Fourier Analysis of Sawtooth Signal with To = 1

AI Thread Summary
The discussion focuses on the Fourier analysis of a sawtooth signal defined by specific parameters. The participants verify the coefficients a_k for k=0 and k≠0, with calculations showing a_0 equals 1/2 and a formula for a_k when k is not zero. There are concerns about integration by parts and boundary evaluations, with suggestions to correct sign errors and ensure k is treated as an integer. The conversation emphasizes the importance of proper application of integration techniques to derive the Fourier series accurately. Overall, the thread highlights common challenges in Fourier analysis and the need for careful mathematical manipulation.
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Homework Statement



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

verify:
<br /> <br /> a_{k} = \left\{\begin{matrix}<br /> \frac{1}{2}, for k=0; &amp; \\\frac{j}{2\pi k}, for k \neq 0; <br /> &amp; <br /> \end{matrix}\right.<br /> <br />

Homework Equations



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

The Attempt at a Solution



for k = 0

a_{0} = \int_{0}^{1} t dt

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

for k != 0

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

u = t
du = dt
dv = e^(-j2\pi kt)

v = \frac{-1}{j2\pi k}e^{-j2\pi kt}t * \frac{-1}{j2\pi k}e^{-j2\pi kt} - \int \frac{-1}{j2\pi k}e^{-j2\pi kt} dt

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

-1/j = j

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

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

getting close but not seeing where to go from here.
 
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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)
\int_a^b u(x)v&#039;(x)\,dx = \left[u(x)v(x)\right]_a^b - \int_a^b u&#039;(x)v(x)\,dx,
first term on the right hand side.
 
Päällikkö said:
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)
\int_a^b u(x)v&#039;(x)\,dx = \left[u(x)v(x)\right]_a^b - \int_a^b u&#039;(x)v(x)\,dx,
first term on the right hand side.
\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}
 
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.
 
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).
 
k is an integer. What is \exp(-j2\pi k) for k integer?
 
Thank you Paallikko, I did not have that one in my notes.
 
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