# How DC offset affects the spectrum of a single pulse

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?

• Grade 1 gobbledegook. Jan 20, 2018 at 17:01
• The amplitude at DC does not affect the amplitude of any of the harmonics. But the question improperly defines a square wave lobes,. Jan 20, 2018 at 17:15
• @Andyaka I edited it to more coherent form. Jan 20, 2018 at 19:06
• @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! Jan 20, 2018 at 19:14
• A Square wave has null values at even harmonics for f=1/T at 2f, 4f or 2/T, 4/T etc. Jan 20, 2018 at 19:43

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 $\mathcal{F}\{f(t)\}|_{\omega=0}=-2\cdot\delta(0)$.