# Routh's stability condition

Assume we have a LTI system which has poles in the half left plane of the s domain.

Before I learnt Routh's stability condition I had imagined that this was enough to decide whether a LTI system was stable or not. Is it possible for a LTI system to have its poles only in the half left plane of the s domain but not be stable? In other words, is Routh's stability condition exclusive from not having any poles which their real part is greater than 0?

And if yes what is the intuition behind it?

Most of your problems with this stem from confusion about terms and mixing up "criterion" and "test". Words matter! And in engineering and math, precise wording matters, so maybe you've not been learning from the cleanest references.

The criterion is (usually) called Routh-Hurwitz stability criterion. And that's what's sufficient and necessary. That criterion is that all roots of the the transfer function have negative real part. End of story.

Routh came up with a test for whether the criterion is fulfilled, it's called Routh's Test. It's an algorithm to determine whether the criterion is fulfilled, not a criterion.

No it's not possible. This is because the presence of poles in the right-hand plane implies that the system has a component with exponentially increasing amplitude. On the other hand, poles in the left-hand plane correspond to exponentially decaying components in the system response, which results in a stable and well-behaved system.

Moreover, as Marcus said:

Routh came up with a test for whether the criterion is fulfilled, it's called Routh's Test. It's an algorithm to determine whether the criterion is fulfilled, not a criterion.

so it's merely an algorithm to check for stability. You will learn more ways of checking for stability as you progress through your control theory course if you haven't already.

But the intuition behind poles in the LHP are stable is as I have stated above.