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I would be grateful if you can help me with i) ii) and iii) especially with ii and iii

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  • $\begingroup$ For i, apply the definition. For ii, find the zeroes of the first derivative. For iii, find the zeroes. $\endgroup$ – Yves Daoust Apr 22 '14 at 20:51
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Define $T=2$ms and write down the definition of the Fourier transform:

$$X(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_0^{T}e^{-j\omega t}dt=-\frac{1}{j\omega}(e^{-j\omega T}-1)=\frac{1}{j\omega}e^{-j\omega T/2}(e^{j\omega T/2}-e^{-j\omega T/2})=e^{-j\omega T/2}T\frac{\sin(\omega T/2)}{\omega T/2}$$

The maximum value of $|X(j\omega)|$ is at $\omega=0$ and its value is $T$. The zeros of $X(j\omega)$ are are the zeros of $\sin(\omega T/2)$ apart from $\omega=0$, i.e. $\omega=2\pi k/T$, $k=\pm 1, \pm 2, \ldots$

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  • $\begingroup$ Can you explain why maximum is at ω=0 ? $\endgroup$ – Anarkie Apr 22 '14 at 21:00
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    $\begingroup$ The function $\sin(x)/x$ is called the sinc-function and attains its maximum value 1 at $x=0$. Check this article: en.wikipedia.org/wiki/Sinc_function $\endgroup$ – Matt L. Apr 22 '14 at 21:03
  • $\begingroup$ Thank you very much, would en.wikipedia.org/wiki/Spectrum_(functional_analysis) help for understanding iii)? $\endgroup$ – Anarkie Apr 22 '14 at 21:07
  • $\begingroup$ No, just read the "Properties" section of the article on the sinc function, then you know that it attains its zeros at non-zero multiples of $\pi$. Since in your case $x=\omega T/2$, the zeros are as I indicated in my answer. $\endgroup$ – Matt L. Apr 22 '14 at 21:11
  • $\begingroup$ I would be grateful if you can have a look at this question math.stackexchange.com/questions/801338/… $\endgroup$ – Anarkie May 20 '14 at 12:16

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