Question

Heaviside step functions, convolution, and dirac delta function. Im having trouble with these chapters all and any help is appreciated! Important test on monday

Answer 1

\[s/(s^2+1)^2\]
use convolution theorem to find inverse laplace transform of given function

Answer 2

@rperez36, first, identify which functions you want to set as \(F(s)\) and \(G(s)\).

Both functions have easy inverse transforms:
\[F(s)=\frac{s}{s^2+1}~\Rightarrow~f(t)=\cos t\\
G(s)=\frac{1}{s^2+1}~\Rightarrow~g(t)=\sin t \]
Then, you have
\[\mathcal{L}^{-1}\left\{F(s)G(s)\right\}=\int_0^t f(t-\tau)g(\tau)~d\tau\]

Answer 3

@rperez36, got it? It's pretty easy to compute the integral.

Answer 4

@SithsAndGiggles thanks for helping me start it. I knew I had to separate it but that's what I have most trouble with in identifying it. it was the denominator that threw me off I wasn't sure if I should've separated it the way you did or leave it squared for both

Answer 5

@Loser66 you start doing laplace transforms yet?

Answer 6

yes, and I passed it. now, eigenvalue.

Answer 7

I have to step away to eat something, will be back later. ok?

Answer 8

well good job! I will be tested on Monday over it. Then we just have power series and then Thursday is our final

Answer 9

ok

Answer 10

well I didn't get the ans...I tried taking integral of
\[\int\limits_{0}^{t}\cos(w)\sin(t-w)\]

Answer 11

\[\int_0^t \cos(t-\tau)\sin\tau~d\tau\]
Use the following identity:
\[\cos(a-b)=\cos a\cos b+\sin a\sin b\]
\[\int_0^t \big(\cos t\cos\tau+\sin t\sin\tau\big)\sin\tau~d\tau\\
\cos t\int_0^t\cos\tau\sin\tau~d\tau+\sin t\int_0^t\sin^2\tau~d\tau\]
The first integral takes a simple substitution. The second requires another identity:
\[\sin^2\theta=\frac{1}{2}\left(1-\cos2\theta\right)\]