Answer:
To find the remaining 4 samples of x(n), we need to use the inverse Discrete Fourier Transform (IDFT) to go from X(k) to x(n).
X(k) is the 10-point DFT of x(n), which means it has a length of 10. However, we are given the first 6 samples of x(n), so we will assume the remaining 4 samples are zero. This is because the DFT assumes periodicity, meaning it repeats the signal values in order to create a longer sequence.
Now, let's calculate the remaining 4 samples using the IDFT:
x(n) = (1/N) * Σ(k=0 to N-1) X(k) * e^(j * 2π * kn/N)
where N is the length of x(n), which is 10 in this case.
Since we know the first 6 samples of x(n) and the remaining 4 samples are assumed to be zero, we can simplify the IDFT equation as follows:
x(n) = (1/10) * Σ(k=0 to 9) X(k) * e^(j * 2π * kn/10)
x(0) = (1/10) * [2 * e^(j * 2π * 0 * 0/10) + 5 * e^(j * 2π * 0 * 1/10) + (7 - j3) * e^(j * 2π * 0 * 2/10) -3 * e^(j * 2π * 0 * 3/10) + (2 + j1.3) * e^(j * 2π * 0 * 4/10) + (-2 + j5) * e^(j * 2π * 0 * 5/10) + 7j * e^(j * 2π * 0 * 6/10) + 5 * e^(j * 2π * 0 * 7/10)]
x(1) = (1/10) * [2 * e^(j * 2π * 1 * 0/10) + 5 * e^(j * 2π * 1 * 1/10) + (7 - j3) * e^(j * 2π * 1 * 2/10) -3 * e^(j * 2π * 1 * 3/10) + (2 + j1.3) * e^(j * 2π * 1 * 4/10) + (-2 + j5) * e^(j * 2π * 1 * 5/10) + 7j * e^(j * 2π * 1 * 6/10) + 5 * e^(j * 2π * 1 * 7/10)]
x(2) = (1/10) * [2 * e^(j * 2π * 2 * 0/10) + 5 * e^(j * 2π * 2 * 1/10) + (7 - j3) * e^(j * 2π * 2 * 2/10) -3 * e^(j * 2π * 2 * 3/10) + (2 + j1.3) * e^(j * 2π * 2 * 4/10) + (-2 + j5) * e^(j * 2π * 2 * 5/10) + 7j * e^(j * 2π * 2 * 6/10) + 5 * e^(j * 2π * 2 * 7/10)]
x(3) = (1/10) * [2 * e^(j * 2π * 3 * 0/10) + 5 * e^(j * 2π * 3 * 1/10) + (7 - j3) * e^(j * 2π * 3 * 2/10) -3 * e^(j * 2π * 3 * 3/10) + (2 + j1.3) * e^(j * 2π * 3 * 4/10) + (-2 + j5) * e^(j * 2π * 3 * 5/10) + 7j * e^(j * 2π * 3 * 6/10) + 5 * e^(j * 2π * 3 * 7/10)]
Calculating these equations will give us the remaining 4 samples of x(n).