Sketching Nyquist Plots for Complex Arguments

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SUMMARY

The discussion focuses on sketching Nyquist plots for the transfer function $$\frac{j\omega-1}{j\omega+1}$$. The initial argument calculation of $$\arctan(-\omega) - \arctan(\omega) = -2\arctan(\omega)$$ is identified as incorrect due to the need for an adjustment when the real part is negative. The correct approach involves using the formula $$\mathrm{arg}(x + yj) = \pi + \arctan(y/x)$$ when the real part is negative, leading to the accurate behavior of the Nyquist plot as $$\omega \to 0$$ and $$\omega \to \infty$$.

PREREQUISITES
  • Understanding of complex numbers and their arguments
  • Familiarity with Nyquist plots in control theory
  • Knowledge of the arctangent function and its properties
  • Basic electrical engineering concepts related to transfer functions
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  • Study the derivation of Nyquist plots for various transfer functions
  • Learn about the implications of phase shifts in control systems
  • Explore the use of Wolfram Alpha for complex function analysis
  • Review the properties of the arctangent function in complex analysis
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Electrical engineers, control system designers, and students studying complex analysis and Nyquist stability criteria will benefit from this discussion.

ajtgraves
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I'm trying to sketch the nyquist plot of
$$\frac{j\omega-1}{j\omega+1}$$
but can't seem to calculate the argument correctly. I think it should be $$\arctan(-\omega) - \arctan(\omega) = -2\arctan(\omega)$$ but this doesn't give the correct nyquist plot behaviour for $\omega \to 0$ and $\omega \to \infty$ - surely $-2\arctan(\omega)$ implies that $\lim_{x\to 0} = 0^\circ$ and $\lim_{x\to \infty} = -180^\circ$?

Wolfram Alpha disagrees but I can't see where I'm going wrong. Am I making a glaring error somewhere? Any help would be greatly appreciated.

Thanks very much
 
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You're an electrical engineer, right? So j means the square root of -1?

Assuming it does, I think you are off by 180 degrees. The formula ##\mathrm{arg}(x + yj) = \arctan(y/x)## is valid when ##x > 0##, but if the real part of ##x + yj## is negative, as it is in the numerator, you need to adjust for the fact that arctan only returns angles in ##(-\pi/2, \pi/2)## by using ##\mathrm{arg}(x + yj) = \pi + \arctan(y/x)##.
 

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