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Reducing nominator's higher order.

  1. Aug 22, 2012 #1

    I have a mathematical dilemma. How can the following transfer function be simplified:


    in order to be able to have a 1st order nominator, maximum, compared to the denominator. But what if there is s^2(/<1...∞>) ? Is this possible?

    I know that the following can be done:

    [itex]\frac{s^{2}+2*s+3}{s^{2}+s+1} → \frac{s+2}{s^{2}+s+1}+1[/itex]

    but I don't know for the others above. I hope I managed to transmit the message I wanted, if not, blame my English. Can anyone help?

    Anticipated thanks,
  2. jcsd
  3. Aug 22, 2012 #2


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    Start by "pulling out" the highest powers of "s" first, then work your way down. The numerator degree is one higher than the denominator, so you know you can pull out a factor of "s"

    [tex]\frac{s^{2}+1}{s+1} = \frac{s(s+1)}{s+1} + \frac{-s + 1}{s+1}[/tex]

    Then work on the second term in that expansion to pull out the factor of "1". What you're left with at that point will of course be a "proper" rational fraction, with the degree of the numerator lower than that of the denominator.

    BTW. Do you know how to do polynomial division?
    Last edited: Aug 22, 2012
  4. Aug 22, 2012 #3
    Thank you for the reply. Yes, and it's because I saw that I can't reduce that equation, in particular, to a 1st_order/1st_order+constant form (knowing my weak mathematical skills), that I thought there might be another way.

    What I obtained (after scribbling a page of notebook) was s-1+2/(s+1), apparently it's a good result because wolframalpha showed the same. But, again, I thought (hoped) that it may be another way to allow for a translation in a state-space form. This was (is) for a custom block in LTspice to allow for 2nd order Laplace transfer functions with RLC (transient simulation is very bad and even un-recommended by the help file, ac is OK, though). Testing is done with a custom state-space block and I had problems implementing a random, but plausible, function to test it. So, I got here.

    Thanks for the help, though, a doubt cleared means one way closed and another opened.
  5. Aug 22, 2012 #4
    If I understood correctly, this is a rather important question as it involves partial fractions (of rational functions), something pretty important in integration.

    In the present case, since the numerator's degree is higher than the denominator's , we can directly divide polynomialwise:

    [tex]s^2+1=(as+b)(s+1)+r=as^2+(a+b)s+b+r\Longrightarrow a=1\,,\,a+b=0\Longrightarrow b=-1\,,\,b+r=1\Longrightarrow r=2[/tex]
    Of course, this case is so simple that it can be done almost "by eye", but the general case can be way more involved. Google "Partial fraction"

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