# How does one arrive at the following transfer function for a certain circuit? [duplicate]

I'm trying to understand the derivation of the transfer function for the following electrical system:

Where $k$ is some forward gain element and $G_{s}$ is the transfer function of some plant. According to an instruction video, the transfer function for this whole system is $$T(s) = \frac{Y}{R} = \frac{k G_{s} }{1+kG_{s}} .$$

I am wondering how this expression for $T(s)$ can be derived. It probably has something to do with the control mechanism, but I am not an electrical engineer so I'm not sure how to derive this expression. I do have a background in mathematics though, so maybe you can take that into account in your answer.

## marked as duplicate by Justin, Andy aka, JRE, Voltage Spike, ThreePhaseEelNov 22 '16 at 23:50

• This is a very similar answer to what you are looking for: electronics.stackexchange.com/a/233239/108065 – Big6 Nov 22 '16 at 14:53
• Visual inspection: (R-Y)*K*Gs=Y. Solve for Y/R. – LvW Nov 22 '16 at 15:00

Just write out the control system into an equation and solve for Y/R ([Signal out] / [Signal in]

Sorry I have not figured out the math function yet.

Y = (R - Y)kGs

Y = RkGs - YkGs

Y + YkGs = RkGs

Y*(1 + kGs) = Rk*Gs

Y/R = kGs / (1 + kGs)

• Thank you! How do you see that $Y = (R - Y) k G_{s}$ , though? – Max Muller Nov 22 '16 at 15:25
Walking backwards from the output to determine what Y is, you see that Y is the output of a transfer function Gs. There's an amplification of K before that so you end up with $$<stuff>*K*Gs=Y$$ "stuff" is the output of the adder/subtractor which happens to be adding R and subtracting Y so you end up with $$(R-Y)*K*Gs = Y$$