in Digital Logic
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in Digital Logic
166 views

1 Answer

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answer (c); $2^{n} -x+y$

from  x to 2^n-1 no.of pulses are

p1= $2^{n} -1 -x$

from 2^n-1 to 0

p2=1

from 0 to y

p3=y

total pulses p=p1+p2+p3
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