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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.

masiewpao
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1 Answers1

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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.

schematic

simulate this circuit – Schematic created using CircuitLab

Hearth
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  • Thank you so much, I didn't notice that the scalar gain was just assumed to be one in the derivation! – masiewpao Apr 28 '17 at 02:04
  • This is quite late, but I think there shouldn't be any feedback? The two equations are for the overall transfer function systems already? – masiewpao May 02 '17 at 00:11
  • If they are, then the same thing holds--just get rid of the feedback! – Hearth May 02 '17 at 00:20
  • 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? – masiewpao May 02 '17 at 15:14
  • When you don't have a K in the numerator, that's equivalent to having K=1. – Hearth May 02 '17 at 16:44
  • Hmm, I phrased that badly. What I meant was, I believe the term 1/(1+st) already includes the amplifier gain in its derivation? – masiewpao May 02 '17 at 16:50
  • Unless the gain is 1, it doesn't. – Hearth May 02 '17 at 16:52
  • I've drawn a quick summary of the derivation from my lecture notes here (it's the derivation for a turntable): http://imgur.com/YkKGX8O Ka is the amplifier gain, which I never assume to be 0, but somehow I still end up with 1/(st+1). Could it be because I am assuming Kt=Kp (which is done so there is 0 error when input = output) – masiewpao May 02 '17 at 16:58
  • Are you sure these are the same K in both cases? It's possible a gain is being added on afterward, as a control compensator. In some simple cases like this, you can improve convergence by adding a gain block. – Hearth May 02 '17 at 17:06
  • Absolutely not, I think that's where my confusion may be coming from! I didn't know you could add on a gain block like that; I thought it would just be multiplied into the overall transfer function! – masiewpao May 02 '17 at 17:25
  • Well, it would be, if you were making a truly overall transfer function. But you can always stick a gain block onto something, that's pretty much what control system design is all about! (well, if you allow for gain blocks to be frequency-dependant and yeah there's a lot more to it than that but the scalar gain is the most basic!) – Hearth May 02 '17 at 17:29