The Fourier series expansion for the periodic function,f(t) = |sin t|is defined in its fundamental interval. Taking π = 3.142, calculate the Fourier cosine series approximation of f(t), up to the 6th harmonics when t = 1.09. Give your answer to 3 decimal places. Anybody can help?
TA E IT A. HALE. SO H S A ION
To calculate the Fourier cosine series approximation of the function f(t) = |sin t| up to the 6th harmonics, we first need to find the Fourier coefficients. The Fourier cosine series is used because f(t) is an even function.
The Fourier coefficients for the cosine series can be calculated using the following formula:
a0 = (2/T) * ∫[0,T] f(t) dt
an = (2/T) * ∫[0,T] f(t) * cos((2πnt)/T) dt, for n = 1, 2, 3, ...
where T is the period of the function (in this case, T = 2π).
Let's calculate a0 first. Since f(t) = |sin t|, we have:
a0 = (2/2π) * ∫[0,2π] |sin t| dt
To evaluate this integral, we need to break it into two intervals, [0, π] and [π, 2π], because |sin t| changes sign at π.
For the interval [0, π], we have:
a0(1) = (2/2π) * ∫[0, π] sin t dt
Using the integration formula, we get:
a0(1) = (2/2π) * (-cos t)|[0, π] = -1/π
For the interval [π, 2π], we have:
a0(2) = (2/2π) * ∫[π, 2π] (-sin t) dt
Again, using the integration formula, we get:
a0(2) = (2/2π) * (-cos t)|[π, 2π] = -1/π
Since f(t) is even, a0(1) = a0(2) = -1/π. Therefore, a0 = -1/π.
Now, let's calculate the coefficients an. We have:
an = (2/2π) * ∫[0,2π] |sin t| * cos((2πnt)/(2π)) dt
= (1/π) * ∫[0,2π] |sin t| * cos(nt) dt
Again, we need to break this integral into two intervals, [0, π] and [π, 2π], due to the change in sign of |sin t| at π.
For the interval [0, π], we have:
an(1) = (1/π) * ∫[0, π] sin t * cos(nt) dt
Using integration by parts with u = sin t and dv = cos(nt) dt, we get:
an(1) = (1/π) * (1/n) * (sin(nt))^2 |[0, π] = (1/πn) * (1 - 0) = 1/πn
For the interval [π, 2π], we have:
an(2) = (1/π) * ∫[π, 2π] (-sin t) * cos(nt) dt
Again, using integration by parts with u = -sin t and dv = cos(nt) dt, we get:
an(2) = (1/π) * (-1/n) * (sin(nt))^2 |[π, 2π] = (-1/πn) * (1 - 0) = -1/πn
Since f(t) is even, an(1) = an(2). Therefore, an = 1/πn for n > 0 (excluding n = 0), and an = 0 for n = 0.
Now, let's calculate the Fourier cosine series approximation for t = 1.09, using up to the 6th harmonics:
f(t) ≈ a0/2 + Σ[1,6] an * cos(nt)
Substituting the coefficients a0 = -1/π and an = 1/πn into the formula, we have:
f(t) ≈ (-1/2π) + Σ[1,6] (1/πn) * cos(nt)
Plugging in the values for t = 1.09 and π = 3.142, we can calculate the approximation:
f(1.09) ≈ (-1/2π) + (1/π) * cos(1 * 1.09) + (1/(2π)) * cos(2 * 1.09) + (1/(3π)) * cos(3 * 1.09) + (1/(4π)) * cos(4 * 1.09) + (1/(5π)) * cos(5 * 1.09) + (1/(6π)) * cos(6 * 1.09)
Calculating this expression will give you the Fourier cosine series approximation of f(t) at t = 1.09 up to the 6th harmonics.