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File size: 20 s, 44.1 KHz, 8-bit stereo

Short answer

Multiply 44,100 by 8 bits by 2 channels by 20 seconds, divide by 8: 1,764,000 bytes (about 1.68 MB).

Calculate the file size if we record the 20 second of stereo music at 44.1 KHz with 8 bits.

The size of an uncompressed (PCM) audio file is found from the sampling rate, the resolution, the number of channels and the duration.

File size (bytes) = sampling rate x bits per sample x channels x time / 8

Given

  • Sampling rate = 44.1 kHz = 44,100 samples per second
  • Resolution = 8 bits per sample = 1 byte
  • Channels = 2 (stereo)
  • Time = 20 seconds

Calculation

  1. Samples per second for both channels = 44,100 x 2 = 88,200
  2. Bits per second = 88,200 x 8 = 705,600 bits/s
  3. Bytes per second = 705,600 / 8 = 88,200 bytes/s
  4. Total size = 88,200 x 20 = 1,764,000 bytes

In larger units

  • 1,764,000 / 1024 = 1,722.66 KB
  • 1,764,000 / (1024 x 1024) = 1.68 MB

Check in one line: 44,100 x 8 x 2 x 20 / 8 = 1,764,000 bytes.

The file size is 1,764,000 bytes (about 1.68 MB). A WAV file would also have a small header (about 44 bytes), which is ignored here.

✓ Answer checked by Claude Opus 5.5 (high effort)

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