Question

the inverse Laplace transform of s/s^4+s^2+1

Answer 1

Answer:

Note that you can factor the denominator by noting that:

s^4 + s^2 + 1 = (s^4 + 2s^2 + 1) - s^2

= (s^2 + 1)^2 - s^2

= [(s^2 + 1) + s][(s^2 + 1) - s], via difference of squares

= (s^2 + s + 1)(s^2 - s + 1).

So, we have:

s/(s^4 + s^2 + 1) = s/[(s^2 + s + 1)(s^2 - s + 1)].

Then, applying Partial Fractions yields:

s/[(s^2 + s + 1)(s^2 - s + 1)] = (As + B)/(s^2 + s + 1) + (Cs + D)/(s^2 - s + 1),

for some A, B, C, and D.

Multiplying both sides by (s^2 + s + 1)(s^2 - s + 1) gives:

s = (As + B)(s^2 - s + 1) + (Cs + D)(s^2 - s + 1).

At this point, you can expand all of this out, collect like terms, and compare the coefficients of the s^3, s^2, s, and constant terms on both sides (note that the s^3 coefficient, the s^2 coefficient, and the constant of the left side is zero) to get a system of equations. At this point, you can solve to get:

A = 0, B = -1/2, C = 0, and D = 1/2.

This gives:

s/[(s^2 + s + 1)(s^2 - s + 1)] = (-1/2)/(s^2 + s + 1) + (1/2)/(s^2 - s + 1)

= (1/2)[1/(s^2 - s + 1) - 1/(s^2 + s + 1)].

By completing the squares, we see that:

(a) 1/(s^2 - s + 1) = 1/[(s^2 - s + 1/4) + 3/4] = 1/[(s - 1/2)^2 + (√3/2)^2]

(b) 1/(s^2 + s + 1) = 1/[(s^2 + s + 1/4) + 3/4] = 1/[(s + 1/2)^2 + (√3/2)^2].

So:

s/(s^4 + s^2 + 1) = (1/2){1/[(s - 1/2)^2 + (√3/2)^2] - 1/[(s + 1/2)^2 + (√3/2)^2]}.

we need to get the a term (√3/2) in the numerator. To do this, multiply and divide by √3/2 to get:

(1/2){1/[(s - 1/2)^2 + (√3/2)^2] - 1/[(s + 1/2)^2 + (√3/2)^2]}

= (1/2){(2/√3)(√3/2)/[(s - 1/2)^2 + (√3/2)^2] - (2/√3)(√3/2)/[(s + 1/2)^2 + (√3/2)^2]}.

Applying #19 yields:

L^-1[f(s)] = (1/2)[(2/√3)e^(t/2)sin(t√3/2) - (2/√3)e^(-t/2)cos(t√3/2)]

= (√3/3)e^(t/2)sin(t√3/2) - (√3/3)e^(-t/2)cos(t√3/2).

I hope this helps!

Answer 2

s^4 + s^2 + 1 = (s^4 + 2s^2 + 1) - s^2

= (s^2 + 1)^2 - s^2

= [(s^2 + 1) + s][(s^2 + 1) - s], through contrast of squares

= (s^2 + s + 1)(s^2 - s + 1).

Along these lines, we have:

s/(s^4 + s^2 + 1) = s/[(s^2 + s + 1)(s^2 - s + 1)].

Then, at that point, applying Partial Fractions yields:

s/[(s^2 + s + 1)(s^2 - s + 1)] = (As + B)/(s^2 + s + 1) + (Cs + D)/(s^2 - s + 1),

for somewhere in the range of A, B, C, and D.

Increasing the two sides by (s^2 + s + 1)(s^2 - s + 1) gives:

s = (As + B)(s^2 - s + 1) + (Cs + D)(s^2 - s + 1).

Now, you can extend all of this out, gather like terms, and analyze the coefficients of the s^3, s^2, s, and consistent terms on the two sides (note that the s^3 coefficient, the s^2 coefficient, and the steady of the left side is zero) to get an arrangement of conditions.

Now, you can tackle to get:

A = 0, B = - 1/2, C = 0, and D = 1/2.

This gives:

s/[(s^2 + s + 1)(s^2 - s + 1)] = (- 1/2)/(s^2 + s + 1) + (1/2)/(s^2 - s + 1)

= (1/2)[1/(s^2 - s + 1) - 1/(s^2 + s + 1)].

By finishing the squares, we see that:

(a) 1/(s^2 - s + 1) = 1/[(s^2 - s + 1/4) + 3/4] = 1/[(s - 1/2)^2 + (√3/2)^2]

(b) 1/(s^2 + s + 1) = 1/[(s^2 + s + 1/4) + 3/4] = 1/[(s + 1/2)^2 + (√3/2)^2].

So:

s/(s^4 + s^2 + 1) = (1/2){1/[(s - 1/2)^2 + (√3/2)^2] - 1/[(s + 1/2)^2 + (√3/2)^2]}.

we want to get the a term (√3/2) in the numerator. To do this, duplicate and gap by √3/2 to get:

(1/2){1/[(s - 1/2)^2 + (√3/2)^2] - 1/[(s + 1/2)^2 + (√3/2)^2]}

= (1/2){(2/√3)(√3/2)/[(s - 1/2)^2 + (√3/2)^2] - (2/√3)(√3/2)/[(s + 1/2)^2 + (√3/2)^2]}.

Applying yields:

L^-1[f(s)] = (1/2)[(2/√3)e^(t/2)sin(t√3/2) - (2/√3)e^(- t/2)cos(t√3/2)]

= (√3/3)e^(t/2)sin(t√3/2) - (√3/3)e^(- t/2)cos(t√3/2).