1. Build a length-8 signal
Edit the entries of $\vec f=(f_0,\ldots,f_7)^T$. The tool computes $\widehat{\vec f}=F_8\vec f$.
2. Time domain and frequency domain
Blue bars show the time samples. Orange bars show magnitudes $|\widehat f_k|$.
3. DFT coefficients
A coefficient with large magnitude means the signal has a strong component in that frequency mode.
4. Inverse DFT reconstruction
The inverse formula is $\vec f=\frac{1}{n}F_n^*\widehat{\vec f}$.
5. Even-odd FFT recursion
For $n=8$, split the signal into even and odd samples. Then combine two length-4 DFTs using twiddle factors.
6. Low-pass filtering
Keep frequency modes $k=0,1,7$ and set the others to zero. This keeps slow variation and removes fast oscillation.
7. Complexity comparison
8. Independent-study practice tasks with answers
Task A. Use the impulse preset. What are all DFT coefficients?
Answer: They are all equal to $1$. A time-zero impulse contains all frequencies equally.
Answer: They are all equal to $1$. A time-zero impulse contains all frequencies equally.
Task B. Use the constant preset. Which frequency is nonzero?
Answer: Only $k=0$ is nonzero. A constant signal has only zero-frequency content.
Answer: Only $k=0$ is nonzero. A constant signal has only zero-frequency content.
Task C. Use the high-frequency preset. Which coefficient is largest?
Answer: For the alternating signal, the main frequency is $k=4$.
Answer: For the alternating signal, the main frequency is $k=4$.
Task D. Explain why FFT is faster than direct DFT.
Answer: Direct DFT costs $O(n^2)$, while FFT recursively splits the problem into two half-size DFTs plus $O(n)$ combination work, giving $O(n\log n)$.
Answer: Direct DFT costs $O(n^2)$, while FFT recursively splits the problem into two half-size DFTs plus $O(n)$ combination work, giving $O(n\log n)$.