Mixing
Laplace transform restriction and differentiation
$begingroup$
every one.I have just started learning Laplace transform.However, there are two main conceptual problems I can't convince myself.
The first problem is about the restriction of this integral, I understand that for a real variable s, it needs to be positive so that the integral converges to a value, e.g. f(t)=1 then L(f(t)) = 1/s . s>0
But,if the s is a complex variable, I don't understand why the book says , e.g. for f(t) = exp(-at) ,then Re(s+a)>0.
Why did they simply ignore the imaginary part of variable s, or are they just considering the real part of the function?(which they didn't mention)?
The second question is when I was asked to verify L(tsinat) =3 http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table_files/eq0021M.gif by using
L(cosat) = 
the problem is why do we have to differentiate this formula with respect to a? I thought a was a constant value which can't be differentiated in many problems I encountered while solving ODEs before.
Thank you so much for helping! Greatly appreciated.
laplace-transform
edited Jan 22 at 8:13
Glorfindel
3,42981830
asked Sep 9 '13 at 7:39
el psy Congrooel psy Congroo
168111
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add a comment |
$begingroup$
every one.I have just started learning Laplace transform.However, there are two main conceptual problems I can't convince myself.
The first problem is about the restriction of this integral, I understand that for a real variable s, it needs to be positive so that the integral converges to a value, e.g. f(t)=1 then L(f(t)) = 1/s . s>0
But,if the s is a complex variable, I don't understand why the book says , e.g. for f(t) = exp(-at) ,then Re(s+a)>0.
Why did they simply ignore the imaginary part of variable s, or are they just considering the real part of the function?(which they didn't mention)?
The second question is when I was asked to verify L(tsinat) =3 http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table_files/eq0021M.gif by using
L(cosat) =
the problem is why do we have to differentiate this formula with respect to a? I thought a was a constant value which can't be differentiated in many problems I encountered while solving ODEs before.
Thank you so much for helping! Greatly appreciated.
laplace-transform
edited Jan 22 at 8:13
Glorfindel
3,42981830
asked Sep 9 '13 at 7:39
el psy Congrooel psy Congroo
168111
$endgroup$
add a comment |
$begingroup$
every one.I have just started learning Laplace transform.However, there are two main conceptual problems I can't convince myself.
The first problem is about the restriction of this integral, I understand that for a real variable s, it needs to be positive so that the integral converges to a value, e.g. f(t)=1 then L(f(t)) = 1/s . s>0
But,if the s is a complex variable, I don't understand why the book says , e.g. for f(t) = exp(-at) ,then Re(s+a)>0.
Why did they simply ignore the imaginary part of variable s, or are they just considering the real part of the function?(which they didn't mention)?
The second question is when I was asked to verify L(tsinat) =3 http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table_files/eq0021M.gif by using
L(cosat) =
the problem is why do we have to differentiate this formula with respect to a? I thought a was a constant value which can't be differentiated in many problems I encountered while solving ODEs before.
Thank you so much for helping! Greatly appreciated.
laplace-transform
edited Jan 22 at 8:13
Glorfindel
3,42981830
asked Sep 9 '13 at 7:39
el psy Congrooel psy Congroo
168111
$endgroup$
every one.I have just started learning Laplace transform.However, there are two main conceptual problems I can't convince myself.
The first problem is about the restriction of this integral, I understand that for a real variable s, it needs to be positive so that the integral converges to a value, e.g. f(t)=1 then L(f(t)) = 1/s . s>0
But,if the s is a complex variable, I don't understand why the book says , e.g. for f(t) = exp(-at) ,then Re(s+a)>0.
Why did they simply ignore the imaginary part of variable s, or are they just considering the real part of the function?(which they didn't mention)?
The second question is when I was asked to verify L(tsinat) =3 http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table_files/eq0021M.gif by using
L(cosat) =
the problem is why do we have to differentiate this formula with respect to a? I thought a was a constant value which can't be differentiated in many problems I encountered while solving ODEs before.
Thank you so much for helping! Greatly appreciated.
laplace-transform
laplace-transform
edited Jan 22 at 8:13
Glorfindel
3,42981830
asked Sep 9 '13 at 7:39
el psy Congrooel psy Congroo
168111
edited Jan 22 at 8:13
Glorfindel
3,42981830
asked Sep 9 '13 at 7:39
el psy Congrooel psy Congroo
168111
edited Jan 22 at 8:13
Glorfindel
3,42981830
edited Jan 22 at 8:13
Glorfindel
3,42981830
edited Jan 22 at 8:13
Glorfindel
3,42981830
3,42981830
asked Sep 9 '13 at 7:39
el psy Congrooel psy Congroo
168111
asked Sep 9 '13 at 7:39
el psy Congrooel psy Congroo
168111
asked Sep 9 '13 at 7:39
el psy Congrooel psy Congroo
168111
168111
add a comment |
add a comment |
2 Answers
2
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For the first question: If $s=sigma + iomega$ and $t$ is real, then
$$|e^{-st}|=|e^{-sigma t}e^{-iomega t}| = e^{-sigma t}$$
since $|e^{iomega t}|=1$. In other words it's only the real part of $s$ that determines where the ontegral converges.
answered Sep 9 '13 at 8:00
mrfmrf
37.6k64786
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add a comment |
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Hint for the second part: Use this fact that:
$$mathcal{L}(t^nf(t))=(-1)^nfrac{d^n F(s)}{ds^n}$$ wherein $mathcal{L}(f(t))=F(s).$
answered Sep 9 '13 at 7:46
mrsmrs
1
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- needs a TU! +1
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– Namaste
Sep 9 '13 at 10:36
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the first question: If $s=sigma + iomega$ and $t$ is real, then
$$|e^{-st}|=|e^{-sigma t}e^{-iomega t}| = e^{-sigma t}$$
since $|e^{iomega t}|=1$. In other words it's only the real part of $s$ that determines where the ontegral converges.
answered Sep 9 '13 at 8:00
mrfmrf
37.6k64786
$endgroup$
add a comment |
$begingroup$
For the first question: If $s=sigma + iomega$ and $t$ is real, then
$$|e^{-st}|=|e^{-sigma t}e^{-iomega t}| = e^{-sigma t}$$
since $|e^{iomega t}|=1$. In other words it's only the real part of $s$ that determines where the ontegral converges.
answered Sep 9 '13 at 8:00
mrfmrf
37.6k64786
$endgroup$
add a comment |
$begingroup$
For the first question: If $s=sigma + iomega$ and $t$ is real, then
$$|e^{-st}|=|e^{-sigma t}e^{-iomega t}| = e^{-sigma t}$$
since $|e^{iomega t}|=1$. In other words it's only the real part of $s$ that determines where the ontegral converges.
answered Sep 9 '13 at 8:00
mrfmrf
37.6k64786
$endgroup$
For the first question: If $s=sigma + iomega$ and $t$ is real, then
$$|e^{-st}|=|e^{-sigma t}e^{-iomega t}| = e^{-sigma t}$$
since $|e^{iomega t}|=1$. In other words it's only the real part of $s$ that determines where the ontegral converges.
answered Sep 9 '13 at 8:00
mrfmrf
37.6k64786
answered Sep 9 '13 at 8:00
mrfmrf
37.6k64786
answered Sep 9 '13 at 8:00
mrfmrf
37.6k64786
answered Sep 9 '13 at 8:00
mrfmrf
37.6k64786
37.6k64786
add a comment |
add a comment |
$begingroup$
Hint for the second part: Use this fact that:
$$mathcal{L}(t^nf(t))=(-1)^nfrac{d^n F(s)}{ds^n}$$ wherein $mathcal{L}(f(t))=F(s).$
answered Sep 9 '13 at 7:46
mrsmrs
1
$endgroup$
$begingroup$
- needs a TU! +1
$endgroup$
– Namaste
Sep 9 '13 at 10:36
add a comment |
$begingroup$
Hint for the second part: Use this fact that:
$$mathcal{L}(t^nf(t))=(-1)^nfrac{d^n F(s)}{ds^n}$$ wherein $mathcal{L}(f(t))=F(s).$
answered Sep 9 '13 at 7:46
mrsmrs
1
$endgroup$
$begingroup$
- needs a TU! +1
$endgroup$
– Namaste
Sep 9 '13 at 10:36
add a comment |
$begingroup$
Hint for the second part: Use this fact that:
$$mathcal{L}(t^nf(t))=(-1)^nfrac{d^n F(s)}{ds^n}$$ wherein $mathcal{L}(f(t))=F(s).$
answered Sep 9 '13 at 7:46
mrsmrs
1
$endgroup$
Hint for the second part: Use this fact that:
$$mathcal{L}(t^nf(t))=(-1)^nfrac{d^n F(s)}{ds^n}$$ wherein $mathcal{L}(f(t))=F(s).$
answered Sep 9 '13 at 7:46
mrsmrs
1
answered Sep 9 '13 at 7:46
mrsmrs
1
answered Sep 9 '13 at 7:46
mrsmrs
1
answered Sep 9 '13 at 7:46
mrsmrs
1
1
$begingroup$
- needs a TU! +1
$endgroup$
– Namaste
Sep 9 '13 at 10:36
add a comment |
$begingroup$
- needs a TU! +1
$endgroup$
– Namaste
Sep 9 '13 at 10:36
$begingroup$
- needs a TU! +1
$endgroup$
– Namaste
Sep 9 '13 at 10:36
$begingroup$
- needs a TU! +1
$endgroup$
– Namaste
Sep 9 '13 at 10:36
add a comment |
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