# Fourier Analysis - sawtooth

1. Sep 30, 2013

### freezer

1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

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

3. 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.

Last edited: Sep 30, 2013
2. Oct 1, 2013

### Päällikkö

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)
$\int_a^b u(x)v'(x)\,dx = \left[u(x)v(x)\right]_a^b - \int_a^b u'(x)v(x)\,dx$,
first term on the right hand side.

3. Oct 1, 2013

### freezer

$\frac{je^{-j2\pi k} }{2\pi k} - \frac{e^{-j2\pi k} }{4\pi^2 k^2} - \frac{1}{4\pi^2 k^2}$

4. Oct 1, 2013

### Päällikkö

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. Oct 1, 2013

### freezer

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. Oct 1, 2013

### Päällikkö

k is an integer. What is $\exp(-j2\pi k)$ for k integer?

7. Oct 1, 2013

### freezer

Thank you Paallikko, I did not have that one in my notes.