# Practice Question on Computing the Fourier Series coefficients of a sine wave

Obtain the Fourier series coefficients of the CT signal

$x(t) = \sin \left(3\pi t + \frac{\pi}{2} \right) . \$

for $sin(t)$, the coefficients are $a_1=\frac{1}{2j},a_{-1}=\frac{-1}{2j}, a_k=0 \mbox{ for }k\ne 1,-1$

Time shift property: $x(t-t_0) \to e^{-jkw_0t_0}a_k$

Thus with $w_0=3\pi\,$ and $t_0=\frac{-\pi}{2}$,

$a_1=\frac{e^{j 3 \pi \frac{\pi}{2}}}{2j},a_{-1}=\frac{-e^{-j 3 \pi \frac{\pi}{2}}}{2j}, a_k=0 \mbox{ for }k\ne 1,-1$

Is that right? I'm not sure about the time shift property.

--Cmcmican 21:09, 7 February 2011 (UTC)

Instructor's comment: we will see the time shifting property later. Can you solve the problem without it? Perhaps you could write sin(u) as a sum of two exponentials, and then replace u by what is inside the sine. You should be able to factor out the phase as a separate exponential (a constant) in front of a complex exponential function. -pm

So like this?

$sin(t)=\frac{1}{2j}e^{jkw_0t}-\frac{1}{2j}e^{-jkw_0t}$

$x(t)=\frac{1}{2j}e^{jk3\pi(t+\frac{\pi}{2})}-\frac{1}{2j}e^{-jk3\pi(t+\frac{\pi}{2})}$

therefore,

$a_1=\frac{e^{j 3 \pi \frac{\pi}{2}}}{2j},a_{-1}=\frac{-e^{-j 3 \pi \frac{\pi}{2}}}{2j}, a_k=0 \mbox{ for }k\ne 1,-1$

--Cmcmican 08:23, 8 February 2011 (UTC)

TA's comment: I think you still have a mistake in your answer. As Prof. Boutin noted above, the phase should factor out.
Hint: Euler's formula is: $\sin(\theta)=\frac{1}{2j}e^{j\theta}-\frac{1}{2j}e^{-j\theta}$

$x(t)=\sin(3 \pi t + \frac{\pi}{2}) = \frac{e^{j (3 \pi t + \frac{\pi}{2}) } - e^{-j(3 \pi t + \frac{\pi}{2})}}{2j}$

$= \frac{1}{2j}e^{j3\pi t}e^{j\frac{\pi}{2}} - \frac{1}{2j}e^{-j3\pi t}e^{-j\frac{\pi}{2}}$

Since:

• $e^{j\frac{\pi}{2}} = j$
• $e^{-j\frac{\pi}{2}} = -j$

We have:

$\frac{1}{2j}(j)e^{j3t\pi} - \frac{1}{2j}(-j)e^{-j3t\pi} \rightarrow \frac{1}{2}e^{(1)j3t\pi} + \frac{1}{2}e^{(-1)j3t\pi}$

$a_{-1} = \frac{1}{2}, a_{1} = \frac{1}{2}, a_{k} = 0 \text{ for }k \neq -1,1$

$x(t) = \sin \left(3\pi t + \frac{\pi}{2} \right) = \cos \left(3\pi t \right) . \$.