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

rheisf 2022-01-18 Answered
Reverse Laplace transform of
3s152s24s+10
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Fasaniu
Answered 2022-01-19 Author has 46 answers
Step 1
3s152s24s+10=32s5s22s+5=32s14(s1)2+4
=32s1(s1)2+2232(s1)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} eαtcos(ωt) and eαtsin(ωt), respectively. Hence the inverse transform of 3s152s24s+10 is 32etcos(2t)3etsin(2t).

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lovagwb
Answered 2022-01-20 Author has 50 answers
First of all, get rid of that 2 factor in the denominator: 2s24s+10=2(s22s+5) and observe that the polynomial s22s+5 has no real roots. So, we write it as follows:
s22s+5=(s1)2+(51)=(s1)2+22
Next, we try to decompose our fraction like this:
3s15(s1)2+22=As1(s1)2+22+B2(s1)2+22,
for some constants A and B which youll

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alenahelenash
Answered 2022-01-24 Author has 368 answers
Pulling out a factor of 2 from the denominator and the constant multiple of 3 from the numerator leads to: 3s152s24s+10=32s5s22s+5 Next we do Completing the Square for the polynomial (s22s+5) in the denominator and re-writing the numerator constant 5 as (14) giving us: =32s14(s1)2+4 Let denominator equal K. Separating the fraction into parts with them being 32(s1)K and 32K; after distributing the 32 to the second term −4. This leads to the expression below now. =32(s1)(s1)2+2232(s1)2+22 From the table of transforms, the inverse transform of sa(sa)2+b2 and b(sa)2+b2 are eatcos(bt) and eatsin(bt), respectively. Hence the inverse laplace transform of our original problem is, L1{3152s24s+10}=32etcos(2t)3etsin(2t)

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