What would be the total storage needed for recording sound at 96 kHz with 24-bit resolution for one hour?

Steps to Solve

1. Identify the given values

We are given:

• Sample Rate = 96 kHz = $96,000$ samples/second
• Bit Depth = 24 bits
• Duration = 1 hour = $1 \times 3600$ seconds

2. Convert duration to seconds

Calculate the duration in seconds: $$ \text{Duration in seconds} = 1 \text{ hour} \times 3600 \text{ seconds/hour} = 3600 \text{ seconds} $$

3. Calculate total storage required

Using the formula for storage: $$ \text{Storage} = \text{Sample Rate} \times \text{Bit Depth} \times \text{Duration} $$

Substituting the values we identified: $$ \text{Storage} = 96,000 \text{ samples/second} \times 24 \text{ bits/sample} \times 3600 \text{ seconds} $$

4. Perform the calculation

Now, calculating the total storage in bits: $$ \text{Storage} = 96,000 \times 24 \times 3600 = 8,294,400,000 \text{ bits} $$

5. Convert bits to bytes

Since there are 8 bits in a byte, we convert to bytes: $$ \text{Storage in bytes} = \frac{8,294,400,000 \text{ bits}}{8} = 1,036,800,000 \text{ bytes} $$

6. Convert bytes to gigabytes (GB)

To convert bytes to gigabytes, divide by $1,073,741,824$ (the number of bytes in 1 GB): $$ \text{Storage in GB} = \frac{1,036,800,000 \text{ bytes}}{1,073,741,824} \approx 0.966 \text{ GB} $$