Fourier series - trig and complex not matching?

  • Thread starter haydez98
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  • #1
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fourier series - trig and complex not matching!?!

i am given a signal which can be written as:
s(t) = -1 {-1 < t < 0}, 1 {0 < t < 1}, 0 {1 < t < 2} [it's a pulse train]
the period, T, is 3.
i have calculated the trig. fourier series representation, which in matlab turns out to be correct, yet when i calculate the exponentical fsr, i get a version of the trig. fsr which has its amplitude halved.

for the trig fsr:

s(t) = 2/(pi * n) * (1 - cos((2 * pi * n)/3)) * sin((2 * pi * n * t)/3);


for the exp fsr:

s(t) = -1/(i * pi * n) * (cos((2 * pi * n)/3) - 1) * exp((i * 2 * pi * n * t)/3)


i also tried

c_n = 0.5 (a_n - i * b_n) = -0.5 * i * ( 2/(pi * n) * (1 - cos((2 * pi * n)/3))



either case, my complex fsr was a scaled amplitude version of my trig fsr...when i get rid of the 0.5 it turns out to be right, but why would i get rid of the 0.5?


thanks in advance
 

Answers and Replies

  • #2
525
7


Hint:
[tex] \sin x = \frac{e^x - e^{-x}}{2 i}[/tex]
 
  • #3
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okay i'll show you my working out...
[tex]
c_{n}=\frac{1}{T}\int_{0}^{T}s(t)e^{\frac{-j2\pi nt}{T}}dt[/tex]
[tex]=\frac{1}{T}\left [\int_{-T/3}^{0}-1e^{\frac{-j2\pi nt}{T}}dt + \int_{0}^{T/3}1e^{\frac{-j2\pi nt}{T}}dt \right ][/tex]
[tex]=\frac{1}{T} \left[\left[\frac{T}{j2\pi n}e^{\frac{-j2\pi nt}{T}} \right]_{-T/3}^{0} +\left[\frac{-T}{j2\pi n}e^{\frac{-j2\pi nt}{T}} \right]_{0}^{T/3}\right][/tex]
[tex]=\frac{1}{j2\pi n}\left[1 - e^{\frac{j2\pi n}{3}}+e^{\frac{-j2\pi n}{3}}+1\right][/tex]
[tex]=\frac{1}{j2\pi n}\left[2-2cos(\frac{2\pi n}{3})\right][/tex]
[tex]=\frac{1}{j\pi n}\left[1-cos(\frac{2 \pi n}{3})\right]
[/tex]

(note: i used the cosine identity)

Therefore:
[tex]
s(t)= \frac{1}{j\pi n}(1-cos(\frac{2 \pi n}{3}))e^{\frac{j\pi nt}{3}}
[/tex]

However, when I put this into matlab, it doesn't satisfy the trig fsr.
 
Last edited:
  • #4
525
7


Almost everything looks fine. Except, well, you still gotta sum over n... ;)

So:
[tex]s(t) = \sum_{n=-\infty,\neq 0}^{\infty} c_n e^{j\pi n t / 3}[/tex]
where
[tex] c_n = \frac{1}{j\pi n}(1-\cos \frac{2\pi n}{3})[/tex]
Note also that [tex]c_0[/tex] is left out (in other words, [tex]c_0=0[/tex], do you see why?)

Now we see that
[tex]c_{-n} = -c_n[/tex]
so the sum can be rewritten as:
[tex]s(t) = \sum_{n=1}^{\infty} c_n e^{j\pi n t / 3}+c_{-n}e^{-j\pi n t / 3}
= \sum_{n=1}^{\infty} c_n e^{j\pi n t / 3}-c_{n}e^{-j\pi n t / 3}
=\sum_{n=1}^{\infty} c_n \left(e^{j\pi n t / 3}-e^{-j\pi n t / 3}\right)
=\sum_{n=1}^{\infty} 2i c_n\sin{j\pi n t / 3} [/tex]

Which is the trig fsr.
 
Last edited:
  • #5
9
0


ah coolness. i didn't know about having to put c_-n in it too

thanks so much
 

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