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Fourier Transform (Triangular Pulse)

  1. Jan 6, 2016 #1
    1. The problem statement, all variables and given/known data
    What is the Fourier transform of the function graphed below?

    triangularpulse.jpg

    According to some textbooks the Fourier transform for this function must be:

    $$ab \left( \frac{sin(\omega b/2)}{\omega b /2} \right)^2$$

    2. Relevant equations


    3. The attempt at a solution

    I believe this triangular pulse is given by:

    $$x(t)= \left\{\begin{matrix} a- \frac{|t|}{b} \ \ \ if |t|<b \\ 0 \ \ \ if |t|>b \end{matrix}\right.$$

    So we need to find the sum of the two integrals:

    $$x_1(\omega)= \int^0_{-b} \left(a+ \frac{|t|}{b} \right) e^{-j \omega t} dt$$

    $$x_2(\omega)= \int^b_0 \left(a- \frac{|t|}{b} \right) e^{-j \omega t} dt$$

    So using integration by parts we evaluate the two integrals:

    ##u=a + \frac{t}{b}##, ##\frac{du}{dt}=\frac{1}{b}##, ##dv=e^{-j\omega t}##, ##v=\frac{j}{\omega} e^{-j\omega t}##

    $$x_1 (\omega)=[\frac{j}{\omega} (a+ \frac{t}{b} e^{-j\omega t}) ]^0_{-b} - \int^0_{-b} \frac{j}{\omega b} e^{-j \omega t}$$

    ##\therefore x_1(\omega)= \frac{ja}{\omega}-\frac{ja}{\omega}+\frac{j}{\omega}e^{+j\omega b} + \frac{1}{\omega^2b} - \frac{1}{\omega^2 b} e^{+j \omega b}## (1)

    Second integral:

    $$x_2 (\omega)=[\frac{j}{\omega} (a- \frac{t}{b}) e^{-j \omega t} ]^b_0 - \int^b_0 (-\frac{1}{b}) . \frac{j}{\omega} e^{-j \omega t}$$

    ##\therefore \ x_2(\omega)= \frac{ja}{\omega} e^{-j\omega b} - \frac{j}{\omega} e^{-j \omega b} - \frac{j}{\omega} a - \frac{1}{\omega^2 b} e^{-j \omega b} + \frac{1}{\omega^2 b}## (2)

    Combining (1) and (2):

    $$x(\omega)=\frac{-ja}{\omega}+\frac{2}{\omega^2 b} + \frac{j}{\omega} (e^{+j \omega b} - e^{-j\omega b}) - \frac{1}{\omega^2 b} (e^{+j \omega b} - e^{-j\omega b}) + \frac{ja}{\omega} e^{-j\omega b}$$

    Now writing this in terms of sines and cosines using Euler's formula:

    $$x(\omega)=\frac{-ja}{\omega}+\frac{2}{\omega^2 b} + \frac{j}{\omega} (2j sin (\omega b)) - \frac{1}{\omega^2 b} (2 j sin (\omega b)) + \frac{ja}{\omega} (cos (- \omega b)+ j sin (- \omega b))$$

    $$x(\omega)=\frac{-ja}{\omega}+\frac{2}{\omega^2 b} + \frac{ja}{\omega} cos (\omega b) + \frac{a}{\omega} sin (\omega b) - \frac{2}{\omega} sin (\omega b) - \frac{2j}{\omega^2 b} sin (\omega b)$$

    I'm stuck here. So how can I get from here to ##ab (\frac{sin(\omega b/2)}{\omega b /2})^2##?

    Where did I go wrong? :confused:

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Jan 6, 2016 #2

    Samy_A

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    Your formula for x(t) is wrong.
    I didn't check the details of the rest of your calculation, but you can simplify it by noticing that the function is an even function.
     
    Last edited: Jan 6, 2016
  4. Jan 6, 2016 #3

    Ray Vickson

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    As 'Sammy_A' pointed out, your expression for ##x(t)## is incorrect, so everything after that is suspect. Again, if you heed the hint given by 'Sammy_A' you should be able to write the FT ##X(\omega)## of ##x(t)## easily in terms of the two integrals ##I_1 = \int_0^b 1 \cos(\omega t) \, dt## and ##I_2 = \int_0^b t \cos(\omega t) \, dt##.[Note that I use one letter ##x## for one function and a different letter ##X## for the other. Never, never use the same letter for two different functions in the same problem!]

    You may also find it useful to apply the trigonometric identity ##\cos(u) = \cos^2(u/2) - \sin^2(u/2) = 1 - 2 \sin^2(u/2)## at some point.
     
  5. Jan 7, 2016 #4
    Here is another idea:

    $$x(t) =\left\{\begin{matrix}a(\frac{b+t}{b}) \ \ (-b<t<0)\\ a(\frac{b-t}{b}) \ \ (0<t<b) \\0 \ \ else\end{matrix}\right.$$

    Is this correct now?

    How do I exactly use the property that this is an even function? I am not sure. But I did use the property that cos(-x)=cos(x) and sin(x)=-sin(x) in my calculations...
     
  6. Jan 7, 2016 #5

    Samy_A

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    Yes.



    As Ray Vickson explained.

    In general, the Fourier transform is given by ##\displaystyle \int_{- \infty}^{+ \infty} f(t) e^{-i \omega t} dt##
    Using the Euler formula you can write this as the sum of two integrals, one with ##\cos(\omega t)##, one with ##\sin(\omega t)##.

    If ##f## is an even function, the integral with the sine is ##0## (as ##f.\sin## is an odd function and the integration range is symmetric around t=0).
    You are left with the integral with the cosine. That function (##f.\cos##), is even, so that instead of integrating from -∞ to +∞, you can take the integral from 0 to ∞ twice.

    Conclusion: since your function ##x## is even, ##\displaystyle X(\omega)=2\int_{0}^{b} x(t) \cos(\omega t)dt##.
    (Here I used Ray's notation ##X## for the Fourier transform.)
     
    Last edited: Jan 7, 2016
  7. Jan 8, 2016 #6
    Thank you for the hint. I am still having some difficulty getting to ##\displaystyle X(\omega)=ab(\sin(\omega b/2)/(\omega b /2))^2##. Here's what I did:

    $$\displaystyle X(\omega)=2\int_0^{b} a \left( \frac{b-t}{b} \right) \cos(\omega t)dt =2a \int^b_0 \cos (\omega t) - \frac{t}{b} \cos (\omega t) dt$$

    Using integration by parts for the second half:

    $$=2a [\frac{t}{\omega b} \sin (\omega t) ]^b_0 - \frac{1}{b \omega} \int^b_0 \sin (\omega t) dt = \frac{1}{\omega} \sin (b \omega) + \frac{1}{b \omega^2} [\cos (\omega t)]^b_0$$

    $$\therefore \displaystyle X(\omega)= 2a \left( \frac{1}{\omega} \sin (b \omega) + \frac{1}{b \omega^2} \cos (\omega b) - \frac{1}{b \omega^2} \right)$$

    I used Ray's trig identity to further simplify this:

    $$\displaystyle X(\omega)= \frac{2a}{\omega} \sin(b \omega) + \frac{2a}{b \omega^2} (1- 2 \sin^2(\frac{\omega b}{2}))- \frac{2a}{b \omega^2} + \frac{2a}{\omega} \sin(b \omega)$$

    $$= -2 \sin^2(\frac{\omega b}{2})+ \frac{2a}{\omega} \sin(b \omega)$$

    Did I make a mistake somewhere, or do I need to use some other identities to get to that expression for ##\displaystyle X(\omega)##?
     
  8. Jan 8, 2016 #7

    Samy_A

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    You seem to have lost the first integral:
    ##\displaystyle 2a \int^b_0 \cos (\omega t) dt##

    There is also something wrong with the sign (and the ##a##).
     
    Last edited: Jan 8, 2016
  9. Jan 10, 2016 #8
    Sorry, that was a typo. I didn't miss the first integral, in fact I evaluated both integrals separately:

    $$X(\omega) = 2a \left( \underbrace{ \int^b_0 \cos(\omega t) dt}_{{\sin(\omega b)}} + \underbrace{\int^b_0 \frac{t}{b} \cos(\omega t) dt}_{{\frac{1}{\omega} \sin(\omega b) + \frac{1}{\omega^2 b} \cos(\omega b) - \frac{1}{\omega^2 b}}} \right)$$

    $$\therefore \ X(\omega) =2a \left( \sin(\omega b) + \frac{1}{\omega} \sin(\omega b) + \frac{1}{\omega^2 b} \cos(\omega b) - \frac{1}{\omega^2 b} \right)$$

    What is wrong with the sign? :confused:

    Using Ray's trig identity the above becomes:

    $$X(\omega) =2a \left( \sin(\omega b) + \frac{1}{\omega} \sin(\omega b) - \frac{2}{\omega^2 b} \sin^2(\frac{\omega b}{2}) \right)$$

    So, how do I get from here to ##ab \left( \frac{\sin(\omega b/2)}{\omega b/2} \right)##?
     
  10. Jan 10, 2016 #9

    Samy_A

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    A number of small errors, adding up to a wrong result.
    1) The second integral with the ##t\cos(\omega t)## misses a minus sign (look at the formula for ##x(t)##).
    2) In the result of the first integral, you miss an ##\omega## in the denominator.

    Once you have fixed the errors, you will see that Ray's identity leads right to the desired result.
     
  11. Jan 11, 2016 #10
    Thank you so much, I finally got it. I really thank you for your time.
     
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