Find the inverse Laplace transform?

Question

Y(s)=α/((s^2+2s+2)(s^2+α^2))

Answer 1

Hello Sun and Yuan P.,

While there is nothing wrong with the approach that Yuan takes, he may not be familiar with the line of questions that you have posed recently.  So, I will answer based on the work you have done in the past few questions; namely, you have expressed inverse Laplace transforms in terms of convolution integrals.  Clearly, there is a lot you can learn from Yuan's approach, too, and I recommend that you study both his solution and mine for maximum benefit.

Let us view the expression Y(s) as a product:

Y(s) = G(s)H(s)

where we take

G(s) = 1/((s+1)²+1)

and

H(s) = α/(s²+α²).

Now, you are already (I hope!) familiar with the inverse Laplace transforms of both G(s) and H(s), but in case you have forgotten, they are (with our usual notational conventions)

g(t) = e^-tsin(t)

and

h(t) = sin(αt).

You may now write the inverse Laplace transform of the product, using the "convolution theorem", as either

y(t) = ∫₀^t g(t-τ)h(τ)dτ = ∫₀^t e^(τ-t)sin(t-τ)sin(ατ)dτ

or

y(t) = ∫₀^t g(τ)h(t-τ)dτ = ∫₀^t e^-τsin(τ)sin(α(t-τ))dτ.

In this way, you keep the expressions as tidy integrals, for your computer to handle when needed.

Regards,

Hassan H.

Answer 2

first a is a constant, so Y(s)/a=1/((s^2+2s+2)(s^2+α^2)),   now we need to simplify right hand side,

assume 1/((s^2+2s+2)(s^2+α^2)) = (As+B)/(s^2+2s+2)+(Cs+D)/(s^2+α^2),  A,B,C,D are constants, you see how I did that?  hint:Partial Fraction Expansion

now if you work on the equation you will find something like this ((A+C)s^3+(B+2C+D)s^2+(A+2C+2D)s+(B+2D))=1, thus we obtain that A+C=0, B+2C+D=0,A+2C+2D=0, and B+2D=1,

solve this system of equation you should get A=2/5, B=3/5, C=-2/5, D=1/5

so Y(s)/a=(1/5)( (2s+3)/(s^2+2s+2)+(-2s+1)/(s^2+α^2))=(1/5)[2*(s+1)/[(s+1)^2+1]+1/[(s+1)^2+1]-2*(s/(s^2+α^2))+1/(s^2+α^2)]

so the inverse Laplace transform will be 

y(t)=(1/5)*[2a*Exp(-t)*Cos(t)+a*Exp(-t)*Sin(t)-2a*Cos(at)+Sin(at)]

 

I hope this helped!