VTU exam Question Paper with Solution of 18CS34 Computer organization Dec-2019-Gopika D.pdf

VTU Computer Organization -Solution (18cs34) Ans 1 Q1 Connection between memory and processor (operating steps) with diagram? NOTE: If a source of destination is a register (R), MAR, MDR steps are not required as registers are directly accessible to ALU as both reside inside processor...

VTU Computer Organization -Solution (18cs34) Ans 1 Q1 Connection between memory and processor (operating steps) with diagram? NOTE: If a source of destination is a register (R), MAR, MDR steps are not required as registers are directly accessible to ALU as both reside inside processor. MAR and MDR are required only if we want to access main memory for read or write operation. Q2 Basic performance equation and SPEC rating Q3 Byte addressability (Big-endian and Little-endian assignments with diagram) Q4 Instruction types (one-address, two-address, three-address instructions)  MODULE 2 1. With a neat diagram, explain general 8-bit parallel interface. Ans Vectored Interrupts Module 3 Q1 Internal organization of RAM/memory chip/128bit memory chip? 10- bit address line is needed, but there is only one data line resulting in 15 external connections. 10-bit address is divided into two groups of 5 bits each to form the row and the column addresses for the cell array. A row address selects a row of 32 cells, all of which are accessed in parallel. However, according to the column address, only one of these cells is connected to the external data line by output multiplexer and input demultiplexer. Q3 Asynchronous and Synchronous DRAM Asynchronous DRAM or 2M*8 Asynchronous DRAM FAST PAGE MODE: When DRAM in the above diagram is accessed, the contents of all 4096 cells in the selected row are sensed, but only 8 bits are placed on the data lines D7-0, as selected by A8-0. Fast page mode makes it possible to access the other bytes in the same row without having to reselect the row. A latch is added at the output of the sense amplifier in each column. Q6 Explain three types of mapping functions for cache memory..  Module 4 With a figure, explain circuit arrangement for binary division. An n-bit positive-divisor is loaded into register M. An n-bit positive-dividend is loaded into register Q at the start of the operation. Register A is set to 0 After division operation, the n-bit quotient is in register Q, and the remainder is in register A. Procedure: step 1: Page 11 of 13 Do the following n times i) If the sign of A is 0, shift A and Q left one bit position and subtract M from A; otherwise, shift A and Q left and add M to A. ii) Now, if the sign of A is 0, set q0 to 1; otherwise set q0 to 0. Step 2: If the sign of A is 1, add M to A (restore). bit carry look-ahead adder CARRY-LOOKAHEAD ADDITIONS The logic expression for si(sum) and ci+1(carry-out) of stage i are si= xi+yi+ci ------(1) ci+1=xiyi+xici+yici ------(2) Page 13 of 13 Factoring (2) into ci+1=xiyi+(xi+yi)ci we can write ci+1=Gi+ PiCi where Gi=xiyi and Pi=xi+yi The expressions Gi and Pi are called generate and propagate functions. If Gi=1, then ci+1=1, independent of the input carry ci. This occurs when both xi and yi are 1. Propagate function means that an input-carry will produce an output-carry when either xi=1 or yi=1. All Gi and Pi functions can be formed independently and in parallel in one logic- gate delay. Expanding ci terms of i-1 subscripted variables and substituting into the ci+1 expression, we obtain ci+1= Gi+PiGi-1+PiPi-1Gi-2......+P1G0+PiPi-1... P0c0 Conclusion: Delay through the adder is 3 gate delays for all carry-bits & 4 gate delays for all sum-bits. Consider the design of a 4-bit adder. The carries can be implemented as c1=G0+P0c0 c2=G1+P1G0+P1P0c0 c3=G2+P2G1+P2P1G0+P2P1P0c0 c4=G3+P3G2+P3P2G1+P3P2P1G0+P3P2P1P0c0 The carries are implemented in the block labeled carry-look ahead logic. An adder implemented in this form is called a carry-look ahead adder.  Limitation: If we try to extend the carry-look ahead adder for longer operands, we run into a problem of gate fan-in constraints. Module 5.Give the control sequence for execution of complete instruction ADD (R3), R1. 3 Explain the differences between Hardwired and Micro-programmed control.