jamshaid
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Homework Statement
L{4u(t-ர)cos2t}
Homework Equations
I have used UNIT STEP FUNCTION but could not get the result
Laplace transformation problem
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L{4u(t-ர)cos2t}
I have used UNIT STEP FUNCTION but could not get the result
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What did you get? Show your work, not just your final result from your attempt.
jamshaid
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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
f(t) = cos2t
F(s) = s/ s^2+4
therefore
4s e^-ர(cosர + sinர)/s^2 +4 Ans
________________________________
that is my work
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jamshaid said:
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
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here is my solve, if I have done correctly.
jamshaid
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Please expert help me.
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Unfortunately, I find your handwriting a bit hard to read. Could you type out your final answer?
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jamshaid said:
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.
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I think the exponential term in the front incorrectly has an a in it rather than s.
jamshaid
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here is my final Answer
4e^(-as) s/(s^(2) + 4)
what do you think...
4e^(-as) s/(s^(2) + 4)
what do you think...
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The variable a shouldn't be in the final answer.
jamshaid
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vela said:
can you giving me some hints?