The butterfly in fig. is taken from radix 2 decimation in time FFT with N = 32. Assume that the five stages of the signal flow graph are indexed by m = 1, 2, 3, 4, 5, where 5 is the last stage. Which of the five stages has butterflies of this form?

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  1. m = 3, 4
  2. m = 4, 5
  3. m = 3, 4, 5
  4. m = 2, 4, 5

Answer (Detailed Solution Below)

Option 2 : m = 4, 5
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Concept:

In a radix-2 Decimation-in-Time (DIT) FFT algorithm, the computation is broken into multiple stages of "butterfly" operations. Each butterfly computes a pair of values using inputs and a complex multiplier called a twiddle factor.

The twiddle factor is given by: \( W_N^k = e^{-j \frac{2\pi k}{N}} \)

As the FFT progresses through stages from m = 1 to m = log2(N), the values of k used in the twiddle factors become more specific (larger powers of WN).

Given:

Total number of points: \( N = 32 \Rightarrow \log_2(32) = 5 \) stages

The butterfly shown involves twiddle factor \( W_N^6 \) and \(= W_N^{N/2} \)

Observation:

The usage of \( W_N^6 \)  occurs in later stages where index distances increase, indicating higher-order butterflies.

Such butterflies are observed in the last two stages of a 5-stage FFT flow graph: m = 4 and m = 5.

Hence, the correct answer is option 2

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