# Underlying reasoning behind lead and lag nomenclature of compensation

If $$\0, the Laplace transform transfer function $$\\frac{s+z}{s+p}\$$ constitute two distinct types of compensators.

If we look at this schematic, $$\0 (lead compensation) implies that for $$\s=\omega j\$$ with $$\0<\omega\$$, $$\s+z\$$ clearly lies to the left of $$\s+p\$$ so that $$\0<\arg(\frac{s+z}{s+p})=\tan^{−1}(\frac{p−z}{\omega^2+p^2})<\frac{\pi}{2}\$$.

Similarly, $$\0 (lag compensation) implies that for $$\s=\omega j\$$ with $$\0<\omega\$$, $$\s+z\$$ is to the right of $$\s+p\$$ so that $$\-\frac{\pi}{2}<\arg(\frac{s+z}{s+p})=\tan^{−1}(\frac{p−z}{\omega^2+p^2})<0\$$.

Is the following understanding about the phase with regards to the lead and lag compensation accurate? The relative location of the complex numbers $$\s+z\$$ and $$\s+p\$$ is the underlying reason that the lead compensation adds (and a lag subtracts) phase to the transfer function it multiplies.

• Where did the quoted text question come from? Is it homework? – Andy aka Mar 25 at 9:10
• @Andyaka this is not a homework-and-exercises question. – kbakshi314 Mar 26 at 20:11
• Lead compensator provide phase lead at all the frequencies of input signal and by setting p and z appropriately you can get max phase lead at desired frequency , but you yourself answer the question by showing how phase shift takes place in lead or lag compensator – user215805 Mar 26 at 20:40
• @user215805 thanks for the comment. The intention of the question is to ascertain the geometric explanation of the phase manipulation using lead or lag compensation. – kbakshi314 Mar 26 at 22:55