this question is in two parts. Firstly, can you please tell me what the structure and amplitude distribution is of harmonics generated by a differentiated square wave of 50% duty cycle, where the time constant is very short, producing alternating bipolar spikes of short duration? I know a square wave produces odd order harmonics that decay in amplitude by 1/n, but not sure if the above does - or has even order harmonics as well? Secondly, what change to the harmonic series and amplitudes occurs for duty cycles tending towards 0% or 100%, from 50%, as if the source was PWM'd by a lower freq signal. Thanks.

  • \$\begingroup\$ Symmetrical signals don't have even harmonics \$\endgroup\$
    – user253751
    Mar 31, 2023 at 20:23
  • 1
    \$\begingroup\$ Just combine (multiply) the harmonic structure of the square wave with the frequency response of a differentiator. Hint: the differentiator's frequency response rises with frequency. Alternatively, just compute the harmonic series of the PWM signal directly, using the Fourier Transform. \$\endgroup\$
    – Dave Tweed
    Mar 31, 2023 at 20:24

1 Answer 1


You can model this as a series of positive and negative Dirac delta functions. You'll find that this ideal case gives you the (similarly ideal) result mentioned by Dave Tweed in the comments: the Fourier transform of a square wave multiplied by that of a differentiator. The drop in amplitude of the square wave harmonics and the rise with frequency of the differentiator cancel out, leaving a series of spikes on the fundamental and odd harmonics, all of equal amplitude.

  • \$\begingroup\$ OK, if the spikes / odd harmonics are all of equal amplitude at 50% duty cycle, then what changes when the square wave is PWM'd and the negative spike deviates about from its center position between the two positive spikes? Is there a rule like for pwm where if the duty cycle is 20%, or 1/5, every fifth harmonic would be missing? \$\endgroup\$
    – Jeff
    Mar 31, 2023 at 21:43
  • \$\begingroup\$ No, the even harmonics are missing because the signal is symmetrical in the time domain. If you change the duty cycle, the even harmonics will start kicking in. \$\endgroup\$ Apr 3, 2023 at 15:31
  • \$\begingroup\$ Thanks for the help. I think I have a clear picture of the process now. \$\endgroup\$
    – Jeff
    Apr 6, 2023 at 0:36

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