In my notes from a controls course, we derived the transfer function for a first order system, and it was given as $$\frac{1}{1+sT}$$

However, the notes also state that $$\frac{K}{1+sT}$$

Is the general transfer function for 1st order systems. I am very confused on how they can both represent the same system. I initially thought I had misunderstood the first equation and that it was a transfer function for a particular solution, but this http://ece.gmu.edu/~gbeale/ece_421/xmpl-421-1st-order-01.pdf set of notes and What is the significance of the standard form of 1st and 2nd order transfer functions? this question suggest that it fact a general transfer function.

The question in the second link actually asks why sometimes one is given instead of the other, but I would like to know how it is possible for two transfer functions to be equivalent when they differ by a factor of K. Where am I going wrong? Thanks in advance.


The former is a special case of the latter when the scalar gain is equal to one. It can be simpler to understand a system if you use 1/(1+sT) and K as separate blocks; the following block diagrams are equivalent.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Thank you so much, I didn't notice that the scalar gain was just assumed to be one in the derivation! \$\endgroup\$ – masiewpao Apr 28 '17 at 2:04
  • \$\begingroup\$ This is quite late, but I think there shouldn't be any feedback? The two equations are for the overall transfer function systems already? \$\endgroup\$ – masiewpao May 2 '17 at 0:11
  • \$\begingroup\$ If they are, then the same thing holds--just get rid of the feedback! \$\endgroup\$ – Hearth May 2 '17 at 0:20
  • \$\begingroup\$ Thanks for replying Felthry, but even then I am still having trouble understanding this. the transfer funtion 1/(1+st) comes from the summing up of the individual transfer function blocks of the system. So how come we don't include the scalar gain in this? Meaning if we have 3 blocks, and one of them K, K1 and K2/s, then we can define some new K0 to include the scalar gain value K. So we still end up without a K in the numerator? \$\endgroup\$ – masiewpao May 2 '17 at 15:14
  • \$\begingroup\$ When you don't have a K in the numerator, that's equivalent to having K=1. \$\endgroup\$ – Hearth May 2 '17 at 16:44

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