Laplace transforms - second shift theorem












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Hi, i'm struggling with understanding how this expansion came about for determining the LT of f(t). Is anyone familiar with this?
thank you






laplace-transform







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  • Are you unclear with where the functions that they are applying the Laplace transform are coming from, or the actual step in applying the transform?
    – Theo C.
    Jan 4 at 2:06










  • Hi, thanks for your reply. I know the second shift theorem for the unit step function (u) is being applied but I don't understand where the expansions come from.
    – user564900
    Jan 4 at 2:08
















1














pic 1



pic 2



Hi, i'm struggling with understanding how this expansion came about for determining the LT of f(t). Is anyone familiar with this?
thank you






laplace-transform







share|cite|improve this question






















  • Are you unclear with where the functions that they are applying the Laplace transform are coming from, or the actual step in applying the transform?
    – Theo C.
    Jan 4 at 2:06










  • Hi, thanks for your reply. I know the second shift theorem for the unit step function (u) is being applied but I don't understand where the expansions come from.
    – user564900
    Jan 4 at 2:08














1












1








1







pic 1



pic 2



Hi, i'm struggling with understanding how this expansion came about for determining the LT of f(t). Is anyone familiar with this?
thank you






laplace-transform







share|cite|improve this question













pic 1



pic 2



Hi, i'm struggling with understanding how this expansion came about for determining the LT of f(t). Is anyone familiar with this?
thank you






laplace-transform




laplace-transform






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 1:06









user564900

305




305












  • Are you unclear with where the functions that they are applying the Laplace transform are coming from, or the actual step in applying the transform?
    – Theo C.
    Jan 4 at 2:06










  • Hi, thanks for your reply. I know the second shift theorem for the unit step function (u) is being applied but I don't understand where the expansions come from.
    – user564900
    Jan 4 at 2:08


















  • Are you unclear with where the functions that they are applying the Laplace transform are coming from, or the actual step in applying the transform?
    – Theo C.
    Jan 4 at 2:06










  • Hi, thanks for your reply. I know the second shift theorem for the unit step function (u) is being applied but I don't understand where the expansions come from.
    – user564900
    Jan 4 at 2:08
















Are you unclear with where the functions that they are applying the Laplace transform are coming from, or the actual step in applying the transform?
– Theo C.
Jan 4 at 2:06




Are you unclear with where the functions that they are applying the Laplace transform are coming from, or the actual step in applying the transform?
– Theo C.
Jan 4 at 2:06












Hi, thanks for your reply. I know the second shift theorem for the unit step function (u) is being applied but I don't understand where the expansions come from.
– user564900
Jan 4 at 2:08




Hi, thanks for your reply. I know the second shift theorem for the unit step function (u) is being applied but I don't understand where the expansions come from.
– user564900
Jan 4 at 2:08










1 Answer
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oldest

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If I've understood your comment correctly, then I think I see the confusion.



Recall that the second shifting theorem says that if $mathcal{L}{ f(t)} = F(s)$ then $mathcal{L}{ f(t-a)u(t-a)} = e^{-as}F(s)$



Now, let's dissect taking the Laplace transform of $frac{1}{2}t^2u(t-1)$. Note that our current function is $f(t) = frac{1}{2}t^2$. So then $f(t-1) = frac{1}{2}(t-1)^2$. As such, to get $f(t)$ in the desired form, we will conveniently add $0$ in a number of different ways:
$$f(t)=f(t) + (t-t) + (frac{1}{2}-frac{1}{2}) = frac{1}{2}(t^2-2t+1) + t - frac{1}{2} = f(t-1) + t-frac{1}{2} + (frac{1}{2}-frac{1}{2}) $$
$$= f(t-1) + (t-1) + frac{1}{2} $$
Now we have $f(t)$ in the desired form to apply the shifting theorem.






share|cite|improve this answer





















  • Thank you!! this was very helpful
    – user564900
    2 days ago











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If I've understood your comment correctly, then I think I see the confusion.



Recall that the second shifting theorem says that if $mathcal{L}{ f(t)} = F(s)$ then $mathcal{L}{ f(t-a)u(t-a)} = e^{-as}F(s)$



Now, let's dissect taking the Laplace transform of $frac{1}{2}t^2u(t-1)$. Note that our current function is $f(t) = frac{1}{2}t^2$. So then $f(t-1) = frac{1}{2}(t-1)^2$. As such, to get $f(t)$ in the desired form, we will conveniently add $0$ in a number of different ways:
$$f(t)=f(t) + (t-t) + (frac{1}{2}-frac{1}{2}) = frac{1}{2}(t^2-2t+1) + t - frac{1}{2} = f(t-1) + t-frac{1}{2} + (frac{1}{2}-frac{1}{2}) $$
$$= f(t-1) + (t-1) + frac{1}{2} $$
Now we have $f(t)$ in the desired form to apply the shifting theorem.






share|cite|improve this answer





















  • Thank you!! this was very helpful
    – user564900
    2 days ago
















1














If I've understood your comment correctly, then I think I see the confusion.



Recall that the second shifting theorem says that if $mathcal{L}{ f(t)} = F(s)$ then $mathcal{L}{ f(t-a)u(t-a)} = e^{-as}F(s)$



Now, let's dissect taking the Laplace transform of $frac{1}{2}t^2u(t-1)$. Note that our current function is $f(t) = frac{1}{2}t^2$. So then $f(t-1) = frac{1}{2}(t-1)^2$. As such, to get $f(t)$ in the desired form, we will conveniently add $0$ in a number of different ways:
$$f(t)=f(t) + (t-t) + (frac{1}{2}-frac{1}{2}) = frac{1}{2}(t^2-2t+1) + t - frac{1}{2} = f(t-1) + t-frac{1}{2} + (frac{1}{2}-frac{1}{2}) $$
$$= f(t-1) + (t-1) + frac{1}{2} $$
Now we have $f(t)$ in the desired form to apply the shifting theorem.






share|cite|improve this answer





















  • Thank you!! this was very helpful
    – user564900
    2 days ago














1












1








1






If I've understood your comment correctly, then I think I see the confusion.



Recall that the second shifting theorem says that if $mathcal{L}{ f(t)} = F(s)$ then $mathcal{L}{ f(t-a)u(t-a)} = e^{-as}F(s)$



Now, let's dissect taking the Laplace transform of $frac{1}{2}t^2u(t-1)$. Note that our current function is $f(t) = frac{1}{2}t^2$. So then $f(t-1) = frac{1}{2}(t-1)^2$. As such, to get $f(t)$ in the desired form, we will conveniently add $0$ in a number of different ways:
$$f(t)=f(t) + (t-t) + (frac{1}{2}-frac{1}{2}) = frac{1}{2}(t^2-2t+1) + t - frac{1}{2} = f(t-1) + t-frac{1}{2} + (frac{1}{2}-frac{1}{2}) $$
$$= f(t-1) + (t-1) + frac{1}{2} $$
Now we have $f(t)$ in the desired form to apply the shifting theorem.






share|cite|improve this answer












If I've understood your comment correctly, then I think I see the confusion.



Recall that the second shifting theorem says that if $mathcal{L}{ f(t)} = F(s)$ then $mathcal{L}{ f(t-a)u(t-a)} = e^{-as}F(s)$



Now, let's dissect taking the Laplace transform of $frac{1}{2}t^2u(t-1)$. Note that our current function is $f(t) = frac{1}{2}t^2$. So then $f(t-1) = frac{1}{2}(t-1)^2$. As such, to get $f(t)$ in the desired form, we will conveniently add $0$ in a number of different ways:
$$f(t)=f(t) + (t-t) + (frac{1}{2}-frac{1}{2}) = frac{1}{2}(t^2-2t+1) + t - frac{1}{2} = f(t-1) + t-frac{1}{2} + (frac{1}{2}-frac{1}{2}) $$
$$= f(t-1) + (t-1) + frac{1}{2} $$
Now we have $f(t)$ in the desired form to apply the shifting theorem.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 2:27









Theo C.

25928




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  • Thank you!! this was very helpful
    – user564900
    2 days ago


















  • Thank you!! this was very helpful
    – user564900
    2 days ago
















Thank you!! this was very helpful
– user564900
2 days ago




Thank you!! this was very helpful
– user564900
2 days ago


















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