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You are recording a friend's 3-minute song with 24 bits per sample at 96 kHz sampling rate for a 5.1 surround sound system (6 channels). How much space will you save by recording this song with 16-bit, 44.1 kHz sampling for a stereo + subwoofer system (3 channels)? Note that you are making these recordings without using any compression. Please give you answer in terms of Gibibits and round to the nearest hundredth.

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Answer:

save space = 1.9626 Gibibits

Explanation:

given data

recording song time = 3 minute = 180 seconds

song per sample =  24 bits per sample

frequency = 96 kHz

surround sound system = 5.1

solution

first we get here required space for 24 bits at  96 kHz that is

required space = 24 × 96 × 10³ × 6 × 180

required space = 2.48832 × [tex]10^{9}[/tex]  bits

as 1 Gib bit = [tex]2^{30}[/tex] bits

so required space = 2.48832 × [tex]10^{9}[/tex]  bits ÷  [tex]2^{30}[/tex] bits

required space = 2.3174 Gibibits   ...............1

and

space required to save record 16 bit at 44.1 kHz

space required = 16 × 44.1 × 10³ × 3 × 180

space required = 0.381024 × [tex]10^{9}[/tex]  bits

space required = 0.381024 × [tex]10^{9}[/tex]  bits  ÷  [tex]2^{30}[/tex] bits  

space required = 0.3548 Gibibits    ...........2

so

we get here that save space in 16 bit at 44.1 kHz

save space = 2.3174 Gibibit - 0.3548 Gibibits

save space = 1.9626 Gibibits

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