Question

How much computer memory (in bytes) in minimum would be required to store 10 seconds of a sensor signal sampled by a 12-bit A/D converter operating at a sampling rate of 5 kHz?

Answer 1

73.24 K byte

Step-by-step 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