Fourier Analysis of Sawtooth Signal with To = 1

In summary, the conversation discusses a sawtooth signal with a period of 1 and a value of 0 at T=0 and a value of 1 at T=1. The equation for the Fourier coefficients is given and the attempt at a solution involves using integration by parts, but there is a sign error and the fact that k is an integer is not taken into account.
  • #1
freezer
76
0

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.
 
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  • #2
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
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)
[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
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
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
k is an integer. What is [itex]\exp(-j2\pi k)[/itex] for k integer?
 
  • #7
Thank you Paallikko, I did not have that one in my notes.
 

FAQ: Fourier Analysis of Sawtooth Signal with To = 1

1. What is Fourier Analysis?

Fourier Analysis is a mathematical technique used to break down complex signals or functions into simpler components. It is based on the concept that any periodic signal can be represented as a combination of sine and cosine waves of different frequencies.

2. What is a sawtooth wave?

A sawtooth wave is a type of periodic signal that resembles the teeth of a saw. It has a linearly increasing or decreasing amplitude with time, and the frequency of the wave remains constant.

3. How is Fourier Analysis used to analyze sawtooth waves?

Fourier Analysis is used to decompose a sawtooth wave into its individual sine and cosine components. This allows us to understand the frequency content and the amplitude of each component, which can then be used to reconstruct the original sawtooth wave.

4. What are the applications of Fourier Analysis in sawtooth waves?

Fourier Analysis is widely used in signal processing, communication systems, and image processing. In sawtooth waves, it is particularly useful in music synthesis, where it is used to generate different tones and create musical sounds.

5. Are there any limitations to using Fourier Analysis on sawtooth waves?

Yes, there are limitations to using Fourier Analysis on sawtooth waves. One major limitation is that the wave must be periodic for the analysis to be accurate. Additionally, the wave must have a finite number of discontinuities, otherwise the Fourier series will not converge.

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