Fourier Series of f1(t) and f2(t): Calculating the Shifted Fourier Series

[tommyhakinen]

Homework Statement

Given that the Fourier series of function f₁(t)

= At/π + A , -π < t < 0
= -At/π +A, 0 < t < π

is f(t) = 0.5A +4A/π²(cos t + 1/9 cos 3t + 1/25 cos 5t + ...)

what is the Fourier series of function f₂(t)
= -At/π, -π < t < 0
= At/π, 0 < t < π

The Attempt at a Solution

for a function f₂(t) = f₁(t + π) means the function f₂(t) is the function of f₁(t) shifted by a distance of π to the left. How do I link this with the above problem?

 

[gabbagabbahey]

Hint: what is \frac{A(-t+\pi)}{\pi}?

 

[tommyhakinen]

Thanks for the reply. The function f₁ is a periodic function. The function f₂ is a function f₁ shifted by a distance π. However, which direction the shifting should be the right answer? I found out that if we shifted f₁ either to the right or left by π distance, we still get the same f₂. or both answer are acceptable?

my answer is : f₂(t) = A/2 + 4A/π² {cos(t+π) + 1/9 cos(3(t+π)) + 1/25 cos(5(t+π)) + ...}

please correct me if i am wrong. thank you.

 

[gabbagabbahey]

Shouldn't it be f_2(t)=f_1(-t+\pi) i.e. with a negative sign in front of t

 

[tommyhakinen]

why is it negative t? can you elaborate more on this? from my observation, i can see that the function f₁ is just shifting to the left or right by π. since f₁ is a periodic function, it looks like sin wave on cosine wave which are π distance from each other.

 

[gabbagabbahey]

well, look at the functions...for 0&lt;t&lt;\pi,

f_1(-t+\pi)=\frac{-A(-t+\pi)}{\pi}+A=\frac{At}{\pi}=f_2(t)

The same relationship holds for \pi&lt;t&lt;0.