Why is the Fourier coefficent calculated like this?

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arhzz
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Homework Statement
Find the Fourier coefficients
Relevant Equations
Fourier Analysis
Hello!

I have a function that is periodic. T = 2. Now I need to find a fourier series of that function, so I need to find coefficent ##a_0 a_k b_k ##

for a0 I have this formula (this is the one we used in class)

c.png


Now my x(t) = (-t+1). and when I plug in all of the values I have in the formula I get that ##a_0 = 2 ## but the solution says it should be 1. I checked the solution and what they did is they multiplied the 2/T with 2. I dont understand why they did this? Also they integrated from 0 to 1, I dont understand this either; Shouldnt the Integral be from -1 to 1 ?

For reference here is how the graph of the function looks like.

1706973736879.png


Thanks in advance!
 
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  • #2
arhzz said:
Homework Statement: Find the Fourier coefficients
Relevant Equations: Fourier Analysis
View attachment 339696

Now my x(t) = (-t+1).
This is the formula for only the right half of the sawtooth function. IOW, it is correct only for ##t \in [0, 1]##. You also need a formula for the left half, the part in the interval [-1, 0]. Since the formula changes, you'll need separate integrals for each coefficient.
arhzz said:
and when I plug in all of the values I have in the formula I get that ##a_0 = 2 ## but the solution says it should be 1. I checked the solution and what they did is they multiplied the 2/T with 2. I dont understand why they did this? Also they integrated from 0 to 1, I dont understand this either; Shouldnt the Integral be from -1 to 1 ?

For reference here is how the graph of the function looks like.

View attachment 339697

Thanks in advance!
 
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  • #3
Mark44 said:
This is the formula for only the right half of the sawtooth function. IOW, it is correct only for ##t \in [0, 1]##. You also need a formula for the left half, the part in the interval [-1, 0]. Since the formula changes, you'll need separate integrals for each coefficient.
Oh wait, this describes only the right part. So I need too calculate the coeffient 2 times, one from the intervall from [0,1] and for the intervall [-1 0] again, and than I have the coefficient a0? But is it possible to calculate the coefficient with 1 Integral, because they did it with 1 Integral? Also when I find the 2 different coefficients with two seperate integrals, should I add them to get a0?
 
  • #4
arhzz said:
So I need too calculate the coeffient 2 times, one from the intervall from [0,1] and for the intervall [-1 0] again, and than I have the coefficient a0? But is it possible to calculate the coefficient with 1 Integral, because they did it with 1 Integral?
Because of the symmetry, you can get by with one integral, by just multiplying the integral by 2. IOW, this will work:
$$a_0 = \frac 2 2 \int_0^1 1 - t ~dt$$
In addition, this integral can be evaluated by inspection, since it represents twice the area of a triangle whose base is 1 and whose altitude is also 1.
arhzz said:
Also when I find the 2 different coefficients with two seperate integrals, should I add them to get a0?
For the other coefficients in your Fourier series, you might have to use two separate integrals for each coefficient, and add them together. I haven't worked it out, though. It sometimes happens that half of the coefficients are zero, so that might make the work a bit easier.
 
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Mark44 said:
Because of the symmetry, you can get by with one integral, by just multiplying the integral by 2. IOW, this will work:
$$a_0 = \frac 2 2 \int_0^1 1 - t ~dt$$
In addition, this integral can be evaluated by inspection, since it represents twice the area of a triangle whose base is 1 and whose altitude is also 1.
For the other coefficients in your Fourier series, you might have to use two separate integrals for each coefficient, and add them together. I haven't worked it out, though. It sometimes happens that half of the coefficients are zero, so that might make the work a bit easier.
Just to report I have solved the problem and it was just like you said. For the coefficients you need to split it up with the 2 intervals and simply integrate, thankfully a lot cancels out since you plug in 0 for a lot of the function values, it is a bit more integrating but the only difference is the (-t+1) and (t+1) for the different intervals as well as the limits but you pretty much need to integrate once and just adjust the sign, and plug in the appropriate limit.

Thanks for the great help as always!
 

FAQ: Why is the Fourier coefficent calculated like this?

Why is the Fourier coefficient calculated using an integral?

The Fourier coefficient is calculated using an integral because it allows for the precise determination of the contribution of each frequency component in a continuous function. The integral effectively decomposes the function into its constituent sinusoidal components by averaging the product of the function and the corresponding basis function over a period.

What is the significance of the exponential function in the Fourier coefficient formula?

The exponential function in the Fourier coefficient formula, often expressed as \( e^{-i2\pi nt} \), represents complex sinusoids (combinations of sine and cosine functions). These complex exponentials form an orthogonal basis for the space of periodic functions, making them ideal for decomposing signals into their frequency components.

Why do we divide by the period when calculating Fourier coefficients?

Dividing by the period when calculating Fourier coefficients normalizes the integral, ensuring that the coefficients represent the average contribution of each frequency component over one period of the function. This normalization is crucial for accurately capturing the amplitude of each frequency component.

How does the orthogonality of sine and cosine functions affect the calculation of Fourier coefficients?

The orthogonality of sine and cosine functions means that the integral of the product of two different basis functions over a period is zero. This property simplifies the calculation of Fourier coefficients, as it ensures that each coefficient only captures the contribution of its corresponding frequency component without interference from others.

Why are there different forms of Fourier coefficients for real and complex functions?

There are different forms of Fourier coefficients for real and complex functions because real functions can be decomposed into sines and cosines, while complex functions are decomposed into complex exponentials. The real form typically uses separate coefficients for sine and cosine terms, while the complex form combines these into a single complex coefficient, which can be more convenient for certain analyses and applications.

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