Fourier Series of Sawtooth Wave from Inverse FT

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SUMMARY

The discussion focuses on deriving the Fourier series of the sawtooth wave using the inverse Fourier transform (IFT) and specific equations. The Fourier coefficients are calculated using the formula \(c_k = \frac{1}{T} F \left( \frac{k}{T} \right)\), leading to the expression for the Fourier series \(f_{p}(t)=c_{k}\ e^{j2\pi\frac{k}{T}t}\). The user identifies an inconsistency in the expected sine terms due to the odd nature of the sawtooth function and seeks clarification on the presence of \(j\) terms in the final expression. The discussion emphasizes the importance of consistent notation and clear derivation steps.

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Students and professionals in electrical engineering, applied mathematics, and physics who are working with signal processing and Fourier analysis, particularly those focusing on waveforms like the sawtooth function.

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.
 

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