Calculate a_0 for Fourier Coefficient | Solution

[Oxymoron]

I am working through the following question:

f(t) = { a + 2a&omega;/&pi;*t when -&pi;/&omega; < t < 0
{ a - 2a&omega;/&pi;*t when 0 < t < &pi;/&omega;

I was wondering if someone could check what a_0 was for the Fourier coefficient. I worked it out to be 2a&pi; / &omega;L.

Cheers.

 

[HallsofIvy]

I get $$\frac{2a\pi}{\omega}$$. I don't see any "L" in the problem.

 

[M98Ranger]

To calculate a_0 for this Fourier coefficient, we need to use the following formula:

a_0 = (1/2&pi;) * ∫f(t) dt from -&pi; to &pi;

Substituting the given function into the formula, we get:

a_0 = (1/2&pi;) * ∫(a + 2a&omega;/&pi;*t) dt from -&pi; to 0 + (1/2&pi;) * ∫(a - 2a&omega;/&pi;*t) dt from 0 to &pi;

Simplifying the integrals, we get:

a_0 = (1/2&pi;) * [at + a&omega;/&pi;*t^2] from -&pi; to 0 + (1/2&pi;) * [at - a&omega;/&pi;*t^2] from 0 to &pi;

Evaluating the integrals and substituting the limits, we get:

a_0 = (1/2&pi;) * [a*0 + a&omega;/&pi;*0^2 - a*(-&pi;) - a&omega;/&pi;*(-&pi;)^2] + (1/2&pi;) * [a*(&pi;) + a&omega;/&pi;*(&pi;)^2 - a*0 - a&omega;/&pi;*0^2]

Simplifying further, we get:

a_0 = (1/2&pi;) * [0 + 0 + a&pi; + a&pi;] = a&pi; / &pi; = a

Therefore, the final value of a_0 for this Fourier coefficient is a. This means that the DC component of the given function is equal to the constant value a. I hope this helps!