I understand that poles on the left hand complex plane (LHP) make a continuous linear dynamic system stable. What's so significant about imaginary poles on the LHP that make a system stable? What does it mean to have poles and zeros on the LHP verses the right hand plane (RHP)?

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    $\begingroup$ This is only true for continuous systems... For discrete systems, the stablility condition is such that the poles lie in the complex unit circle. This is because instead of ODE's, we get difference equations whose solutions have a factor of $\lambda^t$, where lambda is a pole. With this factor, solutions only blow up to infinity for large $t$ when a pole has magnitude greater than unity. $\endgroup$
    – Paul
    Jul 27, 2016 at 1:49

2 Answers 2


Only the poles in the LHP are necessary for stability. This is because the transient response of a LTI system will consists of a linear combinations of $e^{p_i t}$. If a pole is complex, $p_i=\rho_i+i \sigma_i$, you can use Euler's formula, such that the contribution to the transient response can be written as, $$ e^{\rho_i t} \left(\cos(\sigma_i t)+i \sin(\sigma_i t)\right). $$ A complex pole will always occur in pairs (complex conjugate) such that their combination will be real. These transient responses will only die out in time if $\rho_i<0$. This is the same as all poles in the LHP.

  • $\begingroup$ Why do complex poles always occur in pairs? Can we model a system that consists of a single pole (no conjugate), or pairs that are slightly asymmetric around the imaginary axis? It seems to me singular complex poles would explain a bunch of of quantum behaviour. $\endgroup$ Nov 1, 2018 at 11:57
  • $\begingroup$ @PetrusTheron Mathematically you can have complex poles/zeros without a conjugate. However such transfer function will have a complex response to a real signal in the time domain. However in the real world you do not have this, for example you can not measure a complex voltage or distance. But if want model something for which it would be acceptable to be not completely real then you can use it. It can be noted that when using complex conjugate pairs you can still have a complex response if the input is also complex. $\endgroup$
    – fibonatic
    Nov 1, 2018 at 12:05
  • $\begingroup$ Thank you, @fibonatic :) Let's say you could build a complex single pole system, what would the combined transfer function look like if you combined two such systems where iα₁ and iα₂ differed by a small delta? My intuition says you get a Gaussian probability distribution in the real-valued part, or interference pattern. $\endgroup$ Nov 1, 2018 at 12:17
  • $\begingroup$ @PetrusTheron You mean $\alpha_1\approx-\alpha_2$? $\endgroup$
    – fibonatic
    Nov 1, 2018 at 12:21
  • $\begingroup$ yes, I meant if the sum of the near-conjugates approached zero i.e. i(α1 + α2) ≈ i0 $\endgroup$ Nov 1, 2018 at 12:43

Briefly, poles in the left half plane (R < 0) symbolize ringing that dies down over time, at R = 0 ringing that stays the same, and in the right half plane (R > 0) ringing that increases in amplitude over time.


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