Sample-Data System Stability

Example:

$$\dot{x}=\lambda x$$

Recall from Continuous-Time Stability that

$$x(t)=e^{\lambda t}x(0)=(e^{\lambda }{)}^{t}x(0)$$

We sample at $t=kT$ for $k\in {\mathbb{Z}}_{\ge 0}$, such that

$$x[k]=x(kT)=(e^{\lambda }{)}^{kT}x[0]$$

• Note that we just have $x[0]=x(0)$

Then:

$$|x[k]|=|e^{\lambda }{|}^{kT}|x[0]|=(|e^{\lambda }{|}^T{)}^K|x[0]|$$

To understand this, we want to understand the $(|e^{\lambda }{|}^T{)}^K$ term. First, note that:

$$|e^{\lambda }|=|e^{\text{Re }\lambda +j\text{ Im }\lambda }|=|e^{\text{Re }\lambda }||e^{j\text{ Im }\lambda }|$$

The final term, $|e^{j\text{ Im }\lambda }|$, is equal to $1$. This is because $e^{j\theta }=\cos \theta +j\sin \theta$, and since the argument $\text{Im }\lambda$ is entirely imaginary, we are at the top of the unit circle. Thus:

$$\begin{array}{rl}|e^{\lambda }|& =|e^{\text{Re }\lambda }|\\ & =e^{\text{Re }\lambda }\end{array}$$

From our definition of stability:

$$T\text{ stable}⟺\text{Re}(\lambda )<0⟺e^{\text{Re}(\lambda )}<1⟺|e^{\lambda }|<1$$

• “T stable” here means that the resulting-discrete time system is stable.

We also have:

$$|e^{\lambda }|<1⟺|e^{\lambda }{|}^T<1⟺(|e^{\lambda }{|}^T{)}^k<1$$

Thus, the system is stable in discrete time too. This means that if we have a stable continuous time system that we sample, the result is also stable.