Determine the minimum number of magnitude bits plus an additional sign bit required in a PCM...

Question

Question

Determine the minimum number of magnitude bits plus an additional sign bit required in a PCM code for a dynamic range of 65 dB.

Answer 1

Answer #1

Dynamic Range (DR) :

Ratio of the largest possible magnitude to the smallest (other than 0) magnitude is called as Dynamic Range.

Formula for Dynamic Range in PCM is : 20 log(Vmax / Vmin)

In a n-bit PCM system, Vmax = 2^n and Vmin = 1

DR(db) = 20 log(Vmax/Vmin)

= 20 log(2^n - 1)

~ 6.02n

To find out the minimum number of bits required in PCM code for dynamic range of 65 dB, we equate 6.02n (db) with 65db.

6.02n = 65

n = 10.797

n ~11 bits

By considering both magnitude bits plus an additional sign bit, total bits required are:

11 bits + 1 sign bit = 12 bits

Therefore, the minimum number of magnitude bits plus an additional sign bit required in a PCM code for a dynamic range of 65 dB are 12 bits.

Answer 2

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