Fourier series for trigonometric absolute value function

In summary: You can use them to express the integrand, do the integration and then interpret the results. In summary, the conversation discusses finding Fourier coefficients for a given function using Euler's formula and the fact that the absolute value function is an even function. The integrand is expressed in terms of sines and cosines, but when using Euler's formula, the cosine terms cancel out and the sine terms do not. Further analysis is needed to determine coefficients for different values of n. The conversation also suggests making a sketch for different values of n to better understand the behavior of the coefficients.
  • #1
agnimusayoti
240
23
Homework Statement
Assume the period is 1/60 s. Given V(0) = `V(1/120) = V(1/60) = 0, and V(1/240) = V(1/80) = 100. Then output of a simple DC Generator will have the shape of absolute value of a sine function's curve. Expand this function in an appropriate Fourier series.
Relevant Equations
Fourier series:
For a function with period ##2l##, can expanded in Fourier series:
$$f(x) = \frac{1}{2} a_0 + a_n \cos{\frac{n\pi x}{l}} + b_n \sin{\frac{n\pi x}{l}} $$
where a_n and b_n is Fourier's coefficient.
For even function (since absolute value is an even function), then $$b_n = 0$$ and $$a_n = \frac{2}{l} \int_0^{l} f(x) \cos{\frac{n\pi x}{l}} dx$$
Capture.JPG

First, I try to define the function in the figure above: ##V(t)=100\left[sin(120\{pi}\right]##.
Then, I use the fact that absolute value function is an even function, so only Fourier series only contain cosine terms. In other words, ##b_n = 0##
Next, I want to determine Fourier coefficient ##a_n##:
$$a_n=\frac{2}{l} \int_{0}^{l} V(t) \cos{120n\pi t} dt$$.
I think, now V(t) only positive, so ##V(t)=100 sin(120\pi t)##
then,
$$a_n=\frac{2}{l} \int_{0}^{l} \left(100 sin(120\pi t)\right) \cos{120n\pi t} dt$$.
At this stage, is my work true?
Can I evaluate the integral with Euler form? Thanks
 
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  • #2
agnimusayoti said:
At this stage, is my work true?
Up to 1/120 s yes. After that you need a minus sign !

(However, if you look at the integrand, perhaps integrating up to 1/120 isn't all that bad ... :wink: )
 
  • #3
but I am my work next, I found the coefficient is zero (using Euler) without minus sign. The argument for sin differs from cos (in my integrand) right? So I can't modify to sin 2A?
 
  • #4
agnimusayoti said:
but I am my work next, I found the coefficient is zero (using Euler) without minus sign. The argument for sin differs from cos (in my integrand) right? So I can't modify to sin 2A?
I don't see your work if you don't post it, so I can't comment.
Advice: make a sketch for a few cases (n = 0, 1, 2, 3 ...) and see when coefficients are zero and when they are not

(PS: post #1 has
1594128884374.png
)
 
  • #5
Here I try to write it down.
$$a_n=2(120)\int_{0}^{1/120} 100 \sin {120 \pi t} \cos {120n \pi t} dt$$
In Euler formula:
$$\sin {120 \pi t} = \frac {e^{120i\pi t} - e^{- i 120 \pi t}}{2i}$$
$$ \cos {120n \pi t} =\frac{e^{120 n i\pi t} + e^{- i 120 n \pi t}}{2}$$
Substituting these formula to integran gives:
$$a_n=2(120)\int_{0}^{1/120} 100 \frac{1}{4i} \left ( e^{120i\pi t} - e^{- 120 i \pi t} \right)\left ( e^{120ni\pi t} - e^{- 120 n i \pi t} \right)dt$$
$$a_n=\frac{2(120)(100)}{4i}\int_{0}^{1/120} \left[e^{i 120 \pi t \left(1+n \right)} - e^{i 120 \pi t \left(1-n \right)} - e^{i 120 \pi t \left(-1+n \right)}+e^{i 120 \pi t \left(-1-n \right)}\right]dt$$.
At this stage, is that true? If I am not mistaken, I continued the integration:
$$a_n={96 000i} \left[\frac {e^{i 120 \pi t \left(1+n \right)}}{(120\pi i)(1+n)} - \frac {e^{i 120 \pi t \left(1-n \right)}}{(120\pi i)(1-n)} - \frac {e^{i 120 \pi t \left(-1+n \right)}}{(120\pi i)(-1+n)}+\frac {e^{i 120 \pi t \left(-1-n \right)}}{(120\pi i)(-1-n)}\right]^{1/120}_{0} dt$$
$$a_n=\frac{96 000i} {120 \pi i}\left[\frac {e^{i 120 \pi t \left(1+n \right)}}{(1+n)} - \frac {e^{i 120 \pi t \left(1-n \right)}}{(1-n)} + \frac {e^{i 120 \pi t \left(-1+n \right)}}{(1-n)}-\frac {e^{i 120 \pi t \left(-1-n \right)}}{(1+n)}\right]^{1/120}_{0} dt$$
After that, I found that cosine terms in euler Formula cancel each other, but sine is not. Unfortunately, the argument for sine is n pi then sine become zero.
 
  • #6
Hmm, But if I use trigonometric formula for sin A cos B = 1/2 [sin (A+B) sin (A-B)] I got different result. therefore somehow my integration bugging
 
  • #7
BvU said:
Advice: make a sketch for a few cases (n = 0, 1, 2, 3 ...) and see when coefficients are zero and when they are not

agnimusayoti said:
After that, I found that cosine terms in euler Formula cancel each other, but sine is not
And how is that for the integral from 1/120 to 1/60 ?
 
  • #8
It should be the same as half period because V(t) is even function right. But, the problem is executing Euler formula. I wonder what should I do in case I forget the trigonometric formula and not allowed to see it in textbook, since there are so many trigonometric formula. Huft
 
  • #9
agnimusayoti said:
It should be the same as half period because V(t) is even function right
You want to check that anyway. Remember that ##V(t) =\ {\bf -} \ \sin(...)## there !

BvU said:
Advice: make a sketch for a few cases (n = 0, 1, 2, 3 ...) and see when coefficients are zero and when they are not
And if you post them, I might point out some isssues you misss.
Don't be fixed on Euler if that doesn't work for you.. It's sines and cosines.
 

FAQ: Fourier series for trigonometric absolute value function

What is a Fourier series?

A Fourier series is a mathematical representation of a periodic function as a sum of sinusoidal functions. It is used to decompose a complex function into simpler components to make it easier to analyze.

How is a Fourier series used for trigonometric absolute value functions?

A trigonometric absolute value function can be represented as a Fourier series by breaking it down into its even and odd components. The even component is represented by cosine terms, while the odd component is represented by sine terms.

What is the significance of using a Fourier series for trigonometric absolute value functions?

Using a Fourier series for trigonometric absolute value functions allows for a more accurate and efficient representation of the function. It also makes it easier to perform calculations and analyze the behavior of the function.

Can a Fourier series accurately represent any trigonometric absolute value function?

Yes, a Fourier series can accurately represent any periodic function, including trigonometric absolute value functions. However, the accuracy of the representation may depend on the number of terms used in the series.

How is the convergence of a Fourier series for a trigonometric absolute value function determined?

The convergence of a Fourier series for a trigonometric absolute value function can be determined by using the Dirichlet conditions, which state that the function must be periodic, continuous, and have a finite number of discontinuities within a given interval.

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