# In Control Systems Engineering, why do imaginary poles and zeros on the LHP indicate stability?

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)?

• 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. – Paul Jul 27 '16 at 1:49

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.
• @PetrusTheron You mean $\alpha_1\approx-\alpha_2$? – fibonatic Nov 1 '18 at 12:21
• yes, I meant if the sum of the near-conjugates approached zero i.e. i(α1 + α2) ≈ i0 – Petrus Theron Nov 1 '18 at 12:43