Laplace Transform with Integral Convolution

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Homework Statement


Determine the Laplace Transform of
∫(from 0 to t) (t-τ)cos(2(t-τ))e-4τ

using Laplace Transform tables.


Homework Equations


I know the basic convolution theorem is
(f*g)(t) = ∫f(τ)g(t-τ)dτ


The Attempt at a Solution


I'm not sure if this is double convolution or if I'm missing an easy substitution.
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


Determine the Laplace Transform of
∫(from 0 to t) (t-τ)cos(2(t-τ))e-4τ

using Laplace Transform tables.


Homework Equations


I know the basic convolution theorem is
(f*g)(t) = ∫f(τ)g(t-τ)dτ


The Attempt at a Solution


I'm not sure if this is double convolution or if I'm missing an easy substitution.

1. What are ##f(t)## and ##g(t)## in this problem?
2. What is the formula for ##\mathcal L(f*g)##?
 
  • #3
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Response:
1. I'm not sure because I've never seen (t-tau) twice in one of these.
2. Laplace(f*g) = F(s)G(s)
 
  • #4
LCKurtz
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Response:
1. I'm not sure because I've never seen (t-tau) twice in one of these.

As an example, what would ##g(t-\tau)## be for ##g(t) = t^2\sin(t)##?
 
  • #5
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In that problem, it would be setup so that f(t) = t^2 and g(t) = sin(t) making the convolution integral

∫ from 0 to t: t2*sin(t-τ) dτ

and then integrated using trig identities
 
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  • #6
LCKurtz
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As an example, what would ##g(t-\tau)## be for ##g(t) = t^2\sin(t)##?

In that problem, it would be setup so that f(t) = t^2 and g(t) = sin(t) making the convolution integral

∫ from 0 to t: t2*sin(t-τ) dτ

and then integrated using trig identities

No. You didn't answer my question. I didn't ask you anything about convolution. Given ##g(t) = t^2\sin(t)## what would ##g(\pi/2)## be? ##g(s)##? ##g(t-\tau)##?
 
  • #7
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[itex]\int_0^t (t-\tau)cos(2(t-\tau))e^{-4\tau} d\tau[/itex]

it looks as if f(t) = t and g(t) = cos(2*f(t))

Would it make sense to do a simple substitution and end up doing the integral

[itex]\int_0^t u\cdot cos(2u) du[/itex]

giving [itex]\frac{1}{2} u\cdot sin(2u) + cos(2u)[/itex]

[itex]\frac{1}{2} t\cdot sin(2t) + cos(2t)[/itex]

and then taking the laplace of those?
 
  • #8
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No. You didn't answer my question. I didn't ask you anything about convolution. Given ##g(t) = t^2\sin(t)## what would ##g(\pi/2)## be? ##g(s)##? ##g(t-\tau)##?

I would say g(t-τ) would be (t-τ)2sin(t-τ)
 
  • #9
LCKurtz
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[itex]\int_0^t (t-\tau)cos(2(t-\tau))e^{-4\tau} d\tau[/itex]

it looks as if f(t) = t and g(t) = cos(2*f(t))

Would it make sense to do a simple substitution and end up doing the integral

[itex]\int_0^t u\cdot cos(2u) du[/itex]

giving [itex]\frac{1}{2} u\cdot sin(2u) + cos(2u)[/itex]

[itex]\frac{1}{2} t\cdot sin(2t) + cos(2t)[/itex]

and then taking the laplace of those?

No. You are missing the point of the problem. Just answer my question in my previous post for now. That will maybe give you an idea.
 
  • #10
LCKurtz
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I would say g(t-τ) would be (t-τ)2sin(t-τ)

OK, so you see how you can have more than one ##t-\tau## for a function. Now can you identify ##f(t)## and ##g(t)## in your original problem?
 
  • #11
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OK, so you see how you can have more than one ##t-\tau## for a function. Now can you identify ##f(t)## and ##g(t)## in your original problem?

Yes, I do. Does that mean my attempt was incorrect?
 
  • #12
LCKurtz
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OK, so you see how you can have more than one ##t-\tau## for a function. Now can you identify ##f(t)## and ##g(t)## in your original problem?

Yes, I do. Does that mean my attempt was incorrect?

If you mean your attempt in post #7, yes that is incorrect. Where did the exponential go in that post? But never mind that. What do you get for ##f(t)## and ##g(t)##? Do you see how to do the problem knowing those and using post #3?
 
  • #13
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ahh, got it. thanks a lot.

(sleep deprivation makes you miss things)
 

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