1
\$\begingroup\$

In many books I've seen the generalized transfer function expressed using:

\$\displaystyle G(s) = K \frac{(s + z_1)(s + z_2)\ldots}{s^n (s + p_1) (s + p_2)\ldots} \$

What I don't understand is where the lonely \$s^n\$ term in the denominator comes from? I've searched high and low but I've not found any explanation for this term.

\$\endgroup\$
0

3 Answers 3

1
\$\begingroup\$

The s^n term is a pure integrator and the function may be integrated multiple times and that's where the n comes from.

\$\endgroup\$
7
  • \$\begingroup\$ If an integrator is added why not also include a differentiator? It seems a bit arbitary to me, or is it that integrators are common enough that they should be included? \$\endgroup\$
    – rhody
    Aug 29, 2015 at 0:59
  • \$\begingroup\$ Even if a pure differentiator included there is still a constant K out front. That constant ends up changing and the differentiator gets absorbed into the top equation. \$\endgroup\$
    – vini_i
    Aug 29, 2015 at 1:05
  • \$\begingroup\$ Adding additional numerator terms for differentiating does not change the function because you always can combine these s^n terms with the denominator terms. Furthermore, if the degree of the numerator s^n terms dominates, the transfer function is not realistic and makes no sense. There is no real circuit approaching infinity for rising frequencies. \$\endgroup\$
    – LvW
    Aug 29, 2015 at 7:53
  • \$\begingroup\$ @vini_i, how does changing K work? \$\endgroup\$
    – Chu
    Aug 29, 2015 at 8:45
  • \$\begingroup\$ K is the proportional compensation constant. It's generally adjusted while the function is being compensated. \$\endgroup\$
    – vini_i
    Aug 29, 2015 at 10:33
0
\$\begingroup\$

There's no reason that n cannot be negative. Pure integrators are more common than pure differentiators in control systems, so that's probably why it sits in the denominator.

\$\endgroup\$
0
\$\begingroup\$

Real systems are causal. That means that the order of the denominator is higher or equal compared to the nominator. Or in other words, the number of poles is equal or higher than the number of zeros. So if you are to model a system, then the transfer function will be of the form you have shown.

Note that for a controller this is not necessarily true as Chu said. Then you can have more zeros than poles

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.