Question:

In order to compute an N-point DFT, the number of complex multiplications and complex additions required respectively are

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Always distinguish between direct DFT computation and FFT computation.
Direct computation has \( O(N^2) \) complexity.
FFT algorithms have \( O(N \log N) \) complexity.
This difference highlights the immense computational savings provided by FFT.
Updated On: Jul 6, 2026
  • $N^2$ complex multiplications and $(N^2-1)$ complex additions.
  • $N^2$ complex multiplications and $(N^2-N)$ complex additions.
  • $N^2$ complex multiplications and $N^2$ complex additions.
  • $\frac{N}{2}\log_2 N$ complex multiplications and $N\log_2 N$ complex additions.
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
The question asks for the computational complexity (number of complex additions and complex multiplications) required to compute an \( N \)-point Discrete Fourier Transform (DFT) directly using its definition.

Step 2: Key Formula or Approach:

The definition of an \( N \)-point DFT for a sequence \( x(n) \) is:
\[ X(k) = \sum_{n=0}^{N-1} x(n) W_N^{kn}, \quad \text{for } k = 0, 1, \dots, N-1 \] where \( W_N = e^{-j 2\pi / N} \) is the twiddle factor.

Step 3: Detailed Explanation:


• Analyze the operations required to compute a single DFT coefficient \( X(k) \) for a fixed value of \( k \):
- There are \( N \) terms in the summation.
- Each term \( x(n) W_N^{kn} \) requires 1 complex multiplication. Thus, computing \( X(k) \) requires \( N \) complex multiplications.
- Summing these \( N \) terms requires \( N - 1 \) complex additions.

• Calculate the total operations for all \( N \) coefficients (for \( k = 0, 1, \dots, N-1 \)):
- Since we must calculate this for \( N \) different values of \( k \), we scale the individual requirements by \( N \).
- Total Complex Multiplications = \( N \times N = N^2 \).
- Total Complex Additions = \( N \times (N - 1) = N^2 - N \).

• Contrast this with FFT algorithms (like Cooley-Tukey):
- FFT reduces the multiplication complexity to approximately \( \frac{N}{2} \log_2 N \) and additions to \( N \log_2 N \).
- However, for direct computation, the requirements are strictly \( N^2 \) and \( N^2 - N \).
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