Control systems question involving laplace transforms

In summary, the problem was that the residue function of MATLAB wasn't interpreting the r in rpk as the answers to the partial fraction analysis.
  • #1
FOIWATER
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I have narrowed down a control systems question I am doing to the following function in the S-domain.. numerator = (5s+1) denominator = (s+3)(3s+2)(s^2)

Using partial fraction expansion - I compute A = 14/63... D = 1/6... B = -2.25 And I can go no further - because both MATLAB and MAPLE disagree with me on B. They say I have the correct A and C, but they have B listed as -3/4. I have NO idea how they are achieving this result... any ideas?

When I expanded I got this, A/(S+3) B(3S+2) C/S D/S^2 all added... and when i cross multiply and try to solve for B... I get all terms except for B having the (3S+2) (It got canceled out of the num and den in B when I cross multiplied). Can anyone agree with me so far?...

Now, I proceed to make S = -2/3, since this will drive A, C, and D to zero, leaving me with:
5S+1 = B(S+3)S^2 right?

and when I put in -2/3 no matter what I do I get -2.25 I have even tried to expand and use systems of equations, in which case I still get the -2.25... not sure why

The MATLAB function I used with this equation was the [r,p,k]= function to get the residue and poles. in maple i used invlaplace and they both give -3/4 for B, and 19/36 for C, of which I get neither.. However we both agree on A and B what am I doing wrong, anyone?
 
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  • #2
Every system I compute this with gives me 3/4 for B, it must have to do with the way I am dividing B by 3s+2, although I don't see where it is incorrect..

I am so sorry for asking a question that I immediately found the answer to.

The problem was - the residue function of MATLAB apparently - the r in rpk the residue itself is not the answers to the partial fraction analysis, it is the answers to the time domain coefficients of each term? because when I now multiply -2.25 by 1/3 I am going to get the 3/4. I just should of kept going!.
 
Last edited:

1. What is a Laplace transform and how is it used in control systems?

A Laplace transform is a mathematical tool used to convert a function from the time domain to the frequency domain. In control systems, it is used to analyze the behavior of a system and design controllers to achieve desired performance.

2. What is the Laplace transform of a unit step function?

The Laplace transform of a unit step function, also known as the Heaviside function, is 1/s, where s is the complex frequency variable.

3. How do Laplace transforms simplify the analysis of control systems?

Laplace transforms simplify the analysis of control systems by converting differential equations into algebraic equations, making it easier to solve and analyze the behavior of the system in the frequency domain.

4. Can Laplace transforms be used for nonlinear control systems?

Yes, Laplace transforms can be used for nonlinear control systems, but they may require more advanced techniques such as partial fraction expansion and inverse Laplace transforms to analyze and design controllers.

5. What is the relationship between Laplace transforms and transfer functions?

Laplace transforms and transfer functions are closely related, as the Laplace transform of a system's input and output is the system's transfer function. This allows for easy analysis and design of control systems using transfer function techniques.

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