Manually trying to calculate output of an transfer function.

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If your transfer function is $1/(s+1)$ then the Laplace transform of the step response is

$$Y(s)=\frac{1}{s(s+1)}=\frac{1}{s}-\frac{1}{s+1}\tag{1}$$

Using basic Laplace transform relations, the inverse of (1) is

$$y(t)=(1-e^{-t})u(t)$$

EDIT:

If you want to use convolution then you must compute the following integral

$$y(t)=\mathcal{L}^{-1}\{\frac{1}{s}\}*\mathcal{L}^{-1}\{\frac{1}{s+1}\}=\int_0^te^{-\tau}d\tau,\quad t>0$$

which of course results in the same solution.

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newbiemath
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newbiemath

Updated on August 01, 2022

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  • newbiemath
    newbiemath over 1 year

    I am trying to calculate the output of an transfer function due to the input of an step, But some weird reason, I am only getting the inverse output, what Matlab says it should be.

    My transfer function is $ \frac{1}{s+1}$ which i Apply an step to $1/s$ which in time domain gives an function $e^{-t}$... which is inverse of what matlab gives me..

    Why am i getting this???

  • newbiemath
    newbiemath over 9 years
    how come is $\frac{1}{s(s+1)} = \frac{1}{s} -\frac{1}{s+1} $
  • Matt L.
    Matt L. over 9 years
    @control: That's a simple case of partial fraction decomposition: en.wikipedia.org/wiki/Partial_fraction_decomposition I think you can easily prove the equality.
  • newbiemath
    newbiemath over 9 years
    Well.. I don't understand how come such simple expression need to be decomposed.. I mean how come isn't my way correct
  • Matt L.
    Matt L. over 9 years
    @control: It can be decomposed, and in this way the solution is very easy (just basic identities). That's why it is standard to decompose fractions because then the inverse transform is straightforward. Since you didn't share "your way" of doing it, I can't tell you why it doesn't work.
  • newbiemath
    newbiemath over 9 years
    It would just be to rewrite to to $\frac{1}{s} \frac{1}{s+1} $ and then inverseLaplaceTransform each part individually. It will then result in $1 * e^{-t}$ .
  • Matt L.
    Matt L. over 9 years
    @control: OK, you can use convolution if you like. See my edited answer on how to do that.