Question
Download Solution PDFA micro instruction format has micro operation field which is divided into 2 subfields F1 and F2, each having 15 distinct microoperations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field AD. The address space is of 128 memory words. The size of micro instruction is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFMicro instruction format: It consists of two fields of micro-operation. One field from conditions, one for branch field and one is address field.
|
F1 |
F2 |
Condition |
Branch |
Address |
Explanation:
Micro operation field into two subfields containing 15 micro operations each.
Condition field: 4 status bits
Branch: 4 options in conjunction with address
So, bits required for microoperations = log215 + log215 = 4 + 4 = 8
Condition = log24 = 2
Branch in conjunction with address field = log2(4 × 128) = log2 512 = 9
|
F1 (4 bits) |
F2( 4 bits) |
Condition( 2 bits) |
Branch with address (9 bits) |
Size of micro – instruction = 8 + 2 + 9 = 19 bits
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