Answer:
73.24 K byte
Explanation:
Assuming that
N = total number of samples
N = 10 * 5kHz
N = 50*10^3
Also, the total number of bits, T
T = 12 * N
T = 12 * 50*10^3
T = 600 * 10^3
And then, finally, the total number of byte,
B = 600*10^(3/8)
B = 75*10^3 byte
75*10^3 byte = 75*10^3/1024 kilo byte
And on converting to decimal, we will have
= 73.24 K byte
Therefore, the memory required = 73.24 K byte
