Let's revisit the computation to ensure correctness. We'll carefully follow the steps to see if there were any errors.

### Step 1: Input Sequence
The sequence given is:
\[
x(n) = \{1, 2, 3, 4\}
\]
This sequence has a length \( N = 4 \). 

### Step 2: Bit-Reversal Permutation
For \( N = 4 \), we reverse the bits of indices \( n = 0, 1, 2, 3 \):

- \( 00_2 \rightarrow 00_2 \) (n=0)
- \( 01_2 \rightarrow 10_2 \) (n=1)
- \( 10_2 \rightarrow 01_2 \) (n=2)
- \( 11_2 \rightarrow 11_2 \) (n=3)

So the bit-reversed sequence:
\[
\{1, 3, 2, 4\}
\]

### Step 3: Perform FFT Computation

#### Stage 1:
- Compute the 2-point DFTs for the pairs \( (1, 3) \) and \( (2, 4) \).

For pair \( \{1, 3\} \):
\[
\text{DFT}_2 = \begin{cases}
1 + 3 = 4 \quad & \text{(Sum)} \\
1 - 3 = -2 \quad & \text{(Difference)}
\end{cases}
\]

For pair \( \{2, 4\} \):
\[
\text{DFT}_2 = \begin{cases}
2 + 4 = 6 \quad & \text{(Sum)} \\
2 - 4 = -2 \quad & \text{(Difference)}
\end{cases}
\]

After this stage, the sequence is:
\[
\{4, -2, 6, -2\}
\]

#### Stage 2:
- Compute the 4-point DFT using the results from Stage 1.

The combination process involves twiddle factors \( W_N^k \), where \( W_4 = e^{-j\frac{2\pi}{4}} = j \) (since \( j^2 = -1 \)).

\[
X(0) = 4 + 6 = 10
\]

\[
X(1) = -2 + (-2 \times j) = -2 - 2j
\]

\[
X(2) = 4 - 6 = -2
\]

\[
X(3) = -2 - (-2 \times j) = -2 + 2j
\]

### Final Output:
The DFT of the sequence \( x(n) = \{1, 2, 3, 4\} \) using DIT-FFT is:
\[
X(k) = \{10, -2 - 2j, -2, -2 + 2j\}
\]

### Verification of Results:

Your calculated sequence was \( \{10, 2, 2 - 2j, -2j - 2\} \). If we compare:

- \( X(0) = 10 \) (Correct)
- \( X(1) = 2 - 2j \) (It should be \(-2 - 2j\))
- \( X(2) = 2 \) (It should be \(-2\))
- \( X(3) = -2j - 2 \) (It should be \(-2 + 2j\))

The difference seems to come from an incorrect sign in the imaginary component during the second stage of FFT computation. 

If you still get different results, double-check the twiddle factors and the combination step.

Would you like more details or have any questions?

Here are 5 related questions:
1. Why is the twiddle factor \( W_N^k \) crucial in the FFT computation?
2. How does a mistake in the twiddle factor application affect the final result in FFT?
3. What are the specific properties of FFT that make it more efficient than DFT?
4. How do we interpret the complex results obtained from FFT in real-world applications?
5. Can you demonstrate the process for a sequence with a different length, say \( N = 8 \)?

**Tip:** While applying FFT, careful handling of the signs and twiddle factors is crucial to ensure accurate results.

Explore the detailed step-by-step process of computing Fast Fourier Transform (FFT), including bit-reversal permutation, FFT stages, and twiddle factors. This guide includes examples and explanations to clarify complex number handling in FFT computations.

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