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5 votes

TotaL MICROINSTRUCTIONS=2580
SIZE OF ADDRESS FIELD=CEIL(LOG(2580))=12
CONTROL SIGNALS=129
TOTAL CONTROL BITS IN ENCODED CONTROL FIELD=CEIL(LOG(129))=8
SIZE OF CONTROL WORD=20 BIT
TOTAL SIZE OF CONTROL MEMORY=20*2580 BITS= 6450 BYTE

1 votes
1 votes
12 bits for 2580 microinstructions

8 bits for 129 ctrl signals

so 1 ctrl word is 20 bits

so ctrl mem size = 20*2^12 bits

if your ctrl word is using 12 bits for microinstructions then number of control word is 2^12...and not 2580..

because WHATEVER IT IS YOU HAVE 2^12 MEMORY PLACES IN YOUR HARDWARE, WHETHER YOU USE IT OR NOT...

so  20bits*2^12 = 20*2^10 B= 20480 Bytes

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