[ \fracC(s)R(s) = \fracG(s)1 + G(s)H(s) ] This is for a negative feedback system (most common in Kuo).

Below is an example of how a "solid" solution paper addresses specific high-value problems.

Suppose: [ G(s) = \frac10s(s+2), \quad H(s) = 1 ] Find the closed-loop transfer function.

Solution: [ \fracC(s)R(s) = \frac\frac10s(s+2)1 + \frac10s(s+2) ] Multiply numerator and denominator by ( s(s+2) ): [ = \frac10s(s+2) + 10 = \frac10s^2 + 2s + 10 ]