# Digital control stability for more than impulse response stability?

I have a digital control system which is described by the transfer function

$$Y(z) = \frac{-\beta}{z - 1 + k \beta} X(z)$$

where $$\\beta\$$ is a real parameter I can adjust to modify the characteristics of the system, and $$\k\$$ is a positive real constant.

Using the BIBO stability criteria that the poles must be inside the unit circle, this means that for BIBO stability I need $$0 < k \beta < 2$$ which is exactly what I observe when $$\x[n]\$$ is an impulse response: $$x[n] = \begin{cases}0 & n < 0\\ x_0 & n \ge 0\end{cases}$$

However, I am trying to design my system to handle "arbitrary" inputs. For example, in the above example suppose I have an input which jumps between two values: $$x[n] = \begin{cases} x_0 & n~\text{is even}\\ -x_0 & n~\text{is odd} \end{cases}$$ The z-transform for this particular input is $$X(z) = x_0 \frac{1 - \frac{\cos(\pi)}{z}}{1 - \frac{2 \cos(\pi)}{z} + \frac{1}{z^2}} = x_0 \frac{z (z + 1)}{z^2 + 2 z + 1}$$

If I try to simulate the system with this input, for $$\k \beta > 1\$$, $$\y[n]\$$ actually tends towards $$\\pm \infty\$$ (oscillating), which I would consider unstable, or at the very least unacceptable.

How would I go about analyzing (and designing) a control system when I have to consider other possible input responses than just the basic impulse response? Are there any useful search terms I could use for solving this type of problem?

• This may be beter suited for dsp.ee. That said, $-x_0$ falls outside BIBO, so you can't expect it to behave. Which leaves you with two choices: either change the input, or change the transfer function. Commented Aug 11, 2020 at 7:24
• @OP can you add the z transform of the input signal also into your question ? Does it form a double pole (not a complex pair, but 2 poles at same location) on the unit circle ? That may give a clue.
– AJN
Commented Aug 11, 2020 at 7:37
• @aconcernedcitizen I'm a bit surprised that negative inputs are not consider "bounded". Most definitions I've found for BIBO stability only require that $\|x\|_{\infty}$ be "finite". Commented Aug 11, 2020 at 7:41
• @AJN I added the z-transform for the oscillatory input, and it does indeed have a repeat pole at $z=-1$, though I'm not entirely sure why that matters? Commented Aug 11, 2020 at 7:48
• I couldn't reproduce unbounded output with this Octave code ? Is it same as or different from your question ? How did you simulate the system ? I have deleted my two earlier comments.
– AJN
Commented Aug 11, 2020 at 9:25

There might be a problem with your simulation, because for an oscillating input signal (with constant amplitude), the output is also an oscillating signal with constant amplitude, as long as the system is stable, i.e., as long as $$\0 is satisfied.
$$y[n]=\frac{\beta x_0}{2-k\beta}(-1)^n,\qquad 0
Of course, if the input is switched on at some finite point in time, there will be transients before the output approaches $$\(1)\$$, but these transients will decay as long as the stability condition $$\0 is satisfied.
For values of $$\k\beta\$$ close to $$\2\$$ you might be observing the transient, which will increase in amplitude for quite a while, but the output will finally settle at its maximum amplitude $$\\beta/(2-k\beta)\$$, which can be quite large for values of $$\k\beta\$$ close to $$\2\$$.
The figure below shows the output for an oscillating input signal ($$\x_0=1\$$) starting at $$\n=0\$$, and for two different values of $$\k\beta\$$: