Made easy workbook 2015 (CONTROL UNIT DESIGN)

Question

A micro instruction is to be designed to specify none/one of the three micro operations of one kind and none or upto 6 micro operation of another kind. The minimum number of bits in the micro instruction is -

(a) 9          (b)  5        (c) 8        (d) none of these

Answer 1

None/one = encoded binary format = log₂n bits /control signal 
None/more than 1 = decoded binary format = 1bit /control signal 
here none /1 = 3 micro operation so log₂3 = 2 bits  
other i.e none/more than 1 = 6 micro operation so 6 bits   
 
so total 6+2 = 8 bits

Comments

Please explain why we need "encoded binary format" for one category and "decoded binary format" for other category?

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None/one means vertical programming. It is encoded binary formate( log(N) bit).

None/more then one horizontal programming it is in decoded formatemeans (1 bit/ cs)

Answer 2

To specify none or one of 3 operations no of bits required=2

To specify none or upto 6 operations no of bits required=6

So total bits required=2+6=8 bits(added as there was and mention in question)

Comments

To specify none or upto 6 operation.. won't it require log₂6 = 3 bits?

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No, none or upto 6 indicates high degree of parellelism...which furthur indicates it is horizontal CPU which takes 1 bit per control signal...as here we have atmost 6 signals so for horizontal at max 6 bits are required not 3 bits....

Answer 3

i think.The answer could be 4 .because if we use 4 bit b0,b1, b2,b4. Now we can use b4 as special bit to denote the type of operation it is.b4=0 then its a op of 1st kind and b4=1 then its operation of 2nd kind. Now other 3 bit(3 bit is sufficient to denote 6 Mcop) could be use to denote the Microop of corresponding type.