Laplace transform integral problem

[TyErd]

Homework Statement

Hey guys, I've attached the question that's troubling me. I've also attached the table and formulas of Laplace transform for you convenience.

Homework Equations

attached

The Attempt at a Solution

Right now, I am thinking the best way to do this problem is by working backwards. I'm using the convolution integral formula.
I've already established that G(s) is 1/(s+4) using inverse Laplace transforms but I am not sure if that will get me an answer for f(t) rather F(s).

thats my inverse laplace------->invL
and laplace is ------> L

 

[RoshanBBQ]

When you take the Laplace transform of both sides of the equation, what are you getting? Where are you getting stuck?

 

[TyErd]

I think I'm stuck because I'm not sure how to approach a problem like this. anyway, Laplace transform of both sides just makes it L{f(t)}= F(s)G(s) right? no that's not right coz there the 6t-5

 

[RoshanBBQ]

edit:
I am incorrect. Sorry.

 

[TyErd]

so that means h(t)= 6t so H(t) = 6/s^2

therefore
F(s) = 6/s^2 + 5/(s+4) F(s),
rearrange and F(s) = 6(s+4) / s^2(s-1)
and inverse Laplace it and f(t) = 6*(-4t + 5 e^t - 5)

is that right, I am not sure if my calculations are correct.

 

[TyErd]

i just saw your last post...r u sure ur wrong coz it actually made sense to me...

 

[RoshanBBQ]

i just saw your last post...r u sure ur wrong coz it actually made sense to me...

The limits of integration in the problem are from 0 to t, which makes me feel uncomfortable about my answer. If it were convolution, the limits would be -infinity to infinity. If f(u) is nonzero only for u > 0, then the limits could be from 0 to infinity. I'm just unsure why the top limit is t.

 

[TyErd]

in the formula table I've provided, for convolution the limits says 0 to t.

 

[RoshanBBQ]

in the formula table I've provided, for convolution the limits says 0 to t.

If that's the case, then what I said made sense. I suppose G(i) is also zero for i < 0. So G(t-u) is zero outside u > t.

 

[RoshanBBQ]

so that means h(t)= 6t so H(t) = 6/s^2

therefore
F(s) = 6/s^2 + 5/(s+4) F(s),
rearrange and F(s) = 6(s+4) / s^2(s-1)
and inverse Laplace it and f(t) = 6*(-4t + 5 e^t - 5)

is that right, I am not sure if my calculations are correct.

$$F(s) = \frac{6}{s^2} - 5 \frac{1}{s+4} F(s)$$
Note the minus sign. Your next step would have been right if it had been addition (assuming a/bc is atually a/b/c). It is actually:
$$F(s) = 6\frac{s+4}{s^2(s+9)}$$

Your final answer also WOULD have been right without the sign error. Give it a try again with the proper minus sign this time.

 

[TyErd]

ah i see so then the value of f(t) is 6 ((4t)/9 - 5e^(-9 t))/81 + 5/81 correct?

 

[TyErd]

or (8*t)/3 - (10*exp(-9*t)/27) + 10/27

 

[RoshanBBQ]

ah i see so then the value of f(t) is 6 ((4t)/9 - 5e^(-9 t))/81 + 5/81 correct?

This has a lot of right terms in it, but it looks like you messed up on your parentheses.

$$6(\frac{4t}{9} - \frac{5e^{-9t}}{81}+\frac{5}{81})$$