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You got the term $X \left( e^{j (\omega - 2 \pi)/2} \right)$ wrong. It is centered at $2\pi$ and it is non-zero in the interval $(2\pi-2\omega_0,2\pi+2\omega_0)$. So the results obtained in the frequency domain and in the time domain, respectively, are identical. Note that the term $X \left( e^{j \omega/2} \right)$ is $4\pi$-periodic, so the term $X \left( e^... 1 Here's how I like to think about the expression $$\displaystyle \frac{1}{D} \sum_{k=0}^{D-1} e^{j 2 \pi k m / D}.$$ For any value of$D$, the first term is 1, and all of the terms lie on the unit circle. Assuming$D > 1$, each term after the first is found by rotating the previous term through an angle of$2 \pi m / D$radians. If$m$is not a multiple ... 1 Sample rate conversation is easy in theory but tricky in practice. Assuming you want to convert to the standard rate of 44.1 kHz (not 44 kHz), you have an awkward conversion ratio.$3800 =2^3 \cdot 5^2 \cdot 19$and$441 = 3^2 \cdot 7^2$are mutually prime that means that rational sample rate conversion is impractical,so you need irrational sample rate ... 1 Your digital signal is originally sampled at$380KHz$and you want to downsample it to sampling rate$44KHz$. Therefore, you will require fractional sampling rate change. You cannot just downsample to$44KHz$because$\frac{380}{44}$is not an integer. First upsample by a factor of$11$and then downsample by a factor of$95$to reach your goal. Since you ... 0 It's not derived, it's just chosen in a smart way such that the relationship between the decimated and the original sequences becomes obvious. It's just a rearrangement of the terms of the sum. As a simple example, take an infinite sum of numbers$a_r\$: $$S=\sum_{r=-\infty}^{\infty}a_r\tag{1}$$ Under certain conditions that we don't need to bother with now ...