Question

Determine the Fourier transform of the function :
{ 1-t 0≤t≤1
f(t)= { 1+t -1≤t≤0
{ 0 otherwise

Answer 1

Answer on Question #84501 – Math – Trigonometry

Question

Determine the Fourier transform of the function:

$$f(t) = \begin{cases} 1 - t, & 0 \leq t \leq 1 \\ 1 + t, & -1 \leq t \leq 0 \\ 0, & \text{otherwise}. \end{cases}$$

Solution

Find the Fourier transform according to the formula

$$F(\omega) = \int_{-\infty}^{+\infty} f(t) e^{-i\omega t} dt, -\infty \leq \omega \leq +\infty.$$

Substitute in (2) the function (1):

$$F(\omega) = \int_{0}^{1} (1 - t) e^{-i\omega t} dt + \int_{-1}^{0} (1 + t) e^{-i\omega t} dt + 0 = \int_{0}^{1} e^{-i\omega t} dt - \int_{0}^{1} t e^{-i\omega t} dt + \int_{-1}^{0} e^{-i\omega t} dt + \int_{-1}^{0} t e^{-i\omega t} dt = - \frac{e^{-i\omega}}{i\omega} + \frac{1}{i\omega} - \frac{i\omega e^{-i\omega} + e^{-i\omega} - 1}{\omega^2} - \frac{1}{i\omega} + \frac{e^{i\omega}}{i\omega} + \frac{i\omega e^{i\omega} - e^{i\omega} + 1}{\omega^2}$$ $$= \frac{2i\omega \sinh(i\omega) - 2i \sinh(i\omega) - \cosh(i\omega) + 2}{\omega^2}.$$

Answer: $F(\omega) = \frac{2i\omega \sinh(i\omega) - 2i \sinh(i\omega) - \cosh(i\omega) + 2}{\omega^2}$.

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