$mathcalL ( (t+5)U(t-1) ) $
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$mathcalL ( (t+5)U(t-1) ) = U(t-a)f(t-a) = e^-as F(s) $
a= 1
I am having trouble dealing with $f(t-a)$ and finding $F(s)$
$f(t-1) = e^t-3 = e^(t-1)-3+1 = e^(t-1)-2 $
$f(t) = e^t-2 $
How do I Laplace transform this ?
laplace-transform
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up vote
0
down vote
favorite
$mathcalL ( (t+5)U(t-1) ) = U(t-a)f(t-a) = e^-as F(s) $
a= 1
I am having trouble dealing with $f(t-a)$ and finding $F(s)$
$f(t-1) = e^t-3 = e^(t-1)-3+1 = e^(t-1)-2 $
$f(t) = e^t-2 $
How do I Laplace transform this ?
laplace-transform
Are you looking for Laplace transform of $(t+5)U(t-1)$?
â Alla Tarighati
Aug 29 at 6:31
@AllaTarighati yes
â user185692
Aug 29 at 6:37
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$mathcalL ( (t+5)U(t-1) ) = U(t-a)f(t-a) = e^-as F(s) $
a= 1
I am having trouble dealing with $f(t-a)$ and finding $F(s)$
$f(t-1) = e^t-3 = e^(t-1)-3+1 = e^(t-1)-2 $
$f(t) = e^t-2 $
How do I Laplace transform this ?
laplace-transform
$mathcalL ( (t+5)U(t-1) ) = U(t-a)f(t-a) = e^-as F(s) $
a= 1
I am having trouble dealing with $f(t-a)$ and finding $F(s)$
$f(t-1) = e^t-3 = e^(t-1)-3+1 = e^(t-1)-2 $
$f(t) = e^t-2 $
How do I Laplace transform this ?
laplace-transform
asked Aug 29 at 6:21
user185692
1536
1536
Are you looking for Laplace transform of $(t+5)U(t-1)$?
â Alla Tarighati
Aug 29 at 6:31
@AllaTarighati yes
â user185692
Aug 29 at 6:37
add a comment |Â
Are you looking for Laplace transform of $(t+5)U(t-1)$?
â Alla Tarighati
Aug 29 at 6:31
@AllaTarighati yes
â user185692
Aug 29 at 6:37
Are you looking for Laplace transform of $(t+5)U(t-1)$?
â Alla Tarighati
Aug 29 at 6:31
Are you looking for Laplace transform of $(t+5)U(t-1)$?
â Alla Tarighati
Aug 29 at 6:31
@AllaTarighati yes
â user185692
Aug 29 at 6:37
@AllaTarighati yes
â user185692
Aug 29 at 6:37
add a comment |Â
1 Answer
1
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0
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Hint:
$$ mathcalLleft[ (t+5)U(t-1)right]= mathcalL left[(t-1+6)U(t-1)right] = mathcalLleft[ (t-1)U(t-1) + 6U(t-1)right] = mathcalL left[(t-1)U(t-1)right] + 6mathcalLleft[ U(t-1)right]$$
Where we know:
$$ mathcalLU(t) = frac1s,$$
and
$$mathcalLtU(t)=frac1s^2.$$
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
Hint:
$$ mathcalLleft[ (t+5)U(t-1)right]= mathcalL left[(t-1+6)U(t-1)right] = mathcalLleft[ (t-1)U(t-1) + 6U(t-1)right] = mathcalL left[(t-1)U(t-1)right] + 6mathcalLleft[ U(t-1)right]$$
Where we know:
$$ mathcalLU(t) = frac1s,$$
and
$$mathcalLtU(t)=frac1s^2.$$
add a comment |Â
up vote
0
down vote
accepted
Hint:
$$ mathcalLleft[ (t+5)U(t-1)right]= mathcalL left[(t-1+6)U(t-1)right] = mathcalLleft[ (t-1)U(t-1) + 6U(t-1)right] = mathcalL left[(t-1)U(t-1)right] + 6mathcalLleft[ U(t-1)right]$$
Where we know:
$$ mathcalLU(t) = frac1s,$$
and
$$mathcalLtU(t)=frac1s^2.$$
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Hint:
$$ mathcalLleft[ (t+5)U(t-1)right]= mathcalL left[(t-1+6)U(t-1)right] = mathcalLleft[ (t-1)U(t-1) + 6U(t-1)right] = mathcalL left[(t-1)U(t-1)right] + 6mathcalLleft[ U(t-1)right]$$
Where we know:
$$ mathcalLU(t) = frac1s,$$
and
$$mathcalLtU(t)=frac1s^2.$$
Hint:
$$ mathcalLleft[ (t+5)U(t-1)right]= mathcalL left[(t-1+6)U(t-1)right] = mathcalLleft[ (t-1)U(t-1) + 6U(t-1)right] = mathcalL left[(t-1)U(t-1)right] + 6mathcalLleft[ U(t-1)right]$$
Where we know:
$$ mathcalLU(t) = frac1s,$$
and
$$mathcalLtU(t)=frac1s^2.$$
answered Aug 29 at 6:40
Alla Tarighati
2623
2623
add a comment |Â
add a comment |Â
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Are you looking for Laplace transform of $(t+5)U(t-1)$?
â Alla Tarighati
Aug 29 at 6:31
@AllaTarighati yes
â user185692
Aug 29 at 6:37