I know that in the magnitude spectrum of a single square DC pulse (0-8V) the lobes are zero at frequencies $$\frac{1}{t},\frac{2}{t},\frac{3}{t},...etc. $$ where t=pulse duration.

If the square pulse goes from -2V to 6V, is the formula for the zero points of the spectrum still the same, ie. $$\frac{n}{t}$$ where n=1,2,3... etc. and t=pulse duration?

What is the difference?

  • 1
    \$\begingroup\$ Grade 1 gobbledegook. \$\endgroup\$
    – Andy aka
    Jan 20, 2018 at 17:01
  • \$\begingroup\$ The amplitude at DC does not affect the amplitude of any of the harmonics. But the question improperly defines a square wave lobes,. \$\endgroup\$ Jan 20, 2018 at 17:15
  • \$\begingroup\$ @Andyaka I edited it to more coherent form. \$\endgroup\$
    – user287001
    Jan 20, 2018 at 19:06
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    \$\begingroup\$ @TonyStewart.EEsince'75 I edited the question to more coherent form. OP: Please revert the question to it original form if you wanted something else and add a image of your signals. You should show some own attempt to solve the problem, or you will get voted out! \$\endgroup\$
    – user287001
    Jan 20, 2018 at 19:14
  • \$\begingroup\$ A Square wave has null values at even harmonics for f=1/T at 2f, 4f or 2/T, 4/T etc. \$\endgroup\$ Jan 20, 2018 at 19:43

1 Answer 1


Applying an offset to a signal will only change a spectrum at DC (frequency of 0Hz).

If the single square pulse goes from 0V to 8V, the value of DC would be \$\mathcal{F}\{ f(t)\}|_{\omega=0}=0\$. If the square wave goes from -2V to 6V, then \$\mathcal{F}\{f(t)\}|_{\omega=0}=-2\cdot\delta(0)\$.

  • \$\begingroup\$ Your original question was about square waves, so I assumed it was periodic. I'll match my answer to your question. \$\endgroup\$
    – Sven B
    Jan 20, 2018 at 22:55

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