Sept 30 IOP Tutorial

Example: Consider the following standard negative feedback system:

where we have

$$\begin{array}{rl}D[z]& =\frac{(z-1)\left(z-\frac{1}{2}\right)}{(2z+1)(5z-2)}\\ G[z]& =\frac{z+1}{(4z-3)\left(z-\frac{1}{2}\right)}\end{array}$$

To check closed-loop stability, we need to check the closed-loop transfer functions.

$$T_{re}=T_{du}=\frac{1}{1+GD}=\frac{(4z-3)(2z+1)(5z-2)}{40z^3-25z^2-11z+5}$$

• Roots: -0.48, 0.34, 0.77

$$T_{ru}=\frac{D}{1+GD}=\frac{(4z-3)(z-1)\left(z-\frac{1}{2}\right)}{40z^3-25z^2-11z+5}$$

• Roots: -0.48, 0.34, 0.77

$$T_{de}=\frac{-G}{1+GD}=\frac{-(z+1)(2z+1)(5z-2)}{\left(z-\frac{1}{2}\right)(40z^3-25z^2-11z+5)}$$

• Roots: -0.48, 0.34, 0.5, 0.77

Then:

• $T_{re},T_{du},T_{ru},T_{de}$ are real, rational and proper.
• Since the poles of these lie in the open unit disk the TFs are stable [definition of stability].
• These TFs are BIBO stable [theorem from class]
• The system is closed-loop stable [definition of closed-loop stability]