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

If we look at this schematic, \$0<z<p\$ (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<z<p\$ (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.

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  • \$\begingroup\$ Where did the quoted text question come from? Is it homework? \$\endgroup\$ – Andy aka Mar 25 at 9:10
  • \$\begingroup\$ @Andyaka this is not a homework-and-exercises question. \$\endgroup\$ – kbakshi314 Mar 26 at 20:11
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    \$\begingroup\$ 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 \$\endgroup\$ – user215805 Mar 26 at 20:40
  • \$\begingroup\$ @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. \$\endgroup\$ – kbakshi314 Mar 26 at 22:55

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