Answer:
the minimum number of bits needed for quantization is 7
Step-by-step explanation:
Given the data in the question;
if quantization is Δv, then the maximum quantization Error is Δv/2
now if we split up the whole range from -[tex]m_{p}[/tex] to [tex]m_{p}[/tex] evenly into L levels, so
Δv = [tex]2m_{p}[/tex] / L
Δv / 2 = [tex]m_{p}[/tex] / L
given that the error must be at most 1% of [tex]m_{p}[/tex],
1%[tex]m_{p}[/tex] = [tex]m_{p}[/tex] / L
1/100 × [tex]m_{p}[/tex] = [tex]m_{p}[/tex] / L
100 = L
now L must be a factor of 2 to be binary encoded,
so lets consider;
L = 128
2ⁿ = 128¹
2ⁿ = 2⁷
n = 7
Therefore, the minimum number of bits needed for quantization is 7