Why Doesn't My Fourier Series Expansion Look Like a Square Wave?

[back2square1]

for a square wave function,

f(x)= { -1, -∞ ≤ x ≤ 0; +1, 0 ≤ x ≤ ∞

Expanding it in Fourier series gives a function like,

f(x) = (4/π) * Ʃ_n=0^∞( (sin ((2n+1)x) / (2n+) )

Plotting a graph of the equation gives something like this, http://goo.gl/vFJhL
which obviously doesn't look like a square wave. Can anyone tell me where have I gone wrong? What am I missing?

P.S
Fourier co-efficients
a_n=0
a₀=0

 

[Mute]

for a square wave function,

f(x)= { -1, -∞ ≤ x ≤ 0; +1, 0 ≤ x ≤ ∞

The way you have written this, f(x) is not a square wave. It's a signum function.

A Fourier series is a series representation of a periodic function. If you take the Fourier series of a non-periodic function on a finite interval [a,b], then the Fourier series matches your function on that domain, but repeats the function shape with period b-a.

What you have written, however, has an infinite domain, so a Fourier series cannot be used - you would have to use a Fourier transform instead. However, because you mention a square wave, what you probably want to do is define your function on a finite domain, such as

$$f(x) = \left\{\begin{array}{c}{-1,~-1 \leq x < 0 \\ +1,~0 < x \leq 1}\end{array}\right.$$

Then, calculate the Fourier series for that function on the domain [-1,1]. The resulting Fourier series should be a square wave. Note also that it is odd about x = 0, so I would expect only sine terms in the series, and in fact you should get the series you quoted.

As for why your plots aren't working, it looks to me like an order of operations issue. When you write the sum you are writing terms like "4sin(5x)/5pi". The plotter is interpreting this as "(4*sin(5x)/5)*pi". Write it like "4sin(5x)/5/pi" or "4sin(5x)/(5pi)" and you will get the result you want.

 

[back2square1]

Thank you very much, Mute. It worked really nice. You're a wonderful guy. As you said, I added some extra parenthesis to my function and I can see a nice square wave, like this: http://goo.gl/9nu8V
Thank you once again. And yeah, I meant to write it like this,
f(x)= { -1, -π ≤ x ≤ 0; +1, 0 ≤ x ≤ π
I wrongly clicked on ∞ instead of π on the side bar.
And hey, please tell me how did you write that function in such a nice layout?

 

[Bacle2]

Thank you very much, Mute. It worked really nice. You're a wonderful guy. As you said, I added some extra parenthesis to my function and I can see a nice square wave, like this: http://goo.gl/9nu8V
Thank you once again. And yeah, I meant to write it like this,
f(x)= { -1, -π ≤ x ≤ 0; +1, 0 ≤ x ≤ π
I wrongly clicked on ∞ instead of π on the side bar.
And hey, please tell me how did you write that function in such a nice layout?

Use the 'Quote' button on Mute's post, and it will show you the formatting used.

 

[CellCoree]

First of all, it is important to note that the Fourier series expansion of a square wave is an infinite sum of sine functions. This means that the graph of the equation will only be an approximation of a square wave, as it is impossible to have an infinite number of terms in a real-life scenario.

One possible explanation for the discrepancy in the graph could be the number of terms used in the Fourier series expansion. The more terms that are included in the series, the closer the graph will be to a square wave. However, as mentioned earlier, it is impossible to have an infinite number of terms, so there will always be some level of approximation.

Another factor that could affect the accuracy of the graph is the range of values used for x. In the given equation, the range is from -∞ to ∞, which is not practical in real-life situations. It is more common to use a finite range of values, which could also contribute to the discrepancy in the graph.

Additionally, there could be errors in the calculations of the Fourier coefficients, which could also affect the accuracy of the graph. It is important to double-check the calculations and make sure they are accurate.

In conclusion, while the Fourier series expansion is a useful tool for approximating a square wave, it is not a perfect representation. Factors such as the number of terms used, the range of values, and potential errors in calculations can all contribute to discrepancies in the resulting graph. It is important to keep these limitations in mind when using Fourier series to model real-life phenomena.