Question

audio is being sampled at the rate of 44.1 khz using 8 bits. two channels are being used, calculate. (a) the size of one sample in bits (b) the size of a 30 second audio recording in MIB​

Answer 1

Answer:

(a) The size of one sample in bits is 8 bits * 2 channels = 16 bits.

(b) To calculate the size of a 30 second audio recording in MIB (megabytes), we need to know the number of samples in the recording. To find this, we can use the formula:

number of samples = sample rate * duration

Substituting the given values, we get:

number of samples = 44100 samples/second * 30 seconds = 1323000 samples

The size of the audio recording in bits is the number of samples * the size of each sample in bits, which is 1323000 samples * 16 bits/sample = 21120000 bits.

To convert this to megabytes, we need to divide by 8 bits/byte and then by 1024 bytes/kilobyte and then by 1024 kilobytes/megabyte, which gives us a final result of 21120000 bits / (8 bits/byte) / (1024 bytes/kilobyte) / (1024 kilobytes/megabyte) = 2.96 megabytes.

Thus, the size of a 30 second audio recording in MIB is approximately 2.96 MIB.