Sept 23 Tutorial

Proofs:

• Given
• WTS
• Reason

Problem

Suppose $G[z]$ is rational, proper, and BIBO stable. For some fixed positive integer $n$, show $(G[z]{)}^n$ is stable.

Given: $G[z]$ is real, rational, proper, BIBO stable. WTS: $(G[z]{)}^n$ is stable.

Strategy A:

• $G[z]$ is real, rational, proper, and BIBO stable
• $⟹G[z]$ is stable
  • (theorem from class)
• Let $s$ be the set of poles of $G[z]$
• $⟹G[z]=\frac{b_m{z}^m+\cdots +b_{1}z+b_0}{(z-p_1)(z-p_2)\dots (z-p_j)}$ with $\{p_1,\dots p_j\}\in s$
  • (since $G[z]$ is rational)
• $⟹$ Since $G[z]$ is stable, the $s$ lies on the open unit disk
  • (definition of stability)
• $⟹(G[z]{)}^n=\frac{(b_m{s}^m+\cdots +b_{1}s+b_0{)}^n}{(z-p_1{)}^n(z-p_2{)}^n\dots (z-p_j{)}^n}$
  • ${\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n}$
• $⟹$ From this expression, we see the multiplicity of the poles above, but pole locations have not changed
• Thus, $\text{poles}(G[z]{)}^n\subseteq s$
• $⟹\text{poles}(G[z]{)}^n$ lie in open unit disk
  • (since s lies in the open unit disk)
• $⟹(G[z]{)}^n$ is stable
  • (definition of stability)

Strategy B:

• $G[z]$ is real, rational, proper, and BIBO stable
• Let $u[k]$ be bounded. Say $|u[k]|\le \overline{u}\forall k\ge 0$ for some $\overline{u}\ge 0$.
• Let $y[k]=(g^n\ast u)[k]$
• We have:

$$\begin{array}{rl}|y[k]|& =|\sum _{m=0}^k(g[m]{)}^{n}u[k-m]|\text{[discrete convolution]}\\ & =|\sum _{m=0}^k(g[m]{)}^{n-1}g[m]u[k-m]|[a^n=a^{n-1}a]\\ & \le \sum _{m=0}^k|(g[m]{)}^{n-1}g[m]u[k-m]|[\text{triangle inequality}]\\ & =\sum _{m=0}^k|(g[m]{)}^{n-1}|\cdot |g[m]u[k-m]|[|ab|=|a||b|]\end{array}$$

Steady-State Value Example

What is the steady-state value of $Y[k]$ when $r[k]=5\cdot 1[k]$, $d[k]=0.1^k$?

$$D[z]=1,G[z]=\frac{0.25(z+1)}{(z-1)}$$

We have:

$$T_{ry}=\frac{0.25(z+1)}{1.25z-0.75}$$ $$T_{dy}=\frac{0.25(z+1)}{1.25-0.75}$$

So:

$$\begin{array}{r}Y[z]=T_{ry}R[z]+T_{dy}d[z]\\ Yr[z]=\frac{1.25z(z+1)}{(1.25z+0.75)(z-1)}\end{array}$$

Case b:

$$\underset{k\to \infty }{\lim }{y}_r[k]=\underset{z\to 1}{\lim }(z-1)\cdot 1.25$$