Fourier Transfer of sawtooth function

In summary, the Fourier transform of h(t)=t over the interval 0,1 is given by H(s)=1/(s^2)*(i*(s*cos(s)-sin(s))+(s*sin(s)+cos(s)-1)), where s=-i*2*pi*f=-i*w.
  • #1
hotvette
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Homework Statement


Find the magnitude and phase of the Fourier transform of [itex]h(t)=t[/itex] over the interval 0,1

Homework Equations


[itex]H(s) = \int^{1}_{0} h(t) e^{-i 2 \pi f t} dt[/itex]

The Attempt at a Solution


I found this thread: https://www.physicsforums.com/threads/fourier-transform-of-the-sawtooth-function.502113/ but can't seem to come close to the same result. Letting [itex] a = -i 2 \pi f = -i \omega [/itex] we get:
##H(s) = \int^{1}_{0} t e^{a t} dt \\
= \frac{e^{at}}{a^2}(at-1)|^{1}_{0} \\
= \frac{1}{a^2} \left( a t e^{at} - e^{at}\right)|^{1}_{0} \\
= \frac{1}{a^2}(ae^{a} - e^{a} + 1) \\
= \frac{1}{a^2}\left[(a-1) e^{a} + 1 \right] \\
= \frac{1}{a^2}(ae^{a} - e^{a} + 1) \\
= \frac{1}{(-i \omega)^2}\left[(-i \omega-1) e^{-i \omega} + 1 \right] \\
= \frac{-1}{\omega^2}\left[(-i \omega-1) e^{-i \omega} + 1 \right] \\
= \frac{1}{\omega^2}\left[(i \omega+1) e^{-i \omega} - 1 \right] \\
= \frac{1}{\omega^2}\left[(i \omega+1) (\cos \omega - i \sin \omega) - 1 \right] \\
\vdots \\
= \frac{1}{\omega^2} \left[ i (\omega\cos\omega-\sin \omega) + (\omega \sin \omega + \cos \omega -1)\right]##
I just don't see how this is converging to the solution in the referenced thread. Have I gone wrong somewhere?
 
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  • #2
hotvette said:

Homework Statement


Find the magnitude and phase of the Fourier transform of [itex]h(t)=t[/itex] over the interval 0,1

Homework Equations


[itex]H(s) = \int^{1}_{0} h(t) e^{-i 2 \pi f t} dt[/itex]

The Attempt at a Solution


I found this thread: https://www.physicsforums.com/threads/fourier-transform-of-the-sawtooth-function.502113/ but can't seem to come close to the same result. Letting [itex] a = -i 2 \pi f = -i \omega [/itex] we get:
##H(s) = \int^{1}_{0} t e^{a t} dt \\
= \frac{e^{at}}{a^2}(at-1)|^{1}_{0} \\
= \frac{1}{a^2} \left( a t e^{at} - e^{at}\right)|^{1}_{0} \\
= \frac{1}{a^2}(ae^{a} - e^{a} + 1) \\
= \frac{1}{a^2}\left[(a-1) e^{a} + 1 \right] \\
= \frac{1}{a^2}(ae^{a} - e^{a} + 1) \\
= \frac{1}{(-i \omega)^2}\left[(-i \omega-1) e^{-i \omega} + 1 \right] \\
= \frac{-1}{\omega^2}\left[(-i \omega-1) e^{-i \omega} + 1 \right] \\
= \frac{1}{\omega^2}\left[(i \omega+1) e^{-i \omega} - 1 \right] \\
= \frac{1}{\omega^2}\left[(i \omega+1) (\cos \omega - i \sin \omega) - 1 \right] \\
\vdots \\
= \frac{1}{\omega^2} \left[ i (\omega\cos\omega-\sin \omega) + (\omega \sin \omega + \cos \omega -1)\right]##
I just don't see how this is converging to the solution in the referenced thread. Have I gone wrong somewhere?

Your answer is correct; you have not gone wrong anywhere that I can see.
 

Related to Fourier Transfer of sawtooth function

1. What is a sawtooth function?

A sawtooth function is a mathematical function that resembles the shape of a saw blade, with a linear increase followed by a sudden drop to its starting point. It is a periodic function, meaning it repeats itself after a certain interval.

2. What is the Fourier Transform of a sawtooth function?

The Fourier Transform of a sawtooth function is a continuous function that represents the amplitude and phase of the different frequency components that make up the original sawtooth function. It is a way to decompose a complex function into simpler sinusoidal components.

3. How is the Fourier Transform of a sawtooth function calculated?

The Fourier Transform of a sawtooth function can be calculated using the Fourier Transform formula, which involves integrating the original function with respect to frequency. This process is often carried out using specialized software or mathematical tools such as Fourier series or Laplace transforms.

4. What are the applications of the Fourier Transform of a sawtooth function?

The Fourier Transform of a sawtooth function has various applications in signal processing, communications, and image processing. It is used to analyze and manipulate signals in the frequency domain, which can help with tasks such as filtering, noise reduction, and compression.

5. Are there any limitations to the Fourier Transform of a sawtooth function?

While the Fourier Transform is a powerful tool for analyzing signals, it has some limitations when applied to sawtooth functions. For example, the Fourier Transform assumes that the function is continuous and infinite, which may not always be the case for real-world signals. Additionally, the Fourier Transform may not accurately represent sharp changes or discontinuities in the original sawtooth function.

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