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# Find the inverse laplace transform?

For 1/((s^2+1)(s^2+2s+3)).

I would first break the original expression into partial fractions.

1/((s2+1)*(s2+2s+3))=(A+Bs)/(s2+1)+(C+Ds)/(s2+2s+3). Now we need to find A, B, C, and D. Since two expressions must be identical, numerators shall be equal on the left and right parts (denominators are equal, since common denominator on the right is the same as on the left side). Therefore, (A+Bs)(s2+2s+3)+(C+Ds)*(s2+1)=1 or Bs3+(A+2B)s2+(2A+3B)s+3A+Ds3+Cs2+Ds+C=1. Thus we have the following system of equations:

B+D=0; A+2B+C=0; 2A+3B+D=0; 3A+C=1; B=-D from the first, then we are down to three equations:

A+2B+C=0; 2A+3B-B=0; 3A+C=1. From the second of these three A=-B; Therefore, -B+2B+C=0 and -3B+C=1;

B+C=0 and -3B+C=1; From the first of two B=-C, therefore the second equation gives 4C=1 or C=¼. Therefore, B=-¼; A=¼, and D=¼. So we obtain:

1/((s2+1)*(s2+2s+3))=¼[(s-1)/(s2+1)+(s+1)/(s2+2s+3)]

Now we consider each term in brackets separately.

(s-1)/(s2+1)=s/(s2+1)-1/(s2+1). Using linearity of Laplace transform, we can apply inverse transform to each term separately and obtain: F[s/(s2+1)](t)=cos(t); F[1/(s2+1)](t)=sin(t); therefore, F[(s-1)/(s2+1)](t)=cos(t)-sin(t).

(s+1)/(s2+2s+3)=(s+1)/(s2+2s+1+2)=(s+1)/((s+1)2+2); F[(s+1)/((s+1)2+2)](t)=et*cos(t√2)

Combining everything together, we obtain:

F[1/((s2+1)*(s2+2s+3))]=¼[cos(t)-sin(t)+et*cos(t√2)]

To attack such problems, you generally need a good memory or some reference like Schaum's, as well as some properties of Laplace transform.

As for this specific problem, first factor it as follows

1/(s^2+1)  * 1/(s^2+2s+3)

Now recall the convolution property of the Laplace transform (can you?): a product of the Laplace images of two functions correspond to a convolution of the two functions themselves. So suppose function

f has Laplace transform of 1/(s^2+1), and function

g has Laplace transform of 1/(s^2+2s+3),

then the result you are looking for would be the convolution of "f" and "g". Finding function f is easy: it's just sin(t). To find function g, you need to go one step further:

factor 1/(s^2+2s+3) = 1/((s+1)^2+2).

Now using your good memory or consulting a reference, you can figure out that its inverse Laplace transform is

g = e^(-t) sin(sqrt(2)*t)/sqrt(2),

where sqrt(a) means the square root of a. Now I suppose you know how to perform the convolution, which is essentially an integration.