Laplace transform and delay

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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" ?







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  • What is $u$ here?
    – Kavi Rama Murthy
    Aug 29 at 10:04










  • is the step function
    – xmaionx
    Aug 29 at 10:25














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" ?







share|cite|improve this question




















  • What is $u$ here?
    – Kavi Rama Murthy
    Aug 29 at 10:04










  • is the step function
    – xmaionx
    Aug 29 at 10:25












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" ?







share|cite|improve this question












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" ?









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 29 at 9:53









xmaionx

44




44











  • What is $u$ here?
    – Kavi Rama Murthy
    Aug 29 at 10:04










  • is the step function
    – xmaionx
    Aug 29 at 10:25
















  • What is $u$ here?
    – Kavi Rama Murthy
    Aug 29 at 10:04










  • is the step function
    – xmaionx
    Aug 29 at 10:25















What is $u$ here?
– Kavi Rama Murthy
Aug 29 at 10:04




What is $u$ here?
– Kavi Rama Murthy
Aug 29 at 10:04












is the step function
– xmaionx
Aug 29 at 10:25




is the step function
– xmaionx
Aug 29 at 10:25










1 Answer
1






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$$cal Lleft(u(t-c)f(t-c)right)=e^-cscal L(f)$$
now let $c=1$:
$$cal Lleft(u(t-1)e^t-1right)=e^-sdfrac1s-1$$






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  • 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 $e^-1s$ ?
    – xmaionx
    Aug 29 at 10:31










  • first statement is 'yes'. I don't understand the second. sorry :)
    – Nosrati
    Aug 29 at 10:35










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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 $c=1$:
$$cal Lleft(u(t-1)e^t-1right)=e^-sdfrac1s-1$$






share|cite|improve this answer




















  • 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 $e^-1s$ ?
    – xmaionx
    Aug 29 at 10:31










  • first statement is 'yes'. I don't understand the second. sorry :)
    – Nosrati
    Aug 29 at 10:35














up vote
0
down vote



accepted










$$cal Lleft(u(t-c)f(t-c)right)=e^-cscal L(f)$$
now let $c=1$:
$$cal Lleft(u(t-1)e^t-1right)=e^-sdfrac1s-1$$






share|cite|improve this answer




















  • 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 $e^-1s$ ?
    – xmaionx
    Aug 29 at 10:31










  • first statement is 'yes'. I don't understand the second. sorry :)
    – Nosrati
    Aug 29 at 10:35












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 $c=1$:
$$cal Lleft(u(t-1)e^t-1right)=e^-sdfrac1s-1$$






share|cite|improve this answer












$$cal Lleft(u(t-c)f(t-c)right)=e^-cscal L(f)$$
now let $c=1$:
$$cal Lleft(u(t-1)e^t-1right)=e^-sdfrac1s-1$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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 $e^-1s$ ?
    – xmaionx
    Aug 29 at 10:31










  • first statement is 'yes'. I don't understand the second. sorry :)
    – 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 $e^-1s$ ?
    – xmaionx
    Aug 29 at 10:31










  • first statement is 'yes'. I don't understand the second. sorry :)
    – 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 $e^-1s$ ?
– 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 $e^-1s$ ?
– xmaionx
Aug 29 at 10:31












first statement is 'yes'. I don't understand the second. sorry :)
– Nosrati
Aug 29 at 10:35




first statement is 'yes'. I don't understand the second. sorry :)
– Nosrati
Aug 29 at 10:35

















 

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