8.9 Questions for the reader

8.9 Questions for the reader#

Exercise 42

Bin spacing. You compute the DFT of exactly 1 second of audio.

  1. What is the spacing in Hz between adjacent DFT bins in the output?

  2. More generally, what is the bin spacing for an input of duration \(T\) seconds?

  3. If you wanted a frequency resolution of \(1\) Hz, how long a segment would you need to analyze?

Reveal solution

  1. \(1\) Hz for a \(1\) second input

  2. \(1/T\) Hz in general

  3. A \(1\) Hz resolution needs a \(1\) second segment.

Exercise 43

Counting bins. You take the DFT of a \(1024\)-sample window of audio recorded at \(f_s = 44{,}100\) Hz.

  1. How many complex bins does the full DFT produce?

  2. How many non-redundant bins does np.fft.rfft return for this real signal?

  3. What frequency, in Hz, does bin \(k = 100\) correspond to?

Reveal solution

  1. \(1024\) complex bins

  2. \(513\) non-redundant bins

  3. \(\approx 4307\) Hz

Exercise 44

Reading a spectrum. The amplitude spectrum of a periodic tone has strong peaks at 200, 600, and 1000 Hz, with the peaks at 400 and 800 Hz nearly absent.

  1. What is the fundamental frequency?

  2. Which harmonics are present, and what does the pattern of odd-only harmonics suggest about the sound?

Reveal solution

  1. Fundamental \(200\) Hz

  2. The 1st, 3rd, and 5th (odd) harmonics are present, suggesting a square-wave-like sound.

Exercise 45

Why \(O(N \log N)\) matters. A DFT is applied to a window of \(N = 4096\) samples. Estimate the ratio between the number of operations for a naive \(O(N^2)\) DFT and an \(O(N \log N)\) FFT. If a real-time system must compute such a transform hundreds of times per second, why is this difference decisive?

Reveal solution

About \(N / \log_2 N = 4096 / 12 \approx 340\) times fewer operations.

Exercise 46

Redundant bins. For a real input of length \(N = 16\), the DFT bin \(R[k]\) satisfies \(R[k] = R[N-k]\).

  1. Given \(R[3] = 0.7\), what is \(R[13]\)?

  2. Which two bins are guaranteed to have a zero imaginary part, and why?

Reveal solution

  1. \(R[13] = R[3] = 0.7\).

  2. Bins \(k = 0\) and \(k = 8\) (the \(N/2\) bin) have zero imaginary part.

Exercise 47

Windowing before the DFT. You cut a short fragment out of a longer recording and take its DFT. Because the fragment rarely contains a whole number of cycles of every frequency present, its spectrum exhibits spectral leakage.

  1. In one or two sentences, explain what spectral leakage is and why abruptly slicing a fragment out of a longer signal produces it.

  2. What do we typically multiply the fragment by before taking the DFT to reduce leakage, and what property of that function makes it help?

  3. Does a plain rectangular window (i.e., taking the fragment as-is) reduce leakage relative to a tapered window such as a Hann window? Why or why not?

Reveal solution

  1. Omitted (read the chapter to find out!)

  2. Multiply the fragment by a tapered window (such as a Hann window) that falls smoothly to zero at both ends

  3. A plain rectangular window does not reduce leakage.