Final answer:
To solve for the DFTs of the given sequences using the properties of the Discrete Fourier Transform (DFT), we can apply the appropriate properties to each sequence. For sequence y[n] = x[n] cos(π/2n), the DFT is [0, 0, 0, 0]. For y[n] = [0, 0, 1, 0] ∘ x[n], the DFT is [0, 0, 1, 0]. For y[n] = g[n]x[n], where g[n] = [0, 0, 1, 0], the DFT is [0, 0, 1, 0]. For y[n] = x[n-1], the DFT is [0, 0, 0, 0].
Step-by-step explanation:
To solve for the DFTs of the given sequences, we can use the properties of the Discrete Fourier Transform (DFT).
(a) For y[n] = x[n] cos(π/2n), we can use the property that the DFT of a sequence multiplied by a constant is equal to the DFT of the original sequence scaled by that constant. Therefore, the DFT of y[n] would be [0, 0, 0, 0] multiplied element-wise with DFT(x[n]), which results in [0, 0, 0, 0].
(b) For y[n] = [0, 0, 1, 0] ∘ x[n], where ∘ denotes element-wise multiplication, we can use the property that the DFT of a sequence multiplied element-wise with another sequence is equal to the element-wise multiplication of their DFTs. Therefore, the DFT of y[n] would be [0, 0, 1, 0] multiplied element-wise with DFT(x[n]), which results in [0, 0, 1, 0].
(c) For y[n] = g[n]x[n], where g[n] = [0, 0, 1, 0], we can again use the property of element-wise multiplication of sequences under DFT. The DFT of y[n] would be the element-wise multiplication of DFT(g[n]) and DFT(x[n]), which results in [0, 0, 1, 0] multiplied element-wise with DFT(x[n]), resulting in [0, 0, 1, 0].
(d) For y[n] = x[n-1], we can use the property that the DFT of a sequence shifted to the right by k positions is equal to the DFT of the original sequence multiplied by a complex exponential term. Therefore, the DFT of y[n] would be the DFT of x[n] multiplied element-wise with [1, 1 - j, 1 + j, 0], which results in [0, 0, 0, 0].