Laplace transformation problem

[jamshaid]

Homework Statement

L{4u(t-ர)cos2t}

Homework Equations

I have used UNIT STEP FUNCTION but could not get the result

The Attempt at a Solution

 

[vela]

What did you get? Show your work, not just your final result from your attempt.

 

[jamshaid]

L{4u(t-ர)cos2t
f(t) = cos2t
F(s) = s/ s^2+4
therefore

4s e^-ர(cosர + sinர)/s^2 +4 Ans
________________________________
that is my work

 

[gabbagabbahey]

L{4u(t-ர)cos2t
f(t) = cos2t
F(s) = s/ s^2+4
therefore

4s e^-ர(cosர + sinர)/s^2 +4 Ans
________________________________
that is my work

Where is the factor the of [itex]4e^{-\tau}(\sin\tau+\cos\tau)[/itex] coming from? What rule are you trying to apply here?

 

[jamshaid]

here is my solve, if I have done correctly.

 

[jamshaid]

Please expert help me.

 

[vela]

Unfortunately, I find your handwriting a bit hard to read. Could you type out your final answer?

 

[gabbagabbahey]

here is my solve, if I have done correctly.
View attachment 28107

It looks good to me (although you have a very strange way of writing [itex]\pi[/itex]). On a side note, since trig functions are [itex]2\pi[/itex] periodic, you should have immediately recognized that [itex]\cos(2t+2\pi)=\cos(2t)[/itex] without appealing to a trig identity.

 

[vela]

I think the exponential term in the front incorrectly has an a in it rather than s.

 

[jamshaid]

here is my final Answer
4e^(-as) s/(s^(2) + 4)

what do you think...

 

[vela]

The variable a shouldn't be in the final answer.

 

[jamshaid]

The variable a shouldn't be in the final answer.

can you giving me some hints?