"Tony Helton" wrote in message
...

If the switching signal (PWM) is run into an inductive load in a poorly
designed circuit you could have tens or even hundreds of volts as a
transient signal. It's not the current but dI/dt that makes the voltage.
One does not need an antenna to radiate. A plain old wire can work just
fine. Since this is all so variable, it points right at the installation
factors. By the way, radio amateur QRP distances show over 1,000 miles
range with a single NAND gate as the output driver. Figure PWM waveform is
5V @ 5mA = 25mW-- 14 dBm into 50 ohm. VHF radio sensitivity typically
0.25 uV into 50 ohm ---1.25e-15 Watt or -120 dBm. Say the PWM operates at
100 kHz so the 1500th harmonic has to be down -134 dB to be undetectable
on the VHF. The Fourier expansion of a square wave PWM shows the harmonic
coefficients decreasing as 1/n, n being the harmonic number so the VHF
in-band harmonic is 1/1500 of the original signal or about 33 dB down.
That leaves about 100 dB of required attenuation (if all the DC input
power were radiated) not to interfere with the VHF. FCC requires the
harmonics to be much lower, which they are in order to comply. Automotive
testing is rigorous and thorough and it takes a real screw up to get the
thing to radiate and cause interference in the boat. The lesson here is if
you don't understand electrical installations, hire an experienced (even
licensed) technician to do the work It may save your life.

Hey Tony,

The Fourier coefficients for the Power Spectra decrease as 1/(n^2), not 1/n
(that is for voltage). Power is proportional to voltage squared. Hence, your
1500th harmonic is down 66 dB, not 33 dB.

Amen!

Bob Crantz PhD