Finding Exponential Fourier Series for a Periodic Signal

[VinnyCee]

Homework Statement

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.

Homework 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]

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}\,=\,1Now, 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?

 

[vela]

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

I think you meant "least common multiple."

Anyway, your approach seems unnecessarily complicated. Other than the constant term, the Fourier series has terms of the form $\cos n\omega_0 t$ and $\sin n\omega_0 t$. You can see, by inspection, that $\omega_0=1~\rm rad/s$ for the signal you were given.

Now, I use Euler’s formula to convert the cos and sin to exponentials…
x(t)=2+\frac 12 \left[e^{j(t+45^\circ)}+e^{-j(t+45^\circ)}\right]+2\left[e^{j(3t)}+e^{-j(3t)}\right]-2\left[e^{j(4t-60^\circ)}+e^{-j(4t-60^\circ)}\right]I don’t know if the last term (sin) is supposed to be kept as $4t+30^\circ$ OR changed to a cosine to fit Euler’s formula by subtracting ninety degrees: $4t-60^\circ$. I assumed the latter. Is that correct?

Yes, that's correct.

Alternatively, you could use the expression for sine in terms of complex exponentials:
$$\sin \theta = \frac{e^{j\theta}-e^{-j\theta}}{2j}.$$