The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Partial fractions for Laplace transform

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 :smile: Thank you
Reply 1
Avatar for mqb2766
mqb2766
21
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 :smile: 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.
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 :smile: Thank you


8s(s2+1)(s2+4s+5)=As+Bs2+1+Cs+Ds2+4s+5\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)(s2+4s+5)+(Cs+D)(s2+1)8s = (As+B)(s^2 + 4s + 5) + (Cs+D)(s^2 + 1)


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

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

hence 0=B−A0 = B-A and 2=A+B2 = A+B. Hence B=1B= 1 and A=1A=1.

Hence (*) implies D=−5D = -5.

Visually (or, just by inspection), the coeff of s3s^3 on the RHS will be A+CA+C which must be zero when compared to the LHS. Hence C=−1C = -1.


So 8s(s2+1)(s2+4s+5)=s+1s2+1+−s−5s2+4s+5\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.
(edited 5 years ago)
Reply 3
Avatar for Gibbo_
Gibbo_
OP
6
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
8s(s2+1)(s2+4s+5)=As+Bs2+1+Cs+Ds2+4s+5\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)(s2+4s+5)+(Cs+D)(s2+1)8s = (As+B)(s^2 + 4s + 5) + (Cs+D)(s^2 + 1)


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

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

hence 0=B−A0 = B-A and 2=A+B2 = A+B. Hence B=1B= 1 and A=1A=1.

Hence (*) implies D=−5D = -5.

Visually (or, just by inspection), the coeff of s3s^3 on the RHS will be A+CA+C which must be zero when compared to the LHS. Hence C=−1C = -1.


So 8s(s2+1)(s2+4s+5)=s+1s2+1+−s−5s2+4s+5\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!

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you