1. ## Laplace transform

Q1) L {4cost sin2t}

Q2) L { f 0- infinite(the big f sign) e pwt-3t ( cos4t - t^2)dt}

Q3) Given iR + 1/C f 0- infinite(the big f sign) i dt + Ldi/dt = 4, where R,C,L are constants. show that if i(0) = 0, then

L {i} = 4C/LCs^2+RCs+1

Ans:
1) (4/(s-1)^2+4) +(4/s+1)^2 +4

2) 1/s [ (s+3/(s+3)^2 + 16) - 2/(s+3)^2]

3) I(s) = 4s/ (s+3)^2(s^2+4)

Help... i just started learning laplace transform. i am struggling because i am like direct to 2yr to my school and i have never done anything related before.. anyone can refer me to any good online materials that teaches the basics(if any)....also provide me with the workings/expl for the above question. Thanks for helping.

2. Originally Posted by mathsnerd
Q1) L {4cost sin2t}
Use the trig identity:

sin(A+B) - sin(A-B) = 2 sin(A)cos(B)

to reduce this to:

L {4 cos(t) sin(2t)} = L 4[sin(3t) - sin(t)] = 2 [Lsin(3t) + Lsin(t)]

Then look up the Laplace transform of sin(at) in a table to complete

2 [Lsin(3t) + Lsin(t)] = 2[3/(s^2+9) + 1/(s^2+1)]

................ .....=[8 (s^2 + 3) / ((s^2 + 1)·(s^2 + 9))]

Q2) L { f 0- infinite(the big f sign) e pwt-3t ( cos4t - t^2)dt}

Q3) Given iR + 1/C f 0- infinite(the big f sign) i dt + Ldi/dt = 4, where R,C,L are constants. show that if i(0) = 0, then

L {i} = 4C/LCs^2+RCs+1
You will have to improve the way you are typing these questions to make them easier
for us to understand, Q2 and 3 are incomprehensible as they are

Ans:
1) (4/(s-1)^2+4) +(4/s+1)^2 +4

2) 1/s [ (s+3/(s+3)^2 + 16) - 2/(s+3)^2]

3) I(s) = 4s/ (s+3)^2(s^2+4)

Help... i just started learning laplace transform. i am struggling because i am like direct to 2yr to my school and i have never done anything related before.. anyone can refer me to any good online materials that teaches the basics(if any)....also provide me with the workings/expl for the above question. Thanks for helping.
Look here (but I suspect you are lacking a lot of background that
you need to follow this, you might need to revise single variable
calculus first).

RonL

3. Originally Posted by CaptainBlack
Use the trig identity:

sin(A+B) - sin(A-B) = 2 sin(A)cos(B)

to reduce this to:

L {4 cos(t) sin(2t)} = L 4[sin(3t) - sin(t)] = 2 [Lsin(3t) + Lsin(t)]

Then look up the Laplace transform of sin(at) in a table to complete

2 [Lsin(3t) + Lsin(t)] = 2[3/(s^2+9) + 1/(s^2+1)]

................ .....=[8 (s^2 + 3) / ((s^2 + 1)·(s^2 + 9))]
You can have a bit of fun with this using the online Fourier-Laplace calculator here.
This gives for the LT of 4 cos(t) sin(2t):

-2i (s-2i)/[(s-2i)^2+1] + 2i (s+2i)/[(s+2i)^2+1]

which you can have a lot of fun proving is the same as what I gave in the
previous post.

RonL