Stability of Approximation Methods

Left side rule: $s=\frac{1}{T}(z-1)$. Stable poles in CT can get mapped to unstable poles in DT:

$$\frac{1}{s-p}⟶\frac{1}{\frac{1}{T}(z-1)-p}\times \frac{T}{T}=\frac{T}{z-1-Tp}$$

• If $p<-\frac{2}{T}$, we can see that the pole $\notin \mathbb{D}$, and thus unstable.

Right side rule: $s=\frac{1}{T}\frac{z-1}{z}$. Unstable poles in CT can get mapped to stable in DT

$$\frac{1}{s-p}⟶\frac{1}{\frac{1}{T}\frac{(z-1)}{z}-p}\times \frac{zT}{zT}=\frac{zT}{z-1-pzT}⟶\text{pole:}\left\{\frac{1}{1-pT}\right\}$$

• If $p>\frac{2}{T}$, pole $\in \mathbb{D}$, and thus stable.

Trapezoidal rule: $s=\frac{2}{T}\frac{z-1}{z+1}$ Stable poles in CT get mapped to stable poles in DT (note: does not depend on $T$).

$$\begin{array}{rl}\frac{1}{s-p}⟶& \frac{1}{\frac{2}{T}\frac{z-1}{z+1}-p}\times \frac{T(z+1)}{T(z+1)}\\ & =\frac{T(z+1)}{2(z-1)-pT(z+1)}\\ & =\frac{T(z+1)}{(2-pT)z-(2+pT)}\\ ⟶& \text{pole:}\left\{\frac{2+pT}{2-pT}\right\}\end{array}$$

Example 9.1.

Find the discretized version of $C(s)$ using the 3 discretization methods and compare stability:

$$C(s)=\frac{s-2}{s+25},T=0.1$$

We have a stable pole at $-25$.

Left side rule:

$$\begin{array}{rl}D[z]& =C(s){|}_{s=\frac{1}{T}(z-1)}=C(s){|}_{s=10(z-1)}\\ & =\frac{10(z-1)-2}{10(z-1)+25}\\ & =\frac{10z-12}{10z+15}\end{array}$$

which has a pole at $-\frac{3}{2}$, not in $\mathbb{D}$ (unstable).

Right side rule:

$$\begin{array}{rl}D[z]& =C(s){|}_{s=\frac{1}{T}\frac{z-1}{z}}=C(s){|}_{s=10\frac{z-1}{z}}\\ & =\frac{10\frac{z-1}{z}-2}{10\frac{z-1}{z}+25}\times \frac{z}{z}\\ & =\frac{8z-10}{35z-10}\end{array}$$

which has a pole at $\frac{2}{7}$, which is in $\mathbb{D}$, so stable.

Trapezoidal rule:

$$\begin{array}{rl}D[z]& =C(s){|}_{s=\frac{2}{T}\frac{z-1}{z+1}}=C(s){|}_{20\frac{z-1}{z+1}}\\ & =\frac{20\frac{z-1}{z+1}-2}{20\frac{z-1}{z+1}+25}\times \frac{z+1}{z+1}\\ & =\frac{18z-22}{45z+5}\end{array}$$

which has a pole at $-\frac{1}{9}$, which is in $\mathbb{D}$, which is stable.