I have a system G(s). It has no zeroes, but some poles.

When applying unity negative feedback, with a static gain K, why does K remain on the numerator of my transfer function T(s) which describes the whole closed loop system?

If I have no zeroes, Z(s), then surely K multiplied by Z(s) is 0? But it is K.

In addition, in the denominator, there is another term that is again, K multiplied by Z(s). No zeroes, but this term simplifies to K, and not 0.

I understand that Z(s) is a function, and normally takes the form (s+3)(s+1) or something, but if there are no zeroes, can someone explain why K remains on the numerator?



  • \$\begingroup\$ en.wikibooks.org/wiki/Control_Systems/…, i'm unsure of your question \$\endgroup\$
    – sstobbe
    May 31, 2017 at 14:06
  • \$\begingroup\$ K doesn't have any zeroes. \$\endgroup\$
    – Chu
    May 31, 2017 at 14:52
  • 3
    \$\begingroup\$ No zeros means \$Z(s)=1\$, hence \$K Z(s)=K\$ \$\endgroup\$ May 31, 2017 at 15:00

1 Answer 1


The order of the Tf goes from its highest power down to s^0 which is one ie theres always a one there


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