# Laplace transform question

1. Jan 7, 2007

### cabellos

how do i go about solving the laplace transform of sint * cost ???

i know the answer becomes 1/(s^2 + 4) but what is the method?

thanks.

2. Jan 7, 2007

### murshid_islam

do you know that
$$\mathcal{L}\{f(t)*g(t)\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\} = F(s) \cdot G(s)$$

and i think the answer comes out to be $$\frac{s}{{\left(s^2+1\right)}^2}$$

Last edited: Jan 7, 2007
3. Jan 7, 2007

### cabellos

yes i did know that but i didnt think it was as simple as that. From reading the certain rules we can apply to problems e.g the shift rule to solve L(sin2t * e^3t) do we HAVE to use the shift rule or can we seperate each part and solve then multiply them together?

4. Jan 7, 2007

### murshid_islam

i am sorry. do you mean convolution by the asterisk? or do you mean multiplication? what i said in my last post is only correct if the asterisk means convolution.

5. Jan 7, 2007

### cabellos

i meant multiplication

6. Jan 7, 2007

### HallsofIvy

Staff Emeritus
Then use the definition
$$\mathcal{L}(sin(x)cos(x)}= \int_0^{\infnty}e^{-st}sin(t)cos(t)dt$$
which can be done by integration by parts.

7. Jan 7, 2007

### d_leet

Mybe it would help to notice that sin(t)cos(t) is equal to $$\frac{1}{2}sin(2t)$$

8. Mar 4, 2007

### I_power2

I wanna ask about a proof of this transform: L{(f(t)}= sF(s) - f(0) - f(a^+) - f(a^-) - exp^(-as) inwhich f(t) is continous except for an ordinary discontinuity (finite jump) at t=a, a>0

Last edited: Mar 4, 2007
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