Final answer:
To encode and transmit high-fidelity two-channel music with 1000 loudness levels and 22kHz bandwidth, 10 bits per sample are necessary. Sampling at a rate of 44kHz per channel, the total transmission speed required is 1.76 Mbps.
Step-by-step explanation:
The question involves calculating the required transmission speed for encoding and transmitting high-fidelity two-channel music. To determine the transmission speed, we must consider both the sampling rate and the bit depth required to encode the audio. Since the music has a bandwidth of 22kHz, the sampling rate must be at least twice this value according to the Nyquist theorem, which is 44kHz per channel. To minimize quantizing errors, we choose a sampling rate that is twice the bandwidth, hence we use a sampling rate of 44kHz.
To represent 1000 loudness levels, we need to determine the number of bits required per sample. 1000 levels can be approximated as 2³10, since 2³9 is 512 and 2³10 is 1024, and we need an integer number of bits. Therefore, we need 10 bits per sample.
Finally, to calculate the transmission speed, we use the formula:
- Transmission speed = Sampling rate × number of bits per sample × number of channels
- Transmission speed = 44,000 samples/s × 10 bits/sample × 2 channels
- Transmission speed = 880,000 bits/s or 880 Kbps per channel
Therefore, the total transmission speed required for a two-channel music audio is 1.76 Mbps (880 Kbps × 2).