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I have no idea how to solve the following exercise, can you help me, please?

Find the periodic function whose Fourier coefficients are

$$C_{k} = \frac{sin^{2}(\pi k T)}{(\pi k T)^{2}}$$

I think that this function is a definition of sinc ...

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  • \$\begingroup\$ I've edited your question to make the formula more accessible. Please do check I haven't screwed up. \$\endgroup\$
    – Andy aka
    Commented Oct 1, 2019 at 15:29

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Note that \$\text{sinc}\,{\pi k T} = \frac{\sin{\pi kT}}{\pi k T} \$. The inverse fourier transform of this is a box.

Note further that your frequency-domain expression is precisely \$\text{sinc}\,{\pi k T} \cdot \text{sinc}\,{\pi k T} \$. Furthermore, note that multiplication in the frequency domain is the same as convolution in the time domain; the convolution of a box with itself is a triangular pulse. This should suffice as a starting point toward solving the problem; computing the box/triangle support and normalization is left to the reader so as to avoid doing your exercise in its entirety for you.

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Ok, let me see. It looks like this is a complex Fourier Series, so as there is no imaginary part, the function $$x(t)$$ that corresponds to this must be even so that $$x(t)=x(-t)$$ Therefore, it must be the case that there are no sine terms, as these are odd. There are only cosine terms. The inverse Fourier transform of a sinc function is a rectangular pulse. $$\int_{0}^{T}1*e^{j\omega t}dt+\int_{-T}^{0}1*e^{j\omega t}dt =\frac{1}{j\omega}[(e^{j\omega T}-1)+(1-e^{-j\omega T})]=\frac{2j\sin(\omega T)}{j\omega}=\frac{2T\sin(\omega T)}{\omega T}$$ Using the convolution integral $$\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau$$ The final answer of the convolution is $$(t+2T)(u(t+2T))+(2T-t)(u(2T-t))$$ The fourier transform of this is $$\frac{4T^2\sin^2(\omega T)}{\omega^2 T^2}$$ So, for the question, we need $$\frac{1}{4T^2}(t+2T)(u(t+2T))+(2T-t)(u(2T-t))$$ u(t) is the unit step.

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