I want to find the fourier transform of this input signal.

enter image description here

Let the unit function unit (t, a, b) have the value 1 on the interval a≤ t < b and the value 0 otherwise. f(t) = (t)unit(t, 0, 0.5) + (-t)unit(t, 0.5, 1.5) + (t)unit(t, 1.5, 2). I am lost on where to go from here. Without being able to write it using euler's formula(e^(it) = cos(t) + isin(t)), how should I proceed? I am not sure which fourier integral to use.

This is the equation I hope to use. enter image description here

  • $\begingroup$ what do you mean by "euler's formula?" $\endgroup$
    – 5xum
    Commented Mar 2, 2016 at 9:05
  • $\begingroup$ Hint: write the Fourier integrals and compute them piecewise. $\endgroup$
    – user65203
    Commented Mar 2, 2016 at 9:09
  • $\begingroup$ But do i use the fourier integral of sin or cos? My intuition is to use sin, since at t = 0, the y value is 0. but I don't really understand the mathematical way to get here(for example, if it was a completely different, obscure graph). $\endgroup$
    – Jonathan
    Commented Mar 2, 2016 at 9:19

2 Answers 2


I think a simpler and more general way (it can be extended to any continuous piecewise linear function) is to write this function as a linear combination of triangle functions $\Lambda$, with $\Lambda(t)=1-|t|$ with support on $[-1,1]$.


and then apply the rules about F. Transform, yielding something like

$$(e^{-i\pi u}-e^{-3i\pi u})sinc(u/2)^2 \ \ (*)$$

(where $sinc$ is the classical cardinal sine, with sinc $u = \dfrac{\sin{\pi u}}{\pi u}$ for $u \neq 0$)

(I say "something like" because there are different cousin definitions of the F.T., of the cardinal sine, of the triangle function, etc.)

Remark: formula (*) can be transformed by factorization of $e^{-2i\pi u}$.


f(t) = {0 <= t < (1/2), 2t, (1/2) <= t < (3/2), 2-2t, (3/2) <= t < 2, 2t}

integral of [f(t) * e^(-iwt)] from 0 to 2 is the fourier transform of the function.

  • $\begingroup$ I am afraid that it'is only the definition of Fourier Transform. I don't understand why the solution I have given which is $2ie^{-2i\pi u}sin{\pi u}\frac{sin(\pi u/2)^2}{(\pi u/2)^2}$, not only has not been considered, but has been downgraded (-15 !) an hour ago. $\endgroup$
    – Jean Marie
    Commented Mar 25, 2016 at 18:10

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