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What is the bit capacity of a memory that has 1024 addresses and can store 8 bits at each address?
1024 bytes is 8192 bits.
What is memory word size required in an 8085 system?
In an 8085 system, the memory word size required is 8 bits. This means that each memory location can store 8 bits or one byte of data. The 8085 processor accesses memory locations using these 8-bit memory addresses to read or write data during program execution. The memory word size of 8 bits allows the 8085 system to handle data in small, manageable chunks efficiently.
If a memory has 512 words how many bits should the address registry have?
9
What is an addressability of a memory of 8 nibbles?
The addressability of a memory refers to the smallest unit of data that can be individually addressed. In the case of memory with 8 nibbles, where each nibble is 4 bits, the total size is 32 bits (8 nibbles × 4 bits/nibble). Each nibble can be addressed separately, so the addressability of this memory is 1 nibble. This allows for 8 unique addresses corresponding to each of the 8 nibbles.
What is the total capacity expressed in Kb of a memory that is organised as 32K words 4 bits?
To calculate the total capacity of a memory organized as 32K words of 4 bits each, first convert 32K to bytes. Since 1K equals 1024, 32K equals 32 × 1024 = 32,768 words. Each word is 4 bits, so the total capacity in bits is 32,768 words × 4 bits/word = 131,072 bits. To convert bits to kilobits (Kb), divide by 1,000, resulting in a total capacity of 131.072 Kb.
How many words are there in an address space specified by 24 bits and the corresponding memory space by 16 bits?
address space=24bits => (2 Power 24)=16M words
How many address bits would be required for a memory that stores 64 4-bit words?
It requires 6 bits to address 64 words. It does not matter what the word size is.
How many bits of mask are required to provide 30 host addresses?
5 bits
How many bits of address bus are required to address 1mb memory?
You need 20 bits of address bus to address 1 Mb of memory.
What does the number of bits used to represent a memory address determine?
The number of bits used to represent a memory address determines the number of different addresses that can be formed. If the number of bits is N, then 2N addresses can be formed.If N is 16, as it is in the 8085, then 216 = 65,536.If N is 20, as it is in the 8086/8088, then 220 = 1,048,576.If N is 24, as it is in the IBM 360/44, then 224 = 16,777,216.If N is 32, as it is in most 32 bit modern processors, then 232 = 4,294,967,296.If N is 64, as it is in most 64 bit modern processors, then 264 = 18,446,744,073,709,551,616.Note that the amount of addressable memory is not the same as the amount of physical memory. As addressability goes up, often physical memory does not match it, which means that effective addressability is limited.
Consider a logical address space of eight pages of 1024words each mapped onto a physical memory of 32 frames how many bits are there in the logical address?
As was given for a 4 Page, 1024 words & 64 frames (shown below) 4 pages -> 2^2 bits 1024 bytes -> 2^10 bits 64 frames -> 2^6 bits Therefore: Logical memory = 2+10=12 bits Physical memory = 10 +6 =16 bits The answer for this problem is 13. 8 pages -> 2^3 bits 1024 bytes -> 2^10 bits 32 frames -> 2^5 bits Therefore: Logical memory = 3+10=13 bits (Page + Word) Physical memory = 10 + 5 =15 bits (Word + Frame)
How long are Ethernet Mac addresses?
Ethernet addresses are 48 bits long - not 32 bits long like IP addresses. Different single network standards have different address lengths. Ethernet addresses are called MAC addresses for other reasons, Media Access Control.
