## Contents

# 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) . \ $

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!

### Answer 1

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} $

### Answer 2

$ 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 $

### Answer 3

I think that problem will get much easier if you notice that

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