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Help in Laplace Transform, not homework

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  • #1
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I am trying to do laplace transform of [itex]y(t)=e^{-at}\cos (bt)\;[/itex]. The answer should be:

[tex]\frac {s+a}{(s+a)^2+b^2}[/tex]

But here is my work, I can't get rid of the -ve sign. I must be too blind to see the obvious, please help:

[tex]\overline{y(s)}=\int^∞_0 e^{-(s+a)t}\cos (bt) dt\;=\; \frac 1 b [e^{-(s+a)t}\sin(bt)]^∞_0\;+\frac {s+a}{b}\int^∞_0\;e^{-(s+a)t}\sin (bt) dt[/tex]

[tex][e^{-(s+a)t}\sin(bt)]^∞_0 =0 \;\Rightarrow\; \overline{y(s)}= \frac {s+a}{b}\int^∞_0\;e^{-(s+a)t}\sin (bt) dt\;=\;-\frac {s+a}{b^2} [e^{-(s+a)t}\cos(bt)]^∞_0\;-\frac {(s+a)^2}{b^2}\int^∞_0\;e^{-(s+a)t}\cos (bt) dt[/tex]

[tex][e^{-(s+a)t}\cos(bt)]^∞_0 =1 \;\Rightarrow\; \overline{y(s)}= -\frac {s+a}{b^2}-\frac {(s+a)^2}{b^2}\int^∞_0\;e^{-(s+a)t}\cos (bt) dt\;=\;-\frac {s+a}{b^2}-\frac {(s+a)^2}{b^2}\overline{y(s)}[/tex]

[tex]\Rightarrow\;\left(1+\frac {(s+a)^2}{b^2}\right)\overline{y(s)}\;=\;-\frac {s+a}{b^2}\;\Rightarrow\; \overline{y(s)}\;=\; -\frac {s+a}{(s+a)^2+b^2}[/tex]

You see the -ve sign? Please tell me what I did wrong.

Thanks
 

Answers and Replies

  • #2
Ray Vickson
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I am trying to do laplace transform of [itex]y(t)=e^{-at}\cos (bt)\;[/itex]. The answer should be:

[tex]\frac {s+a}{(s+a)^2+b^2}[/tex]

But here is my work, I can't get rid of the -ve sign. I must be too blind to see the obvious, please help:

[tex]\overline{y(s)}=\int^∞_0 e^{-(s+a)t}\cos (bt) dt\;=\; \frac 1 b [e^{-(s+a)t}\sin(bt)]^∞_0\;+\frac {s+a}{b}\int^∞_0\;e^{-(s+a)t}\sin (bt) dt[/tex]

[tex][e^{-(s+a)t}\sin(bt)]^∞_0 =0 \;\Rightarrow\; \overline{y(s)}= \frac {s+a}{b}\int^∞_0\;e^{-(s+a)t}\sin (bt) dt\;=\;-\frac {s+a}{b^2} [e^{-(s+a)t}\cos(bt)]^∞_0\;-\frac {(s+a)^2}{b^2}\int^∞_0\;e^{-(s+a)t}\cos (bt) dt[/tex]

[tex][e^{-(s+a)t}\cos(bt)]^∞_0 =1 \;\Rightarrow\; \overline{y(s)}= -\frac {s+a}{b^2}-\frac {(s+a)^2}{b^2}\int^∞_0\;e^{-(s+a)t}\cos (bt) dt\;=\;-\frac {s+a}{b^2}-\frac {(s+a)^2}{b^2}\overline{y(s)}[/tex]

[tex]\Rightarrow\;\left(1+\frac {(s+a)^2}{b^2}\right)\overline{y(s)}\;=\;-\frac {s+a}{b^2}\;\Rightarrow\; \overline{y(s)}\;=\; -\frac {s+a}{(s+a)^2+b^2}[/tex]

You see the -ve sign? Please tell me what I did wrong.

Thanks
On line 2 you should have a + sign in front of the last integral on the right, because you have a negative v = -w in the integration-by-parts formula uv - ∫ v du = -uw + ∫ w du.

One question: why did you not just get the transform of exp(-at)*exp(i*b*t)? That way you would get transforms of both exp(-at)*cos(bt) and exp(-at)*sin(bt) in the same calculation. and that calculation would take just one step---plus extracting real and imaginary parts.

RGV
 
  • #3
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On line 2 you should have a + sign in front of the last integral on the right, because you have a negative v = -w in the integration-by-parts formula uv - ∫ v du = -uw + ∫ w du.

One question: why did you not just get the transform of exp(-at)*exp(i*b*t)? That way you would get transforms of both exp(-at)*cos(bt) and exp(-at)*sin(bt) in the same calculation. and that calculation would take just one step---plus extracting real and imaginary parts.

RGV
Thanks for the reply. But on the second integral it is
[tex]-\int wdu\;\hbox { where }\; u= e^{-(s+a)t} \Rightarrow du=-(s+a)e^{-(s+a)t} dt\;,\; dw=\sin(bt)dt\;\Rightarrow \;w=-\frac 1 b \cos (bt)[/tex]

[tex]\Rightarrow\; -\int wdu\;=\; -\frac {-(s+a)}{-b}\int^∞_0 e^{-(s+a)t} \cos(bt)dt[/tex]

You see the -ve sign holds?

I just did it the first way come to mind to prove the formula, not working as an exercise. Now I got stuck in this!!! I gone through so many times and just can't see what I did wrong!!!
 
Last edited:
  • #4
You could take two derivatives and find that

x''= -2ax' + (b^2-a^2)x (I hope I did that right)

And take the laplace of both sides, use the rule for the laplace of a derivative, then solve for the laplace of x
 
  • #5
Ray Vickson
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Thanks for the reply. But on the second integral it is
[tex]-\int wdu\;\hbox { where }\; u= e^{-(s+a)t} \Rightarrow du=-(s+a)e^{-(s+a)t} dt\;,\; dw=\sin(bt)dt\;\Rightarrow \;w=-\frac 1 b \cos (bt)[/tex]

[tex]\Rightarrow\; -\int wdu\;=\; -\frac {(s+a)}{b}\int^∞_0 e^{-(s+a)t} \cos(bt)dt[/tex]

You see the -ve sign holds?

I just did it the first way come to mind to prove the formula, not working as an exercise. Now I got stuck in this!!! I gone through so many times and just can't see what I did wrong!!!
Yes, I see them, and I shouldn't, because they do not belong. Go back and look again!

RGV
 
  • #6
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Yes, I see them, and I shouldn't, because they do not belong. Go back and look again!

RGV
I don't see it. I look at this for a long time. I modified the last post to reflect the -ve signs and still the same. Please point it out. I know once a while, I get into this and I just cannot see it. This is supposed to be an easy proof of the formula.
 
  • #7
Ray Vickson
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I don't see it. I look at this for a long time. I modified the last post to reflect the -ve signs and still the same. Please point it out. I know once a while, I get into this and I just cannot see it. This is supposed to be an easy proof of the formula.
I have already pointed out your error, and suggested to you that you do the whole thing again, using exp(i*b*x) instead of sin(bx) or cos(bx). If you do not wish to use my hints, that is OK by me, but I am all out of suggestions.

RGV
 
Last edited:
  • #8
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I have already pointed out your error, and suggested to you that you do the whole thing again, using exp(i*b*x) instead of sin(bx) or cos(bx). If you do not wish to use my hints, that is OK by me, but I am all out of suggestions.

RGV
Yes, I actually did the problem using complex Euler formula already. But that does not answer the question why I can't get the right sign. I check it so many times. Whether the way I do this is the most efficient way is besides the point. The point is why don't I get the right answer using simple integrate by parts. No matter what, I should get the correct answer. Right?
 
  • #9
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Anyone please?
 

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