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Integrate / Convolution of a dirac Delta function from 0 to t

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0

$begingroup$

I have a question about the convolution integral resulting due to an inverse Laplace transform.
Considering one has a multiplication of 2 functions in the laplace Domain

$G(s) F(s)= exp(-a s) F(s)$

To transform that into the time domain one can simply transform $G(s)$ and $F(s)$ separately, which results in the convolution integral

beginalign
int_0^t delta(t-a)f(t) dt &=int_0^t delta(t-tau-a) f(tau) dtau
endalign

I am a bit confused with the boundaries. I know the identity of

$$int_-infty^infty delta(t-a) f(t) dt = f(a)$$

In that case the boundaries are from $-infty$ to $infty$. Unfortunately in my case I integrate from $0$ to $t$, does that still results to the same solution of $f(a)$ or does that result in $f(t-a)$?

laplace-transform convolution dirac-delta

share|cite|improve this question

edited Mar 29 at 11:30

Mattos

2,83121321

asked Mar 29 at 10:43

user654189user654189

112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To avoid confusion, you might want to use cdot (without the gap between the forward slash and the string, which gives $cdot$) for multiplication, and use $*$ only for convolutions.
  $endgroup$
  – Mattos
  Mar 29 at 10:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, I changed it
  $endgroup$
  – user654189
  Mar 29 at 11:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure how you got your result with an integral from $0$ to $t$ or where your delta function is coming from. Could you explain a little bit more?
  $endgroup$
  – Mattos
  Mar 29 at 11:24
  

  
  
  
  

add a comment | 

0

$begingroup$

I have a question about the convolution integral resulting due to an inverse Laplace transform.
Considering one has a multiplication of 2 functions in the laplace Domain

$G(s) F(s)= exp(-a s) F(s)$

To transform that into the time domain one can simply transform $G(s)$ and $F(s)$ separately, which results in the convolution integral

beginalign
int_0^t delta(t-a)f(t) dt &=int_0^t delta(t-tau-a) f(tau) dtau
endalign

I am a bit confused with the boundaries. I know the identity of

$$int_-infty^infty delta(t-a) f(t) dt = f(a)$$

In that case the boundaries are from $-infty$ to $infty$. Unfortunately in my case I integrate from $0$ to $t$, does that still results to the same solution of $f(a)$ or does that result in $f(t-a)$?

laplace-transform convolution dirac-delta

share|cite|improve this question

edited Mar 29 at 11:30

Mattos

2,83121321

asked Mar 29 at 10:43

user654189user654189

112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To avoid confusion, you might want to use cdot (without the gap between the forward slash and the string, which gives $cdot$) for multiplication, and use $*$ only for convolutions.
  $endgroup$
  – Mattos
  Mar 29 at 10:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, I changed it
  $endgroup$
  – user654189
  Mar 29 at 11:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure how you got your result with an integral from $0$ to $t$ or where your delta function is coming from. Could you explain a little bit more?
  $endgroup$
  – Mattos
  Mar 29 at 11:24
  

  
  
  
  

add a comment | 

$begingroup$

I have a question about the convolution integral resulting due to an inverse Laplace transform.
Considering one has a multiplication of 2 functions in the laplace Domain

$G(s) F(s)= exp(-a s) F(s)$

To transform that into the time domain one can simply transform $G(s)$ and $F(s)$ separately, which results in the convolution integral

beginalign
int_0^t delta(t-a)f(t) dt &=int_0^t delta(t-tau-a) f(tau) dtau
endalign

I am a bit confused with the boundaries. I know the identity of

$$int_-infty^infty delta(t-a) f(t) dt = f(a)$$

In that case the boundaries are from $-infty$ to $infty$. Unfortunately in my case I integrate from $0$ to $t$, does that still results to the same solution of $f(a)$ or does that result in $f(t-a)$?

laplace-transform convolution dirac-delta

share|cite|improve this question

edited Mar 29 at 11:30

Mattos

2,83121321

asked Mar 29 at 10:43

user654189user654189

112

$endgroup$

share|cite|improve this question

edited Mar 29 at 11:30

Mattos

2,83121321

asked Mar 29 at 10:43

user654189user654189

112

share|cite|improve this question

edited Mar 29 at 11:30

Mattos

2,83121321

asked Mar 29 at 10:43

user654189user654189

112

edited Mar 29 at 11:30

Mattos

2,83121321

edited Mar 29 at 11:30

Mattos

2,83121321

asked Mar 29 at 10:43

user654189user654189

112

asked Mar 29 at 10:43

user654189user654189

112