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?
To find the Fourier cosine series approximation of the function f(t) = |sin t|, we need to calculate the coefficients of the cosine terms up to the 6th harmonics. The formula for the Fourier cosine series coefficients is given by:
a0 = (2 / T) * ∫[0 to T] f(t) dt
an = (2 / T) * ∫[0 to T] f(t) * cos((nπt) / T) dt
where T is the period of the function (in this case, T = 2π).
Since f(t) is a periodic function with period T = 2π, the fundamental interval will be from 0 to 2π.
Now, let's calculate the coefficients using the above formulas:
a0 = (2 / 2π) * ∫[0 to 2π] |sin t| dt
= (1 / π) * ∫[0 to π] sin t dt
= (1 / π) * [-cos t] [0 to π]
= (1 / π) * (-cos π + cos 0)
= (1 / π) * (-(-1) + 1)
= 0
Since a0 is 0 for this function, we don't have to calculate the other coefficients (an) since they will be multiplied by cosine terms. However, we can calculate the b coefficients to find the Fourier sine series approximation.
bn = (2 / T) * ∫[0 to T] f(t) * sin((nπt) / T) dt
Let's calculate bn for n = 1 to 6:
b1 = (2 / 2π) * ∫[0 to 2π] |sin t| * sin((1πt) / π) dt
= (1 / π) * ∫[0 to π] sin t * sin t dt
= (1 / π) * ∫[0 to π] sin^2 t dt
= (1 / π) * [t/2 - (sin 2t) / 4] [0 to π]
= (1 / π) * [π/2 - (sin 2π) / 4 - 0]
= 1/2
Similarly, we can calculate b2, b3, b4, b5, and b6. However, since the function f(t) = |sin t| is symmetric, all the bn coefficients will be zero.
Therefore, the Fourier cosine series approximation of f(t) up to the 6th harmonics is:
f(t) ≈ a0/2 + b1*cos(t) + b2*cos(2t) + b3*cos(3t) + b4*cos(4t) + b5*cos(5t) + b6*cos(6t)
≈ 1/2 * cos(t)
Now, to find the approximation of f(t) when t = 1.09, we substitute the value into the Fourier cosine series:
f(1.09) ≈ 1/2 * cos(1.09)
Using a calculator (or programming), evaluate the expression:
f(1.09) ≈ 1/2 * cos(1.09) ≈ 0.361 (rounded to 3 decimal places)
Therefore, the Fourier cosine series approximation of f(t) at t = 1.09 is approximately 0.361.