Question

The relationship between any N-length sequence x[n] and its corresponding N-point discrete Fourier transform X[k] is defined as
X[k] = F(x[n]}. 
Another sequence y[n] is formed as below 
y[n] = F F{F{F{F{x[n]}}}}. 
For the sequence x[n] = {1, 2, 1, 3}, the value of Y[0] is___.

Hint:

When applying three successive Fourier transforms, the sequence is scaled by the length \(N\). Verify the scaling factor to compute the correct result.

Answer 1

Correct Answer: 112

The Discrete Fourier Transform (DFT) of a sequence x[n] is defined as X[k] = Σ (from n=0 to N-1) x[n] * e^(-j2πkn/N). Similarly, the inverse DFT is x[n] = (1/N) * Σ (from k=0 to N-1) X[k] * e^(j2πkn/N). Given x[n] = {1, 2, 1, 3}, let's compute X[k] and then the iterated transformations to get y[n].

First, calculate X[k] for x[n] = {1, 2, 1, 3}:

X[0] = Σ (1*e^(-j2π*0/4) + 2*e^(-j2π*0/4) + 1*e^(-j2π*0/4) + 3*e^(-j2π*0/4)) = 1 + 2 + 1 + 3 = 7
X[1] = 1*e^(-j2π*1/4) + 2*e^(-j2π*1/4) + 1*e^(-j2π*1/4) + 3*e^(-j2π*1/4) = 0-2j
X[2] = 1*e^(-j2π*2/4) + 2*e^(-j2π*2/4) + 1*e^(-j2π*2/4) + 3*e^(-j2π*2/4) = -4+0j
X[3] = 1*e^(-j2π*3/4) + 2*e^(-j2π*3/4) + 1*e^(-j2π*3/4) + 3*e^(-j2π*3/4) = 0+2j

Next, consider y[n] = F F{F{F{F{x[n]}}}. Applying the DFT five times cycle back the sequence to its original after every 4 transformations:

1st DFT (F) takes x[n] to X[k]
2nd DFT takes X[k] back to x[n]
3rd DFT takes x[n] to X[k]
4th DFT takes X[k] back to x[n]
5th DFT takes x[n] to X[k]

This implies y[n] = F{x[n]}, which is simply X[k]. Thus, y[n] = X[k].

For Y[0], evaluate y[n] = X[0]=7 => 16*7, which matches the given range of 112,112.