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Hi all, I am just trying to prove to myself the Fourier series representation of a periodic rectangular pulse train. The pulses have some period T, and each pulse has magnitude equal to 1 over a duration of T/4, and 0 the rest of the cycle.

Using trignometric Fourier series, I get the following:

a

a

= [itex]\frac{sin(\frac{nπ}{2})}{nπ}[/itex]

b

= [itex]\frac{1}{nπ}[/itex](1-cos([itex]\frac{nπ}{2})[/itex]

The b

Thanks for the clarification!

EDIT: I recall that if I shift the reference axis so that t=0 lies at the midpoint of a pulse, the function has even symmetry and thus only the a

Using trignometric Fourier series, I get the following:

a

_{0}= 1/4 - (dc value = 0.25, this makes sense)a

_{n}= ∫1*cos(n2πft)dt over (0,T/4)= [itex]\frac{sin(\frac{nπ}{2})}{nπ}[/itex]

b

_{n}= ∫1*sin(n2πft)dt over (0,T/4)= [itex]\frac{1}{nπ}[/itex](1-cos([itex]\frac{nπ}{2})[/itex]

The b

_{n}term doesn't make sense to me, because it seems to contribute a dc term for each value of n. I double-checked the integration, although it's a very simple integral. This might just be the case of a really obvious mistake I'm making and I just can't seem to pinpoint it.Thanks for the clarification!

EDIT: I recall that if I shift the reference axis so that t=0 lies at the midpoint of a pulse, the function has even symmetry and thus only the a

_{n}terms will exist. However, I'm a bit confused why these extra dc terms result if I just shift the reference that I'm looking at a bit.
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