I have encountered the following theorem in a book.enter image description here

According to me, I can get an asymptotically stable system even if cancellations exist for the unstable poles. my example, enter image description here Then Why does this book definition state "no hidden cancellations of unstable modes" ? It is implying necessity I understand

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    \$\begingroup\$ Theoretically your diagram is stable but in practise, this wouldn't necessarily be regarded as a fuilly stable system due to delays between blocks and temperature effects making the first block s-1.99999. \$\endgroup\$ – Andy aka Mar 27 '18 at 9:02
  • \$\begingroup\$ Andy's spot on. Also a step input (which is bounded) to the (s-2) block would give rise to an infinite output at the output of the (s-2) block at t=0, due to the differentiator. \$\endgroup\$ – Chu Mar 27 '18 at 11:34
  • \$\begingroup\$ Perfect andy! Please add it to the answer so I can accept it as the answer \$\endgroup\$ – aadil095 Mar 27 '18 at 15:41

Asymptotic stability and BIBO stability are entirely different. System response depends on both zero state and zero input conditions. Thus we have two forms of stability criterions, one that concerns with input and other concerns only with characteristic modes of a system.When a system is observable and controllable, its external and internal descriptions are same. So BIBO and asymptotic stabilities are same.

The example that you have provided has impulse response,


which is absolutely integrable and hence BIBO stable.

But the system 1 has characteristic root at 2 while system 2 has at 2 and -3. Since these two systems do not load each other, the characteristic modes are also independent. The characteristic mode of system 1 does not converge thus making entire system asymptotically unstable.

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  • \$\begingroup\$ what is bibo? basic input bo? \$\endgroup\$ – Dan Powers May 13 '19 at 7:33
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    \$\begingroup\$ bounded input bounded output \$\endgroup\$ – Parth May 14 '19 at 10:28

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