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## Main Question or Discussion Point

I hope this is OK to post here. I thought it would be better here than in the math questions forum, since you are EEs, and probably have more experience dealing with things related to the delta function.

Let

[tex]\hat{x}(t) = \sum_{k=-\infty}^{\infty}\delta(t-2k).[/tex]

Now let

[tex]x(t) = 2\hat{x}(t) + \hat{x}(t-1).[/tex]

Find the Fourier Transform of [itex]x(t)[/itex].

Here is the official solution given by Oppenheim:

From Table 4.2,

[tex]\hat{X}(j\omega) = \pi \sum_{k=-\infty}^{\infty}\delta(\omega - \pi k)[/tex]

Therefore,

[tex]X(j\omega)=\hat{X}(j\omega)[2+e^{-\omega}] = \pi \sum_{k=-\infty}^{\infty}\delta(\omega-\pi k)[2 + (-1)^k].[/tex]

Thank you!

**Problem**Let

[tex]\hat{x}(t) = \sum_{k=-\infty}^{\infty}\delta(t-2k).[/tex]

Now let

[tex]x(t) = 2\hat{x}(t) + \hat{x}(t-1).[/tex]

Find the Fourier Transform of [itex]x(t)[/itex].

**Given Solution**Here is the official solution given by Oppenheim:

From Table 4.2,

[tex]\hat{X}(j\omega) = \pi \sum_{k=-\infty}^{\infty}\delta(\omega - \pi k)[/tex]

Therefore,

[tex]X(j\omega)=\hat{X}(j\omega)[2+e^{-\omega}] = \pi \sum_{k=-\infty}^{\infty}\delta(\omega-\pi k)[2 + (-1)^k].[/tex]

**My Two Questions****Question 1.**When applying the time-shifting property of Fourier series, why did they multiply by [itex]e^{-\omega}[/itex] and*not*[itex]e^{-j\omega}[/itex]??**Question 2.**How did the [itex]e^{-\omega}[/itex] become [itex](-1)^k[/itex] in the final answer?Thank you!

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