Question

hi guys can you solve this using inverse laplace
"2 over (s+1)^2 times (s^2 +4)

Answer 1

\[\frac{2}{(s+1)^2(s^2+4)}\]
We need to find the partial fractions for this. Do you know how to do that?

Answer 2

i just did but i cant find the abcd help me pls

Answer 3

I never use the A, B, C method. Let me show you how I do this.

Answer 4

teach me master

Answer 5

Subtract the two terms in the denominator,
\[(s^2+4)-(s+1)^2=\underline{-2s+3}\]
I'll create the underlined term in the numerator
\[\frac{2}{3}\frac{3}{(s+1)^2(s^2+4)}\]
\[\frac{2}{3}(\frac{3-2s+2s}{(s+1)^2(s^2+4)})\]
Do you get this part?

Answer 6

where did you get 2/3

Answer 7

I had a 2 in the numerator, I took it out then I multiplied and divided by 3
\[\frac{2}{1}\]
\[\frac{2\times 3}{3}=\frac{2}{3} \times 3\]

Answer 8

ahh ok get it just continue to answer ur own step

Answer 9

Cool :)

Answer 10

Wait a minute, the steps I took, aren't helping. Just a moment. I'll think for sometime

Answer 11

awts

Answer 12

OK, Let's use the partial fraction method here. My method is failing here
\[\frac 2 {(s+1)^2(s^2+4)}=\frac A{(s+1)}+\frac B {(s+1)^2}+\frac{Cs+D}{s^2+4}\]

Answer 13

Let's rearrange this, we'd get
\[2=A(s+1)(s^2+4)+B(s^2+4)+(Cs+D)(s+1)^2\]
First put s=-1
\[2=B(1^2+4)\]or
\[B=\frac 2 5\]
Do you get it @coldi3845 ?

Answer 14

but 2Cs right? bcoz the derivative of s^2 +4 is 2S?

Answer 15

We can do that way also, here C will come twice of what we get if we use (2Cs)

Answer 16

how did you get=-1

Answer 17

I didn't get, I put s=-1, which makes the first and the last term on right side 0, then I can find the value of B

Answer 18

@coldi3845 do you get this?

Answer 19

not really

Answer 20

OK, I'll explain you with this
\[\frac 1{(s+1)(s+2)}=\frac A{s+1}+\frac B{s+2}\]
We need to find the value of A and B.

Answer 21

@coldi3845 ???

Answer 22

why it becomes A over s+1 and B over s+2

Answer 23

We are splitting the fraction on the right to 2 partial fractions.
It'd be easy to fine the inverse laplace transform that way.
isn't it?

Answer 24

ok ok just continue

Answer 25

Should I continue, with our question?
Do you get how I find the value of B?

Answer 26

yeah u assume s=-1 to find it

Answer 27

\[2=A(s+1)(s^2+4)+B(s^2+4)+(Cs+D)(s+1)^2\]
now Let's put s=+2i that will make
\[s^2+4=-4+4=0\]
\[2=(C\times 2i+D)(-4+1+4i)\]
\[2=(C\times 2i+D)(-3+4i)\]
\[2=-6iC-8C-3D+4iD\]
Equate imaginary and real parts
\[-8C-3D=2\]
\[-6C+4D=0\]
Do you get this?

Answer 28

@coldi3845 ???

Answer 29

ok ok and then?

Answer 30

Can you find the value of C and D?
\[-8C-3D=2\]
\[-6C+4D=0\]

Answer 31

@coldi3845

Answer 32

no

Answer 33

You have two equations, two variables. It's simultaneous linear equation
@coldi3845

Answer 34

just finish it sir bcoz i will compare it to my convolution if it is right

Answer 35

@coldi3845 I can't you the full solution, just like that. I want your active involvement in the solution

Answer 36

*give

Answer 37

i don't know

Answer 38

just solved it pls