Laplace transform of complex exponential

  • Thread starter EvLer
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I just want to be sure I understand this correctly, usually L[f(t)] = 1/(s-a), where [tex]f(t) = e^{at}[/tex], but if it is a complex number would still be 1/(s - complex_number)?
techinically, i think it should be, since every number can be reprsented as complex number. Just want to be sure about this with Laplace transform.
thanks much.
 

Answers and Replies

  • #2
Tide
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Definitely! You can demonstrate it by direct integration.
 
  • #3
lurflurf
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EvLer said:
I just want to be sure I understand this correctly, usually L[f(t)] = 1/(s-a), where [tex]f(t) = e^{at}[/tex], but if it is a complex number would still be 1/(s - complex_number)?
techinically, i think it should be, since every number can be reprsented as complex number. Just want to be sure about this with Laplace transform.
thanks much.
Yes and it is a nice way to find the laplace transforms of sin and cos.
L[cos(a x)+i sin(a x)]=L[exp(i x)]=1/(s-a i)=(s+a i)/(s^2+a^2)
hence (equating real and imaginary parts)
L[cos(a x)]=s/(s^2+a^2)
L[sin(a x)]=a/(s^2+a^2)
 

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