Question

Find the inverse transform of s/[(s^2+a^2)(s^2+b^2)], where a^2 is not equal to b^2 and ab does not equal 0. Please show two different ways to find the inverse transform.

Answer 1

have you tried partial fractions ?

Answer 2

No I haven't but I'm sure that is one of the ways to solve it. I fear it will get very tricky very quickly lol

Answer 3

not really

Answer 4

if partial fractions is one way....could convolution be another way?

Answer 5

\[\large \dfrac{s}{(s^2+a^2)(s^2+b^2)} = \dfrac{s(s^2+b^2-s^2-a^2)}{(s^2+a^2)(s^2+b^2)}\]

Answer 6

that simplifies nicely but its not correct, it needs some more tweaking

Answer 7

looks there is an extra (b^2-a^2) in the numerator ?

Answer 8

yeah I was just wondering about that lol

Answer 9

can we say \[\large \dfrac{s}{(s^2+a^2)(s^2+b^2)} = \dfrac{1}{b^2-a^2}\dfrac{s(s^2+b^2-s^2-a^2)}{(s^2+a^2)(s^2+b^2)}\]

Answer 10

\[\begin{align} \dfrac{s}{(s^2+a^2)(s^2+b^2)} &= \dfrac{1}{b^2-a^2}\dfrac{s(s^2+b^2-s^2-a^2)}{(s^2+a^2)(s^2+b^2)}\\~\\

&= \dfrac{1}{b^2-a^2}\left[\dfrac{s}{s^2+a^2}-\dfrac{s}{s^2+b^2}\right]\\~\\

\end{align}\]

Answer 11

does that look okay

Answer 12

that looks correct from the step above it if that's what you mean. so from there would it be....something times cos(at)-cos(bt)...?

Answer 13

Yep! thats it

Answer 14

so my final answer is (1/b^2-a^2)*(cos(at)-cos(bt))? I leave it like that?

Answer 15

will I get the same answer by convolution....I have to be able to solve it two ways for my class lol

Answer 16

yes convolution will produce same result
then u have to take
\[F_1(s)=\frac{ s }{ s^2+b^2}\\F_2(s)=\frac{ 1 }{ s^2+a^2 }\\then ~simply\\ \int\limits_{0}^{t}f_1(u)*f_2(t-u)du\]

Answer 17

\[where\\f_1(u)=L^{-1}F_1(t)|_{s=u}\\f_2(u)=L^{-1}F_2(t)|_{s=t-u}\]

Answer 18

awesome thanks!

Answer 19

@sidsiddhartha I am a bit confused on the convolution part. How would I take the inverse laplace of 1/(s^2+a^2) so I could do the rest of the problem?

Answer 20

ok we have\[F_2(s)=\frac{ 1 }{ s^2+a^2 }\\so\\f_2(t)=L^{-1}[F_2(s)]=L^{-1}[\frac{ 1 }{ a }*\frac{ a }{ a^2+s^2 }]\\=\frac{ 1 }{ a }*L^{-1}\frac{ a }{ a^2+s^2 }\]
make sense?

Answer 21

I think so. So would the inverse be 1/a*cos(at)?

Answer 22

sin

Answer 23

no wait that would be sin instead of cos I think

Answer 24

yup right :)

Answer 25

ok! So from here I just turn the t's into u's and put them into the formula to solve the convolution?

Answer 26

yes
and \[f_1(t)=\cos(bt)\]
then turn the t's into u's and put them into the formula to solve the convolution
as u said :)

Answer 27

I know how to do the convolution when finding the laplace transform....is it the same process for finding the inverse laplace transform?

Answer 28

yes basically tha same thing

Answer 29

ok thank you!

Answer 30

ok sorry one more question...the formula becomes integral from 0 to t cos(bu)*(1/a)sin(t-au)du? that last part is throwing me off