Sawtooth function Fourier transform

  • #1
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Homework Statement


For a periodic sawtooth function ##f_p (t) = t## of period ##T## defined over the interval ##[0, T]##, calculate the Fourier transform of a function made up of only a single period of ##f_p (t),## i.e.

$$f(t)=\left\{\begin{matrix}f_p (t) \ \ 0<t<T\\0 \ \ elsewhere \end{matrix}\right.$$

Use the result that

$$FT \Big[ f(t) = t \Big] = \frac{j \delta' (\nu)}{2 \pi}$$

Homework Equations




The Attempt at a Solution



I am not sure how to approach this problem. I think somehow we need to evaluate ##f(\nu) = \frac{j \delta' (\nu)}{2 \pi}## over the interval ##[0, T]##, but it's not possible since we are in Fourier space. If this is the case, do we need to write the interval in terms of the corresponding frequency (##[\nu = 0, \ 1/T ]##)?

Any help would be greatly appreciated.
 

Answers and Replies

  • #2
blue_leaf77
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Note that ##f(t)## can be obtained by multiplying a function ##y_1(t)=t## and a top hat function
$$
y_2(t)=\left\{\begin{matrix}1 \ \ 0<t<T\\0 \ \ elsewhere \end{matrix}\right.
$$
i.e. ##f(t)=y_1(t)y_2(t)##. The Fourier transform of ##f(t)## is then the convolution between ##\tilde{y}_1(\nu)## and ##\tilde{y}_2(\nu)##. Now, you have been given ##\tilde{y}_1(\nu)##, which is equal to ##\frac{j \delta' (\nu)}{2 \pi}##, and your task now is to find ##\tilde{y}_2(\nu)##.
 
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  • #3
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Thank you very much for this hint. So if I write the answer as:

$$F(\nu)=\left\{\begin{matrix}\frac{j \delta'(\nu)}{2 \pi} \ \ 0<t<T\\0 \ \ elsewhere \end{matrix}\right.$$

is that alright?
 
  • #4
blue_leaf77
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As I said, the Fourier transform of ##f(t)## will be a convolution between the transform of the individual functions in the time domain. The one you proposed above doesn't seem to be one. Have you calculated the transform of ##y_1## and ##y_2##?
 
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  • #5
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I see. So for the case ##0<t<T## the convolution would be:

$$\tilde{y_1} * \tilde{y_2} = \int^T_0 \tilde{y_1} (\tau) \tilde{y_2} (t- \tau) \ d\tau = \frac{j \delta' (\nu)}{2 \pi} \int^T_0 1 \ d \tau = \frac{j \delta' (\nu)}{2 \pi} T$$

So, we have

$$F(\nu)=\left\{\begin{matrix}\frac{j \delta'(\nu)}{2 \pi} T \ \ 0<t<T\\0 \ \ elsewhere \end{matrix}\right.$$

Is this correct now?
 
  • #6
vela
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No, it's not correct. The Fourier transforms are functions of frequency, not time.
 
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  • #7
blue_leaf77
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I see. So for the case ##0<t<T## the convolution would be:

$$\tilde{y_1} * \tilde{y_2} = \int^T_0 \tilde{y_1} (\tau) \tilde{y_2} (t- \tau) \ d\tau = \frac{j \delta' (\nu)}{2 \pi} \int^T_0 1 \ d \tau = \frac{j \delta' (\nu)}{2 \pi} T$$

So, we have

$$F(\nu)=\left\{\begin{matrix}\frac{j \delta'(\nu)}{2 \pi} T \ \ 0<t<T\\0 \ \ elsewhere \end{matrix}\right.$$

Is this correct now?
No, that's wrong. What the convolution theorem states is that when you have two functions ##y_1(t)## and ##y_2(t)## multiplied in the time domain ##f(t)=y_1(t)y_2(t)##, then Fourier transform of the product function in frequency domain will be a convolution of the form
$$
\tilde{f}(\nu) = FT[f(t)] = \int_{-\infty}^\infty \tilde{y}_1(\nu')\tilde{y}_2(\nu-\nu') d\nu'
$$
 
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  • #8
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Since ##\tilde{y}_2## is equal to ##1## on (0, T) the convolution would become

$$\int_{-\infty}^\infty \tilde{y}_1(\nu')\tilde{y}_2(\nu-\nu') d\nu' = \frac{j}{2 \pi} \int_{-\infty}^\infty \delta' (\nu') . 1 \ d \nu'=- \frac{j}{2 \pi} \Big[ \delta(\nu') \Big]^{\infty}_{-\infty}$$

But the last expression is equal to zero since the Dirac delta is zero everywhere other than on ##0## (or alternatively it equals to ##1## if 0 is contained within the integration limits).

So ##\tilde{f} (\nu) = 0## in this case? Is that right?
 
  • #9
blue_leaf77
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Since ~y2y~2\tilde{y}_2 is equal to 111 on (0, T) the convolution would become
No. It's ##y_2(t)## which is unity in the interval [0,T], not ##\tilde{y}_2(\nu)## which is the Fourier transform of ##y_2(t)##.
 
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  • #10
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No. It's ##y_2(t)## which is unity in the interval [0,T], not ##\tilde{y}_2(\nu)## which is the Fourier transform of ##y_2(t)##.
Thanks, so I know that the Fourier transform of the uniform function is equal to ##\delta(\nu)##. So the convolution integral becomes:

$$\frac{j}{2 \pi} \int^\infty_{-\infty} \delta ' (\nu ') \ \delta (\nu - \nu') d \nu '$$

I know that there is a property that says: ##\int^\infty_{-\infty} \delta' (\nu) \phi (\nu ') d \nu' = - \phi' (0)##. Is it possible to use this? Or do I need to proceed with integration by parts?
 
  • #11
blue_leaf77
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so I know that the Fourier transform of the uniform function is equal to δ(ν)
##y_2(t)## is not a constant function, it has top hat form and hence its Fourier transform is not equal to a delta function.
 
  • #12
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##y_2(t)## is not a constant function, it has top hat form and hence its Fourier transform is not equal to a delta function.
What do you mean? Because you said earlier that it is equal to unity in the interval [0, T], so the Fourier transform of 1 is just a delta function.
 
  • #13
blue_leaf77
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Look at the form of ##y_2(t)## in post#2. Does that look like a constant function over the entire time axis?
 
  • #14
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In post #2 you have ##y_2(t) =1## which is a constant over the interval under consideration (##0<t<T##). So I looked the FT only in this region.

But if we consider the time axis outside this region, the function looks like a rectangular pulse function of duration ##T## (in our course it is denoted by ##\Pi##). In this case it would be ##\Pi(\frac{t}{T})## whose FT is ##T sinc(T \nu)##. Is this what you meant?
 
  • #15
blue_leaf77
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Is this what you meant?
Yes. However, there is a small difference. The rectangular pulse we have in this problem is not centered with respect to the origin. If you know the shifting theorem of Fourier transform, this can help you find the corresponding transform.
 
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  • #16
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Yes. However, there is a small difference. The rectangular pulse we have in this problem is not centered with respect to the origin. If you know the shifting theorem of Fourier transform, this can help you find the corresponding transform.
Okay so if we rewrite the rectangular function as centered at T/2:

$$\Pi \left( \frac{t- (\frac{T}{2})}{T} \right)$$

Using the shifting property (##f(t-t_0) \iff e^{-j 2 \pi \nu t_0} F(\nu)##) the FT becomes: ##T \ sinc (T \nu) e^{-j 2 \pi (T/2)}.## So

$$\tilde{f} = \frac{jT}{2 \pi} \int^{\infty}_{- \infty} \delta ' (\nu ') \ sinc (T \nu) e^{-j 2 \pi \nu (T/2)} \ d \nu'$$

I am not sure how to proceed from here. The delta function still becomes zero over this integration limit. Are there any special properties we could use?
 
  • #17
blue_leaf77
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It should be
$$
\tilde{f} = \frac{jT}{2 \pi} \int^{\infty}_{- \infty} \delta ' (\nu ') \ sinc (T (\nu-\nu')) e^{-j 2 \pi (\nu-\nu') (T/2)} \ d \nu'
$$
writing ##\delta ' (\nu ')d \nu' = d\delta(\nu ')##, that integral becomes
$$
\tilde{f} = \frac{jT}{2 \pi} \int^{\infty}_{- \infty} \ sinc (T (\nu-\nu')) e^{-j 2 \pi (\nu-\nu') (T/2)} \ d\delta(\nu ')
$$
At this point, consider using integration by parts.
 
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  • #18
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Thank you so much for the help.
 

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