Fourier Series of Sawtooth Wave from Inverse FT

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The discussion focuses on deriving the Fourier series of a sawtooth wave using inverse Fourier transform techniques. The user attempts to calculate the Fourier coefficients but finds discrepancies in their results, particularly regarding the presence of sine terms due to the odd nature of the sawtooth function. There is confusion about the application of a low pass rectangular filter and the role of the delta function in the Fourier transform. The user seeks clarification on eliminating the imaginary unit in their final expression and requests visual aids for better understanding. The conversation emphasizes the importance of consistent notation and clear mathematical derivations in solving the problem.
roam
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Homework Statement


I want to find the Fourier series of the sawtooth function in terms of real sine and cosine functions by using the formula:

$$f_p (t)=\sum^\infty_{k=-\infty} c_k \exp \left(j2\pi \frac{k}{T}t \right) \tag{1}$$

This gives the Fourier series of a periodic function, with the coefficients:

$$c_k = \frac{1}{T} F \left( \frac{k}{T} \right) \tag{2}$$

where the capital letter denotes the Fourier transform.

Homework Equations



Equation (1) and (2) above are found by considering the IFT which recovers ##f_p## from ##F_p##:

$$\intop^\infty_{-\infty} F_p (\nu) e^{j2\pi \nu t} \ d\nu = \intop^\infty_{-\infty} \frac{1}{T} \sum^\infty_{-\infty} F \left( \frac{k}{T} \right) \delta(\nu - \frac{k}{T}) e^{j2\pi \nu t} \ d\nu=\sum^\infty_{-\infty} \frac{1}{T} F \left( \frac{k}{T} \right) e^{j2\pi \frac{k}{T} t}$$

The Attempt at a Solution


[/B]
By using a low pass rectangular filter, a single period of the sawtooth function is given by

$$f(t)=t\Pi\left(\frac{t}{T}\right)$$

Since we have the following Fourier transform pair:

$$t \leftrightarrow \frac{j \delta'(\nu)}{2 \pi}$$

We can write the FT of a single period of the sawtooth wave as:

$$F(\nu)=\frac{j\delta'\left(\nu\right)}{2\pi}*T\ sinc\left(\nu T\right)=\frac{jT}{2\pi}\ sinc^{\prime}(\nu T)=\left(\frac{jT}{2\pi}\right)\left(\frac{\cos(\pi\nu T)}{\nu T}-\frac{\sin(\pi\nu T)}{\pi\nu^{2}T^{2}}\right)$$

Using equation (2), we get the coefficients:

$$c_{k}=\frac{1}{T}\left(\frac{jT}{2\pi}\right)\left(\frac{\cos(\pi\frac{k}{T}T)}{\frac{k}{T}T}-\frac{\sin(\pi\frac{k}{T}T)}{\pi\left(\frac{k}{T}\right)^{AC2}T^{2}}\right)=\frac{j}{2\pi}\left(\frac{\cos(\pi k)}{k}-\frac{\sin(\pi k)}{\pi k^{2}}\right).$$

And therefore, the Fourier series becomes:

$$f_{p}(t)=c_{k}\ e^{j2\pi\frac{k}{T}t}=\frac{j}{2\pi}\left(\frac{\cos(\pi k)}{k}-\frac{\sin(\pi k)}{\pi k^{2}}\right)e^{j2\pi\frac{k}{T}t}$$

But this does not look correct (it is very different than the Fourier series of the sawtooth given here). Since the sawtooth function is odd, I think we must only have the sine terms present. What is wrong here?

Also, because I need to plot this function, how can I get rid of the ##j## terms?

Any help would be appreciated.
 
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roam said:
using a low pass rectangular filter, a single period of the sawtooth function is given by

f(t)=tΠ(t/T)​
Can you explain this ? With some pictures preferably ? It doesn't look to me like ##s(x) ={x\over \pi}## at all ?
 
Last edited:
Where'd the delta function come from?? it isn't part of the expression you derived for the coefficients. Notation, notation, notation, be consistent and explicit. Do the integrals and the answer will pop out quickly, it isn't a hard problem to solve.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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