# Find the exponential Fourier series of x(t) = 2 + 0.5cos(t+45)+2cos(3t)-2sin(4t+30)

1. ### VinnyCee

492
1. The problem statement, all variables and given/known data

For the periodic signal

$$x(t)\,=\,2\,+\,\frac{1}{2}\,cos\left(t\,+\,45^{\circ}\right)\,+\,2\,cos\left(3\,t\right)\,-\,2\,sin\left(4\,t\,+\,30^{\circ}\right)$$

Find the exponential Fourier series.

2. Relevant equations

Euler’s Formula
$$x(t)\,=\,A\,cos\left(\omega_0\,t\,+\,\phi\right)\,=\,A\,\left[e^{j\,\left(\omega_0\,t\,+\,\phi\right)}\,+\, e^{-j\,\left(\omega_0\,t\,+\,\phi\right)}\right]$$

3. The attempt at a solution

To get $\omega_0$, we need to find the least common denominator between the following periods…

$$\frac{2\,\pi}{3},\,2\,\pi,\,\frac{\pi}{2}$$

Which is $2\,\pi$.

So, now I use the formula $\omega_0\,=\,\frac{2\,\pi}{T}$…

$$\omega_0\,=\,\frac{2\,\pi}{2\,\pi}\,=\,1$$

Now, I use Euler’s formula to convert the cos and sin to exponentials…

$$x(t)\,=\,2\,+\,\frac{1}{2}\,\left[e^{j\left(t\,+\,45^{\circ}\right)}\,+\,e^{-j\left(t\,+\,45^{\circ}\right)\right]\,+\,2\,\left[e^{j\left(3\,t\right)}\,+\,e^{-j\left(3\,t\right)}\right]\,-\,2\left[e^{j\left(4\,t\,-\,60^{\circ}\right)}\,+\,e^{-j\left(4\,t\,-\,60^{\circ}\right)}\right]$$

I don’t know if the last term (sin) is supposed to be kept as $$4\,t\,+\,30^{\circ}$$

OR changed to a cosine to fit Euler’s formula by subtracting ninety degrees: $$4\,t\,-\,60^{\circ}$$

I assumed the latter, is that correct?