Fourier Series of cos4t + sin8t

[helderdias]

Hi everyone,

So I was trying to calcule the Fourier Series of x(t) = cos4t + sin8t, but I'm a little bit confused. What would be ω₀ in this case since I have a combination of two functions with different frequencies?

Thank you in advance.

 

[HallsofIvy]

I don't know what you mean by $\omega_0$ here. A Fourier series is a sum
$$\sum_{n=0}^\infty A_n sin(nt)+ B_n cos(nt)$$
Here, obviously $A_8= 1$, $B_4= 1$ and all other coefficients are 0.

 

[helderdias]

I thought the series was the sum of An*cos(nw0t) + Bn*sen(nw0t)

 

[theorem4.5.9]

Typically Fourier series are presented as HallsofIvy posted. However, there's no reason not to include a frequency term $\omega_0$. This can simplify some expressions, and you are free to choose any $\omega_0$ you like. It's important to remember that any choice of $\omega_0$ changes the periodicity to $\frac{2\pi}{\omega_0}$.

For your example, it's clearly best to choose $\omega_0=1$

 

[blue_raver22]

Hello there,

I can provide some insight into your question. In this case, ω0 would be the fundamental frequency, which is the lowest common multiple of the individual frequencies (4 and 8 in this case). This means that ω0 = 4*8 = 32.

To calculate the Fourier Series for this function, you can use the formula:

x(t) = a0 + ∑(an*cos(nω0t) + bn*sin(nω0t))

Where a0 is the DC component, an and bn are the coefficients for the cosine and sine terms respectively, and n is the harmonic number (1, 2, 3...).

In this case, the coefficients can be calculated as follows:

a0 = (1/π) * ∫[0,2π] x(t) dt = 0 (since x(t) is an odd function)

an = (1/π) * ∫[0,2π] x(t)*cos(nω0t) dt = 0 (since x(t) is an odd function)

bn = (1/π) * ∫[0,2π] x(t)*sin(nω0t) dt = (1/π) * ∫[0,2π] (cos4t + sin8t)*sin(nω0t) dt

= (1/π) * ∫[0,2π] (cos4t*sin(nω0t) + sin8t*sin(nω0t)) dt

Using trigonometric identities, we can simplify this to:

bn = (1/2π) * ∫[0,2π] (cos(4-n)ω0t - cos(4+n)ω0t + cos(8-n)ω0t - cos(8+n)ω0t) dt

Since nω0 = n*32 and the integral is over one period (2π), we can further simplify this to:

bn = (1/2) * (δ(n-4) - δ(n+4) + δ(n-8) - δ(n+8))

Where δ is the Dirac delta function.

Therefore, the Fourier Series for x(t) = cos4t + sin8t is:

x(t) = ∑(bn*sin(nω0t))

= (1/2) *