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In motor-drives textbooks (e.g Ned Mohan's Electric Motor Drives), the text states that the stator windings on ac machines (e.g. PMAC three-phase motors and induction machines) are, in theory, sinusoidally distributed in the air gap. That is, the conductor density as a function of the mechanical angle (for a two-pole motor) can be written $$ n_s(\theta)=\frac{N_s}{2} \sin(\theta) $$ where \$N_s\$ is the number of turns. This is not possible in practice due to the slots, but it is approximated.

Is this true for modern PMAC motors ? If we consider one of the three phases, are there more conductors per angle at 90 and 270 degree angles from the magnetic axis than there are at f.ex 60 and 120 degrees ? I tried to find photos, but could not so easily see this...

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Yes:

A PMAC motor has an (approximately) sinusoidally distributed stator winding to produce sinusoidal back EMF waveforms. (Similarly a 3-phase AC induction motor does as well).

A "brushless DC" motor intended for six-step commutation may have trapazoidally distributed windings.

A good summary is Here

An interesting graphical representation with animation is Here

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  • \$\begingroup\$ I asked an experienced motor designer this question (do PMAC motors come with sinusoidally distributed stator windings). The answer was surprisingly "NO"... and from 3D drawing of one of their motors, the windings looked more uniformly distributed. Is it so that this is true only for most PMAC motors, e.g. for medium and lower size motors? \$\endgroup\$ – Ronny Landsverk Oct 18 '17 at 18:12
  • \$\begingroup\$ It's actually the overall properties of the magnetic circuit that influence the back emf waveforms (and therefore the ideal drive waveforms). So if the magnetic circuit is designed to produce roughly sinusoidal back-emf with somewhat uniformly distributed windings then that's a valid alternate way to design the motor. However, lots of PM brushless DC motors have evenly distributed winding which gives rise to trapezoidal back-emf waveforms. These are usually driven with six-step commutation. \$\endgroup\$ – John D Oct 18 '17 at 20:14

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