am i applying the property correctly??
$begingroup$
The time shifting property states that:
$$ mathcal{L}(u(t-t_0)f(t-t_0))=e^{-st_0}F(s) $$
Now I am given a question which is:
$$ mathcal{L}(e^{-4t}u(t-3)) $$
Can I do like I have done here? That the time shift is of 3 seconds, so using the property I get final answer as:
$$ frac{e^{-3s}}{s+4} $$
Edit: if we are asked to find unilateral Laplace Transform then and only then do I get the above answer?
laplace-transform
edited Dec 30 '18 at 22:10
Paul Enta
5,18111334
asked Dec 20 '18 at 15:52
Ahmad QayyumAhmad Qayyum
677
$endgroup$
add a comment |
$begingroup$
The time shifting property states that:
$$ mathcal{L}(u(t-t_0)f(t-t_0))=e^{-st_0}F(s) $$
Now I am given a question which is:
$$ mathcal{L}(e^{-4t}u(t-3)) $$
Can I do like I have done here? That the time shift is of 3 seconds, so using the property I get final answer as:
$$ frac{e^{-3s}}{s+4} $$
Edit: if we are asked to find unilateral Laplace Transform then and only then do I get the above answer?
laplace-transform
edited Dec 30 '18 at 22:10
Paul Enta
5,18111334
asked Dec 20 '18 at 15:52
Ahmad QayyumAhmad Qayyum
677
$endgroup$
add a comment |
$begingroup$
The time shifting property states that:
$$ mathcal{L}(u(t-t_0)f(t-t_0))=e^{-st_0}F(s) $$
Now I am given a question which is:
$$ mathcal{L}(e^{-4t}u(t-3)) $$
Can I do like I have done here? That the time shift is of 3 seconds, so using the property I get final answer as:
$$ frac{e^{-3s}}{s+4} $$
Edit: if we are asked to find unilateral Laplace Transform then and only then do I get the above answer?
laplace-transform
edited Dec 30 '18 at 22:10
Paul Enta
5,18111334
asked Dec 20 '18 at 15:52
Ahmad QayyumAhmad Qayyum
677
$endgroup$
The time shifting property states that:
$$ mathcal{L}(u(t-t_0)f(t-t_0))=e^{-st_0}F(s) $$
Now I am given a question which is:
$$ mathcal{L}(e^{-4t}u(t-3)) $$
Can I do like I have done here? That the time shift is of 3 seconds, so using the property I get final answer as:
$$ frac{e^{-3s}}{s+4} $$
Edit: if we are asked to find unilateral Laplace Transform then and only then do I get the above answer?
laplace-transform
laplace-transform
edited Dec 30 '18 at 22:10
Paul Enta
5,18111334
asked Dec 20 '18 at 15:52
Ahmad QayyumAhmad Qayyum
677
edited Dec 30 '18 at 22:10
Paul Enta
5,18111334
asked Dec 20 '18 at 15:52
Ahmad QayyumAhmad Qayyum
677
edited Dec 30 '18 at 22:10
Paul Enta
5,18111334
edited Dec 30 '18 at 22:10
Paul Enta
5,18111334
edited Dec 30 '18 at 22:10
Paul Enta
5,18111334
5,18111334
asked Dec 20 '18 at 15:52
Ahmad QayyumAhmad Qayyum
677
asked Dec 20 '18 at 15:52
Ahmad QayyumAhmad Qayyum
677
asked Dec 20 '18 at 15:52
Ahmad QayyumAhmad Qayyum
677
677
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's wrong!
Go step by step.
Step $1$: Find frequency shift due to $e^{-4t}$ using:
If $mathcal{L}(f(t))=F(s)$, then
$mathcal{L}(e^{-4t}f(t))=F(s+4)$. (which is the answer)
Step $2$: Find $mathcal{L}(f(t))$ by the property of time shift as follows:
$$mathcal{L}(u(t-3).1)=frac{e^{-3s}}{s}$$
where $f(t)=u(t-3).1$
So that finally $$F(s+4)=frac{e^{-3(s+4)}}{s+4}$$
answered Dec 20 '18 at 17:54
Sameer BahetiSameer Baheti
5718
$endgroup$
$begingroup$
can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
$endgroup$
– Ahmad Qayyum
Dec 25 '18 at 18:19
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's wrong!
Go step by step.
Step $1$: Find frequency shift due to $e^{-4t}$ using:
If $mathcal{L}(f(t))=F(s)$, then
$mathcal{L}(e^{-4t}f(t))=F(s+4)$. (which is the answer)
Step $2$: Find $mathcal{L}(f(t))$ by the property of time shift as follows:
$$mathcal{L}(u(t-3).1)=frac{e^{-3s}}{s}$$
where $f(t)=u(t-3).1$
So that finally $$F(s+4)=frac{e^{-3(s+4)}}{s+4}$$
answered Dec 20 '18 at 17:54
Sameer BahetiSameer Baheti
5718
$endgroup$
$begingroup$
can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
$endgroup$
– Ahmad Qayyum
Dec 25 '18 at 18:19
add a comment |
$begingroup$
It's wrong!
Go step by step.
Step $1$: Find frequency shift due to $e^{-4t}$ using:
If $mathcal{L}(f(t))=F(s)$, then
$mathcal{L}(e^{-4t}f(t))=F(s+4)$. (which is the answer)
Step $2$: Find $mathcal{L}(f(t))$ by the property of time shift as follows:
$$mathcal{L}(u(t-3).1)=frac{e^{-3s}}{s}$$
where $f(t)=u(t-3).1$
So that finally $$F(s+4)=frac{e^{-3(s+4)}}{s+4}$$
answered Dec 20 '18 at 17:54
Sameer BahetiSameer Baheti
5718
$endgroup$
$begingroup$
can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
$endgroup$
– Ahmad Qayyum
Dec 25 '18 at 18:19
add a comment |
$begingroup$
It's wrong!
Go step by step.
Step $1$: Find frequency shift due to $e^{-4t}$ using:
If $mathcal{L}(f(t))=F(s)$, then
$mathcal{L}(e^{-4t}f(t))=F(s+4)$. (which is the answer)
Step $2$: Find $mathcal{L}(f(t))$ by the property of time shift as follows:
$$mathcal{L}(u(t-3).1)=frac{e^{-3s}}{s}$$
where $f(t)=u(t-3).1$
So that finally $$F(s+4)=frac{e^{-3(s+4)}}{s+4}$$
answered Dec 20 '18 at 17:54
Sameer BahetiSameer Baheti
5718
$endgroup$
It's wrong!
Go step by step.
Step $1$: Find frequency shift due to $e^{-4t}$ using:
If $mathcal{L}(f(t))=F(s)$, then
$mathcal{L}(e^{-4t}f(t))=F(s+4)$. (which is the answer)
Step $2$: Find $mathcal{L}(f(t))$ by the property of time shift as follows:
$$mathcal{L}(u(t-3).1)=frac{e^{-3s}}{s}$$
where $f(t)=u(t-3).1$
So that finally $$F(s+4)=frac{e^{-3(s+4)}}{s+4}$$
answered Dec 20 '18 at 17:54
Sameer BahetiSameer Baheti
5718
edited Dec 20 '18 at 18:19
answered Dec 20 '18 at 17:54
Sameer BahetiSameer Baheti
5718
answered Dec 20 '18 at 17:54
Sameer BahetiSameer Baheti
5718
answered Dec 20 '18 at 17:54
Sameer BahetiSameer Baheti
5718
5718
$begingroup$
can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
$endgroup$
– Ahmad Qayyum
Dec 25 '18 at 18:19
add a comment |
$begingroup$
can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
$endgroup$
– Ahmad Qayyum
Dec 25 '18 at 18:19
$begingroup$
can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
$endgroup$
– Ahmad Qayyum
Dec 25 '18 at 18:19
$begingroup$
can you tell me why can't apply the property over here? what are the cases where we can apply this property? does those cases involve the unshifted version of unit step functions?
$endgroup$
– Ahmad Qayyum
Dec 25 '18 at 18:19
add a comment |
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