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What is the transfer function of a line, can i ignore it?

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After simplification i got the diagram:

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I don't know what to do with the empty feedback

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    $\begingroup$ It's an ideal connection, i.e. its 'transfer function' is 1. In this case the output is simply subtracted from the input. $\endgroup$ – Matt L. May 28 '14 at 11:02
  • $\begingroup$ Can i use the rule G(s)/1+ G(s)H(s) ? , where H(s) = 1) $\endgroup$ – Giannis Foulidis May 28 '14 at 11:21
  • $\begingroup$ Not sure which 'rule' you mean, but the Laplace transform of the signal at the input to the first block (K) is simply $R(s)-C(s)$. I guess you need to find the transfer function of the whole system. If you added to the question what you've got so far, people could help you to obtain the final result. $\endgroup$ – Matt L. May 28 '14 at 11:27
  • $\begingroup$ @MattL.: The rule he's talking about is $$\frac{C(s)}{R(s)} = \frac{1}{1+G(s)H(s)}$$, where $H(s)$ is the transfer function of the plant (the system in the middle of the diagram in his case) and $G(s)$ is the transfer function of the feedback path ($G(s) = 1$ in his diagram. This is the closed-loop transfer function for a system surrounded by a feedback loop. $\endgroup$ – Jason R May 28 '14 at 17:33
  • $\begingroup$ @JasonR: Oh yes, I see it now, thanks. A quick look wasn't enough for me to correctly parse the OP's formula ... $\endgroup$ – Matt L. May 28 '14 at 18:34
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The transfer function of a line is 1.

And hence you can reduce your blocks into a single block with transfer function $$\frac{C(s)}{R(s)} = \frac{G(s)}{1+G(s)}$$

where, $$G(s) = \frac{K}{s(s+2)(s+3)+s}$$

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