1. ## Laplace to time

Is this correct:? $$\frac{A}{C+B+sD} = A \frac{1} {s \frac{C+B}{D}}$$ using $$1 = \frac{1}{s}$$ so: $$f(t) = \frac{A}{\frac{C+B}{D}}$$?

2. ## Re: Laplace to time

Originally Posted by Nforce
Is this correct:? $$\frac{A}{C+B+sD} = A \frac{1} {s \frac{C+B}{D}}$$ using $$1 = \frac{1}{s}$$ so: $$f(t) = \frac{A}{\frac{C+B}{D}}$$?
No...

$\dfrac{A}{C + B + sD} = \dfrac{A}{C+B} \cdot \dfrac{1}{1+\frac{D}{C+B}s}$

Then use

$\mathscr{L}(f(c t) ) = \dfrac 1 c F\left(\dfrac s c \right)$

and

$\mathscr{L}(a f(t)) = a F(s)$

with $F(s) = \mathscr{L}(f(t))$

to complete the solution

3. ## Re: Laplace to time

Sorry, wrong problem. No wonder it get's really strange.

The right expression is:

$$\frac{A}{s(B+C+sD)}$$

I should get something like this: K1e^something + K2e^something ...

Because physics never lie. Thanks.

4. ## Re: Laplace to time

Originally Posted by Nforce
Sorry, wrong problem. No wonder it get's really strange.

The right expression is:

$$\frac{A}{s(B+C+sD)}$$

I should get something like this: K1e^something + K2e^something ...

Because physics never lie. Thanks.
it's time to teach yourself about partial fractions.

$\dfrac{A}{s(B+C+sD)} = \dfrac{\alpha}{s} + \dfrac{\beta}{B+C+sD}$

and there is a method for determining $\alpha$ and $\beta$

Once separated into partial fractions it's much easier to compute the inverse transforms of the terms.

5. ## Re: Laplace to time

We can rewrite the fraction like this:

$$\frac{A}{s(B+C+sD)} =$$

$$\frac{A}{s}\frac{1}{(B+C+sD)}$$

6. ## Re: Laplace to time

Originally Posted by Nforce
We can rewrite the fraction like this:

$$\frac{A}{s(B+C+sD)} =$$

$$\frac{A}{s}\frac{1}{(B+C+sD)}$$
and what does this get you?

7. ## Re: Laplace to time

$$\frac{A}{s} = A \frac{1}{s} = A$$

$$\frac{1}{B+C+sD} = \frac{1}{\frac{B+C}{D}+ s} = e^{-\frac{B+C}{D}t}$$

=?

8. ## Re: Laplace to time

Originally Posted by Nforce
$$\frac{A}{s} = A \frac{1}{s} = A$$

$$\frac{1}{B+C+sD} = \frac{1}{\frac{B+C}{D}+ s} = e^{-\frac{B+C}{D}t}$$

=?
that would work if the two terms were being added but the way you have it they are being multiplied.

What the method of partial fractions gets you is

$\dfrac{A}{s (B+C+D s)}=\dfrac{A}{s (B+C)}-\dfrac{A D}{(B+C) (B+C+D s)}$

Now you can mangle these terms to match the common transform templates and invert them as usual.

9. ## Re: Laplace to time

Sorry, maybe you are right about learning the partial fraction decomposition.

$$\alpha (B+C+sD) + s\beta = A$$

Now I solve for $\alpha$ and $\beta$. But how, there is just A on the right hand side?

10. ## Re: Laplace to time

Originally Posted by Nforce
Sorry, maybe you are right about learning the partial fraction decomposition.

$$\alpha (B+C+sD) + s\beta = A$$

Now I solve for $\alpha$ and $\beta$. But how, there is just A on the right hand side?
I'm not checking that what you've done to reach above is correct. But assuming it is what you do is break the equation up into powers of $s$

$\alpha(B+C) = A$ (equation for constant terms)

$(\alpha D + \beta) = 0$ (equation for terms of $s$ )

solve this system for $\alpha$ and $\beta$

IMPORTANT NOTE FOR THE FUTURE

The rules for the partial fraction decomposition get more complicated than this as powers of $s$ go up.