1
$\begingroup$

I'm studying for an exam and this is an old exam question that I don't understand:

Is the following system non-minimum phase?

$$G(s) = \frac{e^{-2s}}{s+2}$$

I can see that the real part of the pole is on the left half plane, but regarding the zeros I don't know what to do. If plotting $e^{-2s}$ it seem to never cross the x-axis, so I thought it didn't have a zero at all, and thus no zero/pole is in the right half plane $\rightarrow$ the system is minimum phase. But the correct answer was apparently that the system is non-minimum phase, and the explanation was "contains a time-lag". How was I supposed to think?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

A linear time-invariant system is said to be minimum-phase if the system and its inverse are 1) causal and 2) stable.

In your case the inverse of the time-delay $e^{-2s}$ makes the inverse non-causal; hence not minimum phase.

Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.