100kHz PWM Motor driver PCB/cable consideations
The analog signals will be trashed, if there is strong coupling via electrostatic fields and the analog node impedance (not resistance but impedance) is HIGH. The coupling is by high-pass-filter; any resistance on the analog signal is sufficient to implement a DC_response.

Consider capacitive coupling, of 1mm by 1mm overlap of your perpendicular traces, with 1/3 of 1/16" spacing (0.5mm). Er of FR4 is 5. The capacitance is

Eo * Er * Area/Distance = 9e-12 * 5 * 1mm * 1mm  0.5mm = 9e-11 * 1mm = 9e-14 farad. Or ~~ 1e-13 Farad

If your node has 10pF capacitance, your injected voltage level is 1% of the PWM signal (plus any switching spikes).

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Regarding magnetic fields, I use

Vinduce = 2e-7 * Area/Distance * dI/dT

where Area is the length * width of a vulnerable analog signal/RTN loop and Distance is the closest spacing between PWM lines and that analog loop.

Assuming 1cm loop (0.01 meter * 0.01 meter) at 1cm distance, the PWM switching 1amp in 100 nanoseconds, the Vinduce is

2e-7 * 0.01 * 0.01 / 0.01 * 10^7 amp/second

2e-7 * 0.01 * 10^7 = 0.02 volts

At 100 nanosecond switching edge, any planes will have a few dB of Skin Effect shielding. Slower edges will be even less attenuated, but also will induce smaller voltages into the analog loop.