Answer:
The number of required bits in our counter is 9
Explanation:
Given:
Incoming frequency [tex]f_{1} = 32.768[/tex] kHz
Output frequency [tex]f_{2} = 64[/tex] Hz
From the formula of finding the number of required bits of a counter,
[tex]f_{2} = \frac{f_{1} }{2^{n} }[/tex]
Where [tex]n =[/tex] number of bits,
Put all required values in above equation,
[tex]64 = \frac{32768}{2^{n} }[/tex]
[tex]2^{n} = \frac{32768}{64}[/tex]
[tex]2^{n} = 512[/tex]
Take natural log on both side,
[tex]2 \ln n= \ln (512)[/tex]
[tex]n \times 0.693 = 6.2383[/tex]
[tex]n = \frac{6.2383}{0.693}[/tex]
[tex]n[/tex] ≅ [tex]9[/tex]
Therefore, the number of required bits in our counter is 9
