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I am having trouble figuring out this step in an example question I've been given. I'm supposed to be finding the 'transfer function in the forward path'.

Here is the step I don't understand

This image shows the block diagram (top) and what I have been told is the correct forward path transfer function (circled with red).

But based off of my understanding of that a forward path transfer function is, I would have said Gfp = 5 * s+4/(s+5)(s+8)

Where am I going wrong?

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You have missed the feedback path.
(R(s) - 2*C(S))*5(s+4)/ ((s+5)(s+8)) = C(s), and so on

Also, the closed loop transfer function vs open in case the feedback is 1

H(s) = F(s) / (1+F(s)); in your case the feedback is 2 so, you have to move it in the loop.

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Your forward path transfer function Gfp is correct. The circled expression is the transfer function of the whole system with feedback. However, the denominator should be D(s)=[1+2(Gfp)] because the loop gain is LG=-2(Gfp).

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I can't understand why it's (2-1). For negative feedback it would be just 1+10(s+4)/[(s+5)(s+8)] in denominator. Thus, you get 5(s+4)/(s^2+23s+80).or didnt I understand you right?

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Thank you everyone for your answers. In the end I got so confused I had to go and hunt down my lecturer in his office and ask. He said that because there is a component in the feedback loop (so it's not unity gain) I had to use this formula to convert the transfer function into the 'equivalent unity gain transfer function':

Ge(s) = G(s) / (1 + H(s)G(s) - G(s))

where G(s) is the top part of the block diagram and H(s) is the feedback loop.

btw the e in Ge(s) stands for 'equivalent'.

This gave me the right answer :)

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  • \$\begingroup\$ The equation Ge(s)=.... is false. According to the well-known formula from H. Black for feedback systems we have Ge(s)=G(s)/(1+G(s)H(s)). It is very easy to proof this function. Instead of asking again your lecturer you should consult a good textbook. \$\endgroup\$
    – LvW
    Oct 5 '15 at 8:02

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