# Tag Info

## Hot answers tagged stability

5

For continuous-time systems, a pole at location $s_0=\sigma_0+j\omega_0$ will create a time-domain contribution of the form $$e^{s_0t}=e^{\sigma_0t}e^{j\omega_0t}\tag{1}$$ which is a damped oscillation if the pole is in the left half-plane (i.e., $\sigma_0<0$), and if the pole is not on the real axis (i.e., $\omega_0\neq 0)$. For $\omega_0=0$ there is ...

4

Although what @Fat32 wrote is correct, I think the potential instability of IIR filters is not the main reason for the instability of an adaptive IIR filter. After all, we can calculate the poles in each iteration and put a hard constraint to avoid poles out of the unit circle. Even within the case of the FIR filters - which are unconditionally stable- we ...

4

The IIR filter doesn't have to be unstable, but it has the potential of being so; unlike the FIR case which doesn't have even the potential. One reason for the (potential) unstability of an IIR (adaptive) filter is the numerical issues due to coefficient quantization. When the poles are closer to unit circle this will be critical. This is especially ...

4

They are independent of each other. Continuous systems: For stability, the ROC (region of convergence) must include the jw-axis of the s-plane. Causal systems have a ROC which is a right-sided plane, with $Re(s)>\alpha$. Here $\alpha$ is the real part of the "most to the right" pole. Due to this, for a continuous system to be causal and stable, all its ...

3

A system with simple distinct poles on the imaginary axis (and note that the origin is on the imaginary axis) and no poles in the right half-plane is called marginally stable. If you have poles with multiplicity greater than $1$ on the imaginary axis, or if there are poles in the right half-plane, then the system is unstable. For discrete-time systems, the ...

3

Repeated poles simply means there are more than one pole at the same location. If a pole is not repeated then it is a Distinct Pole. Consider the simple case of a cascade of two integrators (in s) which is similar to the cascade of two accumulators (in z). The Laplace transform of an integrator Is $\frac{1}{s}$, which is one pole at the origin. Two ...

2

For the biquad section that is cascaded, the quantization issues regarding the pole locations are well understood. For a biquad transfer function: \begin{align} H(z) &= \frac{b_0+b_1z^{-1}+b_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}} \\ \\ &= \frac{b_0z^2+b_1z+b_2}{z^2+a_1z+a_2} \\ \\ &= b_0\frac{z^2+\frac{b_1}{b_0}z+\frac{b_2}{b_0}}{z^2+a_1z+a_2} \...

2

With infinite-precision arithmetic they will be stable. However, since you don't have infinite-precision arithmetic, you will have quantization issues even if you use 64-bit precision. These quantization issues can make your filter unstable. Even if your filter is stable, perhaps you will not get the frequency response wanted because of these quantization ...

2

The problem here is that for the given functions, the usual Nyquist contour (see figure below) results in Nyquist plots from which no decision can be made about stability. Nyquist contour ($M\to\infty$) (from Signals and Systems by A.V.Oppenheim) The reason is the fact that for all three functions the limit for $|s|\to\infty$ equals zero. So the function ...

2

How can a stable system have a RHP pole? It can't. Now, I have that if I do the root locus I have: (image left out of quote) so I have a pole in the RHP. This is where you misunderstand. The word "locus" means "The set of all points whose coordinates satisfy a given equation or condition". The traces in that root locus show all possible locations ...

1

Like you mentionned, you cannot cancel a right-half-plane zero (or a zero outside the unit circle) by placing a pole on it. A unstable pole in your compensator will make the command of your controller unbounded (i.e. it will reach infinity). There are no ways to cancel a right-half-plane zero. It's sometimes possible to "remove" a right-half-plane zero by ...

1

Marginally stable means that an otherwise stable system has one or more simple poles on the unit circle (in discrete time), or on the imaginary axis (in continuous time). The consequence of that is that transients don't decay, but they also don't grow without bounds. Marginally stable systems are unstable in the bounded-input bounded-output (BIBO) sense.

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In the first case you have a pole at the location (-0.7). In the second case, your pole is at 0.7. Having a pole at -0.7 means that the natural frequency of your system is fs/2, that's why you have an oscillation at fs/2. Since the pole is stable, i.e. inside the unit circle, the oscillation eventually dies down. Edit : You happen to have a pole at fs/2, ...

1

We don't ignore poles on the contour. As mentioned in a comment, poles are avoided by modifying the contour as shown in the figure below, where a contour appropriate for a pole at $s=0$ is shown. Fig. 1: Nyquist contour for a pole at $s=0$ (from "Modern Control Engineering" by K. Ogata). The contour moves around the pole along a semi-circle centered at the ...

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