# Very simple issue on Fourier series

• DWill
In summary, the conversation discusses the Fourier series representation of a periodic rectangular pulse train and the calculation of the coefficients using trigonometric integrals. The bn term is found to be incorrect and it is determined that a 1/2 scaling factor is missing. The conversation also mentions the natural period for Fourier series and adjustments that need to be made for different periods.
DWill
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:

a0 = 1/4 - (dc value = 0.25, this makes sense)

an = ∫1*cos(n2πft)dt over (0,T/4)
= $\frac{sin(\frac{nπ}{2})}{nπ}$

bn = ∫1*sin(n2πft)dt over (0,T/4)
= $\frac{1}{nπ}$(1-cos($\frac{nπ}{2})$

The bn 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 an 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.

Last edited:
Hey DWill.

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/Fourie...general_interval_.5Ba.2C.C2.A0a_.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.

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

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## 1. What is a Fourier series?

A Fourier series is a mathematical tool used to represent periodic functions as a sum of sine and cosine waves. This allows for the analysis and manipulation of complex periodic functions.

## 2. How is a Fourier series calculated?

A Fourier series is calculated by taking the coefficients of the sine and cosine waves that make up the function over one period. These coefficients are found using integration or other methods, depending on the complexity of the function.

## 3. What is the significance of the Fourier series in science?

The Fourier series has numerous applications in various fields of science, including physics, engineering, and signal processing. It allows for the analysis of complex periodic phenomena and the prediction of future behavior.

## 4. Can any function be represented as a Fourier series?

No, not all functions can be accurately represented by a Fourier series. The function must be periodic and satisfy certain mathematical conditions for the series to converge and accurately represent the function.

## 5. Are there any limitations to using a Fourier series?

While the Fourier series is a powerful tool, it does have some limitations. For example, it may not accurately represent functions with sharp discontinuities or functions that are not periodic. Additionally, the convergence of the series may be slow for certain functions.

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