I understand that stability for an LTI system is defined with respect to Bounded input bounded output condition. However I'm not clear on why non repeated poles on the imaginary axis makes the system marginally stable. For a unit step input, a single pole at origin produces an unbounded ramp output [Unbounded Response] and non repeated conjugate poles on the imaginary axis produces a bounded sinusoidal output [Bounded response]. Then why are both these Marginally stable systems?
\$\begingroup\$
\$\endgroup\$
1
-
\$\begingroup\$ yeah, it's not BIBO stable. if you banged it with an impulse, the output would ring forever and be bounded. personally i don't ever call a system stable if there are any poles on or to the right of the \$j\omega\$ axis of the \$s\$-plane or any poles on or outside the unit circle in the \$z\$-plane. \$\endgroup\$– robert bristow-johnsonJun 13, 2016 at 6:36
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
6
Complex conjugate poles on the \$j\omega\$ axis can also produce an unbounded output, just like a pole at \$s=0\$. It just depends on the input signal. If you excite a system with a single pole at \$s=0\$ with an impulse, the output is bounded (it's a step). If you excite if with a step, the output is unbounded (it's a ramp). If you excite a system with complex conjugate poles at \$\pm j\omega_0\$ with a sinusoidal input signal with frequency \$\omega_0\$, then you'll get a sinusoidal output signal with linearly increasing amplitude (a ramped sinusoid), i.e., an unbounded signal.
-
\$\begingroup\$ I rather think, that a system with a pole pair at ±jωo is an oscillator. Why should the amplitude increase? This happens only if the pole pair has a positive real part. \$\endgroup\$– LvWJun 13, 2016 at 6:59
-
\$\begingroup\$ @LvW: No, with a positive real part the amplitude would increase exponentially. A pole pair at \$j\omega_0\$ will lead to a linearly increasing output signal, if it is excited with an input signal of the appropriate frequency. Hence marginally stable. It's exactly the same mechanism as with a pole at \$s=0\$. Would you agree that the output of such a system would increase linearly when excited by a step? \$\endgroup\$– Matt L.Jun 13, 2016 at 7:04
-
\$\begingroup\$ Of course, an integrator has a linearly increasing step response. No doubt about it. But - would YOU agree that an (ideal) oscillator has a pole pair directly on the jw axis? \$\endgroup\$– LvWJun 13, 2016 at 8:07
-
\$\begingroup\$ @LvW: Well, OK, so if you excite such a system with an impulse you get a sinusoidal output that continues forever. However, if you excite it with a sinusoid of its own oscillation frequency, you get a sinusoid with linearly increasing amplitude. As an analogy, call a system with a pole at \$s=0\$ an "oscillator" with zero frequency. Excite it with an impulse, and you get its zero frequency output (a step); excite it with its (zero) frequency (a step), and you'll get an unbounded (linearly increasing) output. Exactly the same thing. \$\endgroup\$– Matt L.Jun 13, 2016 at 8:19
-
\$\begingroup\$ OK - you speak of external excitement of a self-oscillating system with a frequency identical to the oscillation frequency, correct? A rather "uncommon" situation, I think. And - yes - in such a (idealized) system the output will (theoretically) be unbounded. \$\endgroup\$– LvWJun 13, 2016 at 10:35