Partial fractions for Laplace transform Watch

Gibbo_
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Hi,

Is there any other way to find the coefficients for the partial fraction of 8s/((s^+1)(s^2+4s+5)) to find the Laplace transform other than having to plug in 4 values for s and use matrix methods to solve the simultaneous equations? It's the first question in an exercise so I feel that I am maybe missing something quite obvious.

Any help would be appreciated Thank you
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mqb2766
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(Original post by Gibbo_)
Hi,

Is there any other way to find the coefficients for the partial fraction of 8s/((s^+1)(s^2+4s+5)) to find the Laplace transform other than having to plug in 4 values for s and use matrix methods to solve the simultaneous equations? It's the first question in an exercise so I feel that I am maybe missing something quite obvious.

Any help would be appreciated Thank you
Nothing appears obvious. Looks like you'll get a couple of sinusoidal terms, an e^-t and an e^-4t.

If you don't factorize the (s^2+4s+5) immediately, you get
(As+B)/(s^2+1) + (Cs+D)/(s^2+4s+5)
You can spot that
D = -5B (constant coefficient)
A = -C (s^3 coefficient
That will reduce it to a couple of simultaneous equations with two variables for the s and s^2, which isn't too bad. Then factorize the (Cs+D)/(s^2+4s+5) part. Note that you're equating coefficients here, rather than subbing values for s.
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RDKGames
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(Original post by Gibbo_)
Hi,

Is there any other way to find the coefficients for the partial fraction of 8s/((s^+1)(s^2+4s+5)) to find the Laplace transform other than having to plug in 4 values for s and use matrix methods to solve the simultaneous equations? It's the first question in an exercise so I feel that I am maybe missing something quite obvious.

Any help would be appreciated Thank you
\dfrac{8s}{(s^2+1)(s^2+4s+5)} = \dfrac{As+B}{s^2+1} + \dfrac{Cs+D}{s^2 + 4s + 5}

so 8s = (As+B)(s^2 + 4s + 5) + (Cs+D)(s^2 + 1)


When s= 0 we get:  0 = 5B + D (*)

When s= \sqrt{-1} = i, we get: 8i = (B+iA)(4+4i)= 4[(B-A) + i(A+B)]

hence 0 = B-A and 2 = A+B. Hence B= 1 and A=1.

Hence (*) implies D = -5.

Visually (or, just by inspection), the coeff of s^3 on the RHS will be A+C which must be zero when compared to the LHS. Hence C = -1.


So \dfrac{8s}{(s^2+1)(s^2+4s+5)} = \dfrac{s+1}{s^2+1} + \dfrac{-s-5}{s^2 + 4s + 5} without too much work.
Last edited by RDKGames; 4 weeks ago
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Gibbo_
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(Original post by mqb2766)
Nothing appears obvious. Looks like you'll get a couple of sinusoidal terms, an e^-t and an e^-4t.

If you don't factorize the (s^2+4s+5) immediately, you get
(As+B)/(s^2+1) + (Cs+D)/(s^2+4s+5)
You can spot that
D = -5B (constant coefficient)
A = -C (s^3 coefficient
That will reduce it to a couple of simultaneous equations with two variables for the s and s^2, which isn't too bad. Then factorize the (Cs+D)/(s^2+4s+5) part. Note that you're equating coefficients here, rather than subbing values for s.
(Original post by RDKGames)
\dfrac{8s}{(s^2+1)(s^2+4s+5)} = \dfrac{As+B}{s^2+1} + \dfrac{Cs+D}{s^2 + 4s + 5}

so 8s = (As+B)(s^2 + 4s + 5) + (Cs+D)(s^2 + 1)


When s= 0 we get:  0 = 5B + D (*)

When s= \sqrt{-1} = i, we get: 8i = (B+iA)(4+4i)= 4[(B-A) + i(A+B)]

hence 0 = B-A and 2 = A+B. Hence B= 1 and A=1.

Hence (*) implies D = -5.

Visually (or, just by inspection), the coeff of s^3 on the RHS will be A+C which must be zero when compared to the LHS. Hence C = -1.


So \dfrac{8s}{(s^2+1)(s^2+4s+5)} = \dfrac{s+1}{s^2+1} + \dfrac{-s-5}{s^2 + 4s + 5} without too much work.
Thank you both!
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