1s$ ?
â xmaionx
Aug 29 at 10:31
first statement is 'yes'. I don't understand the second. sorry :)
â Nosrati
Aug 29 at 10:35
add a comment |Â
so after i have multiplied by $e^-cs$ i only have to make the transorm as it was f(t) and not f(t-c) ? so in my example i have to make the transform of $e^tu(t)$ as if this was my transform and after all multiply by $e^-1s$ ?
â xmaionx
Aug 29 at 10:31
first statement is 'yes'. I don't understand the second. sorry :)
â Nosrati
Aug 29 at 10:35
so after i have multiplied by $e^-cs$ i only have to make the transorm as it was f(t) and not f(t-c) ? so in my example i have to make the transform of $e^tu(t)$ as if this was my transform and after all multiply by $e^-1s$ ?
â xmaionx
Aug 29 at 10:31
so after i have multiplied by $e^-cs$ i only have to make the transorm as it was f(t) and not f(t-c) ? so in my example i have to make the transform of $e^tu(t)$ as if this was my transform and after all multiply by $e^-1s$ ?
â xmaionx
Aug 29 at 10:31
first statement is 'yes'. I don't understand the second. sorry :)
â Nosrati
Aug 29 at 10:35
first statement is 'yes'. I don't understand the second. sorry :)
â Nosrati
Aug 29 at 10:35
add a comment |Â
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What is $u$ here?
â Kavi Rama Murthy
Aug 29 at 10:04
is the step function
â xmaionx
Aug 29 at 10:25