1. Inverse Laplace transform

Could anybody please point me in the right direction with this inverse laplace transform?

L-1 {(e^(-0.165s)) / ((1+0.63s)s)

i think its a heaviside function

so H(t-a)= H(t-0.165)

which leaves me with

1 / ((1+0.63s)s)

and this where i'm stuck (No doubt the solution is staring me in the face!)

2. Originally Posted by al2308
Could anybody please point me in the right direction with this inverse laplace transform?

L-1 {(e^(-0.165s)) / ((1+0.63s)s)

i think its a heaviside function

so H(t-a)= H(t-0.165)

which leaves me with

1 / ((1+0.63s)s)

and this where i'm stuck (No doubt the solution is staring me in the face!)
Let $LT^{-1}[F(s)] = f(t)$.

Then $LT^{-1}[e^{-as} F(s)] = f(t - a) H(t - a)$.

In your case you have the form $F(s) = \frac{1}{s(bs + 1)}$ so you need to find $LT^{-1} \left[\, \frac{1}{s(bs + 1)} \, \right]$.

To do this, express $\frac{1}{s(bs + 1)}$ in the partial fraction form $\frac{A_1}{s} + \frac{A_2}{bs + 1}$.

Note that $\frac{A_2}{bs + 1} = \frac{A_2/b}{s - \left( -\frac{1}{b} \right)}$.

3. I thought this one looked simple but i was wrong again!

I've split it down into the partial fractions

A/s + B/(bs+1)

so

1=A(bs+1) + Bs

let S=0

A=1

expanding the equation

1=Abs+A+Bs

equating the coefficients

0=Ab+B

as A=1

then B=-b

i think this is right so far (but i'm often wrong )

this leaves me with the following to transform

(L-1) 1/s - b/(bs+1)

or

(L-1) 1/s - 1/(s-(-1/b))

= 1- e^-((1/b)t)

is this near the mark?

4. Originally Posted by al2308
I thought this one looked simple but i was wrong again!

I've split it down into the partial fractions

A/s + B/(bs+1)

so

1=A(bs+1) + Bs

let S=0

A=1

expanding the equation

1=Abs+A+Bs

equating the coefficients

0=Ab+B

as A=1

then B=-b

i think this is right so far (but i'm often wrong )

this leaves me with the following to transform

(L-1) 1/s - b/(bs+1)

or

(L-1) 1/s - 1/(s-(-1/b))

= 1- e^-((1/b)t)

is this near the mark?