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:Hint: Euler's formula is: <math class="inline">\sin(\theta)=\frac{1}{2j}e^{j\theta}-\frac{1}{2j}e^{-j\theta}</math> | :Hint: Euler's formula is: <math class="inline">\sin(\theta)=\frac{1}{2j}e^{j\theta}-\frac{1}{2j}e^{-j\theta}</math> | ||
===Answer 2=== | ===Answer 2=== | ||
| − | + | <math> 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} </math> | |
| + | |||
| + | <math> = \frac{1}{2j}e^{j3\pi t}e^{j\frac{\pi}{2}} - \frac{1}{2j}e^{-j3\pi t}e^{-j\frac{\pi}{2}}</math> | ||
| + | |||
| + | Since: | ||
| + | |||
| + | *<math> e^{j\frac{\pi}{2}} = j</math> | ||
| + | *<math> e^{-j\frac{\pi}{2}} = -j</math> | ||
| + | |||
| + | We have: | ||
| + | <math> \frac{1}{2}e^{j3t\pi} + \frac{1}{2}e^{-j3t\pi} \rightarrow \frac{1}{2}e^{(1)j3t\pi} + \frac{1}{2}e^{(-1)j3t\pi}</math> | ||
| + | |||
| + | <math> a_{-1} = \frac{1}{2}, a_{1} = \frac{1}{2}, a_{k} = 0 \text{ for }k \neq -1,1 </math> | ||
===Answer 3=== | ===Answer 3=== | ||
Write it here. | Write it here. | ||
---- | ---- | ||
[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | [[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
Revision as of 17:57, 15 February 2011
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}{2}e^{j3t\pi} + \frac{1}{2}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
Write it here.
