Reverse Laplace transform of 3s−152s2−4s+10

Name: Reverse Laplace transform of 3s−152s2−4s+10
Uploaded: 2022-02-23T17:22:57+00:00
Description: Reverse Laplace transform of 3s−152s2−4s+10

Question

Reverse Laplace transform of  3s−152s2−4s+10

Fasaniu · Accepted Answer

Step 1  3s−152s2−4s+10=32s−5s2−2s+5=32s−1−4(s−1)2+4  =32s−1(s−1)2+22−32(s−1)2+22  Now let α=−1 and ω=2.  The inverse transforms of s+α(s+α)2+ω2 and ω(s+α)2+ω2 tetx{are} e−αtcos⁡(ωt) and e−αtsin⁡(ωt), respectively. Hence the inverse transform of 3s−152s2−4s+10 is 32etcos⁡(2t)−3etsin⁡(2t).

lovagwb · Accepted Answer

First of all, get rid of that 2 factor in the denominator:

2 s^{2} - 4 s + 10 = 2 (s^{2} - 2 s + 5)

and observe that the polynomial

s^{2} - 2 s + 5

has no real roots. So, we write it as follows:

s^{2} - 2 s + 5 = {(s - 1)}^{2} + (5 - 1) = {(s - 1)}^{2} + 2^{2}

Next, we try to decompose our fraction like this:

\frac{3 s - 15}{{(s - 1)}^{2} + 2^{2}} = A \frac{s - 1}{{(s - 1)}^{2} + 2^{2}} + B \frac{2}{{(s - 1)}^{2} + 2^{2}}

,
for some constants A and B which youll

alenahelenash · Accepted Answer

Pulling out a factor of 2 from the denominator and the constant multiple of 3 from the numerator leads to: 3s−152s2−4s+10=32s−5s2−2s+5 Next we do Completing the Square for the polynomial (s2−2s+5) in the denominator and re-writing the numerator constant −5 as (−1−4) giving us: =32⋅s−1−4(s−1)2+4 Let denominator equal K. Separating the fraction into parts with them being 32⋅(s−1)K and −3⋅2K; after distributing the 32 to the second term −4. This leads to the expression below now. =32⋅(s−1)(s−1)2+22−3⋅2(s−1)2+22 From the table of transforms, the inverse transform of s−a(s−a)2+b2 and b(s−a)2+b2 are eatcos⁡(bt) and eatsin⁡(bt), respectively. Hence the inverse laplace transform of our original problem is, L−1{3−152s2−4s+10}=32etcos⁡(2t)−3etsin⁡(2t)

Reverse Laplace transform of \frac{3s-15}{2s^{2}-4s+10}

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