If there are m input lines and n output lines for a decoder that is used to uniquely address a byte addressable 4 KB RAM, then the minimum value of m + n is
Q: n our 32-bit MIPS architecture, what is the minimum number of byte e allocated to store the…
A: Lets understand how is MIPS architecture and what are its data types --->MIPS uses a 32-bit…
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A: Answer:- (1)C codeA = B + CMIPS Codelw $t1, 0($s4) #$t1 = Blw $t2, 0($s5) #$t2 =…
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A: Introduction: Here we are required to determine the minimum number of address bits required, also we…
Q: If there are m input lines and n output lines for a decoder that is used to uniquely address a byte…
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A: Given: Total number of rows of memory cell in the DRAM is = 225 48 -bit width main memory.…
Q: The write operation in Dynamic Random Access Memory (DRAM) is using a voltage signal to represent…
A: I have provided solution in step2
Q: The write operation in Dynamic Random Access Memory (DRAM) is using a voltage signal to represent…
A: During a write operation, a voltage (high=1, low=0) is applied to the DQ. This voltage is translated…
Q: The amount of ROM needed to implement at 4 bit multiplier is
A: The answer can be found by using the architecture of the Random Access Memory (RAM) as described in…
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A: The answer is
Q: rue/False During the 'READ' of the SRAM structure studied in the class, the bit and bit_bar are…
A: let us see the answer:- The correct answer is true:-
Q: Draw a diagram of a static ROM in the form of 4 bit NMOS NOR array for two words 1010 and 1001.
A:
Q: 10101001; DR = 10101010 Determine the 8-bit values in each register after the execution of the…
A:
Q: If the size of each 4-bit memory component is 4 x n cells where n = 1G (i.e., 4 xn uniquely…
A: The answer is
Q: A message of 200 character stored in external ROM is to be transferred serially at baud rate 2400bps…
A: Message length: 200 Baud rate: 2400 Asked:- Transfer Characters stored in external ROM serially at…
Q: A10 0 A9 A8 1 2 3 4 LS 5 138 6 7 The lowest hex address decoded by the circuit above is C B A A12…
A: The solution for the above give question is given below:
Q: With drawing Compute the Average access time for memory system when the time for Main Memory is 2000…
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A: MOV AX,#89000H; ( Transferred bits from memory location to Accumulator register) AND AX,0xff00;…
Q: QUESTION 21 Give the maximum hex value that can be brought into the CPU at a time for a 20-bit data…
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A: Dear Student, ADDWF instruction is used to add values stored at a location with Wreg and store the…
Q: Consider a 32-bit machine where four-level paging scheme is used. If the hit ratio to TLB is 98%,…
A: Introduction :Given ,4 level paging architecture hit rate of TLB = 98%TLB time = 20 ns main memory…
Q: The write operation in Dynamic Random Access Memory (DRAM) is using a voltage signal to represent…
A: Actually, DRAM stands for Dynamic Random Access Memory.
Q: If the size of each 4-bit memory component is 4 x n cells
A: The answer is
Q: For a multiplexer based bus system in an 8 bit computer system with 4 registers: a- What is the MUX…
A: NOTE: As per Barlteby guideline, if there multiple part then we are allowed to solve only first…
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A: 1. Data Hazards in RISC Architecture: Data hazards occur when instructions that exhibit data…
Q: Calculate the number of entries in page table for a virtual memory of MIPS with 20 KB page size.
A: The page table entry consists of various information about the page. The information consists in the…
Q: In a dynamic random access memory (DRAM) computer chip, each memory cell chiefly consists of a…
A: Introduction: We are requested to solve for surplus electrons on the -ve plate of a cell capacitor.…
Q: If the size of each 4-bit memory component is 4 xn cells where n = 1G (i.e., 4 xn uniquely…
A: The answer is
Q: What is the highest address in MIPS memory architecture referring to a word in hex? The lowest…
A: H . in MIPS 32bit the address can go from 0x0 to 0xffff i.e. 232 - 1 in MIPS 64 bit the address…
Q: Address Content 50 10 51 57 52 21 53 OA 54 52 55 01 56 32 57 58 59 Со 00 Suppose the memory cells at…
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Q: 3. Show the inputs and outputs for a 128 MByte memory with 32-bit read/write operations and byte…
A: The diagram is drawn in the next step :
Q: Draw the complete design for constructing a SRAM memory of a total capacity 512KX32 using a SRAM…
A: Draw the complete design for constructing a SRAM memory of a total capacity 512k * 32 using a SRAM…
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A: Given ROM capacity 16K x 8 ROM capacity as 2^m x n Where m is no of address lines n is no of data…
Q: How many 256K × 16 RAM chips are needed to provide a memory capacity of 16 MB? How many address…
A: 1). Ans Capacity of each chip = 256K * 16 = 4096K bits Memory capacity = 16M bytes = 16 * 1024 * 8K…
Q: When designing a new computer system, is it better to have a large or a small TLB (translation look…
A: Solution is in Step 2.
Q: Show a schematic diagram for interfacing an 8KB ROM starting at 0000H and 4KB RAM with 8085. The…
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Q: instructions
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Q: Given the binary format of an instruction as follows 0000 0011 0000 1000 1000 1000 0110 0011 a. What…
A:
Q: Calculate the number of entries in page table for a virtual memory of MIPS with 10 KB page size.
A: Solution Plan There are two types of MIPS architecture 32 bits and 64 bits. Calculate the number…
Q: Consider a system in which the 32kb memory space is implemented using (4 *4 )kb EPROM memory type.…
A: Consider a system in which 32kb memory space is implemented using (4*4)kb EPROM memory type. The…
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A:
Q: The contents of memory location B0007H are FFH and those at B000AH are O0H. What is the data word…
A: The contents of the memory location B0000H are FFH, and those atB0001H are 00H, what is data word…
Q: In a 128x64 bit ROM the number of the gates in the row address decoder is gates, and the number of…
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Q: 4 bytes.) ow assume the same 1kB of memory but now, word size is 64 bits. The starting address of…
A: Here in the given diagram, Memory size is 1KB Word size is 64bit
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- The write operation in Dynamic Random Access Memory (DRAM) is using a voltage signal to represent bit 0 and bit 1. The high voltage represent bit 1 and low voltage represent bit 0? O a. false O b. TrueThere is an application that requires the following hardware: an Intel 8031, a Program ROM of 8Kx8, a Data ROM of 4Kx8 for look-up tables and a Data RAMs of 8Kx8. The memory map of the design: Program ROM should start at address 0000H. Then, the Data ROM should come above the Program ROM. Finally the Data RAM must go to the top of the memory map. There should be no gaps between the memory addresses of the external ROMs. Show the design’s address space on a memory map, starting with 0000H at the bottom and FFFFH at the top.micro 8051 Given that two arrays X1 and X2 are stored in ROM where each array consists of ten byte elements. Produce array Z of the same size where: Z[i]= 9*X1[i] + X2[i] where i=0,1,….9 Store Z in internal RAM starting at address 30H. Run your program with the following data (shown in Hex): X1 : 35, A3, 7B, 90, 5C, 4F, 38, D5, 39, 7A X2: B0, 36, 2A,75, 82, 9B,C4, 7D,18,B4
- There is an application that requires the following hardware: an Intel 8031, a Program ROM of 8Kx8, a Data ROM of 4Kx8 for look-up tables and a Data RAMs of 8Kx8. The memory map of the design: Program ROM should start at address 0000H. Then, the Data ROM should come above the Program ROM. Finally the Data RAM must go to the top of the memory map. There should be no gaps between the memory addresses of the external ROMs. Calculate the address space of the ROMs and RAMs of your design.In MATLAB, a grey-scale image is stored a a matrix of numbers, all between 0 (black) and 255 (white). Since there are 256 possible values, 8 bits are needed or each pixel, in contrast to a normal integer, which is 64 bits.To save memory, grey-scale images are stored with a special data type, uint8 (unsigned 8-bit integer). In computer terminology, 8 bits = 1 byte. Even at one byte per pixel, a large image can consume a lot of memory. For example, a 1000x1000-pixel image takes 1 million bytes (which is a little less than 1 megabyte, since by convention 1 Mbyte = 2^20 bytes). One technique for reducing the size of an image is quantization - reducing the number of allowable levels from 256 to something smaller. Surprisingly, most images can be quantized by a large factor without losing much information. In this problem, you will quantize a grey-scale image to only 4 levels, so that it could (in principle) be stored as 2 bits per pixel. The template will read in an image file. Your job…A CPU that supports little endian format reads two integer (4-byte) values from address 0x1000 and 0x2000. The values read are 55 and 6850 respectively. Please show the memory contents (byte-wise) at address 0x1000 and 0x2000?
- The table below shows a segment of primary memory from a Von Neumann model computer Address Data 10101000 10001000 11001000 10011001 10100000 10101010 10110100 10111011 10001100 11001100 The program counter (PC) contains a value of 11001000. Find the value (in binary) that will be placed in MAR (memory address register)? MAR (in binray) %3D Find the value (in binary) that will be placed in MBR (memory buffer register)? MBR = (binray) %3DSuppose a program segment consists of a purely sequential part which takes 100 cycles to execute, and an iterated loop which takes 400 cycles per iteration. Assume that the loop is dependent on the sequential part, i.e., both parts cannot run in parallel. Also assume that the loop iterations are independent, and cannot be further parallelized. If the loop is to be executed 100 times, what is the maximum speedup possible using an infinite number of processors (as many processors as you could possibly need) compared to a single processor?The associative memory formed by the four column vectors x1, x2, X3, and x4 is given by (a) X1X2X3X4 (b) x1x1* + x2x2* + X3X3'+ X4X4 (c) x1*x1 + x2*x2 + x3%3 + X4x4* (d) x1x1 - x2*x2 - x3'x3 - x4x4*
- A disk has a capacity of two tera-byte size. Say the file system uses a multi-level inode structure for locating the data blocks of afile.The inode stores pointers to data blocks, including a single indirect block, a double indirect block, and several direct blocks in the 64 B of available space.Now, it is given that the disk has a block size of 512 B. The maximum file size that can be stored in such a file system in MB (round off upto 2 decimal places).8Gbx32 ROM element is given. a) Specify the address line and the data number line. b) How many bits is the total storage capacity of the memory? c) The total storage capacity of the memory can be specified in Megabytes.d) If we have two 4Gbx16 ROMs, two 2Gbx16 and 2Gbx32 modules, use these elements to design the 8Gbx32 bit memory unit as block diagrams. Express it in a descriptive way.Given a predefined array named arr consisting of N elements of type double, where N is already stored in $a1. The instructions needed to store in sa3 the memory address of the first byte after arr are la Sa3,arr sil Sa1,sa1,2 addi sa1,Sat,4 add sa3,sa3,sa1 none la $a3,orr addi sa1,Sa1,-1 sill Sa1,Sa1,3 add Sa3,Sa3,Sat la Sa3,arr sil Sat,Sa1,3 add $a3,Sa3,Sal la Sa3,orr E sl Sat,Sa1,2 add $a3,Sa3,Sa1