
#1
Mar712, 01:18 AM

P: 70

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_{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](1cos([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 doublechecked 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. 



#2
Mar712, 02:57 AM

P: 4,570

Hey DWill.
Just a few comments. As far as I remember, the natural period for fourier series is going to be 2*pi units. So if you have a period of T, then this implies that we have to adjust the terms inside the sines and cosines as well as the normalization terms for the projections. What this translate to is that if T is our period then we use the following formula: http://en.wikipedia.org/wiki/Fourier..._.2B_.CF.84.5D This means that we have to figure out the following integrals: an = 1/T∫1*cos(n2πt/T)dt over (0,T/4) = 1/T x T/(2πn) [sin(2πnt/T)] t=(0,T/4) = 1/(2πn) x sin(πn/2) bn = 1/T∫1*sin(n2πt/T)dt over (0,T/4) = 1/T * T/(2πn) [cos(2πnt/T)] t = (0,T/4) = 1/(2πn) [1 cos(πn/2)] So if my calculations are right, you are missing a 1/2 scaling factor term. Now I had a bit of a mess around with deriving the the final function which is meant to match the wiki definition of the fourier series but I made a mistake when using some trig substitutions so I haven't posted it. 



#3
Mar712, 03:36 PM

Sci Advisor
P: 5,942

The bn terms are not dc. They multiply sines, just as the an terms multiply cosines.



Register to reply 
Related Discussions  
Simple Fourier Series Question  Calculus & Beyond Homework  14  
How is this possible? (simple fourier series)  General Math  3  
Simple Fourier series problem  Calculus & Beyond Homework  1  
Fourier Series of simple function.  Calculus & Beyond Homework  1  
Fourier Series / Fourier Transform Question  Electrical Engineering  6 