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4

Expand the denominator into a geometric series like this: $$\frac{1}{1 + \alpha e^{-2\pi i f T}} = \frac{1}{1 - (- \alpha e^{-2\pi i f T})} = \sum_{n=0}^{\infty} (-\alpha \cdot e^{-2\pi i f T})^n,$$ where the series converges uniformly as long as $|\alpha|<1$ (this is the case in your question). Hence, we can interchange summation and integration. ...

2

The vectors $u_j^n$ are powers of the primitive $n$-th root of unity $\zeta_n$ in the complex plane. The collection of vectors $S_n$ you are looking at for each fixed $n$ consists of all combinations of sign assignments of $$\pm\zeta_n^1 \pm \zeta_n^2 \pm \cdots \pm \zeta_n^n.$$ This follows from the fact that $\left\lfloor{(k-1)/2^{j-1}}\right\rfloor$ ...

2

The Fourier transform is simply the $z$ transform evaluated at $z = e^{i\omega}$. Thus, the Fourier transform exists iff $z = e^{i\omega}$ is in the region of convergence for all $\omega \in \mathbb{R}$. This means that we just need the unit circle to be in the region of convergence.

1

You can also solve this problem with the standard tools of complex analysis. If you make a substitution $z=\exp(2\pi i fT)$, then the integral becomes $$p_\ell=\frac{1}{2\pi i}\int dz\ \frac{z^\ell}{z+\alpha}$$ where the contour of integration is the unit circle oriented counter-clockwise. Because $\alpha\in(0,\tfrac{1}{2})$, there is a pole at $z=-\alpha$ ...

1

After your recent edit, your result $A_0 = 2/(5\pi)$ is correct. Also, you are correct that all of the $B_n$'s are zero because the integrand of $$B_n = \frac{1}{\pi}\int_{-\pi}^{\pi} \cos(5x/2)\sin(nx) dx$$ is an even function times an odd function, hence odd. So that leaves the $A_n$ integral for $n \neq 0$. We have A_n = \frac{1}{\pi}\int_{-\pi}^{\pi} ...

1

Hint: $\cos^2(2x) = \frac{1+\cos(4x)}{2}$, and you don't need to do any integration.

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