(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

For the periodic signal

[tex]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)[/tex]

Find the exponential Fourier series.

2. Relevant equations

Euler’s Formula

[tex]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][/tex]

3. The attempt at a solution

To get [itex]\omega_0[/itex], we need to find the least common denominator between the following periods…

[tex]\frac{2\,\pi}{3},\,2\,\pi,\,\frac{\pi}{2}[/tex]

Which is [itex]2\,\pi[/itex].

So, now I use the formula [itex]\omega_0\,=\,\frac{2\,\pi}{T}[/itex]…

[tex]\omega_0\,=\,\frac{2\,\pi}{2\,\pi}\,=\,1[/tex]

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

[tex]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][/tex]

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

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

I assumed the latter, is that correct?

**Physics Forums - The Fusion of Science and Community**

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

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

**Physics Forums - The Fusion of Science and Community**