# Effect of RHP zeros on stability

Why do zeros in the right half of the $$\s\$$ plane reduce stability (i.e., gain / phase margin)?

• What research have you made that needs clarifying? – Peter Smith Feb 23 at 17:26
• A screenshot of some notes link – TVV Feb 23 at 17:47
• I don't understand why boosting the gain and lagging the phase is bad for stability. – TVV Feb 23 at 17:52

Instability issues concern us when a system is used in a negative-feedback connection. Remember a negative-feedback system may oscillate at frequency $$\\omega\$$ if the phase shift around the loop at this frequency is more negative than -180 that the feedback becomes positive (because negative feedback itself introduces 180 degrees of phase shift) while the loop gain is still greater than one.

Therefore as a measure of stability we must have the gain crossover frequency occur well before the phase crossover. This condition guarantees that phase margin stays positive and the system will avoid oscillation. Now the problem with a RHP zero is that it adds extra phase shift when the magnitude of loop gain reaches 0dB (as indicated in above plot). This interprets as a negative phase margin or instability (a total phase shift of >360) in a negative-feedback system.

To ensure that any feedback system is stable, the loop gain needs to have a phase difference less than 180 degrees when its magnitude is unity, in other words it needs to ave sufficient phase margin.
Having a right half plane zero is non-favorable in this sense. It increases your loop gain and increases your phase difference. Consequently, your phase margin is definitely going to reduce because of the RHP zero and hence your stability is degraded.

Because zeros attract root locus branches - remember that each root locus branch starts at a pole and ends at a zero.

By the same token, a left plane zero can counteract the destabilising effect of a right plane pole. For example, an open loop TF, $$\\frac{1}{s-1}\$$, can be stabilised by adding a zero at, say, $$\\small s=-2\$$, i.e OLTF = $$\\frac{s+2}{s-1}\$$ gives a CLTF = $$\\frac{s+2}{2s+1}\$$. So the unstable pole at $$\\small s=1\$$ is dragged into the left plane; becoming a stable pole at $$\\small s=-0.5\$$