Derivision of a transfer function from a block diagram with gain blocks

Question

I have derived the transfer function G(s) shown in the image.

I now need to work out the closed loop transfer function for the entire system and I can select my own values of k but I am doubting my method.

Is it as simple as this;

$$\frac{k(s)G(s)*2.51}{1+k(s)(G(s)*2.51}$$

Or am I mixing up concepts?

Answer 1

The general rule for SISO systems is the following:

Let $G_1(s)$ be the transfer function of the feedforward path and $G_2(s)$ be the transfer function of the feedback path. Then the closed loop transfer function for negative feedback is:

$$G_{cl}(s) = \frac{G_1(s)}{1+G_1(s)G_2(s)}$$

In your case, it is possible to move the 2.51 into the loop since it is present in both routes before the sum. This means that $$G_1(s) = \frac{2.51*0.36*K(s)}{s(s+1.95)}$$ and $$G_2(s) = 1$$.

In this case it is often helpful to apply the following formula: $$G_1(s) = \frac{N(s)}{D(s)} $$ $$G_{cl}(s) = \frac{N(s)}{N(s) + D(s)}$$