Answer:
1. A parallel full-duplex synchronized data transmission is the parallel bidirectional transmission which is used for short distances. The parallel transmission of the data needs a lot of wires, and its 1 wire per bit, as well as it is expensive. The data over long distances might become skewed. And it can be used inside LAN, where the computer and server can talk to each other.
2. The serial full-duplex asynchronized transmission is a bidirectional serial transmission of the data over a full-duplex connection. And the messages are sent by both sides serially and can be sent anytime as well as at irregular intervals. This requires a single wire. A perfect example is while we talk through a telephone.
3. The synchronized full-duplex serial transmission is a bidirectional data transmission one by one in a serial manner and at regular intervals. And it requires a single wire. The speed of the data transmission depends upon the bandwidth of the full-duplex channel. The perfect example is full-duplex walky-talky. In this, the sender sends serially messages in a synchronized manner, and the same thing is done by the receiver, and at the same time, and as a response to each of the messages, the receiver receives. This can create a little confusion in real-time.
Explanation:
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Answer:
31 bits.
Explanation:
Given, total number of registers = 55
Total instructions = 60
Size of memory = 16 KB
Now, no of registers are 55. We find the next greater or equal power of 2 which is 64 = 26. Hence, 6 bits are required to represent a register operand.
Number of instructions = 60. We find the next greater or equal power of 2 which is 64 = 26. Hence, 6 bits are required to represent a instruction.
Size of memory = 64 KB = 26 * 210 * 23 bits = 219 bits. Hence, 19 bits are required to represent a memory location.
Now, an instruction has 2 parts, opcode and operand. As given there are only two address instructions which are memory operand and register operand.
Hence, total bits would be: 6 bits (opcode) + 6 bits (register operand) + 19 (memory operand) = 31 bits.