# Thread: Inverse Laplace transform problem 2

1. ## Inverse Laplace transform problem 2

Dear all,

The following problem I have is this one. The transfer function of a closed loop system is
$10 \over {s^2 + 2s + 20}$

In this case the unit step response is needed. Allright, just multiply by $1 \over s$ and obtain the inverse, of course.

However, I have the solution but not the way it is obtained. The solution is stated as
$Y(s) = {0.12 \over s} - {0.24 \over {s + 10}} + {0.12 \over {s + 20}}$

and I see no way how to split up the above equation in the partial fractions over (s+10) and (s+20). Again.

Regards,

Jurgen

2. Hm, either everybody who browses this thread either understands the solution while I do not, or everybody is equally anxious to know what the solution is.

3. That transfer function has a negative discriminant for the denominator (that is, $b^{2}-4ac=4-4(20)=-76$), which means that it doesn't factor over the real numbers. Hence, you cannot possibly separate it out via partial fractions into the solution you stated. If you're sure the solution you've stated is correct, I would conclude that the transfer function is incorrect. Could you please state the complete original problem?

Thanks!

4. The problem is E2.29 from 'Modern Control Systems', P125 (Eleventh Edition, 2008).

The problem is given as the transfer function
$G(s) = {10 \over {{s^2} + {2s} + 10}}$
to be used in a unity feedback loop, which computes to the expression in the original question.

I have found the PDF containing the solutions on the internet, and the authors state that Y(s) for unity step is the answer also given in the original question. Which would mean that is possible to transform
${1 \over s} \cdot {10 \over {s^2 + 2s + 10}}$

Regards,

Jurgen

5. Originally Posted by chthon
Dear all,

The following problem I have is this one. The transfer function of a closed loop system is
$10 \over {s^2 + 2s + 20}$

In this case the unit step response is needed. Allright, just multiply by $1 \over s$ and obtain the inverse, of course.

However, I have the solution but not the way it is obtained. The solution is stated as
$Y(s) = {0.12 \over s} - {0.24 \over {s + 10}} + {0.12 \over {s + 20}}$

and I see no way how to split up the above equation in the partial fractions over (s+10) and (s+20). Again.

Regards,

Jurgen
That is not the step response for that transfer function, check for typos in the transfer function, the wording of the question and that you are looking at the answer for the question you are working on.

CB

6. Originally Posted by chthon
The problem is E2.29 from 'Modern Control Systems', P125 (Eleventh Edition, 2008).

The problem is given as the transfer function
$G(s) = {10 \over {{s^2} + {2s} + 10}}$
to be used in a unity feedback loop, which computes to the expression in the original question.

I have found the PDF containing the solutions on the internet, and the authors state that Y(s) for unity step is the answer also given in the original question. Which would mean that is possible to transform
${1 \over s} \cdot {10 \over {s^2 + 2s + 10}}$

Regards,

Jurgen
If you have a block with transfer function $H(s)$ inside a unity gain negative feedback loop the transfer function of the loop is:

$G(s)=\dfrac{H(s)}{1+H(s)}$

If:

$H(s)=\dfrac{10}{{s^2} + {2s} + 10}}$

then:

$G(s)=\dfrac{10}{s^2+2s+20}$

which is what you said in the original post.

The if $x(t)$ is the input and $y(t)$ the output:

$Y(s)=G(s)X(s)$

and if $x(t)=u(t)$ we have $X(s)=1/s$ and so:

$Y(s)=\dfrac{10}{s(s^2+2s+20)}$

But this does not give the partial fraction expansion given.

CB

7. I am glad that someone confirms for me that it is not possible. Which probably means that there is a big and nasty error in the book.
I needed this confirmation because I am not very confident in my math skills, and things such as these make me doubt myself.