Approximately how much bandwidth is utilized by a unicast audio flow with 4 channels at 24-bit, 48kHz, and 1 ms latency?

Approximately how much bandwidth is utilized by a unicast audio flow with 4 channels at 24-bit, 48kHz, and 1 ms latency?

• A. 1 Mbps
• B. 6 Mbps
• C. 20 Mbps
• D. 45 Mbps

Answer: (B)

To determine the bandwidth utilized by a unicast audio flow with 4 channels at 24-bit depth and a sample rate of 48kHz, you can calculate the audio data rate using the following formula: 1. **Calculate the number of samples per second for all channels**: - Each channel outputs samples at 48,000 times per second (48kHz), so for four channels, this is: \[ 48,000 \, \text{samples/second} \times 4 \, \text{channels} = 192,000 \, \text{samples/second} \] 2. **Calculate the number of bits per second**: - Each sample is 24 bits, so the total number of bits per second is: \[ 192,000 \, \text{samples/second} \times 24 \, \text{bits/sample} = 4,608,000 \, \text{bits/second} \] 3. **Convert bits per second into megabits per second (Mbps)**: - Now convert 4,608,000 bits/second to Mbps: \[ \frac{4

To determine the bandwidth utilized by a unicast audio flow with 4 channels at 24-bit depth and a sample rate of 48kHz, you can calculate the audio data rate using the following formula:

1. Calculate the number of samples per second for all channels:

• Each channel outputs samples at 48,000 times per second (48kHz), so for four channels, this is:

[

48,000 , \text{samples/second} \times 4 , \text{channels} = 192,000 , \text{samples/second}

]

2. Calculate the number of bits per second:

• Each sample is 24 bits, so the total number of bits per second is:

[

192,000 , \text{samples/second} \times 24 , \text{bits/sample} = 4,608,000 , \text{bits/second}

]

3. Convert bits per second into megabits per second (Mbps):

• Now convert 4,608,000 bits/second to Mbps:

[

\frac{4