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Fourier Analysis - sawtooth

  • Thread starter freezer
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  • #1
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Homework Statement



Sawtooth signal with To = 1, at T=0, x = 0, at T=1, x =1

verify:
[itex]

a_{k} = \left\{\begin{matrix}
\frac{1}{2}, for k=0; & \\\frac{j}{2\pi k}, for k \neq 0;
&
\end{matrix}\right.

[/itex]


Homework Equations



[itex]\frac{1}{T_{0}} \int_{0}^{T_{0}} te^{-j(2\pi/T_{0}))kt}dt[/itex]

The Attempt at a Solution



for k = 0

[itex] a_{0} = \int_{0}^{1} t dt[/itex]

[itex]a_{0} = \frac{1}{2} t^{2} [/itex] from 0 to 1 = 1/2

for k != 0

[itex]\int_{0}^{1} te^{-j(2\pi) kt}dt[/itex]

u = t
du = dt
dv = [itex]e^(-j2\pi kt)[/itex]

[itex] v = \frac{-1}{j2\pi k}e^{-j2\pi kt} [/itex]


[itex] t * \frac{-1}{j2\pi k}e^{-j2\pi kt} - \int \frac{-1}{j2\pi k}e^{-j2\pi kt} dt[/itex]

[itex] t * \frac{-1}{j2\pi k}e^{-j2\pi kt} - \frac{e^{-j2\pi kt}}{4\pi^2k^2} [/itex]

-1/j = j

[itex]t * \frac{j}{2\pi k}e^{-j2\pi kt} - \frac{e^{-j2\pi kt}}{4\pi^2k^2}[/itex]

[itex]e^{-j2\pi kt} (t \frac{j}{2\pi k} - \frac{1}{4\pi^2 k^2})[/itex]

getting close but not seeing where to go from here.
 
Last edited:

Answers and Replies

  • #2
Päällikkö
Homework Helper
517
10
Check the integration by parts rules: You seem to have forgotten to evaluate the first part at the boundaries (in particular, if you integrate over t from 0 to 1, there is no way t should remain in the final expression)
[itex] \int_a^b u(x)v'(x)\,dx = \left[u(x)v(x)\right]_a^b - \int_a^b u'(x)v(x)\,dx[/itex],
first term on the right hand side.
 
  • #3
76
0
Check the integration by parts rules: You seem to have forgotten to evaluate the first part at the boundaries (in particular, if you integrate over t from 0 to 1, there is no way t should remain in the final expression)
[itex] \int_a^b u(x)v'(x)\,dx = \left[u(x)v(x)\right]_a^b - \int_a^b u'(x)v(x)\,dx[/itex],
first term on the right hand side.


[itex] \frac{je^{-j2\pi k} }{2\pi k} - \frac{e^{-j2\pi k} }{4\pi^2 k^2} - \frac{1}{4\pi^2 k^2}[/itex]
 
  • #4
Päällikkö
Homework Helper
517
10
You seem to have a sign error. Also, remember that k is an integer (a periodic function is mapped into a series in fourier space), and you should be able to arrive at the result.
 
  • #5
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Okay, see the sign error but still not seeing how that is going to get
the other terms to fall out leaving just j/(2pik).
 
  • #6
Päällikkö
Homework Helper
517
10
k is an integer. What is [itex]\exp(-j2\pi k)[/itex] for k integer?
 
  • #7
76
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Thank you Paallikko, I did not have that one in my notes.
 

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